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\title{Sets, Classes, and Families}

\author{David Pierce}
\date{\today}
\usepackage[dvips]{graphicx} % for the German script picture
\usepackage{verbatim}  % allows a comment environment:

\address{Mathematics Department\\
Middle East Technical University\\
Ankara 06531\\
Turkey}
\email{dpierce@metu.edu.tr}
\urladdr{www.math.metu.edu.tr/~dpierce/}
\subjclass[2000]{03Exx}

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\newtheorem{lemdefn}[theorem]{Lemma and Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{axiomscheme}[theorem]{Axiom Scheme}
\newtheorem{porism}[theorem]{Porism}

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\newtheorem{definition}[theorem]{Definition}
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\newcommand{\Forall}[1]{\forall{#1}\qsep }
\newcommand{\Exists}[1]{\exists{#1}\qsep }
\newcommand{\existsunique}{\exists!}
\newcommand{\Existsunique}[1]{\exists!\,{#1}\qsep }
\newcommand{\subst}[3]{(#3)_{#1}^{#2}} % substitution of #1 for #2 in #3

\newcommand{\setim}[1][]{''{#1}} % image of set under preceding function 
\newcommand{\setimb}[1]{[\,#1\,]}    % image of set under preceding function 
\newcommand{\conv}[1]{\breve{#1}}       % converse of a relation
\newcommand{\To}{\longrightarrow}
\newcommand{\inv}{^{-1}}                % mult. inverse
\newcommand{\rest}[1]{\restriction{#1}}% restriction of function to #1
\newcommand{\id}[1]{\operatorname{id}_{#1}}
\newcommand{\mapset}[2]{{}^{#1}#2}     % set of functions from #1 to #2

\newcommand{\Iff}{}
\let\Iff\iff
%\renewcommand{\iff}{\leftrightarrow}
\renewcommand{\iff}{\Leftrightarrow}
%\newcommand{\lto}{\to}
\newcommand{\lto}{\Rightarrow}
\renewcommand{\land}{\mathrel{\binampersand}}

%%%%%%%%  sets:

\newcommand{\set}[2]{\{#1\colon #2\}}
\newcommand{\cls}[2]{\{#1\colon #2\}}
\newcommand{\class}[1]{\bm{#1}}
\newcommand{\cA}{\class A}
\newcommand{\cB}{\class B}
\newcommand{\cC}{\class C}
\newcommand{\cD}{\class D}
\newcommand{\cE}{\class E}
\newcommand{\cF}{\class F}
\newcommand{\cG}{\class G}
\newcommand{\cH}{\class H}
\newcommand{\cK}{\class K}
\newcommand{\cM}{\class M}
\newcommand{\cR}{\class R}
\newcommand{\cS}{\class S}
\newcommand{\cT}{\class T}

\newcommand{\universe}{\mathrm{\mathbf V}}
\newcommand{\cOm}{\bm{\mathit{\Omega}}}
\newcommand{\on}{\mathbf{ON}}           % the class of ordinals
\newcommand{\cn}{\mathbf{CN}}           % the class of cardinals
\newcommand{\wf}{\mathbf{WF}}           % class of well-founded sets
\newcommand{\rwf}{\mathrm{\mathbf R}}   % R used to define WF
\newcommand{\defr}[2]{\mathscr{D}_{#1}(#2)} % definable n-ary relations
\newcommand{\cu}{\mathbf{L}}            % constructible universe L
\newcommand{\const}{\mathrm{\mathbf L}} % C used to define L
\newcommand{\tc}[1]{\operatorname{tc}(#1)}  % transitive closure
\newcommand{\cl}[2][]{\operatorname{cl}_{#1}(#2)}

\newcommand{\str}[1]{\mathfrak{#1}}     % structure
\newcommand{\nmodels}{\nvDash}
\newcommand{\proves}{\vdash}
%\newcommand{\lang}{\mathcal{L}}        % a language or signature
\newcommand{\End}[1]{\operatorname{End}(#1)} % endomorphisms

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\renewcommand{\emptyset}{\varnothing}  % [I like the other symbol better]

% unary:

\newcommand{\comp}{^{\mathrm{c}}}      % set-theoretic complement
\newcommand{\pow}[1]{\mathscr{P}(#1)}  % power set

% binary:
\renewcommand{\setminus}{\smallsetminus} % I don't like the standard symbol
\newcommand{\symdiff}{\vartriangle}    % symmetric difference

%%%%%%%%%%%%  Cardinality

\newcommand{\injects}{\preccurlyeq}    % injects in
\newcommand{\pinjects}{\prec}
\newcommand{\equip}{\approx}           % equipollent (is in bijection with)
\newcommand{\nequip}{\not\approx}      % not equipollent
\newcommand{\vscr}[1]{#1'}        % successor ordinal (v for von Neumann)
\newcommand{\ord}[1]{\operatorname{ord}(#1)} % order-type (ordinality)
\newcommand{\supr}[1]{\sup(#1)}        % supremum
%\newcommand{\size}[1]{\lvert#1\rvert}  % cardinality
\newcommand{\card}[1]{\operatorname{card}(#1)} % cardinality
\newcommand{\pl}{^{+}}                 % successor cardinal
\newcommand{\cntm}{\mathfrak c}        % cardinality of the continuum
\newcommand{\pred}[1]{\operatorname{pred}(#1)}
\newcommand{\rank}[1]{\operatorname{rank}(#1)}
\newcommand{\rankl}[1]{\operatorname{rank}_{\const}(#1)}

%%%%%%%%%%%%%  Functions

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\newcommand{\ran}[1]{\operatorname{rng}(#1)} % range
\newcommand{\fld}[1]{\operatorname{f{}ld}(#1)} % field
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\newcommand{\surjection}{\twoheadrightarrow}
%\newcommand{\bijection}{\rightarrowtail\mkern-17mu\shortrightarrow}
\newcommand{\bijection}{\rightarrowtail\mkern-18mu\twoheadrightarrow}
\newcommand{\supp}[1]{\operatorname{supp}(#1)} % support

%%%%%%%%%%%%%

\newcommand{\included}{\subseteq}      % [the name suggests the meaning here]
\newcommand{\nincluded}{\not\subseteq} % not included
\newcommand{\pincluded}{\subset}       % proper inclusion    
\newcommand{\npincluded}{\not\subset}       % proper inclusion    
\newcommand{\includes}{\supseteq}

%%%%%%%%%%%%%%%

\newcommand{\axioms}[1]{\mathrm{#1}}   % set-theoretic axiom or set of axioms
\newcommand{\zf}{\axioms{ZF}}          % Zermelo-Frankel axioms
\newcommand{\ac}{\axioms{AC}}	       % Axiom of Choice
\newcommand{\zfc}{\axioms{ZFC}}        % ZF + AC
%\newcommand{\ch}{\axioms{CH}}          % Continuum Hypothesis
\newcommand{\gch}{\axioms{GCH}}          % Generalized Continuum Hypothesis
%\newcommand{\pa}{\axioms{PA}}          % first-order Peano arithmetic

%%%%%%%%%%%%%%%%

% Standard structures

\newcommand{\vnn}{\upomega}       % von Neumann natural numbers

\newcommand{\stnd}[1]{\mathbb{#1}}
\newcommand{\N}{\stnd{N}}         % natural numbers
\newcommand{\scr}[1]{#1^{+}}           % successor
\newcommand{\Zp}{\Z_{+}}          % positive integers
\newcommand{\Z}{\stnd{Z}}         % integers
\newcommand{\Q}{\stnd{Q}}         % rationals
\newcommand{\R}{\stnd{R}}         % reals
\newcommand{\C}{\stnd{C}}         % complex numbers


%\usepackage[BCOR22pt]{typearea}
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\begin{document}
\frontmatter

\mainmatter

\thispagestyle{empty}

\mbox{}\vfill

\noindent\hspace{87pt}
\begin{minipage}{300pt}
\begin{center}
{\Huge Sets, Classes, and Families}

\mbox{}
\newcommand{\mygap}{1ex}
\vspace{\mygap}
\vspace{\mygap}

{\huge Notes on set theory}

\mbox{}
\vspace{\mygap}

{\LARGE David Pierce}

\mbox{}

{\Large\today}
\end{center}

\mbox{}

%\includegraphics[width=300pt,height=400pt]{isabey.eps}
%\includegraphics[width=300pt,height=400pt]{zinciriye.eps}
\includegraphics[width=300pt,height=400pt]{zinciriye.eps}
%{\small \.Isa Bey Mosque, Sel\c cuk, Turkey}
\end{minipage}
\vfill
\tableofcontents

%\setcounter{chapter}{-1}
\chapter*{Preface}
%\addcontentsline{toc}{chapter}{Preface}
%\markboth{\it PREFACE}{\it PREFACE}
%\input{preface}

%THESE NOTES ARE BEING EDITED.


These notes are for use in a course called Set Theory, given in the
mathematics department of Middle East Technical University, Ankara,
under the designation Math~320.  The catalogue description of the
course is:
\begin{quote}
Language and axioms of set theory. Ordered pairs, relations and
functions. Order relation[s] and well ordered sets. Ordinal numbers,
transfinite induction, arithmetic of ordinal numbers. Cardinality and
arithmetic of cardinal numbers. Axiom of choice, generalized continuum
hypothesis. 
\end{quote}
The notes cover these topics and more.
There have been three previous editions:
\begin{enumerate}
  \item
In 2001, I wrote notes on ordinals and cardinals for Fundamentals of
Mathematics (Math~111); a good part of what I wrote went beyond what
that course had time for.  
\item
I revised those notes for use in Math~320 as
taught by Ay\c se Berkman in
the spring semester of the 2004/5 academic year.  
\item
When I was to teach
that course myself, in the spring of 2006/7, I
completely rewrote the notes, using
Ay\c se's copy of the second edition, with her comments.
\end{enumerate}
All of these editions must be considered as rough drafts.  The same is
true of the present edition, since it has many differences with the
third edition.  
Starting with that third edition, I place more
emphasis on the following picture of set theory:
\begin{enumerate}
\item
There are things called 
\tech{set}{s,}
with certain properties.  Sets compose a so-called 
\techplain{universal class},%
\tindexsub{universal}{--- class}%
\tindexsub{class}{universal ---} 
$\universe$ (\parref{par:class=}).  
\item
There is a logical language for talking about sets (\parref{par:1});
the one non-logical symbol of this language is $\in$, to express
\tech{membership}{} of one set in another (\parref{par:truth}).
\techplain{Equality}{}%
\tindexsub{equal}{---ity of sets} 
of sets can be \emph{defined} in terms of membership
(\parref{par:=def}). 
\item
In the language of sets, a formula $\phi$ with one free variable $x$
defines the class $\cls x{\phi(x)}$ (\parref{par:class=}), which is
a \techplain{subclass}{}%
\tindex{subclass}%
\tindexsub{class}{sub---}
of the universal class (\parref{par:subclass}).  
\item
Sets are also classes (\parref{par:=def}).  Most of things
that one does with sets in mathematics---like taking unions or
intersections or power sets---can be done with classes.
\item
There is no reason to assume that all classes are sets.  Indeed, there
is a class of all sets that are not members of themselves, namely
$\cls x{x\notin x}$, but it is not a set.  Thus, the so-called
Russell Paradox is simply a basic theorem of set theory (\parref{thm:R}).
\end{enumerate}
This picture of set theory can be seen in Levy~\cite{MR1924429}; I
have found his
book a useful reference, though it is dense with detail.  Levy is also
a good source for historical
references.  
Other useful books (at different levels) have been
Suppes~\cite{MR0349389}, Kunen~\cite{MR85e:03003}, and
Moschovakis~\cite{MR95a:04001}; also Enderton~\cite{MR0337470} for
Chapter~\ref{ch:logic}, and Shoenfield \cite{MR0225631} and
Cohen~\cite{MR0232676} for Ch.~\ref{ch:models}.  
I have used Fraenkel \emph{et al.}~\cite{MR0345816} for a review
of the development of the individual axioms that generate what is
called Zermelo--Fraenkel set theory.  Those axioms are among those listed on
p.~\pageref{axiomlist} below.  In that list, weaker forms of the
Union Axiom (called Augmentation and Binary Union respectively) are
included, along with the usual form.  In the text, I try to introduce
axioms only when they are
needed for something interesting; the full Union Axiom is not
needed for a long while; meanwhile, weaker versions
suffice.\footnote{In the 2007 edition I remarked: `Possibly I have
  overlooked some earlier implicit use of Union or other 
axioms; I remind the reader of my comment that these notes are still a
rough draft.'}   

Unlike many set-theory textbooks, I aim, in Chapter~\ref{ch:logic}, to
give a precise formal account of
the logic behind set theory.  It may be pointed out that the
Zermelo--Fraenkel axioms were worked out \emph{before} the formalism
of the logic was worked out.  However, the consistency results of
Chapter~\ref{ch:models} are all about the logic.

I try to work with the idea that most of the Zermelo--Fraenkel axioms
amount to assertions that certain classes are sets.  I have not seen
the Infinity Axiom treated explicitly in this way, except in these
notes.  Here I take some trouble, in Chapter~\ref{ch:size-order}, to see
how the \emph{class} of
natural numbers can be obtained, \emph{without} the assumption that it
already exists as a set.

In studying the natural numbers, Dedekind~\cite{MR0159773} was clear
about the distinction between
\techplain{proof by induction}{}%
\tindexsub{proof}{--- by induction}%
\tindex{induction}
and
\techplain{definition by recursion}{}%
\tindex{definition by recursion, recursive ---}%
\tindex{recursion}.
Peano~\cite{Peano} was not, though he was aware of Dedekind's work;
and many writers today seem not to be clear about the distinction.  In
Chapters~\ref{ch:size-order} and~\ref{ch:N}, I try to be clear.

Since I avoid treating classes as sets, it is useful to
have a name for something that contains classes as such.  I introduce
the word \tech{family}{} for this.

These notes are not intended for use in isolation from the classroom.
Points presented here in outline may be elaborated more fully in
lectures, or else simply omitted there.  
Lectures may take matters in a different order from the notes.
I take Euclid's \emph{Elements} \cite{MR1932864} as a
model.  Euclid simply presents propositions and proofs, with no
explanation of \emph{why} one would want to prove these propositions.
The explanation, if needed, is left to the living teacher.

The notes are divided into eight chapters; these are divided into
sections (\S), and these into paragraphs (\P).  Some paragraphs are
labelled as Axioms, Lemmas, Theorems, or (in one case) Porism.
Certain displayed formulas are assigned arabic numerals in a single
sequence through the notes.
Terms being 
\defnplain{defined}{}%
\dindexsub{define}{terms being ---d}
 in the notes are printed in boldface;
technical terms being emphasized, but not properly defined, are
\techplain{slanted}.  There is an index of these terms at the back.  All
but the first chapter have exercises at
the end; often the exercises ask the reader to supply the proofs of
propositions (lemmas, theorems) given in the chapter.

On the front and back pages are photographs I took of
Zinciriye Madrasah (built in 1385) and Mor
Gabriel Monastery (founded 397), in Mardin.



\listoffigures
%\addcontentsline{toc}{chapter}{List of Figures}

%\listofaxioms

\chapter*{List of Axioms}
%\addcontentsline{toc}{chapter}{List of Axioms}
\label{axiomlist}

Here, $a$ and $b$ are arbitrary sets, $\cC$ is an
        arbitrary class, $\cF$ is an arbitrary function, and $\vnn$,
        $\universe$, and $\wf$ are 
        the classes defined on pp.~\pageref{par:omega-as-class},
        \pageref{par:class=}, and~\pageref{par:wf-universe} respectively.
\begin{enumerate} 
\item{}
[p.~\pageref{ax:extension}]
Extension: \[a=b\iff a\included b\land b\included a.\]
\item{}
[p.~\pageref{ax:comprehension}]
Separation Scheme: \[\Exists xx=a\cap\cC.\]
\item{}
[p.~\pageref{ax:pairing}]
Pairing: \[\Exists xx=\{a,b\}.\]
\item{}
[p.~\pageref{ax:replacement}]
Replacement Scheme: \[a\included\dom{\cF}\lto\Exists xx=\cF\setimb{a}.\]
\item{}
[p.~\pageref{ax:power-set}]
Power Set: \[\Exists xx=\pow a.\]
\item{}
[p.~\pageref{ax:augmentation}]
Augmentation: \[\Exists xx=a\cup\{b\}.\]
\item{}
[p.~\pageref{ax:binary-union}]
Binary union: \[\Exists xx=a\cup b.\]
\item{}
[p.~\pageref{ax:union}]
Union: \[\Exists xx=\bigcup a.\]
\item{}
[p.~\pageref{ax:infinity}]
Infinity: \[\Exists xx=\vnn.\]
\item{}
[p.~\pageref{ax:choice}]
Choice: \[\text{Every set has a choice function.}\]
\item{}
[p.~\pageref{ax:foundation}]
Foundation: \[\universe=\wf.\]
\end{enumerate}
The axioms of Extension, Pairing, Power Set, Union, Infinity, and
Foundation, along with the axiom schemes of Separation and
Replacement, are known collectively as $\zf$ (pp.~\pageref{par:incomp}
and~\pageref{par:zf}); but Separation is redundant in $\zf$
(p.~\pageref{ex:sep-from-rep}).  The Augmentation and Weak Union axioms
are my addition, but they are consequences of $\zf$.  When the Axiom
of Choice is added, $\zf$ becomes $\zfc$.  Some results on consistency
and independence are worked out in Chapter~\ref{ch:models}.


\chapter{Introduction}

%\setcounter{equation}{9} % not needed---no equations

\section{What a set is}

\begin{parag}\label{par:set}
A \defn{set}{} is a thing that 
\defnplain{contain}s%
\dindex{contain: a set ---s things}
 other things.  Those other
  things are called \defn{member}{s} or
  \defn{element}{s} of the set.  But the set cannot be separated 
  from its elements the way a box can be emptied of its contents.  The 
  set 
\defnplain{comprise}{s}%
\dindexsub{comprise}{set ---s its members}
 its members, and the
  members 
\defnplain{compose}{}%
\dindexsub{compose}{members --- a set}
the set.
  A set \emph{is} its elements, considered as one thing.  It is a
  \tech{multitude}{} that is also a \tech{unity}.
\end{parag}

\begin{parag}\label{par:collective}
In ordinary speech, one may speak of a flock of pigeons, a pair
  of socks, a deck of cards, or a number of days.
Words like \Eng{flock, pair, deck,} and \Eng{number} are (or can be)
\defn{collective noun}{s.}
In English, such nouns can be used as 
subjects of singular or plural verbs:
\begin{quote}
  \Eng{A flock of pigeons is attacking that crust of bread.}\\
  \Eng{A flock of pigeons are attacking that crust of bread.}
\end{quote}
In mathematics, the word \techplain{set}{} is introduced as the
most abstract%
\footnote{I originally called \Eng{set} the most \emph{generally
  applicable} abstract noun (after \Eng{class}), but
  this was shown by a student, Sajjad Haider, to be a mistake: our
  universe of sets will not contain flocks of pigeons.}
 collective noun.  But now we have something new, \techplain{set}{s,} that
 can belong to something else: that something else is called a
 \tech{class}.  In turn, a class may belong to a \tech{family}{} of classes
 (\parref{par:aRRa}). 
Sets are classes (\parref{par:=def}), and classes are families, but not
every class is a set (\parref{thm:R}; in
  \parref{sect:e-wf}, we shall see how to understand a family as a
  class). 
\end{parag}

\begin{parag}\label{par:10}
  A set contains other things (\parref{par:set}).  As it happens, we
  may modify the 
  expression \Eng{other things} in three ways:  We might ignore the
  \Eng{other,} allowing a set to contain \emph{itself.}\footnote{We
  ultimately rule out this possibility with the Foundation Axiom in
  Ch.~\ref{ch:models}.}  Or, a set might
  contain, not other \emph{things,} but just \emph{one} other thing.  Or
  possibly a set contains \emph{nothing} at all.
\end{parag}

\section{Why study sets}

\begin{parag}\label{par:set-math}
  Sets are a foundation for
  mathematics, in that the objects of mathematics can be
  understood as sets.  A function $f$ can be understood
  as the set of 
\techplain{ordered pair}s%
\tindexsub{order}{---ed pair}%
\index{pair|seealso{ordered pair}}
 $(a,b)$ such
  that $f(a)=b$; and the
  ordered pair $(a,b)$ can be understood as the set
  $\{\{a\},\{a,b\}\}$.  Perhaps $a$ and $b$ 
  here are 
\techplain{real number}{s.}%
\index{real|see{number}}%
\tindexsub{number}{real ---}
A real number can be
  understood as a certain
  set of 
\techplain{rational number}{s.}%
\index{rational|see{number}}%
\tindexsub{number}{rational ---}
A  rational number can be understood 
  as a certain set of 
\techplain{integer}{s.}%
\index{integer|see{number}}%
\tindexsub{number}{integer}
An integer can be 
  understood as a certain set of 
\techplain{natural number}{s.}%
\index{natural|see{number}}%
\tindexsub{number}{natural ---} 
A natural number can be understood as the set of its
\tech{predecessor}{s.}  There is one natural number with no
predecessors; so this is the empty set. 
\end{parag}

\begin{parag}
  Sets form a connection between mathematics and logic,\index{logic}
  the science of reasoning.  See Figure~\ref{fig:lms}.  
  \begin{figure}[t]
    \begin{pspicture}(-3,-1)(3,5.6)
%\psgrid
\psline[linestyle=dashed](-3,0)(-3,5.5)
\psline[linestyle=dashed](3,-1)(3,5.2)
\psset{plotpoints=200}
      \parametricplot[arrows=>->, arrowsize=2mm 2]{-45}{945}{
t cos 3 mul 
t sin t 360 div 2 mul add}
%      \parametricplot[linestyle=dashed]{0}{360}{
%t cos 3 mul 
%t sin}
\pstextpath[c]{
\parametricplot[linestyle=none]{540}{720}{t cos 3 mul 
t sin t 360 div 2 mul add 2.3 sub} 
}
{set theory} 
\psdots(3,0)(3,2)(3,4)(-3,1)(-3,3)(-3,5)
\uput[r](3,1){mathematics}
\uput[l](-3,2){logic}
    \end{pspicture}
    \caption{Logic and mathematics, connected by set theory}\label{fig:lms}
  \end{figure}
There is, for
  example, a correspondence between the 
  \tech{union}, $a\cup b$, of two sets $a$ and $b$ and the
\tech{disjunction}, $P\lor Q$, of two propositions
$P$ and $Q$.  Indeed, if $a$ is the set of \tech{possible
  situation}s\footnote{I borrow this term from Wilfrid Hodges
  \cite[\S~1, p.~1]{Hodges-Logic}, who writes:  `A set of beliefs is
  called \tech{consistent}{} if these beliefs could all be true
  together in some possible situation.'} in which
$P$ is true, and $b$ is the set of possible situations in which $Q$ is
true, then $a\cup b$ is the set of possible situations in which $P\lor
Q$ is true.\footnote{However, if one is in a situation where one knows
  that $P$ is true, but $Q$ is false, then one is not likely to assert
  that $P\lor Q$ is true; one will probably just say that $P$ is true
  and $Q$ is false.}
\end{parag}

\begin{parag}\label{par:axiomatic}
Sets may serve both in the definition of, and in an
example of, the logical notion of an
\defnplain{axiomatic system}.%
\dindexsub{axiom}{---atic system}%
\dindexsub{system}{axiomatic ---}
In such a
system, one 
\defn{postulate}s certain truths, called
\defn{axiom}{s,} 
which are held to be self\-{}-evident; 
from the axioms, by means of logic, one derives other truths, which
are not so evident.  Thus, set theory is a modern example of a
method as old
as Euclid~\cite{MR1932864} for organizing and developing a body of
mathematical work. 
\end{parag}

\section{Cardinals and ordinals}\label{sect:co}

\begin{parag}
As young
children, we learn to chant a sequence of numerals: \Eng{one, two,
  three} in English; \Tur{bir, iki, \"u\c c} in Turkish.  We
learn to use these numerals as 
\techplain{cardinal number}{s,}%
\tindexsub{number}{cardinal ---}% 
\tindexsub{cardinal}{--- (number)} 
to indicate the
\emph{sizes} of sets.  Later 
we learn that \Eng{zero}/\Tur{s\i f\i r} is also a size.
But the sequence of numerals also has an \emph{order,} like the order
of letters in the alphabet.  So we can use numerals as
\tech{ordinal number}{s,}%
\tindexsub{number}{ordinal ---} to indicate
\emph{position} of elements
\emph{within} a set.  We also learn special words for ordinal numbers:
\Eng{first, second, third} in English; \Tur{birinci, ikinci,
  \"u\c c\"unc\"u} in Turkish.
\end{parag}

\begin{parag}
So numerals have two uses, as cardinals and as ordinals; these uses are
completely different; and yet the \emph{same} underlying numerals are
used in each case.  For example, if a book starts on page 1 and ends
on page~\pageref{LastPage}, 
then we know two things:  the book has~\pageref{LastPage} pages,
and the last page of the book is 
the~\pageref{LastPage}%
\whiledo{\value{mynum}<\pageref{end}}{\stepcounter{mynum}}%
\setcounter{modnum}{\value{mynum}-\value{mynum}/10*10}%
  \ifthenelse{\value{modnum}=1}{st}{%
   \ifthenelse{\value{modnum}=2}{nd}{%
    \ifthenelse{\value{modnum}=3}{rd}{th}}} page of the book.
\end{parag}

\begin{parag}
Once I saw a cargo of small cardboard boxes being loaded on an
airplane.  Armed guards were standing by.  As the boxes ascended the
ramp into the cargo hold, a man counted them by writing
  numerals on them: 1, 2, 3, and so on.
  I did not see how the boxes were unloaded at the end of the flight.
  But no matter the order with which the boxes came off the plane,
  they could have been counted by the same procedure used when they
  were loaded.  If no box was missing, then the same number of boxes
  would have been found.
\end{parag}

\begin{parag}
When we count a \emph{finite} set, then the order in which we count
the elements in
the set does not matter.  This is a fact so basic that
we do not need to learn it in school.  But it fails for
\emph{infinite} sets. 
\end{parag}

\begin{parag}\label{par:omega}
  What is a number?  As suggested in~\parref{par:set-math}, we
  can identify certain \emph{sets} to be called $0$, $1$, $2$, $3$,
  and so forth.  These sets will be the 
  \tech{natural number}{s.}  Each natural number
  $n$ has a
  \tech{successor}, which can be called $n+1$.  We
  \emph{assume} that 
  all of the natural numbers compose a set, called%
\footnote{The
  letter $\vnn$ is not~\lett w, the so-called double \lett u; it is the
  Greek minuscule \emph{omega,} the last
  letter of the alphabet.  See Appendix~\ref{app:Greek}.  Omega is a
  \emph{large} or
  \emph{long} (mega)
  \lett o, to be contrasted with the small or short (micro) \lett o,
  omicron.  One might even call omega a double \lett o, and indeed its 
  written minuscule form seems to come from \lett{oo.}} 
$\vnn$.  We treat
  $\vnn$ as a new number.  Then we can form the new numbers $\vnn+1$,
  $\vnn+2$, and so forth.  Beyond all of these new numbers, there
  is $\vnn+\vnn$ or $\vnn\cdot2$, then $\vnn\cdot2+1$, and so forth;
  then $\vnn\cdot3$ and so forth; then $\vnn\cdot\vnn$ or $\vnn^2$ and
  so forth, then $\vnn^{\vnn}$, $\vnn^{\vnn^{\vnn}}$, and so
  forth.  All of these new numbers are examples of 
\techplain{transfinite ordinal number}{s.}%
\tindexsub{transfinite}{--- ordinal}%
\tindexsub{ordinal}{transfinite ---}%
\tindexsub{number}{transfinite ordinal ---}
We may be able to use these transfinite ordinals to count a set.
  Perhaps we can assign \emph{each} natural number to some different
  element of the set, only to find that some elements of the set are
  left over.  So we assign $\vnn$ itself to one of these, and
  $\vnn+1$ to another, and so forth.  Perhaps, in this way, all of
  the ordinals that come before $\vnn^3\cdot 7+\vnn^2\cdot
  4+\vnn\cdot 5$ are assigned to elements, but there is no element
  left to which we can assign
 $\vnn^3\cdot 7+\vnn^2\cdot 4+\vnn\cdot
  5$ itself.  Then how many elements has the set?  The same that
  $\vnn$ has.  We can number the set alternatively so that \emph{each} 
  element is assigned a natural number.
\end{parag}

\begin{parag}\label{par:aleph}
There \emph{are} sets larger than $\vnn$; these too can be `counted'
  with ordinals, but with ambiguity as before.
  The size of $\vnn$ is given the name $\aleph_0$;%
\footnote{The letter
  $\aleph$ is \emph{aleph,} the first letter of the Hebrew alphabet.}
 this is the first
  \techplain{transfinite cardinal number}.%
\tindexsub{transfinite}{--- cardinal}%
\tindexsub{cardinal}{transfinite ---}%
\tindexsub{number}{transfinite cardinal ---}
  There is a next larger cardinal number, $\aleph_1$; then we have
  $\aleph_2$, $\aleph_3$, \dots, $\aleph_{\vnn}$,
  $\aleph_{\vnn+1}$, \dots, $\aleph_{\vnn\cdot2}$, and so on.  Thus the
  transfinite cardinal numbers are \tech{indexed}{} by the ordinal
  numbers.  
But while every cardinal is an ordinal, not every ordinal is a
cardinal.  The size of an infinite set is not obtained or defined
  directly by a process of counting.
\end{parag}

\section{Sets as numbers}

\begin{parag}\label{par:pick}
  I suggested in \parref{par:collective} that \emph{number} can be a
collective noun.  It is not always so.  If I ask you, 
\begin{quote}
\Eng{Pick a number between one and ten,}
\end{quote}
you will probably not think of a set or a number \emph{of things;} you
will just pick one abstract thing, called \emph{five} perhaps, or
\emph{eight.}  But if we observe, 
\begin{quote}
\Eng{A number of people are gathering in the street,} 
\end{quote}
then the
emphasis is on the \emph{people} as much as on how many there are.
In this latter sentence, the sense of
\Eng{number} would seem to be that of the Greek \Gk{>arijm'oc.}
(See Appendix~\ref{app:Greek}
%Table~\vref{table:Greek} 
for the Greek alphabet.)  The
word \Gk{>arijm'oc} is the origin of the word
\Eng{arithmetic,} and it is commonly translated as \Eng{number;} but
note how Euclid defines it \cite[Book VII, Definitions]{MR1932864}: 
\begin{quote}
  \begin{enumerate}
\item
\Gk{{Mon'ac} >estin, kaj'' <`hn <'ekaston t~wn >'ontwn <`en l'egetai.}\\
A \defn{unit}{} is that by virtue of which each thing
is called \emph{one} (\Gk{<'en}).
    \item
  \Gk{>Arijm`oc d`e t`o >ek mon'adwn sugke'imenon pl~hjoc.} \\
A \defn{number}{} is a multitude\index{multitude} (\Gk{pl~hjoc})
composed of units.
  \end{enumerate}
\end{quote}
This account of \Eng{number} bears some resemblance to the account of
\Eng{set} in
\parref{par:set}.  However, Euclid does not allow the exceptional cases
discussed in \parref{par:10}; in particular, for Euclid, \emph{one} is
not a number.\footnote{This can be inferred from some other
  definitions in Book VII of the \emph{Elements:} `A 
\defn{prime number}{}%
\dindexsub{number}{prime ---}
 is that which is measured by a unit alone.  A
  \defnplain{composite number}{}%
\dindexsub{number}{composite ---}%
\dindexsub{composite}{--- number}
 is that which is measured by some other
  number.'}
\end{parag}

\begin{parag}
If we take \Eng{number} seriously as a collective
noun roughly equivalent to \Eng{set,} then a certain passage by Plato,
in the work commonly called the \emph{Republic,} becomes
an argument in favor of studying sets.
Numbers and sets are worth studying, because they somehow combine
opposites like \emph{many} and \emph{one,} \emph{multiplicity} and
\emph{unity.} 
  The \emph{Republic} is written as if by Plato's teacher Socrates; in it,
  Socrates recounts a long 
  conversation in which he describes an ideal city, as an analogy for
  the ideal person.  Certain citizens
  of the ideal city will be \emph{guardians;} Socrates describes their
  education. 
The following translation from Book VII (524d--525b) is mine, but
depends on the translations of Shorey \cite{Shorey} and Waterfield
\cite{Waterfield}.  I have inserted some of the original Greek words,
especially\footnote{I have also included certain derivatives of the
  present participle \Gk{>'ont-} corresponding to the English
  \Eng{being} and the Turkish \Tur{olan} or \Tur{olur.}  Addition of
  the abstract\-{}-noun suffix \Gk{-'ia} to the feminine form of
  \Gk{>'ont-} yields
  \Gk{o>us'ia}; the corresponding Turkish might be \Tur{olurluk.}  The
  Greek \Gk{o>us'ia} is sometimes translated as \Eng{substance}, and
  indeed both words can connote wealth.
  Putting the definite article in front of the nominative neuter form
  of \Gk{>'ont-} creates \Gk{t`o >'on.}}
those that are origins of English words.
\begin{quotation}
`So this is what I %[Socrates] 
was just trying to explain:  Some things
  are \emph{thought\-{}-provoking} (\Gk{paraklhtik`a t~hc diano'iac}), and
  some are not.  Those things are called
  \defn{thought-provoking}{} that strike our sense together with their
  opposites.  Those that do not, do not
tend to awaken reflection.'

`Ah, now I understand' he [Glaucon] said.  `It seems that way to me,
too.'

`Okay then. Which of these do \emph{multiplicity}
  (\Gk{>arijm'oc}) and \emph{unity} (\Gk{t`o <'en}) seem to be?' 

`I can't imagine' he said. 

`Well,' I said `reason it out from
what we said.  If unity is fully grasped alone, in itself,  by sight
or some other sense, then it must be [an object] like a finger, as we
were explaining: it does not draw us towards \emph{being\-{}-ness}
(\Gk{o>us'ia}).  But if some discrepancy is always seen with it, so as
to appear not rather \emph{one} (\Gk{<'en}) than its opposite, then a
decision is needed---indeed, the \emph{soul} (\Gk{yuq'h}) in itself is
compelled to be puzzled, and to cast about, arousing thought within
itself, and to ask:
What then is unity as such?  And so the \emph{study}
  (\Gk{m'ajhsic}) of unity must be among those that lead and guide
  [the soul] to the sight of \emph{that which is} (\Gk{t`o >'on}).'

 `But certainly' he
  said `vision is especially like that.  For, the same thing is seen
  as one and as \emph{indefinite multitude}\index{indefinite
    multitude}\index{multitude}  
  (\Gk{>'apeira t`o pl~hjoc}).'

`If it is so with unity,' I said `is it not so with every  \emph{number}
  (\Gk{>arijm'oc})?'

`How could it not be?'

`But \emph{calculation}
  (\Gk{logistik'h}) and \emph{number\-{}-theory} (\Gk{>arijmhtik'h}) are
entirely about number.'

`Absolutely.'

`And these things appear to lead to truth.'

`Yes, and extremely well.'

`So it seems that these must be some of the \emph{studies}
  (\Gk{majhm'ata}) that
  we are looking for.  Indeed, the \emph{military} (\Gk{polemik'on}) needs to
  learn them for deployment [of troops],---and the philosopher,
  because he has to rise out of [the world of] \emph{becoming}
  (\Gk{g'enesic}) in order to take hold of being\-{}-ness, or else he will
  never \emph{become a calculator} (\Gk{logistik\~w| gen'esjai}).'

`Just so' he said.

`And our guardian happens to be both military man and philosopher.' 

`Of course.'

`So, Glaucon, it is appropriate to require this study by law and to
  persuade those who intend to take part in the greatest affairs of
  the city to go into calculation and to engage in it not \emph{as a pastime}
  (\Gk{>idiwtik\~wc}), but until they have attained, by thought
  itself, the vision of the nature of numbers, not [for the sake of]
  buying and selling, as if they were preparing to be merchants or
  shopkeepers, but for the sake of war\footnote{One can hardly be sure
that Socrates is not pulling Glaucon's leg.  Socrates previously
(369b--372c) described a primitive, peaceful,
vegetarian city, which Glaucon rejected (372c--d) as
being fit only for pigs.} and an easy turning of the soul
  itself from becoming towards truth and being\-{}-ness.'

`You speak superbly' he said.
\end{quotation}
\end{parag}

\chapter{Logic}\label{ch:logic}

%\setcounter{equation}{9}

\section{Propositional logic}

\begin{parag}
For present purposes, a \defn{logic}{} consists of
\begin{enumerate}
\item
an \defnplain{alphabet}{}%
\dindex{alphabet for a logic}
 of symbols;
\item
%a \defn{language}, namely
a way to combine the symbols of the alphabet into \defn{formula}{s;}
\item
a correspondence between the formulas and something else that we want
to understand.  
\end{enumerate}
To understand sets, we shall develop a
\techplain{predicate logic}.%
\index{predicate|seealso{logic}}
\tindexsub{logic}{predicate ---}
This will be based on a 
\defnplain{propositional logic}.%
\index{proposition|seealso{logic}}
\dindexsub{proposition}{---al logic}%
\dindexsub{logic}{propositional ---}
\end{parag}



\begin{parag}\label{par:prop}
The alphabet of our propositional logic will comprise
  \begin{enumerate}
    \item
\defnplain{(propositional) variable}{s}
\dindexsub{proposition}{---al variable}%
\dindexsub{variable}{propositional ---} $P$, $P'$, $P''$, \dots;
\item
the
\defnplain{propositional connective}{s}%
\dindexsub{proposition}{---al connective}%
\dindex{connective}%
\footnote{In place of the double-shafted arrow $\lto$,
  the single-shafted arrow $\to$ is often used; but this might be
  confused with the arrow used in denoting functions
  (\parref{par:functions}).}
$\lnot$ and $\lto$;
\item
the \defn{bracket}s $($ and $)$.
  \end{enumerate}
From the variables, connectives, and brackets,
\defnplain{propositional formula}s%
\dindexsub{proposition}{---al formula}%
\dindexsub{formula}{propositional ---}
are built up 
\defn{recursive}{ly} as follows.
\begin{enumerate}
  \item\label{item:P}
Every propositional variable is a propositional formula;
\item\label{item:not}
if $F$ is a propositional formula, then so is its
\defn{negation}, $\lnot F$;
\item\label{item:to}
If $F$ and $G$ are propositional formulas, then so is the
\defn{implication}{}%
\footnote{One might read the
  formula also as the implication of $G$ \emph{in} $F$.  One normally
  refers to a
  formula $(F\lto G)$ merely as an \emph{implication,} without
  specifying how the sub\-{}-formulas $F$ and $G$ are involved in the
  implication.  But we may read the formula as `$F$ implies $G$.'  The
verb \Eng{imply} is from the Latin for \Eng{fold in;} so the formula
$(F\lto G)$ suggests that $G$ is `folded into' $F$, so that, when one
`has' $F$, then one also has $G$.}
of $G$ by $F$, namely $(F\lto G)$.
\end{enumerate}
If a certain string of symbols is a propositional formula, this can be
shown by means of a 
\defn{tree}{} as in Figure~\ref{fig:tree}.
\begin{figure}[t]
\begin{center}
\pstree[levelsep=30pt,treemode=U]
{\Treebox{$((P\lto Q)\lto(\lnot Q\lto\lnot P))$}}
         {\pstree{\Treebox{$(P\lto Q)$}}
                 {\Treebox{$P$}\Treebox{$Q$}}
          \pstree{\Treebox{$(\lnot Q\lto\lnot P)$}}
                 {\pstree{\Treebox{$\lnot Q$}}
                         {\Treebox{$Q$}}
                  \pstree{\Treebox{$\lnot P$}}{\Treebox{$P$}}}
         }
\end{center}
\caption{A propositional formula with its tree}\label{fig:tree}
\end{figure}
The \defn{node}s
of the tree are the
\defn{subformula}s%
\dindexsub{formula}{sub---}
of the formula; these are just the formulas that are created during
the construction of the formula.
\end{parag}

\begin{parag}\label{par:prop-abbrevs}
For our convenience, we may write a formula in abbreviated form:
\begin{asparaenum}
  \item
a formula $(F\lto G)$ can be written as $F\lto G$, with
outer brackets removed;
\item
a subformula
$(\dotsb\lto(H\lto(G\lto F))\dotsb)$ can be written as 
\begin{equation*}
(\dotsb\lto H\lto G\lto F), 
\end{equation*}
so that repeated signs of implication are applied
from the right.  Note then that $P\lto P'$ is \emph{not} a subformula
of $P\lto P'\lto P''$.
\end{asparaenum}
We may also use $Q$ and $R$ as variables, instead of $P'$, $P''$, \dots
\end{parag}

\begin{parag}\label{par:pur}
  It is an important fact that a given propositional formula can be
  constructed in only one way.  Obviously a propositional variable by
  itself is neither a negation nor an implication; and a negation is
  not an implication.  Obviously a particular negation takes the form
  $\lnot F$ for some \emph{unique} subformula $F$.  Not so obviously,
  a particular implication takes the form $(F\lto G)$ for some
  \emph{unique} subformulas $F$ and $G$.
\end{parag}

\begin{parag}\label{par:prop-rules}
A
\defnplain{truth-assignment}{}%
\dindexsub{truth}{---{}-assignment}%
\index{assignment|see{truth-assignment}}
is an assignment of a
\defnplain{truth-value}{}%
\dindexsub{truth}{---{}-value}%
\index{value|see{truth-value}}%
---true or false---to each propositional variable.
Under a truth-assignment, by~\parref{par:pur}, a formula takes on a
unique truth-value according to the following rules:
  \begin{enumerate}
    \item
If a formula is a variable, then it takes the truth\-{}-value assigned
to that variable. 
\item
The formula $\lnot F$ is false just in case $F$ is true.
\item
The formula $F\lto G$ is false just in case $F$ is true and $G$ is
false. 
  \end{enumerate}
The truth-values of a formula under all
possible truth-assignments can be given in a
\defnplain{truth-table}.%
\dindexsub{truth}{---{}-table}%
\dindex{table, truth-}
Usually when one computes a truth-table for a formula, one includes
the possible truth-values of the subformulas.  A convenient way to do
this is in a table where
\begin{asparaenum}
\item 
false appears
as $0$, and true as $1$;%
\footnote{Some writers, as
  Stoll~\cite[Ch.~4, Exercise 3.7]{MR83e:04002}, use $0$ and $1$ in
  the opposite sense.} 
\item
the value of a variable is written beneath it;
\item
the value of $\lnot F$ is written beneath the $\lnot$;
\item
the value of $F\lto G$ is written beneath the $\lto$.
\end{asparaenum}
These rules are expressed by three tables:
\begin{align*}
&
  \begin{array}{c}
    P\\\hline 0\\1
  \end{array}&
&  \begin{array}{cc}
\lnot & F\\\hline
1&0\\
0&1
\end{array}&
&
\begin{array}{ccc}
  F&\lto&G\\\hline
0&1&0\\
1&0&0\\
0&1&1\\
1&1&1
\end{array}
\end{align*}
A full example is worked out in stages in
  Figure~\ref{figure:truth}.  The formula in that table happens to be a
  \defnplain{(propositional) tautology}:%
\dindexsub{proposition}{---al tautology}%
\dindexsub{tautology}{propositional ---}
that is, it is true under every
  truth\-{}-assignment. 
  \begin{figure}[t]
      \begin{equation*}
    \begin{array}{ccccccccc}
P&\lto&\lnot&Q&\lto&\lnot&(P&\lto&Q)\\\hline
0&   &      &0&   &     & 0&   &0\\      
1&   &      &0&   &     & 1&   &0\\      
0&   &      &1&   &     & 0&   &1\\      
1&   &      &1&   &     & 1&   &1\\\hline
0&   &     1&0&   &     & 0&  1&0\\      
1&   &     1&0&   &     & 1&  0&0\\      
0&   &     0&1&   &     & 0&  1&1\\      
1&   &     0&1&   &     & 1&  1&1\\\hline 
0&   &     1&0&   &    0& 0&  1&0\\      
1&   &     1&0&   &    1& 1&  0&0\\      
0&   &     0&1&   &    0& 0&  1&1\\      
1&   &     0&1&   &    1& 1&  1&1\\\hline
0&   &     1&0&  0&    0& 0&  1&0\\      
1&   &     1&0&  1&    1& 1&  0&0\\      
0&   &     0&1&  1&    0& 0&  1&1\\      
1&   &     0&1&  1&    1& 1&  1&1\\\hline
0&  1&     1&0&  0&    0& 0&  1&0\\      
1&  1&     1&0&  1&    1& 1&  0&0\\      
0&  1&     0&1&  1&    0& 0&  1&1\\      
1&  1&     0&1&  1&    1& 1&  1&1      
    \end{array}
  \end{equation*}
\caption{The filling-out of a truth-table}\label{figure:truth}
  \end{figure}
\end{parag}

\begin{parag}\label{par:prop-more}
It is convenient to use the following abbreviations for certain
formulas:
\begin{enumerate}
\item 
$F\lor G$ stands for $\lnot F\lto G$;
\item
$F\land G$ stands for $\lnot(F\lto \lnot G)$;
\item
$F\iff G$ stands for $\lnot((F\lto G)\lto\lnot(G\lto F))$;
\item
$0$ stands for $\lnot(P\lto P)$;
\item
$1$ stands for $P\lto P$.
\end{enumerate}
In writing with these abbreviations, we may follow the convention
whereby $\land$ and $\lor$ are
applied before $\lto$ and $\iff$, so that, for example, $F\lto G\land
G\lto F$ stands
for $F\lto(G\land G)\lto F$ (which stands for $F\lto((G\land G)\lto
F)$ by~\parref{par:prop-abbrevs}).
\end{parag}

\begin{parag}
  Two propositional formulas $F$ and $G$ are 
\defnplain{equivalent}{}%
\dindex{equivalent formulas}
if,
  under every truth\-{}-assignment, $F$ and $G$ take the same truth\-{}-value;
  in this case, we may write
  \begin{equation*}
    F\sim G.
  \end{equation*}
Hence for example
%\begin{gather*}
  \begin{align*}
   \lnot\lnot F&\sim F,& F\lor G&\sim\lnot(\lnot F\land\lnot G),&
    F\land G&\sim\lnot(\lnot F\lor\lnot G),
  \end{align*}
\begin{align*}
F\iff G&\sim(F\lto G)\land(G\lto F),& F\iff G&\sim(F\land G)\lor(\lnot
F\land\lnot G). 
\end{align*}
%\end{gather*}
Two formulas $F$ and $G$ are equivalent if 
and only if the formula $F\iff G$ is a tautology.
\end{parag}

\begin{parag}
  Every truth-table is the truth-table of some formula, in the
  following sense.  Suppose we are asked to find a formula $F$ whose
  variables are $P$, $Q$, and $R$ only and which takes on values as in
  the following table.
  \begin{equation*}
    \begin{array}{cccc}
      P&Q&R&F\\\hline
0&0&0&0\\
1&0&0&1\\
0&1&0&1\\
1&1&0&0\\
0&0&1&1\\
1&0&1&0\\
0&1&1&0\\
1&1&1&0
    \end{array}
  \end{equation*}
Then we can let $F$ be $(\lnot R\lto G)\land(R\lto H)$, where $G$ and
$H$ are as in the following table derived from the one above.
\begin{equation*}
  \begin{array}{cccc}
    P&Q&G&H\\\hline
0&0&0&1\\
1&0&1&0\\
0&1&1&0\\
1&1&0&0
  \end{array}
\end{equation*}
That is, $G$ takes the value of $F$ when $R$ is $0$, and $H$ takes the
value of $F$ when $R$ is $1$.  Analyzing $G$ and $H$ as we analyzed
$F$, we get
\begin{align*}
  G&\sim(\lnot Q\lto P)\land(Q\lto\lnot P),&
  H&\sim(\lnot Q\lto \lnot P)\land(Q\lto0),
\end{align*}
so that
\begin{equation*}
  F\sim(\lnot R\lto(\lnot Q\lto P)\land(Q\lto\lnot P))\land
(R\lto(\lnot Q\lto \lnot P)\land(Q\lto0)).
\end{equation*}
Alternatively, looking at where $F$ is $1$ in the original table, we
obtain the \defn{disjunctive normal form}:
\begin{equation*}
  F\sim(P\land\lnot Q\land\lnot R)\lor(\lnot P\land Q\land\lnot R)\lor
(\lnot P\land\lnot Q\land R).
\end{equation*}
\end{parag}

\begin{parag}
  When we say that $F$ is a propositional formula, we do not mean that
  the \emph{letter} $F$ is
  itself a formula.  The letter itself merely \emph{stands for} a
  formula; the letter is thus a kind of variable.  It is not a
  variable of our propositional logic; it is a variable of the
  language that we are using to talk \emph{about} the
  logic.  If one wants to give it a name, such a variable can be
  called a \defn{syntactical variable}{}%
\dindexsub{variable}{syntactical ---}
  \cite[\S08]{MR18:631a}.  I
  shall not worry further about identifying syntactical variables as
  such. 
\end{parag}

\section{Predicate logic}\label{sect:1}

\begin{parag}\label{par:1}
Our logic for talking about sets will be a so\-{}-called
  \defnplain{predicate logic}.%
\dindexsub{logic}{predicate ---}  
Its alphabet will comprise 
\begin{enumerate}
\item
the \defnplain{binary predicate}{}%
\dindexsub{binary}{--- predicate}%
\dindexsub{predicate}{binary ---}
$\in$,
the sign of \defn{membership}{} in a set;\footnote{The $\in$ can be
  understood as a form of the Greek letter \Gk{e} (epsilon), standing
  for \Gk{>est'i} \emph{is.}}
  \item
\defnplain{(individual) variable}{s}%
\dindexsub{individual}{--- variable}%
\dindexsub{variable}{individual ---} 
$x$, $x'$, $x''$, \dots;
\item
\defnplain{(individual) constant}{s}%
\dindexsub{individual}{--- constant}%
\dindexsub{constant}{individual ---} $a$, $a'$, $a''$, \dots;
\item
the propositional connectives $\lnot$ and $\lto$;
\item
the 
\defn{existential quantifier},%
\dindex{existential quantifier}%
\dindexsub{quantifier}{existential ---} 
$\exists$; 
\item
the brackets $($ and $)$.
\end{enumerate}
We may use $x$, $y$, $z$, \dots to stand for variables; and $a$, $b$, $c$,
\dots, for constants.
A \defn{term}{} in our logic is an individual variable or constant.
In talking about terms, we may symbolize them with letters like $t$
and $s$.
An \defn{atomic formula}{}\dindexsub{formula}{atomic ---} is an
expression of the form
\begin{equation*}
  s\in t
\end{equation*}
(where $s$ and $t$ are terms).  
  The \defn{formula}s in general are
defined recursively, as in \parref{par:prop}, but with an additional
possibility:
\begin{enumerate}
  \item
Atomic formulas are formulas;
\item
if $\phi$ is a formula, then so is $\lnot\phi$;
\item
if $\phi$ and $\psi$ are formulas, then so is $(\phi\lto\psi)$;
\item
if $\phi$ is a formula, and $x$ is a variable, then $\Exists x\phi$ is
a formula.
\end{enumerate}
We may use additional propositional connectives, as in
\parref{par:prop-more}. 
\end{parag}

\begin{parag}
The \defn{subformula}{s}%
\dindexsub{formula}{sub---}
of a formula $\phi$ are the formulas obtained during the construction
of $\phi$.  There is a recursive definition:
\begin{enumerate}
  \item
Every formula is a subformula of itself;
\item
$\phi$ is a subformula of $\lnot\phi$ and of $\Exists x\phi$;
\item
$\phi$ and $\psi$ are subformulas of $(\phi\lto\psi)$;
\item
if $\phi$ is a subformula of $\psi$, and $\psi$ is a subformula of
$\chi$, then $\phi$ is a subformula of $\chi$.
\end{enumerate}
\end{parag}

\begin{parag}
  If $\phi$ is a formula and $x$ is a variable, then $x$ may or may
  not occur in $\phi$.  Indeed, $x$ occurs once in $x\in y$ (if $y$ is
  not $x$), twice in $x\in x$, and three times in $\Exists xx\in x$.
  If $\phi$ is a subformula of $\psi$, then every occurrence of $x$ in
  $\phi$ is also an occurrence in $\psi$.  Some occurrences of $x$ are
  \defnplain{free},%
\dindex{free variable}%
\dindexsub{variable}{free ---}
 while all others are 
\defnplain{bound}.%
\defnplain{bound variable}{}%
\dindexsub{variable}{bound ---}%
\dindexsub{bound}{--- variable}
The definition of free occurrences is recursive:
  \begin{enumerate}
  \item
All occurrences of $x$ in an atomic formula are free.
\item
The free occurrences of $x$ in $\lnot\phi$ are just those in $\phi$.
\item
The free occurrences of $x$ in $\phi\lto\psi$ are just those in $\phi$
and $\psi$.
\item
If $y$ is a variable different from $x$, then the free occurrences of
$x$ in $\Exists y\phi$ are just those in $\phi$.
\item
There are no free occurrences of $x$ in $\Exists x\phi$.
  \end{enumerate}
If some occurrence of a variable in a formula is free, then that
variable is one of the
\defn{free variable}{s}
of the formula.  
So $x$ is a free variable of $(\Exists xx\in
x)\lto x\in y$, although, in this formula, $x$ has three bound
occurrences, but only one free occurrence (if $y$ is not $x$).
If $t$ is a term, then let the expression
  \begin{equation*}
    \subst tx{\phi}%(\phi)^x_t
  \end{equation*}
denote the result of replacing each \emph{free} occurrence of $x$ in
$\phi$ with $t$.  
For example, if $\phi$ is $x\in y\lto\Exists xx\in y$, then $\subst ax{\phi}$
is $a\in y\lto\Exists xx\in y$.
A formula with at most one free variable is
a \defn{singulary}{}%
\dindexsub{formula}{singulary ---}
 or \defnplain{unary}{}%
\dindexsub{unary}{--- formula}%
\dindexsub{formula}{unary ---}
    formula.%
\footnote{Following Quine, Church \cite[\S~02, p.~12,
    n.~29]{MR18:631a} suggests 
  \Eng{singulary} as a more etymologically correct word than
  \Eng{unary.}  Indeed, whereas the first five Latin cardinal numbers are
  \Lat{un-}, \Lat{du-}, \Lat{tri-}, \Lat{quattuor}, \Lat{quinque}, the
  first five Latin \emph{distributive} numbers---corresponding to the
  Turkish \Tur{birer}, \Tur{iki\c ser}, \Tur{\"u\c cer},
  \Tur{d\"order}, \Tur{be\c ser} \cite{LatinDili}---are \Lat{singul-},
  \Lat{bin-}, \Lat{tern-}, \Lat{quatern-}, \Lat{quin-.}  The
  latter sequence gives us \Eng{binary, ternary, quaternary,} and
    \Eng{quinary.}  So 
  \Eng{singulary} appears to be a better word than \Eng{unary.}  In
  fact, \Eng{singulary} does not appear in the original \emph{Oxford English
Dictionary} \cite{OED}.  The word \Eng{unary} \emph{does} appear in
this dictionary, but it is considered
obsolete: only one use of the word, from 1576, was discovered in
English literature.  There, \Eng{unary} meant \emph{unit}, although
the word \Eng{unit} was not actually invented until 1570, when it was
introduced by [John] Dee to correspond to the Greek \Gk{mon'ac,
    monad-} (for which see \parref{par:pick}).  The on-line
    \emph{OED} (2nd ed., 1989; \url{http://dictionary.oed.com},
    accessed January 26, 2009) does have
    \Eng{singulary,} for which Quine is quoted from 1940.}
If that variable is $x$, and the formula is
$\phi$, then we may write $\phi$ also as
\begin{equation*}
  \phi(x).
\end{equation*}
A formula with no free variables is a
\defn{sentence}.%
\dindexsub{formula}{sentence}
Sentences are also singulary formulas.  
\end{parag}

\begin{parag}\label{par:truth}
A sentence 
  with no constants is either true or false; a sentence \emph{with}
  constants \emph{becomes} true or
  false when those constants are 
\defnplain{interpreted}{}%
\dindex{interpretation of (individual) constants}%
\dindexsub{constant}{interpretation of (individual) ---s}
 as particular
  sets.  The rules are as follows.
  \begin{enumerate}
    \item
If $a$ and $b$ are constants, then the sentence
\begin{equation*}
  a\in b
\end{equation*}
is true whenever $a$ and $b$ denote sets such that
the set (denoted by) $b$ contains the set (denoted by) $a$.
\item
If $\sigma$ and $\tau$ are sentences, then the truth\-{}-values of
$\lnot\sigma$ and $(\sigma\lto\tau)$ follow from those of $\sigma$ and
$\tau$ according to the rules of propositional logic in
\parref{par:prop-rules}. 
\item\label{item:exists}
Suppose $\Exists x\phi$ is a sentence, and the constant $a$ does not
appear in~$\phi$.  Then $\Exists x\phi$ is true just in case the sentence
$\subst ax{\phi}$ is true under \emph{some} interpretation of the constant $a$
(as a set).
  \end{enumerate}
The qualification about $a$ in \eqref{item:exists} is needed to guard
against examples like the following.  If $\phi$ is $x\in a$, then
$\subst ax{\phi}$ is $a\in a$; this will ultimately be false, by the
Foundation Axiom (\parref{ax:foundation}); but $\Exists x\phi$
\emph{is} true, unless $a$ has no members (that are sets) at all.  If
$a$ does have a set as a member,
then we can call it $b$, so that $\subst bx{\phi}$ is true.
\end{parag}

\begin{parag}
  Suppose $\phi$ is $\Exists yx\in y$, where $y$ is not $x$.  Then
  $\phi$ is $\phi(x)$, that
  is, $\phi$ has no free variable other than $x$.  In fact, $x$ is
  free in $\phi$.  But $y$ is not free in $\subst yx{\phi}$, since this
  formula is the sentence $\Exists yy\in y$.  We may say then that $y$
  is not \defn{substitutable}{} for $x$ in $\phi$.  As usual, there is
  a recursive definition:
  \begin{enumerate}
  \item
In every atomic formula, $y$ is substitutable for $x$.
\item
If $y$ is substitutable for $x$ in $\phi$ and $\psi$, then it is so in
$\lnot\phi$ and $\phi\lto\psi$. 
\item
If $y$ is substitutable for $x$ in $\phi$, and $z$ is not $y$, then
$y$ is substitutable for $x$ in $\Exists z\phi$.
\item
If $x$ is not free in $\phi$, then $y$ is substitutable for $x$ in
$\Exists y\phi$. 
  \end{enumerate}
If $x$ is free in $\phi$, then $y$ is \emph{not} substitutable for $x$
in $\Exists y\phi$.  Trivially, a variable is always substitutable for
itself.  A constant is always substitutable for a variable. 
If $\phi$ is $\phi(x)$, and $t$ is substitutable for
$x$ in $\phi$, then
we can write $\subst tx{\phi}$ as
\begin{equation*}
  \phi(t).
\end{equation*}
If $\Exists y\psi$ is known to be a sentence, then $\psi$ is
$\psi(y)$, so
$\subst ay{\psi}$ can be written as $\psi(a)$.
\end{parag}



\begin{parag}
As implied in \parref{par:truth}, in our logic, variables and
constants refer
\emph{only} to sets.  This means that the only elements of sets that
we can talk about are other sets.  Indeed, we shall restrict our
attention to the sets whose \emph{only} members are other sets.  These
are the 
\defnplain{pure}{}%
\dindexsub{set}{pure ---} or the 
\defn{hereditary set}s%
\dindexsub{set}{hereditary ---}.  (So a set is hereditary if and only
if all its members are hereditary.)
We shall see that this is not a real limitation, mathematically
speaking. 
\end{parag}

\begin{parag}\label{par:forall}
We may write
  \begin{equation*}
    s\notin t
  \end{equation*}
instead of ${}\mathrel{\lnot} s\in t$.
  Also, we may write
  \begin{equation*}
    \Forall x\phi
  \end{equation*}
instead of $\lnot\Exists x\lnot\phi$; here $\forall$ is the
\defnplain{universal quantifier}.%
\dindexsub{universal}{--- quantifier}%
\dindexsub{quantifier}{universal ---} 
If $\Forall x\phi$ is a \emph{sentence,}
then it is true just in case $\phi(a)$ is true under \emph{every}
interpretation of $a$ (as a set), assuming that $a$ does not
already appear in $\phi$.  Hence, as an alternative to asserting
$\Forall x\phi$, we may assert $\phi(a)$ simply, when it is understood
that $a$ may be any set.  \emph{I shall often follow this convention.}
\end{parag}

\section{Classes}\label{sect:classes}

\begin{parag}\label{par:class=}
  If any constants in a singulary formula $\phi(x)$ have been
  interpreted, then the formula may be said
  to \defn{define}{} something, namely the  
  \defn{class}{} of those sets $a$ (assuming $a$ does not appear in
  $\phi$) such that $\phi(a)$ is true.  Then
  the class 
\defnplain{comprise}s%
\dindexsub{comprise}{class ---s its members} such $a$, and 
  the $a$ are 
\defn{member}s or
\defn{element}s of the class, and these
\defnplain{compose}{}%
\dindexsub{compose}{members --- a class}
the class.  The class defined by $\phi$ can be
  denoted by 
  \begin{equation*}
    \cls x{\phi(x)},
  \end{equation*}
or by $\cls y{\phi(y)}$ if $y$ is substitutable for $x$ in $\phi$.
Arbitrary classes can be denoted by boldface capital letters, as
$\cC$, $\cD$, and so on.\footnote{This
  convention is followed by Kunen~\cite[Ch.~1, \S9]{MR85e:03003}, for
  example, 
though not by Moschovakis~\cite[\textbf{3.19}]{MR95a:04001}; Levy
  \cite[I.4.1]{MR1924429} uses plainface capital letters for classes.
  The handwritten version of a
boldface letter is the letter with a wavy line underneath; so $\cC$
can be written by hand as
$\utilde{\mathsf{C}}$.}
Two classes are considered 
\defnplain{equal}{}%
\dindexsub{equal}{--- classes} or the \defn{same}{} if they
  have the same
  members, regardless of whether they are defined by the same formula.
In a word, classes are equal when they have the same
  \defn{extension}. 
  For example, $\cls x{x\in x\lto x\in x}$ is the same as
  the class $\cls x{x\in x\lto x\in x\lto x\in x}$, namely the class
  of all sets, which can be denoted by
  \begin{equation*}\label{eqn:universe}
    \universe;
  \end{equation*}
this is the
\defnplain{universal class}.%
\dindexsub{universal}{--- class}%
\dindexsub{class}{universal ---}
\end{parag}

\begin{parag}\label{par:mistake}
A class appears to be something like a set.  However, we are defining
classes \emph{in terms of} sets.  We should not expect all classes to
be sets.  At the beginning of the study of sets in the nineteenth century,
no possibility of a distinction between classes and sets was
recognized.  This led to problems, as in Theorem~\ref{thm:R} below.  Hence an
\emph{axiomatic} treatment of sets was pursued, as described in
\parref{par:axiomatic}, in an attempt to avoid the problems.  These
notes present set theory as a full\-{}-blown axiomatic
system;%
\footnote{William Wordsworth:
\begin{verse}
To me the meanest flower that blows can give\\	  
Thoughts that do often lie too deep for tears.
\end{verse}
---from the \emph{Ode: Intimations of Immortality from Recollections of Early
  Childhood.}}  
however, its development as such spans several decades of (almost) living
memory.  Euclid's \emph{Elements} is the classical presentation of an
axiomatic system, but it too is given to us full\-{}-blown; we do not
(to my knowledge) have
any earlier texts to tell us how the idea of erecting a mathematical
theory on axioms was discovered.
\end{parag}

\begin{parag}\label{par:*}
If we
let $\cC$ denote a class $\cls x{\phi(x)}$, and $t$ is substitutable
for $x$ in $\phi$, then, in any formula, we may use
  the expression
\begin{equation}\label{eqn:t-in-C}
  t\in\cC
\end{equation}
instead of $\phi(t)$.  If $t$ is a variable $y$ that is perhaps \emph{not}
substitutable for $x$ in $\phi$, we can still write $y\in\cC$; now
this means $\phi^*(y)$, where $\phi*$ is the result of replacing
\emph{each} occurrence of $y$ in $\phi$ with some variable $z$ that
does not appear in $\phi$.
\end{parag}

\begin{theorem}\label{thm:*}
  If $\phi^*$ is as in \parref{par:*}, then $y$ is substituble for $x$
  in $\phi^*$, and 
  \begin{equation}\label{eqn:*}
  \Forall x(\phi\iff\phi^*).
  \end{equation}
\end{theorem}

\begin{proof}
  We use induction on \emph{singulary} formulas, in the following sense.
  Suppose $\phi$ is atomic.  Since we
  assume that $\phi$ has no free variables other than $x$, $\phi$ must
  be $x\in a$ or $x\in x$ or $a\in x$ for some constant $a$.  Then
  $\phi^*$ is the same, so the claim~\eqref{eqn:*} is trivially true.
  Suppose the claim is true when $\phi$ is $\psi$ or $\chi$.  Since
  $(\lnot\psi)^*$ is $\lnot\psi^*$, and $(\psi\lto\chi)^*$ is
  $\psi^*\lto\chi^*$, the claim holds when $\phi$ is $\lnot\psi$ or
  $\psi\lto\chi$.  In the next step, we should like to suppose that
  the claim holds when $\phi$ is a formula $\rho$, and then prove it
  when $\phi$ is $\Exists u\rho$ for some variable $u$.  But then
  $\rho$ may have both $u$ and $x$ free.  So our inductive hypothesis
  should be that~\eqref{eqn:*} holds when $\phi$ is $\subst bu{\rho}$
  for some $b$.  Now say $\phi$ is $\Exists u\rho$.  We may assume
  that $u$ is not $z$.
If $u$ is not $x$ or
  $y$, then $\phi^*$ is $\Exists u\phi^*$, and
  \begin{align*}
    \phi^*(a)
&\text{\quad if and only if\quad}\subst ax{\Exists u\rho^*}\\
&\text{\quad if and only if\quad}\Exists u\subst ax{\rho^*}\\
&\text{\quad if and only if\quad}\subst bu{\subst ax{\rho^*}}\text{
      for some $b$}\\
&\text{\quad if and only if\quad}\subst ax{\subst bu{\rho^*}}\text{
      for some $b$}\\
&\text{\quad if and only if\quad}\subst ax{(\subst bu{\rho})^*}\text{
      for some $b$}\\
&\text{\quad if and only if\quad}\subst ax{\subst bu{\rho}}\text{
      for some $b$\quad[by inductive hypothesis]}\\
&\text{\quad if and only if\quad}\subst bu{\subst ax{\rho}}\text{
      for some $b$}\\
&\text{\quad if and only if\quad}\Exists u{\subst ax{\rho}}\\
&\text{\quad if and only if\quad}\subst ax{\Exists u\rho}\\
&\text{\quad if and only if\quad}\phi(a),
  \end{align*}
and the claim holds.  If $u$ is $y$, then $\phi^*$ is $\Exists
z\rho^*$, and the argument is as before, with $z$ for $u$.  If $u$ is
$x$, then $\phi$ is already a sentence, so the inductive hypothesis
has settled the matter.  This completes the induction.
\end{proof}

\begin{parag}
We can now use the sentence
\begin{equation}\label{eqn:C=D}
  \cC=\cD
\end{equation}
to stand for the sentence
$\Forall x(x\in\cC\iff x\in \cD)$.  For
${}\mathrel{\lnot}\cC=\cD$, we can write
\begin{equation*}
  \cC\neq\cD.
\end{equation*}
\end{parag}

\begin{parag}\label{par:subclass}
  A class $\cC$ is a \defn{subclass}{}%
\dindexsub{class}{sub---} of a class $\cD$ if
  $\cD$ contains all members of $\cC$.  In that case, we may
  say also that $\cD$ \defnplain{includes}{}%
\dindex{includes, included in} $\cC$, or that $\cC$ is
  \defnplain{included in}{}
$\cD$; also, we write 
  \begin{equation*}
    \cC\included\cD;
  \end{equation*}
so this is an abbreviation for $\Forall x(x\in\cC\lto x\in\cD)$.
The definition of equality of classes (\parref{par:class=}) is
now expressed by the sentence
\begin{equation}\label{eqn:ext-classes}
  \cC=\cD\iff\cC\included\cD\land\cD\included\cC.
\end{equation}
Instead of ${}\mathrel{\lnot}\cC\included\cD$, we may write 
\begin{equation*}
  \cC\not\included\cD.
\end{equation*}
The class $\cC$ is a 
\defnplain{proper subclass}{}%
\dindexsub{proper}{--- subclass}%
\dindexsub{subclass}{proper ---}%
\dindexsub{class}{proper sub---} of $\cD$ if
$\cC\included\cD$, but $\cC\neq\cD$; in that case,
we write
\begin{equation*}
  \cC\pincluded\cD
\end{equation*}
and say that $\cD$ 
\defnplain{properly includes}{}%
\dindexsub{proper}{---ly include}
$\cC$, or that $\cC$ is 
\defnplain{properly included}{}
in $\cD$.  
Instead of ${}\mathrel{\lnot}\cC\pincluded\cD$, we may write 
\begin{equation*}
  \cC\npincluded\cD.
\end{equation*}
\end{parag}

\begin{parag}\label{par:class-ops}
  Several 
\defn{operation}s on classes correspond
  to logical operations on formulas:
  \begin{enumerate}
    \item
The 
\defnplain{complement}{}%
\dindex{complement of a class}%
\dindexsub{class}{complement of a ---}
 of $\cC$ is $\cls x{x\notin\cC}$,
denoted by
\begin{equation*}
  \cC\comp.
\end{equation*}
\item
The 
\defnplain{union}{}%
\dindexsub{union}{--- of two classes}%
\dindexsub{class}{union of two ---es}
 of $\cC$ and $\cD$ is $\cls x{x\in\cC\lor x\in\cD}$,
denoted by 
\begin{equation*}
  \cC\cup\cD.
\end{equation*}
\item
The 
\defnplain{intersection}{}%
\dindexsub{intersection}{--- of two classes}%
\dindexsub{class}{intersection of two ---}
of $\cC$ and $\cD$ is $\cls x{x\in\cC\land
  x\in\cD}$, denoted by 
\begin{equation*}
  \cC\cap\cD.
\end{equation*}
\item
The \defnplain{empty class}{}%
\dindexsub{empty}{class}
\dindexsub{class}{empty ---}
 is $\cls x{x\in x\land x\notin x}$, denoted by
\begin{equation*}
  \emptyset.
\end{equation*}
\item
The 
\defnplain{difference}{}%
\dindexsub{difference}{--- of two classes}%
\dindexsub{class}{difference of two ---es}
 of $\cC$ from $\cD$ is $\cls
x{x\in\cC\land x\notin\cD}$ or $\cC\cap\cD\comp$,
denoted also by
\begin{equation*}
  \cC\setminus\cD.
\end{equation*}
\item
The \defnplain{symmetric difference}{}%
\dindexsub{symmetric}{--- difference of two classes}%
\dindexsub{difference}{symmetric --- of two classes}%
\dindexsub{class}{symmetric difference of two ---es}
 of $\cC$ and $\cD$ is $\cls
x{x\in\cC\iff x\notin\cD}$ or
\begin{equation*}
(\cC\setminus\cD)\cup(\cD\setminus\cC), 
\end{equation*}
denoted also by
\begin{equation*}
  \cC\symdiff\cD.
\end{equation*}
  \end{enumerate}
Hence for example (Exercise~\ref{ex:0VCD})
\begin{align}\label{eqn:0VCD}
\emptyset\comp&=\universe,&
  \cC=\cD\iff\cC\symdiff\cD&=\emptyset.
\end{align}
\end{parag}

\begin{parag}\label{par:infinitary}
None of the operations on classes defined in~\parref{par:class-ops}
made any real use of \emph{membership} of sets.
Here are two that do.
  \begin{enumerate}
    \item
The \defn{union}{} of a single class $\cC$ is the class $\cls
x{\Exists y(y\in\cC\land x\in y)}$ of elements of the elements of
$\cC$; it is denoted by
\begin{equation*}
  \bigcup{\cC}.
\end{equation*}
\item
The \defnplain{intersection}{}%
\dindexsub{intersection}{--- of one class}%
\dindexsub{class}{intersection of one ---} of a class $\cC$ is the class
$\cls x{\Forall y(y\in\cC\lto x\in y)}$ of elements common to the
elements of $\cC$; it is denoted by
\begin{equation*}
  \bigcap\cC.
\end{equation*}
  \end{enumerate}
\end{parag}

\section{Relations}\label{sect:relations}

\begin{parag}\label{par:=inc}
  \techplain{Equality}{}%
\tindexsub{equal}{---ity of classes}%
\tindexsub{class}{equality of ---es}
and 
\techplain{inclusion}{}%
\tindexsub{inclusion}{--- of classes}%
\tindexsub{class}{inclusion of ---es}
 of classes are examples of 
\defn{relation}{s.}  In particular,
  equality is a relation that is
  \begin{enumerate}
    \item
\defnplain{reflexive},%
\index{reflexive|see{relation}}%
\dindexsub{relation}{reflexive ---} because $\cC=\cC$ for all classes
$\cC$;
\item
\defn{symmetric},%
\dindexsub{relation}{symmetric ---} because $\cC=\cD\iff\cD=\cC$ for
all $\cC$ and $\cD$;
\item
\defnplain{transitive},%
\index{transitive|seealso{relation}}%
\dindexsub{relation}{transitive ---} because
$\cC=\cD\land\cD=\cE\lto\cC=\cE$ for all $\cC$, $\cD$, and $\cE$.
  \end{enumerate}
Therefore equality is the prototypical example of an
\defnplain{equivalence-relation}.%
\dindexsub{equivalence}{---{}-relation}%
\dindexsub{relation}{equivalence-{}---}
Inclusion, like equality, is
reflexive and transitive; but it is also 
\defnplain{anti-symmetric},%
\index{anti-symmetric|see{relation}}%
\dindexsub{relation}{anti-symmetric ---} because
\begin{equation*}
  \cC\included\cD\land\cD\included\cC\lto\cC=\cD.
\end{equation*}
%Note the use of equality in this definition.
Because it is reflexive, anti-symmetric, and transitive, inclusion is
called an \defn{order}{ing.}%
\dindexsub{relation}{ordering}
\emph{Proper} inclusion is symmetric and transitive, but also
\defnplain{irreflexive},%
\index{irreflexive|see{relation}}%
\dindexsub{relation}{irreflexive ---}
because always
\begin{equation*}
  \cC\npincluded\cC.
\end{equation*}
Still, proper inclusion is refered to as an ordering: it is a
\defn{strict}{}%
\dindexsub{order}{strict ---ing}%
\dindexsub{relation}{strict ordering}%
\dindexsub{strict}{--- ordering} ordering.
\end{parag}

\begin{parag}
  A formula with at most two free variables is a 
\defnplain{binary formula}.%
\dindexsub{formula}{binary ---}%
\dindexsub{binary}{--- formula} 
Suppose $\phi$ is such, with its free
  variables among $x$ and $y$.  Then we can write $\phi$
  as
  \begin{equation*}
    \phi(x,y).
  \end{equation*}
In this case, by writing
\begin{equation*}
  \subst{s\;t}{x\;y}{\phi},
\end{equation*}
we mean  the result of \emph{simultaneously} replacing free occurrences
of $x$ with $s$, and $y$ with $t$.  So it is the same formula as
$\subst{t\;s}{y\;x}{\phi}$, but it need not be the same as
$\subst ty{\subst sx{\phi}}$.  For example, if $\phi$ is $x\in y$, then 
$\subst{y\;x}{x\;y}{\phi}$ is $y\in x$, while $\subst xy{\subst
  yx{\phi}}$ is $x\in x$. 
If $s$ is substitutable for $x$, and $t$ for $y$, in $\phi$, then we
can write $\subst{s\;t}{x\;y}{\phi}$ as
\begin{equation*}
  \phi(s,t).
\end{equation*}
\end{parag}

\begin{parag}\label{par:rel-sets}
If any constants in a binary formula $\phi(x,y)$ have been
interpreted, then the formula may be said to \defn{define}{}
something, namely a
\defnplain{binary relation}{}%
\dindexsub{binary}{--- relation}%
\dindexsub{relation}{binary ---}
between sets.  If this relation is called $\cR$, then in any formula,
we may use the expression
\begin{equation*}
  s\mathrel{\cR}t
\end{equation*}
instead of $\phi(s,t)$, and $\lnot(s\mathrel{\cR}t)$ for
$\lnot\phi(s,t)$.  (We may have to replace $\phi$ with $\phi^*$ as
in~\parref{par:*}.) 
The 
\defnplain{domain}{}%
\dindexsub{domain}{of a relation}
 of $\cR$
is the class
  $\cls x{\Exists yx\mathrel{\cR}y}$;
this can be denoted by
\begin{equation*}
  \dom{\cR}.
\end{equation*}
The \defn{range}{} of $\cR$ is the class
  $\cls y{\Exists xx\mathrel{\cR}y}$;
this can be denoted by
\begin{equation*}
  \ran{\cR}.
\end{equation*}
If $\dom{\cR}\included\cC$, and $\ran{\cR}\included\cD$, then $\cR$
can be called a relation 
\defnplain{from}{}%
\dindexsub{from}{relation --- a class to another}%
\dindexsub{relation}{--- from a class to another}
 $\cC$ 
\defnplain{to}{}%
\dindexsub{to}{relation from a class --- another}
 $\cD$.
If also $\cC=\cD$, then $\cR$ is a relation 
\defnplain{on}{}%
\dindexsub{on}{relation --- a class}
 $\cC$.
\end{parag}

\begin{parag}\label{par:refl-irrefl}
  A relation $\cR$ is 
\defnplain{reflexive}{}%
\dindexsub{relation}{reflexive ---}
if $\Forall xx\mathrel{\cR}x$; 
\defnplain{irreflexive},%
\dindexsub{relation}{irreflexive ---} if
 $\Forall x\lnot(x\mathrel{\cR}x)$.  The relation defined by a formula
$(\phi(x)\iff\phi(y))\lor\psi(x,y)$, for example, is reflexive; by
$\lnot(\phi(x)\iff\phi(y))\land\psi(x,y)$, irreflexive. 
\end{parag}

\begin{parag}\label{par:symm}
A binary relation $\cR$ has a 
\defn{converse},
which can be denoted by
\begin{equation*}
  \conv{\cR};
\end{equation*}
it is defined by the same formula as $\cR$, but with the two free
variables considered in the other order.  That is, if $\phi(x,y)$
defines $\cR$, then $\phi(y,x)$ defines $\conv{\cR}$.
We have
\begin{gather}\label{eqn:dom-ran}
  \dom{\conv{\cR}}=\ran{\cR},\\\label{eqn:ran-dom}
  \ran{\conv{\cR}}=\dom{\cR},\\\label{eqn:conv-conv}
\conv{\conv{\cR}}=\cR
\end{gather}
(Exercise~\ref{ex:conv}).
If $\conv{\cR}=\cR$, then $\cR$ is called
\defn{symmetric}{}%
\dindexsub{relation}{symmetric ---}
(compare~\parref{par:=inc}).  So, the relation defined by $\phi(x,y)$
is symmetric if and only if
\begin{equation*}
  \Forall x\Forall y(\phi(x,y)\iff\phi(y,x)).
\end{equation*}
%More generally, $\cR$ is \defnplain{symmetric on}{}%
%\dindexsub{symmetric}{--- on}%
%\dindexsub{on}{symmetric ---}
%$\cC$ if $\cR\cap(\cC\times\cC)$ is
%symmetric. 
\end{parag}

\begin{parag}\label{par:composite}
If $\cR$ and  $\cS$ are both binary relations, then the
relation defined by 
\begin{equation*}
 \Exists z(x\mathrel{\cR}z\land z\mathrel{\cS}y)
\end{equation*}
is the 
\defnplain{composite}{} %
\dindexsub{composite}{--- of two relations}
of $\cR$ and
$\cS$, denoted by%
\footnote{Tarski~\cite[\S~28, p.~92]{Tarski-Intro} uses
the notation $\cR/\cS$ and refers to the indicated class as
the \tech{relative product}{}%
\tindexsub{product}{relative ---}
 of $\cR$ and $\cS$. 
Suppes~\cite[\S~3.1, Definition 7, p.~63]{MR0349389} also uses the
notation.}  
\begin{equation*}
  \cR/\cS.
\end{equation*}
Logically, if $\cT$ is also a relation, then
\begin{equation}\label{eqn:comp-assoc}
  (\cR/\cS)/\cT=\cR/(\cS/\cT).
\end{equation}
If $\cR/\cR\included\cR$, then $\cR$ is called
\defnplain{transitive}{}%
\dindexsub{relation}{transitive ---}
(again, compare~\parref{par:=inc}).
%this means
%\begin{equation}\label{eqn:transitive}
%  a\mathrel{\cR}b\land b\mathrel{\cR}c\lto a\mathrel{\cR}c.
%\end{equation}
In this case, instead of
$a\mathrel{\cR}b\land b\mathrel{\cR}c$, we may write
\begin{equation*}
  a\mathrel{\cR}b\mathrel{\cR}c.
\end{equation*}
The relation defined by a formula $\phi(x)\lto\phi(y)$ is transitive;
so is the relation defined by $\Forall z(\psi(x,z)\lto\psi(y,z))$, and
so on.  As in~\parref{par:=inc}, a reflexive, symmetric, transitive
relation is an 
\defnplain{equivalence-relation}.%
\dindexsub{equivalence}{---{}-relation}
\dindexsub{relation}{equivalence-{}---}
\end{parag}

\section{Proof}

\begin{parag}\label{par:ta}
  A \defnplain{truth-assignment}{}%
\dindexsub{truth}{---{}-assignment}
in predicate logic is an assignment of truth-values to the atomic
sentences.  Let $\str A$ be such an assignment.  This determines a
truth-value $\sigma^{\str A}$ for each sentence $\sigma$ by the
following rules.
\begin{enumerate}
\item 
If $\sigma$ is atomic, then $\sigma^{\str A}$ is whatever value is
assigned to $\sigma$ by $\str A$.
\item
$(\lnot\sigma)^{\str A}$ is true if and only if $\sigma^{\str A}$ is
  false.
\item
$(\sigma\lto\tau)^{\str A}$ is true if and only if $\sigma^{\str A}$ is
  false or $\tau^{\str A}$ is true.
\item
$(\Exists x\phi)^{\str A}$ is true if and only if $\phi(a)^{\str A}$
  is true for some constant $a$.
\end{enumerate}
If $\sigma^{\str A}$ is true, we may write also
\begin{equation*}
  \str A\models\sigma,
\end{equation*}
saying $\sigma$ is 
\defnplain{true in}{}%
\dindexsub{true}{sentence --- in a truth-assignment} 
$\str A$;
otherwise,
\begin{equation*}
  \str A\nmodels\sigma,
\end{equation*}
and $\sigma$ is
\defnplain{false in}{}%
\dindex{false: sentence --- in a truth-assignment}
$\str A$.  A sentence that is true in every truth-assignment is
\defnplain{logically true}.%
\dindexsub{logic}{---ally true}%
\dindexsub{true}{logically ---}
The proof of Theorem~\ref{thm:*} is really a proof that~\eqref{eqn:*}
is logically true. 
\end{parag}

\begin{parag}
  Suppose $F$ is a tautology of propositional logic, and for each
  variable $P$ appearing in $F$, for some formula $\phi$, we
  replace each instance of $P$ in $F$ with $\phi$.  The resulting
  formula is a 
\defn{tautology}{}%
\dindexsub{formula}{tautology}
of predicate logic.  
If, to an arbitrary formula, we prefix $\forall x$ for some variables $x$,
among which are all of the free variables of the 
formula, then the result, a sentence, is called a  
\defn{generalization}{}%
\dindexsub{formula}{generalization of a ---} 
of the original formula.  Generalizations of tautologies are logically true.
\end{parag}

\begin{parag}
  There follows a list of several kinds of formulas whose
  generalizations are logically true sentences;
  we shall refer to all of these generalizations as 
\defnplain{logical axiom}{s:}%
\dindexsub{logic}{---al axiom}%
\dindexsub{axiom}{logical ---}
\begin{enumerate}
\item\label{item:ax-taut}
tautologies;
\item\label{item:ax-A}
$\theta\lto\Forall x\theta$, where $x$ is not free in $\theta$;
\item\label{item:ax-AAA}
$\Forall x(\phi\lto\psi)\lto\Forall x\phi\lto\Forall x\psi$;
\item\label{item:ax-E}
%$\Forall x\phi\lto\subst tx{\phi}$, 
$\subst tx{\phi}\lto\Exists x\phi$,
where $t$ is a term that is
  substitutable for $x$ in $\phi$.
\end{enumerate}
\end{parag}

\begin{parag}
The 
\defnplain{logical theorem}{s}%
\dindexsub{logic}{---al theorem}%
\dindexsub{theorem}{logical ---}
are obtained from the logical axioms by repeated application of a
\defn{rule of inference},%
\dindex{inference, rule of}
called \defnplain{\emph{Modus Ponens}}{}%
\dindex{modus ponens@\emph{Modus Ponens}}
in Latin and 
\defn{Detachment}{} in English.  This means:
\begin{enumerate}
  \item
the logical axioms are logical theorems;
\item
if $\sigma$ and $\sigma\lto\tau$ are logical theorems, then so is
$\tau$. 
\end{enumerate}
More generally, if $\Gamma$ is a list%
\footnote{I use the word \emph{list} here mainly to avoid using
  \emph{set,} although the word \emph{list} should suggest an
  ordering.}%
\index{list}
of sentences, then sentences are 
\defn{deducible}{} from $\Gamma$ by the following rules:
\begin{enumerate}
\item
logical axioms, and sentences in $\Gamma$, are deducible from
$\Gamma$;
\item
if $\sigma$ and $\sigma\lto\tau$ are deducible from $\Gamma$, then so
is $\tau$.
\end{enumerate}
If $\sigma$ is deducible from $\Gamma$, we may say also that $\Gamma$
\defn{entail}s $\sigma$, writing
\begin{equation*}
  \Gamma\proves\sigma.
\end{equation*}
As a special case, if $\sigma$ is a logical theorem, we may write
\begin{equation*}
  \proves\sigma.
\end{equation*}
The logical axioms, together with the rule of inference, constitute a
\defnplain{proof system}.%
\dindexsub{proof}{--- system}%
\dindexsub{system}{proof system}
In general, if $\Gamma\proves\sigma$, this can be shown with a
\defn{formal proof},%
\dindexsub{proof}{formal ---}
namely a \emph{finite}%
\footnote{A definition of finite sets will come later,
  in~\parref{par:inductive}; meanwhile, we must rely on our
  intuition.}%
\index{finite}  
list of sentences, ending with $\sigma$, such that each
sentence $\tau$ on the list is a logical axiom, is in $\Gamma$, or is
preceded in the list by sentences $\theta$ and $\theta\lto\tau$.  In
general, capital Greek letters will stand for lists of sentences (or
formulas); then an expression like
\begin{equation*}
  \Gamma,\sigma
\end{equation*}
will stand for the list obtained from the list $\Gamma$ by adding
$\sigma$.
\end{parag}

\section{Completeness}

\begin{theorem}[Deduction]
  $\Gamma,\sigma\proves\tau$ if and only if
  \begin{equation*}
  \Gamma\proves\sigma\lto\tau.
  \end{equation*}
\end{theorem}

\begin{proof}
  A formal proof that $\Gamma\proves\sigma\lto\tau$ is also a formal
  proof that $\Gamma,\sigma\proves\sigma\lto\tau$.  If
  $\Delta,\sigma\lto\tau$ is 
  such a proof, then $\Delta,\sigma\lto\tau,\sigma,\tau$ is a formal proof
  that  $\Gamma,\sigma\proves\tau$.

Suppose conversely that $\Gamma,\sigma\proves\tau$.  Suppose $\Delta$
is a proof of this.  We convert $\Delta$ to a proof that
 $\Gamma\proves\sigma\lto\tau$ by replacing the sentences in $\Delta$
one by one.  Suppose $\theta$ is such a sentence.  
\begin{asparaenum}[1.]
  \item
If $\theta$ is a
logical axiom or a sentence of $\Gamma$, then we replace $\theta$ with
the list
\begin{equation*}
  \theta,\ \theta\lto\sigma\lto\theta,\ \sigma\lto\theta,
\end{equation*}
which is itself a formal proof that $\Gamma\proves\sigma\lto\theta$,
since $\theta\lto\sigma\lto\theta$ is a tautology.
\item
If $\theta$ is $\sigma$, we replace it with the logical axiom
$\sigma\lto\sigma$.
\item
If $\theta$ is preceded by sentences $\rho$ and $\rho\lto\theta$, then
we replace $\theta$ with the list
\begin{equation*}
  (\sigma\lto\rho\lto\theta)\lto(\sigma\lto\rho)\lto\sigma\lto\theta,\ 
                               (\sigma\lto\rho)\lto\sigma\lto\theta,\ 
                                                   \sigma\lto\theta,
\end{equation*}
which is a formal proof that
$\sigma\lto\rho,\sigma\lto\rho\lto\theta\proves\sigma\lto\theta$.
\end{asparaenum}

In the end, $\Delta$ is converted to a formal proof that
$\Gamma\proves\sigma\lto\tau$. 
\end{proof}

\begin{lemma}\label{lem:E-to-A}
  If $y$ is substitutable for $x$ in $\phi(x)$, then $\proves\Forall
  y(\Forall x\phi(x)\lto\phi(y))$.
\end{lemma}

\begin{proof}
Here is a formal proof:
  \begin{align*}
    &\Forall y\bigl((\lnot\phi(y)\lto\Exists x\lnot\phi(x))\lto
\Forall x\phi(x)\lto\phi(y)\bigr)&&\text{[Axiom~\eqref{item:ax-taut}]}\\
& \Forall y\bigl((\lnot\phi(y)\lto\Exists x\lnot\phi(x))\lto
\Forall x\phi(x)\lto\phi(y)\bigr)\lto&&\\
&\qquad\qquad\lto \Forall y(\lnot\phi(y)\lto\Exists x\lnot\phi(x))\lto
\Forall y\bigl(\Forall x\phi(x)\lto\phi(y)\bigr)
  &&\text{[Axiom~\eqref{item:ax-AAA}]}\\
&\Forall y(\lnot\phi(y)\lto\Exists x\lnot\phi(x))\lto
\Forall y\bigl(\Forall x\phi(x)\lto\phi(y)\bigr)&&\text{[Detachment]}\\
&\Forall y(\lnot\phi(y)\lto\Exists
x\lnot\phi(x))&&\text{[Axiom~\eqref{item:ax-E}]}\\ 
&\Forall y\bigl(\Forall x\phi(x)\lto\phi(y)\bigr);&&\text{[Detachment]}
  \end{align*}
thus $\proves\Forall y\bigl(\Forall x\phi(x)\lto\phi(y)\bigr)$.
\end{proof}

\begin{theorem}[Generalization]\label{thm:gen}
  If $\phi$ is a formula $\phi(x)$, which does not feature $a$, and
$\Gamma\proves\phi(a)$, where $\Gamma$ is a list of sentences
  that do not feature $a$, 
then 
  \begin{equation*}
  \Gamma\proves\Forall x\phi(x).
  \end{equation*}
\end{theorem}

\begin{proof}
  Under the given conditions, there is a formal proof $\Delta$ that
  $\Gamma\proves\phi(a)$.  Let $y$ be a variable not appearing in
  $\Delta$.  Each sentence in $\Delta$ can be understood as
  $\psi(a)$, where $\psi$ has no variable other than $y$ free (so
  $\psi$ is $\psi(y)$), and $a$ does not appear in $\psi$. 
  \begin{asparaenum}[1.]
    \item
If $\psi(a)$ is a logical axiom, then we replace it with $\Forall
y\psi$, which is also a logical axiom.
\item
If $\psi(a)$ is from $\Gamma$, then $\psi(a)$ must be $\psi$, a
sentence, and we replace it with 
\begin{equation*}
\psi,\ \psi\lto\Forall y\psi,\ \Forall y\psi,
\end{equation*}
which is a formal proof that $\Gamma\proves\Forall y\psi$.
\item
If $\psi(a)$ is preceded in $\Delta$ by $\chi(a)$ and
$\chi(a)\lto\rho(a)$, then we replace $\psi(a)$ in $\Delta$ with
\begin{equation*}
  \Forall y(\chi\lto\rho)\lto\Forall y\chi\lto\Forall y\rho,\   
                             \Forall y\chi\lto\Forall y\rho,\  
                                              \Forall y\rho,
\end{equation*}
which is a formal proof that $\Forall y\chi,\Forall
y(\chi\lto\rho)\proves\Forall y\rho$.
  \end{asparaenum}

In the end, $\Delta$ is converted to a formal proof $\Delta^*$ that
$\Gamma\proves\Forall y\phi(y)$.  Note that $x$ is substitutable for
$y$ in $\phi(y)$.
Now can we continue $\Delta^*$ with
\begin{multline*}
\Forall y\phi(y)\lto\Forall x\Forall y\phi(y),\ 
\Forall x\Forall y\phi(y),\ \dots,\ 
  \Forall x(\Forall y\phi(y)\lto\phi(x)),\\
  \Forall x(\Forall y\phi(y)\lto\phi(x))\lto\Forall x\Forall
  y\phi(y)\lto\Forall x\phi(x),\ \\
\Forall x\Forall
  y\phi(y)\lto\Forall x\phi(x),\ 
\Forall x\phi(x),
\end{multline*}
where the missing steps exist by Lemma~\ref{lem:E-to-A};
thus we get a proof that $\Gamma\proves\Forall x\phi(x)$.
\end{proof}

\begin{parag}
  Let us denote the negation of an arbitrary logically true sentence
  by
  \begin{equation*}
    \bot.
  \end{equation*}
A list of sentences is \defn{inconsistent}{} if it entails every
sentence; otherwise it is \defn{consistent}.
\end{parag}

\begin{lemma}\label{lem:inconsistent}
  Let $\Gamma$ be a list of sentences; and $\sigma$, a sentence.
  \begin{enumerate}
  \item 
$\Gamma$ is inconsistent if and only if $\Gamma\proves\bot$.
\item\label{item:fin-con}
$\Gamma$ is consistent if and only if each finite
sublist is consistent.
\item\label{item:nsinc}
$\Gamma,\lnot\sigma$ is inconsistent if and only if
  $\Gamma\proves\sigma$.
\item\label{item:sinc}
$\Gamma,\sigma$ is inconsistent if and only if
  $\Gamma\proves\lnot\sigma$.
\item\label{item:inc}
$\Gamma,\sigma$ is inconsistent if and only if
  $\Gamma,\sigma\proves\lnot\sigma$. 
\item\label{item:con-2}
If $\Gamma$ is consistent, then at least one of $\Gamma,\sigma$ and
$\Gamma,\lnot\sigma$ is consistent.
  \end{enumerate}
\end{lemma}

\begin{proof}
We prove only~\eqref{item:fin-con}; the rest is
Exercise~\ref{ex:inconsistent}.  If $\Gamma$ is inconsistent, then
there is a formal proof $\Delta$ that $\Gamma\proves\bot$.  Let
$\Gamma_0$ comprise the sentences from $\Gamma$ that actually appear
in $\Delta$.  Then $\Gamma_0$ is finite, and $\Gamma_0\proves\bot$.
\end{proof}

\begin{theorem}[Completeness]%
\dindex{Completeness Theorem}%
\dindexsub{theorem}{(Goedel's) Completeness Th---@(G\"odel's)
  Completeness Th---}%
\dindexsub{Goedel@G\"odel}{---'s Completeness Theorem}
  Every logically true sentence is a logical theorem.
\end{theorem}

\begin{proof}
  Suppose $\sigma$ is not a logical theorem.  
We  shall find a truth-assignment $\str A$ such that $\str
  A\models\lnot\sigma$, so that $\sigma$ is not logically true.

Arrange \emph{all} sentences in an (infinite) list $\Gamma$ that
starts with $\lnot\sigma$ (Exercise~\ref{ex:list}).  We recursively
construct a new list $\Delta_{\theta}$ for each sentence $\theta$ in
$\Gamma$.  To start with, $\Delta_{\lnot\sigma}$ is just
$\lnot\sigma$, which is a finite list that is
consistent by Lemma~\ref{lem:inconsistent}~\eqref{item:nsinc}.  
Suppose
$\Delta_{\theta}$ is finite and consistent, and $\theta$ is followed
by $\theta'$ in $\Gamma$.  We consider several cases.
\begin{asparaenum}[1.]
  \item
If $\Delta_{\theta},\theta'$ is inconsistent, then
$\Delta_{\theta},\lnot\theta'$ is consistent by
Lemma~\ref{lem:inconsistent}~\eqref{item:con-2}, so we let
$\Delta_{\theta'}$ be $\Delta_{\theta},\lnot\theta'$.
\item
If $\Delta_{\theta},\theta'$ is consistent, but $\theta'$ does
\emph{not} have the form $\Exists x\lnot\lnot\phi(x)$ for any formula
$\phi$ and variable $x$, then we let $\Delta_{\theta'}$ be
$\Delta_{\theta},\theta'$.
\item
Suppose $\Delta_{\theta},\theta'$ is consistent, and $\theta'$ has the
form $\Exists x\lnot\lnot\phi(x)$ for some formula $\phi$ and variable
$x$.  Then we let  $\Delta_{\theta'}$ be
$\Delta_{\theta},\theta',\phi(a)$, where $a$ is a constant not
appearing in $\Delta_{\theta}$ or $\phi$.  Note that this list is
consistent.  Indeed, suppose $\Lambda,\theta,\phi(a)$
is inconsistent.  By Lemma~\ref{lem:inconsistent}~\eqref{item:sinc},
we have
\begin{equation*}
  \Lambda,\theta\proves\lnot\phi(a).
\end{equation*}
By the Generalization Theorem,~\ref{thm:gen}, we have
  $\Lambda,\theta\proves\lnot\theta$.  By
Lemma~\ref{lem:inconsistent}~\eqref{item:inc}, we have that
$\Lambda,\theta$ is inconsistent.
\end{asparaenum}

Now let $\Delta$ consist of the sentences belonging to the
$\Delta_{\theta}$.  Then $\Delta$ is consistent by
Lemma~\ref{lem:inconsistent}~\eqref{item:fin-con} because every finite
sublist of $\Delta$ is a sublist of some $\Delta_{\theta}$.  By
construction then, for each sentence $\tau$, the list $\Delta$
features either $\tau$ or $\lnot\tau$, but not both.  Hence we can
define $\str A$ so that, for every atomic sentence $\alpha$, we have
$\str A\models\alpha$ if and only if $\alpha$ is in $\Delta$.  We now
show that $\str A\models\tau$ if and only if $\tau$ is in $\Delta$,
for all sentences $\tau$.
\begin{asparaenum}[1.]
  \item
The claim is true when $\tau$ is atomic, by definition of $\str A$.
\item
If the claim is true when $\tau$ is $\rho$, then it is true when
$\tau$ is $\lnot\rho$, since $\rho$ belongs to $\Delta$ if and only if
$\lnot\rho$ does not.
\item
If the claim is true when $\tau$ is $\rho$ or $\theta$, then it is
true when $\tau$ is $\rho\lto\theta$, since $\rho\lto\theta$ belongs
to $\Delta$ if and only if $\lnot\rho$ or $\theta$ does.
\item
Suppose for some $\phi(x)$ that the claim holds whenever $\rho$ is
$\phi(a)$ for some constant $a$.  If $\Exists x\phi(x)$ belongs to
$\Delta$, then by construction $\phi(a)$ is in $\Delta$ for some $a$,
so $\str A\models\phi(a)$ and therefore $\str A\models\Exists
x\phi(x)$.  Conversely, if  $\str A\models\Exists
x\phi(x)$, then $\str A\models\phi(a)$ for some $a$, so $\phi(a)$ is
in $\Gamma$ and therefore $\Gamma\proves\Exists x\phi(x)$, so $\Exists
x\phi(x)$ must be in $\Gamma$. 
\end{asparaenum}

As a special case, since $\lnot\sigma$ is in $\Delta$, we have
$\str A\models\lnot\sigma$.
\end{proof}

\begin{porism}\label{por}
  For every consistent list of sentences, there is a truth-assignment
  in which all of the sentences are true.
\end{porism}

\begin{proof}
  If $\Sigma$ is a consistent list of sentences, then, in the
  preceding proof, we can start the list $\Gamma$ with $\Sigma$
  instead of $\lnot\sigma$.  Maybe $\Sigma$ is infinite; but the
  finiteness of the lists $\Delta_{\theta}$ in the proof above serve
  only to ensure that there are constants not appearing in
  $\Delta_{\theta}$.  All we really need is that there is an infinite
  list $b$, $b'$, $b''$, \dots of constants not appearing in
  $\Sigma$. 
\end{proof}

\section{Set theory}

\begin{parag}
  A consequence of Porism~\ref{por} is 
\defnplain{Compactness}.%
\dindex{Compactness (Theorem)}%
\dindexsub{theorem}{Compactness Th---}  
A truth-assignment in which every sentence on a given list is true is
  a \defn{model}{} of the list.  Suppose every finite sublist of some
  list $\Gamma$ of sentences has a model.  Then those finite sublists
  are consistent, so $\Gamma$ is consistent by
  Lemma~\ref{lem:inconsistent}~\eqref{item:fin-con}, so it has a
  model.  The interest for us is that if $\sigma$ is true in every
  model of $\Gamma$, then $\Gamma\proves\sigma$, since if $\Gamma$
  does not entail $\sigma$, then $\Gamma,\lnot\sigma$ is consistent by
  Lemma~\ref{lem:inconsistent}~\eqref{item:nsinc}, so by the porism,
  $\Gamma$ has a model in which $\sigma$ is false.  Thus we can
  establish entailment without actually exhibiting a formal proof.
  This is what we shall do from now on.
\end{parag}

\begin{parag}\label{par:incomp}
  We now aim to identify a list of sentences that are \emph{true} in
  the sense of \parref{par:truth} and that entail all true sentences.
  We aim at this; but we must fail, by 
\techplain{G\"odel's Incompleteness Theorem}.%
\tindexsub{Goedel@G\"odel}{---'s Incompleteness Theorem}%
\tindexsub{theorem}{Goedel's Incompleteness Th---@G\"odel's
  Incompleteness Theorem}
  We settle for a list, called $\zfc$, of true sentences that entail
  everything about sets that is useful for mathematics.  The sentences
  of $\zfc$ are not logical axioms; they are not logically true;
  but they are \defn{axiom}s of
  \defn{set theory}.  We say that the sentences in $\zfc$ are true.
  If they \emph{are} true, then $\zfc$ is consistent.  However (also
  by G\"odel) there is no
  proof that $\zfc$ is consistent; we must accept our intuition.  If
  our intuition were wrong, this could be 
  shown by a formal proof of $\bot$ from $\zfc$.  But we cannot prove
  our intuition \emph{correct.}
\end{parag}

\begin{parag}
  We shall sometimes make claims about \emph{all} classes.  Such a
  claim cannot be expressed by a sentence of our predicate logic;
  rather, it is expressed by a \defn{scheme}{} of sentences, one for
  each singulary formula.  Some of the axioms in $\zfc$ are really
  schemes of axioms.
\end{parag}

\section*{Exercises}

\begin{xca}
  Draw the tree showing the construction of 
  \begin{equation*}
  \lnot(\lnot P\lto(Q\lto P))\lto\lnot(R\lto Q)\lto\lnot\lnot R.
  \end{equation*}
\end{xca}

\begin{xca}
In at least two ways, find a formula $F$ with the following
truth-table:
\begin{equation*}
  \begin{array}{c|*{16}c}
P&0&1&0&1&0&1&0&1&0&1&0&1&0&1&0&1\\
Q&0&0&1&1&0&0&1&1&0&0&1&1&0&0&1&1\\
R&0&0&0&0&1&1&1&1&0&0&0&0&1&1&1&1\\
S&0&0&0&0&0&0&0&0&1&1&1&1&1&1&1&1\\
F&0&1&1&0&1&0&0&0&0&1&0&1&1&1&1&0
  \end{array}
\end{equation*}
\end{xca}

\begin{xca}
  Draw a tree showing the construction of the sentence
  \begin{equation*}
    \Forall x(\phi\lto\Exists y(\psi\lto\Forall z(\chi\lto\rho)))
  \end{equation*}
(treating $\phi$, $\psi$, $\chi$, and $\rho$ as atomic formulas).
\end{xca}

\begin{xca}\label{ex:0VCD}
  Prove sentences~\eqref{eqn:0VCD} (p.~\pageref{eqn:0VCD}).
\end{xca}

\begin{xca}
  Determine whether 
  \begin{enumerate}
  \item 
  $\cC\included\cD\lto\bigcap\cD\included\bigcap\cC$;
\item
$\Exists x\Exists y\Forall z(x\in x\iff z\notin y)$.
  \end{enumerate}
\end{xca}

\begin{xca}\label{ex:cop}
What are $\bigcap\emptyset$ and $\bigcup\emptyset$?
\end{xca}

\begin{xca}\label{ex:conv}
Verify~\eqref{eqn:dom-ran}, \eqref{eqn:ran-dom}, and
\eqref{eqn:conv-conv} on p.~\pageref{eqn:dom-ran}.
\end{xca}

\begin{xca}
  Using a singulary formula $\phi(x)$, write down a formula that
  defines an equivalence-relation.
\end{xca}

\begin{xca}\label{ex:inconsistent}
Prove Lemma~\ref{lem:inconsistent}.
\end{xca}

\begin{xca}
  Let $\sigma$ be $\Forall x\phi\lto\Exists x\phi$, where $\phi$ is
  $\phi(x)$.
  \begin{enumerate}
  \item
Prove that $\sigma$ is a logical theorem.
\item
Prove that $\sigma$ is a logical theorem, without using the
Completeness Theorem.
\item
Give a formal proof of $\sigma$.
  \end{enumerate}
\end{xca}

\begin{xca}
  Determine whether $\phi(a)\lto\psi(a)\proves\Forall
  x\phi(x)\lto\Forall x\psi(x)$.
\end{xca}

\begin{xca}\label{ex:list}
Show that all sentences can be arranged in one infinite list, as the
proof of the Completeness Theorem requires.
\end{xca}

\chapter{Functions}

%\setcounter{equation}{10}

\section{Equality of sets}

\begin{parag}\label{par:=def}
We have defined equality of classes (\parref{par:class=}), but not
sets.  This we now do.  We introduce the formula $s=t$ as an
\emph{abbreviation} for 
\begin{equation*}
  \Forall x(s\in x\iff t\in x)
\end{equation*}
where $x$ is neither $s$ nor $t$.  So, by \emph{definition,} two sets
are \defn{equal}{} if (and only if) they are members of the same
sets.  We now \emph{assert} that two sets are equal if (and only if)
they have the same extension, in the sense of~\parref{par:class=}.
This means a set $a$ is equal to the class $\cls x{x\in a}$.  This
assertion is the first axiom of set theory.
\end{parag}

\begin{axiom}[Extension]\label{ax:extension} 
\dindexsub{axiom}{Extension A---}%
\dindexsub{extension}{E--- Axiom}
%\axiomname{Extension: \[a=b\iff a\included b\land b\included a.\]}
  Two sets are equal if and only if they have the same elements:
  \begin{equation}\label{eqn:a=b'}
    a=b\iff\Forall x(x\in a\iff x\in b).
  \end{equation}
\end{axiom}

\begin{parag}\label{par:a=b-subsets}
Since sets are classes, the notions and notations used in
\S\S~\ref{sect:classes} and~\ref{sect:relations} can be used for sets.
A subclass that is also a set can be
called a \defn{subset}.%
\dindexsub{set}{sub---}
We can rewrite \eqref{eqn:a=b'} as
\begin{equation}\label{eqn:ext}
  a=b\iff a\included b\land b\included a.
\end{equation}
Two sets are equal just in case each one is a subset of the
other.  
\end{parag}

\begin{theorem}\label{thm:a=b->f(a)<->f(b)}
  For all formulas $\phi(x)$,
  \begin{equation*}
    a=b\lto(\phi(a)\iff\phi(b)).
  \end{equation*}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:a=b->f(a)<->f(b)}
\end{proof}

\begin{theorem}[Russell Paradox]\label{thm:R}\dindex{Russell Paradox}
  Not all classes are sets; in particular, the class
  \begin{equation*}
    \cls x{x\notin x}
  \end{equation*}
is not a set: symbolically,
  $\lnot\Exists y\Forall x(x\in y\iff x\notin x)$.
\end{theorem}

\begin{proof}
  Let $\cR$ be the given class.  It suffices by
  \parref{par:a=b-subsets} to show that no subset of $\cR$ is
  equal to $\cR$.  Suppose $r\included\cR$, so
  that
  \begin{equation*}
    \Forall x(x\in r\lto x\in\cR).
  \end{equation*}
Then, in particular, if
  $r\in r$, then $r\in\cR$, so $r\notin r$ by definition of
  $\cR$.
Therefore, logically,
\begin{equation}\label{eqn:r-notin-r}
  r\notin r,
\end{equation}
which, by definition of $\cR$, means
\begin{equation}\label{eqn:r-in-R}
  r\in\cR.
\end{equation}
The last two conclusions---\eqref{eqn:r-notin-r}
and~\eqref{eqn:r-in-R}---imply $\cR\not\included r$, so
$r\neq\cR$.
\end{proof}

\begin{parag}\label{par:contradiction}
Often Theorem~\ref{thm:R} is proved by 
\defnplain{contradiction}{}%
\dindex{contradiction, proof by}%
\dindexsub{proof}{--- by contradiction}
as follows:
\begin{quote}
  Suppose $\cR$ is a set.  Then $\cR\in\cR\lto\cR\notin\cR$ and
  $\cR\notin\cR\lto\cR\in\cR$, so $\cR\in\cR\iff\cR\notin\cR$, which
  is absurd.  Therefore $\cR$ is not a set.
\end{quote}
This is a valid argument.  However, I prefer to avoid proofs by
contradiction,
for reasons of style.  In a proof of $P\lto Q$ by contradiction, one
assumes $P$ and $\lnot Q$, and proves an absurdity like $P\land\lnot
P$.  Often in such proofs, however, one does not need the assumption of
$P$; one really just proves $\lnot Q\lto\lnot P$, the
\defn{contrapositive}{} of $P\lto Q$.  Then the needless assumption of $P$
simply prevents anything in the proof from having independent value.
By contrast, in the proof given in
\parref{thm:R}, we happen to learn something more than the truth of
the theorem:
namely that no subset of $\cR$ is a member of itself.\footnote{The
argument for this---$r\included\cR\land r\in r\lto r\notin r$---can be
understood as using the method of contradiction.  However, we still
prove $r\notin r$ directly; we do not have to go back and say that our
original assumption that $r\in r$ is wrong; we simply use the
tautology $(P\lto\lnot P)\lto\lnot P$.}
\end{parag}

\section{New classes}\label{sect:new-classes}

\begin{parag}\label{par:pow}
There is a class of \emph{subsets} of a class $\cC$, namely $\cls
x{\Forall y(y\in x\lto y\in\cC)}$ or $\cls x{x\included\cC}$; we may
call this the 
\defn{power class}{}%
\dindexsub{class}{power ---}
of $\cC$, and denote it by
\begin{equation*}
  \pow{\cC}.
\end{equation*}
The power class of a \emph{set} will later be called the
\tech{power set}{}
of the set; the power set will \emph{be} a set, by
\parref{ax:power-set}; but for now, it is simply a class.
\end{parag}

\begin{parag}\label{par:pair-class}
  Having defined equality of sets, we can put two sets $a$ and
  $b$ into a class, namely the 
  class 
  \begin{equation*}
  \cls x{x=a\lor x=b};
  \end{equation*}
this class is commonly denoted by
\begin{equation*}
  \{a,b\}.
\end{equation*}
This class is going to be a set (\parref{ax:pairing}); but without
assuming this, we can still observe:
\begin{gather}\label{eqn:cup-bigcup}
  a\cup b=\bigcup\{a,b\};\\\label{eqn:cap-bigcap}
  a\cap b=\bigcap\{a,b\}
\end{gather}
(Exercise~\ref{ex:cop-bigcop}).
We do not have such equations for classes in general, since
we do not have a way to put classes, as such, into other classes.
However, the union of a set will be a set; but we shall not need to use
this until after \parref{ax:union}.
Meanwhile, if $\cC\included\pow b$, then
$\bigcup \cC\included b$, so $\bigcup\cC$ will be a set by
\parref{ax:comprehension}
(Exercise~\ref{ex:UC-set}). 
\end{parag}

\section{New sets}

\begin{parag}\label{par:proper}
Classes that are not sets are called 
\defnplain{proper class}{es.}%
\dindexsub{proper}{--- class}%
\footnote{A proper class need not be a proper subclass of any class;
  nor need a proper subclass of some class be a proper class
  (Exercise~\ref{ex:proper}).} 
The proof of the Russell Paradox \eqref{thm:R} suggests that proper
classes are \emph{too big} to be sets.  In the belief that size is the
only bar to being a set, we postulate the following \techplain{scheme}{} of
axioms: it is a \defn{scheme}, because it comprises one
  axiom for each 
singulary formula:
\end{parag}

\begin{axiomscheme}[Separation]\label{ax:comprehension}
%  \axiomname{Comprehension-Scheme: \[\Exists xx=a\cap\cC.\]} 
Every subclass of a set is a set:
  For every singulary formula $\phi$,
  \begin{equation}\label{eqn:comprehension}
    \Exists x\Forall y(y\in x\iff y\in a\land\phi(y)).
  \end{equation}
\end{axiomscheme}

\begin{parag}\label{par:comprehension}
  The set whose existence is expressed by~\eqref{eqn:comprehension} is
  the intersection 
  \begin{equation*}
  a\cap\cls x{\phi(x)}; 
  \end{equation*}
this can be denoted by
  \begin{equation*}
    \set{x\in a}{\phi(x)}.
  \end{equation*}
We may refer to this as \emph{the} set guaranteed
by~\eqref{eqn:comprehension}, 
  because of the Extension Axiom (\parref{ax:extension}).  As a first
  consequence of the
  Separation Scheme, we have that, if $a\in\cC$, then
  $\bigcap\cC$ is the \emph{set} $\set{x\in a}{x\in\bigcap\cC}$
  (Exercise~\ref{ex:cap}).  Likewise, $a\setminus\cD$ is the set
  $\set{x\in a}{x\notin\cD}$.
\end{parag}

\begin{parag}
  There is an assumption so basic that we do not bother to state it
  formally as an axiom.\footnote{Some writers do, as Kunen \cite[I~5,
  p.~10]{MR85e:03003}.}  This assumption is that there \emph{are}
  sets.  Hence, by Separation, the empty class 
  \begin{equation*}
  \emptyset
  \end{equation*}
  is a set, called the \defnplain{empty set}.%
\dindexsub{empty}{set}%
\dindexsub{set}{empty ---}
  (There is only one empty set, by the Extension Axiom.)
\end{parag}

\begin{parag}
  We observed (\parref{par:pair-class}) that, from two sets $a$ and
  $b$, we can form the class denoted by $\{a,b\}$.  But if any class is a
  set, surely this class is:
\end{parag}

\begin{axiom}[Pairing]\label{ax:pairing} 
\dindexsub{axiom}{Pairing A---}%
\dindexsub{pair}{P---ing Axiom}
%\axiomname{Pairing: \[\Exists xx=\{a,b\}.\]}
  Any two sets are contained in a third:
  \begin{equation*}
    \Exists x(a\in x\land b\in x).
  \end{equation*}
\end{axiom}

\begin{parag}
As stated, the axiom is merely that some set contains $a$ and $b$;
  additional members of the set are not excluded.  So the class
  $\{a,b\}$ is a subclass of some set.
  By Separation, $\{a,b\}$ is itself a set.
This set is an \defnplain{(unordered) pair}.%
\dindex{unordered pair}%
\dindexsub{pair}{unordered ---}
  In case $a=b$, the set is a 
\defn{singleton}, denoted by
\begin{equation*}
  \{a\}.
\end{equation*}
\end{parag}

\section{Relations}

\begin{parag}\label{par:in}
Equality and inclusion are relations between classes, but they are not
classes themselves: they are just features of our logic.
In~\parref{par:rel-sets} we defined relations between sets as being
defined by binary formulas; but we did
not establish such relations as classes.  This we now do.
\end{parag}

\begin{parag}\label{par:u-pair}
  A binary formula determines a class of pairs.
  Indeed, $\phi(x,y)$ determines the class
  \begin{equation*}
    \cls z{\Exists x\Exists y(z=\{x,y\}\land\phi(x,y))}.
  \end{equation*}
However, this is also the class determined in the same way by
$\phi(x,y)\lor\phi(y,x)$ (Exercise~\ref{ex:u-pair}).
\end{parag}

\begin{parag}\label{par:op}
  For more control, we want to combine $a$ and $b$ into an
  \defnplain{ordered pair},%
\dindexsub{order}{---ed pair} denoted by
  \begin{equation*}
    (a,b),
  \end{equation*}
such that
\begin{equation}\label{eqn:op-iff}
  (a,b)=(c,d)\iff a=c\land b=d.
\end{equation}
One way (but not the only way) to achieve this is by defining $(a,b)$
as
\begin{equation*}
  \{\{a\},\{a,b\}\}.
\end{equation*}
(See Exercises~\ref{ex:op} and~\ref{ex:pair-alt}.)  Then the class
  $\cls z{\Exists x\Exists
  y(z=(x,y)\land\phi(x,y))}$ can be denoted by
\begin{equation*}
  \cls{(x,y)}{\phi(x,y)};
\end{equation*}
this is the class 
\defnplain{defined by}{}%
\dindexsub{define}{---d by} $\phi(x,y)$.
\end{parag}

\begin{parag}\label{par:cartesian}
In particular, we now we have a new operation on classes:  the
\defn{Cartesian product}{}%
\dindexsub{product}{Cartesian ---}%
\dindexsub{operation}{Cartesian product} 
of $\cC$ and $\cD$ is
\begin{equation*}
  \cls {(x,y)}{x\in\cC\land y\in\cD};
\end{equation*}
this class is denoted by
\begin{equation*}
  \cC\times\cD,
\end{equation*}
and it is a subclass of $\pow{\pow{\cC\cup\cD}}$
(Exercise~\ref{ex:pair-pow}).
A relation from $\cC$ to $\cD$ (\parref{par:rel-sets}) can now be
understood as a subclass of $\cC\times\cD$.
\end{parag}

\section{Kinds of relations}

\begin{parag}\label{par:diagonal}
For any class $\cC$, the relation
$\cls{(x,y)}{x=y\land x\in\cC}$
is the \defn{diagonal}{} on ${\cC}$, denoted by
\begin{equation*}
   \Delta_{\cC}.
\end{equation*}
Then
\begin{gather}\label{eqn:dom}
  \Delta_{\dom{\cR}}\included\cR/\conv{\cR},\\\label{eqn:ran} 
  \Delta_{\ran{\cR}}\included\conv{\cR}/\cR,
\end{gather}
for every binary relation $\cR$ (Exercise~\ref{ex:comp}).  We defined
reflexive and irreflexive relations in~\parref{par:refl-irrefl};
symmetric, in~\parref{par:symm}; transitive,
in~\parref{par:composite}.  Some alternative formulations of
definitions are now possible.
A binary relation $\cR$
is called% (compare~\parref{par:=inc}): 
\begin{enumerate}
  \item
\defnplain{reflexive},%
\dindexsub{relation}{reflexive ---}
 if $\Delta_{\universe}\included \cR$;
\item
\defnplain{irreflexive},%
\dindexsub{relation}{irreflexive ---} 
if $\cR\cap\Delta_{\universe}=\emptyset$;
\item
\defnplain{anti-symmetric},%
\dindexsub{relation}{anti-symmetric ---} 
if $\cR\cap \conv{\cR} \included \Delta_{\universe}$.
\end{enumerate}
There are relative versions.  The relation $\cR$ is:
\begin{enumerate}
\item
\defnplain{reflexive on}{}%
\dindexsub{relation}{--- reflexive --- on a class} 
$\cC$, if  $\Delta_{\cC}\included \cR$;
\item
\defnplain{irreflexive on}{}%
\dindexsub{relation}{--- irreflexive on a class} $\cC$, if
$\cR\cap\Delta_{\cC}=\emptyset$; 
\item
\defnplain{anti-symmetric on}{}%
\dindexsub{relation}{--- anti-symmetric on a class} $\cC$, if $\cR\cap
\conv{\cR}\cap(\cC\times\cC)\included\Delta_{\cC}$
\item
\defnplain{transitive on}{}%
\dindexsub{relation}{---transitive on a class}
$\cC$, if $\cR\cap(\cC\times\cC)$ is transitive.
\end{enumerate}
\end{parag}

\begin{parag}\label{par:equiv-reln}
A relation is an 
\defnplain{equivalence-relation on}{}%
\dindexsub{equivalence}{---{}-relation on a class}
$\cC$ if it is reflexive,
symmetric and transitive on $\cC$.
Then \defn{equality}, understood as
$\Delta_{\universe}$, is an
equivalence\-{}-relation on every class (Exercise~\ref{ex:=-equiv}).
The diagonal 
$\Delta_{\cC}$ is an equivalence-relation on every subclass of~$\cC$. 
\end{parag}

\begin{parag}\label{par:ordering}
The relation $\cR$ is:
\begin{enumerate}
  \item
an \defnplain{ordering of}{}%
\dindexsub{order}{---ing of a class} 
$\cC$, if it is
anti\-{}-symmetric, transitive, and either reflexive or irreflexive,
on $\cC$;
\item
a \defnplain{total ordering of}{}%
\dindex{total ordering of a class}%
\dindexsub{order}{total ---ing of a class}
 $\cC$, if it is an ordering of $\cC$ and
\begin{equation*}
\cC\times\cC\included
  \cR\cup\conv{\cR}\cup\Delta_{\cC}.
\end{equation*}  
\end{enumerate}
If $\cR$ is an ordering of $\cC$, then $\cC$ is 
\defnplain{ordered}{}%
\dindexsub{order}{---ed}
by $\cR$, and we may refer to the ordered pair $(\cC,\cR)$ as an
\defnplain{ordered class}{} or simply an \defn{order}.  Note however
that if $\cC$ is a proper
class, then $(\cC,\cR)$ is not literally an ordered pair in the sense
of~\parref{par:op}.  Note also that we do not assume that $\cR$ is a
relation \emph{on} $\cC$.
An ordering in the present sense is often called a 
\techplain{partial ordering},%
\tindexsub{order}{partial ---ing} even though it might be total.  An
\emph{irreflexive} ordering is 
also called a 
\defn{strict ordering}.%
\dindexsub{ordering}{strict ---}
The converse of an ordering is an ordering (Exercise~\ref{ex:conv-order}).
If $\cR$ is a reflexive ordering of $\cC$, then there is a
corresponding strict ordering of $\cC$, namely
$\cR\setminus\Delta_{\cC}$ (or $\cR\setminus\Delta_{\universe}$, for
example; it doesn't matter what $\cR\setminus(\cC\times\cC)$ is).
A strict ordering~$\cS$ of $\cC$ has
the corresponding reflexive ordering $\cS\cup\Delta_{\cC}$.
\end{parag}

\begin{parag}
There are standard examples.  (Compare~\parref{par:=inc}.)
\begin{asparaenum}
\item
\defnplain{Inclusion}{}
\dindexsub{inclusion}{--- of sets}%
\dindexsub{relation}{inclusion} 
of sets is a reflexive ordering of every class.  (Inclusion of sets is
the class $\cls{(x,y)}{x\included y}$, which can be denoted by
$\included$ alone.)
\item
The converse of inclusion is usually denoted by $\includes$; it is a
reflexive ordering, by the comment in \parref{par:ordering}.
\item
\defnplain{Proper inclusion}{}%
\dindexsub{proper}{--- inclusion}%
\dindexsub{inclusion}{proper ---}%
\dindexsub{relation}{proper inclusion} 
of sets is a strict ordering.
(Proper inclusion of sets is the class
$\cls{(x,y)}{x\pincluded y}$, or
$\pincluded$.)
\end{asparaenum}
It will take more work to define a good example of a total ordering.
Often a reflexive ordering is symbolized by $\leq$; then the
corresponding strict ordering is denoted
by $<$.  The
converse of $\leq$ is $\geq$; the converse of $<$ is $>$.
\defnplain{Membership}{}%
\dindex{membership}%
\dindexsub{relation}{membership} 
of sets (the relation $\cls{(x,y)}{x\in y}$ or
$\in$) is not yet an
example of anything in particular; but it will
be.
\end{parag}

\begin{parag}\label{par:minimum}
Suppose $<$ is a strict ordering of $\cC$.  If $a$ and $b$ are in
$\cC$, and $a<b$, then $a$ is \defn{less}{} than $b$, and $b$ is
\defn{greater}{} than $a$.\footnote{Likewise, if $a\leq b$, then $a$ is
  less than \emph{or equal to} $b$.  This standard language brings out
a point about English.  If the letter $b$ is replaced with the
first-person pronoun, then we should say $a$ is less than \emph{I} am
or is equal to \emph{me.}  The pronoun retains a distinct
objective form.  But a noun---which is what we are using the letter
$b$ as---has no such form.} 
An element $a$ of $\cC$ is \defn{minimal}{} with respect to $<$, or
\defnplain{$<$-minimal},%
\dindex{minimal}
if
\begin{equation*}
  b\in\cC\lto{}\mathrel{\lnot}b<a.
\end{equation*}
A $>$-minimal element is 
\defnplain{$<$-maximal}.%
\dindex{maximal}
An element $a$ of
  $\cC$ is 
\defn{least}{} or 
\defn{minimum}{} with respect to $<$, or
\defnplain{$<$-least},%
\dindex{least}
 if
\begin{equation*}
  b\in\cC\lto a\leq b.
\end{equation*}
A $>$-least
element is 
\defnplain{$<$-greatest}.
\dindex{greatest}
Least elements are minimal elements.
Least elements are unique when they exist
(Exercise~\ref{ex:least-unique}); but they need 
not exist.  With respect to a total ordering, a minimal element is a
least element.
\end{parag}

\begin{parag}\label{par:pred1}
Suppose again that $<$ is a strict ordering of $\cC$.
An 
\defnplain{initial segment}{}%
\dindexsub{initial}{--- segment}%
\dindex{segment, initial} of $\cC$ with respect to $<$ is a
subclass $\cD$ of $\cC$ such that 
\begin{equation*}
  a\in\cD\land b\in\cC\land b< a\lto b\in\cD.
\end{equation*}
A \defn{proper initial segment}{}%
\dindexsub{initial}{proper --- segment}%
\dindexsub{segment}{proper initial ---}
 is an initial segment that is a proper
  subclass.  If $a\in\cC$, then the class $\cls x{x\in\cC\land
    x<a}$ is the class of \defn{predecessor}s of $a$ in $\cC$ with
  respect to $<$; it can be denoted by
  \begin{equation*}
    \pred{a,\cC,<},
  \end{equation*}
or more simply by $\pred a$ if there will be no ambiguity.  Such a
class is a particular kind of proper initial segment, called a
\defnplain{section}.%
\dindex{section of an ordered class}
%If $\cR=\cls{(x,y)\in\cC\times\cC}{x<y}$, then $\pred a=\cR a$.
\end{parag}

\begin{parag}\label{par:least}
  The class $\cC$ is 
\defnplain{well-ordered}{}%
\index{well-ordered|see{order}}
\dindexsub{order}{well-{}---ed class}%
\dindexsub{class}{well-ordered ---}
 by $<$ if:
  \begin{enumerate}
\item
$<$ is a strict total ordering of $\cC$;
\item
every section of $\cC$ with respect to $<$ is a
set;
    \item
every non-empty subset of $\cC$ has a $<$-least element.
  \end{enumerate}
In this case, every non-empty sub\emph{class} of $\cC$ has a least element:
Indeed, if $\cD$ is such a subclass, with an element $a$, then either
$a$ is its least element, or else its least element is the least
element of $\cD\cap\pred{a,\cC,<}$.
If $<$ well-orders $\cC$, then the order $(\cC,<)$ is in particular a
\defnplain{well-ordered set}.%
\dindexsub{order}{well-{}---ed set}%
\dindexsub{set}{well-ordered ---}
\end{parag}

\section{Functions}

\begin{parag}
  The various operations on classes defined in
  \P\P~\ref{par:class-ops}, \ref{par:infinitary},
\ref{par:pow}, \ref{par:pair-class}, 
  and~\ref{par:cartesian} are examples of
  \defn{function}{s.}  The 
  union operation in \parref{par:class-ops} is the function by which
  the class $\cC\cup\cD$ is
  obtained from the classes $\cC$ and $\cD$; the power class
  operation in \parref{par:pow} is a function converting $\cC$ into
  $\pow{\cC}$.  In this sense, a function is not a set or a class; it
  is a feature of our logic.  
\end{parag}

\begin{parag}\label{par:aRRa}
Orderings determine functions as in~\parref{par:pred1}.
More generally, suppose $\cR$ is a binary relation.  Then we can make
the definitions: 
  \begin{align*}
    a\cR&=\cls x{a\mathrel{\cR}x},&
    \cR a&=\cls x{x\mathrel{\cR}a}.
  \end{align*}
If $\cR$ is a relation \emph{on} $\cC$ that strictly orders $\cC$,
then $Ra=\pred a$.  In general,
the classes $a\cR$ and $\cR a$ are functions of the set $a$.  Some
kinds of relations can be understood in terms of these functions.
For example, the
relation $\cR$ is symmetric if and only if $\Forall xx\cR=\cR x$
(Exercise~\ref{ex:sym}).  
\end{parag}

\begin{parag}\label{par:=classes}
Suppose $\cE\included\cC\times\cC$ and is an
equivalence-relation on
$\cC$.  If $a\in\cC$, then $a\cE$ is the
\defnplain{equivalence-class}{}%
\dindexsub{equivalence}{---{}-class}%
\dindexsub{class}{equivalence-{}---}
of $a$ with respect to $\cE$, or the
\defnplain{$\cE$-class}{} of $a$, and $a$ is a
\defn{representative}{} of this class; every other member of the class
is also a representative.  We may understand the $\cE$-classes as
composing a \defn{family}, denoted by
\begin{equation*}
  \cC/\cE.
\end{equation*}
Then $\cD$ belongs to $\cC/\cE$ if and only if $\cD=a\cE$ for some $a$
in $\cC$. 
Since classes contain only sets, and equivalence-classes may be proper
classes, the family $\cC/\cE$ may not be a class.
However, see \sref{sect:e-wf}. 
\end{parag}

\begin{parag}\label{par:functions}
Often it is \emph{sets} that are functions of other sets.
A binary relation $\cF$ is 
\defnplain{functional}{}%
\dindexsub{function}{---al relation} if
\begin{equation*}
  \conv{\cF}/\cF\included\Delta_{\universe},
\end{equation*}
that is,
\begin{equation}\label{eqn:functional}
a\mathrel{\cF}b\land a\mathrel{\cF}c\lto b=c.
\end{equation}
This is equivalent to $\conv{\cF}/\cF=\Delta_{\ran{\cF}}$
(Exercise~\ref{ex:functional}). 
In this case, $\cF$ is a 
\defnplain{function on}{}%
\dindex{function}%
\dindexsub{on}{function ---}
its domain.  Suppose 
that domain is $\cC$, and $\ran{\cF}\included\cD$.
Then we may write either of
\begin{equation*}
  \cF\colon \cC\to\cD,\qquad
\cC\overset{\cF}{\To}\cD; 
\end{equation*}
these are abbreviations
of the sentence
\begin{multline*}
\Forall x\biggl((\Exists yx\mathrel{\cF} y\lto x\in\cC)\land{}\\
{}\land
\Bigl(x\in\cC\lto\Exists y\bigl(x\mathrel{\cF}y\land y\in\cD
  \land\Forall z(x\mathrel{\cF}z\lto y=z)\bigr)\Bigr)\biggr).
\end{multline*}
We may also say that $\cF$ is a function 
\defnplain{from $\cC$ (in)to}{}%
\dindexsub{from}{function --- a class to another}%
\dindexsub{into}{function --- a class}%
\dindexsub{to}{function --- a class} $\cD$.
If $a\mathrel{\cF}b$, then we usually write
\begin{equation*}
  \cF(a)=b.
\end{equation*}
This notation is consistent with our definition of equality of sets,
by which  equality is an equivalence-relation
(\parref{par:equiv-reln}).  Indeed, now the
implication~\eqref{eqn:functional} becomes
\begin{equation*}
  \cF(a)=b\land\cF(a)=c\lto b=c.
\end{equation*}
As an alternative notation for $\cF$ itself, we may write
\begin{equation*}
  x\mapsto\cF(x).
\end{equation*}
One must not confuse $\cF(a)$ with $\cR a$ as defined
in~\parref{par:aRRa}. 
\end{parag}

\begin{parag}\label{par:f}
  We can produce a few examples of functions in the sense of
  \parref{par:functions}:
  \begin{asparaenum}
    \item
  Most basic is the
  \defnplain{identity function},%
\dindexsub{identity}{--- function}%
\dindexsub{function}{identity ---} which is
  $x\mapsto x$ or 
  $\Delta_{\universe}$; considered as a function, this can be denoted
  by 
  \begin{equation*}
  \id{\universe}.
  \end{equation*}   
The \defnplain{identity on}{}%
\dindexsub{identity}{--- on a class}%
\dindexsub{on}{identity ---} $\cC$ is $\Delta_{\cC}$, usually
denoted by
\begin{equation*}
  \id{\cC}.
\end{equation*}
\item
By (a special case of) the Pairing Axiom, we have a function
$x\mapsto\{x\}$ on $\universe$.
\item\label{item:function}
If $\cF$ is a function, then so is
  $x\mapsto(x,\cF(x))$: its domain is $\dom{\cF}$, and its
range is $\cF$. 
\item
If $a$ is a set, then
  $x\mapsto a$ is a \defnplain{constant function},%
\dindexsub{constant}{--- function}%
\dindexsub{function}{constant ---} with domain 
  $\universe$. 
\item
If $\cF\colon \cC\to\cD$, and $\cE\included\cC$,
then $\cF\cap(\cE\times\universe)$ (which is $\cF\cap(\cE\times\cD)$)
is a function with domain
$\cE$ called the \defn{restriction}{}%
\dindexsub{function}{restriction} of $\cF$ to $\cE$
  and denoted by
  \begin{equation*}
    \cF\rest{\cE}.
  \end{equation*}
In particular, $\id{\universe}\rest{\cC}$ is $\id{\cC}$.
\item
By \parref{par:op}, there is a function $\cF$ on
$\universe\times\universe$ such that $\cF(b)=a\iff\Exists
xb=(a,x)$; this function can be denoted by
\begin{equation*}
  (x,y)\mapsto x
\end{equation*}
or $\pi_0$;
it is \defn{projection}{}%
\dindexsub{function}{projection} onto the first coordinate.  Likewise,
there is $(x,y)\mapsto y$ or $\pi_1$.
\item
Hence we have a function $(x,y)\mapsto\{x,y\}$.
  \end{asparaenum}
\end{parag}

\begin{parag}
  If $\cF$ and $\cG$ are  functional relations, and
  $\ran{\cF}\included\dom{\cG}$, then the composite $\cF/\cG$ is a
  function on $\dom{\cF}$; usually this function is denoted by either of
  \begin{align*}
    &(\cG\circ\cF),&
&x\mapsto\cG(\cF(x)).
  \end{align*}
If also $\cH$ is functional, and $\ran{\cG}\included\dom{\cH}$, then,
  as we have~\eqref{eqn:comp-assoc}, so
  \begin{equation}\label{eqn:comp-f-assoc}
    \cH\circ(\cG\circ\cF)=(\cH\circ\cG)\circ\cF. 
  \end{equation}
\end{parag}

\begin{parag}\label{par:inv}
Suppose $\cF\colon \cC\to\cD$, and
  $\cF/\conv{\cF}\included\Delta_{\universe}$, equivalently,
  $\cF/\conv{\cF}=\Delta_{\dom{\cF}}$ (Exercise \ref{ex:inj}).
  This means 
\begin{equation*}
  \cF(a)=\cF(b)\lto a=b.
\end{equation*}
Then $\cF$ is called an 
\defnplain{injective}{}%
\dindex{injection, injective function}
\dindexsub{function}{injective ---, injection} function, or an
  \defnplain{injection}{}
from $\cC$ (in)to $\cD$, or an
  \defn{embedding}{}%
\index{function|seealso{embedding}}
\dindexsub{embedding}{classes} 
of $\cC$ in $\cD$;
we may write
\begin{equation*}
  \cF\colon \cC\injection\cD.
\end{equation*}
(So the tail of the arrow indicates injectivity.)
The converse of an injective function is also a function
  (Exercise~\ref{ex:conv-f-inv}), called the \defn{inverse}{}%
\dindexsub{function}{inverse} of the
  function; when it exists, the inverse of $\cF$ is denoted
  \begin{equation*}
    \cF\inv.
  \end{equation*}
\end{parag}

\begin{parag}
Suppose again $\cF\colon \cC\to\cD$.
If $\cD=\ran{\cF}$, then $\cF$ is \defnplain{surjective}{}%
\index{surjection, surjective function|seealso{function}}
\dindex{surjection, surjective function}
\defnplain{onto}{}%
\dindexsub{onto}{function surjective, surjection, --- a class}%
\dindexsub{function}{surjective ---, surjection} 
$\cD$ (or a \defnplain{surjection onto}{}
$\cD$); we may write
\begin{equation*}
  \cF\colon \cC\surjection\cD.
\end{equation*}
(So the second head of the arrow indicates surjectivity.)
Note well that a function cannot be surjective simply; it is only
surjective with respect to the set that the function is surjective
\emph{onto} (namely its \emph{range}).
If $\cF$ is injective, and surjective onto $\cD$, then
$\cF$ is a \defn{bijection}{}%
\dindexsub{function}{bijective ---, bijection} from $\cC$ to $\cD$,
and we may write
\begin{equation*}
  \cF\colon \cC\bijection\cD.
\end{equation*}
\end{parag}

\begin{theorem}\label{thm:f-inj-bij}
  Suppose $\cF\colon\cC\to\cD$.
  \begin{enumerate}
    \item
$\cF$ is injective if and only if $\cC$ is empty or there is
      a function $\cG$ from $\cD$ to $\cC$ such that
      $\cG\circ\cF=\id{\cC}$.
\item
$\cF$ is a bijection from $\cC$ to $\cD$ if and only if
  there is a function $\cG$ from $\cD$ to $\cC$ such that
      $\cG\circ\cF=\id{\cC}$ and $\cF\circ\cG=\id{\cD}$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:f-inj-bij}.
\end{proof}

\section{Functions from functions}

\begin{parag}
A function $\cF$ induces two functions on classes:
\begin{enumerate}
  \item
If $\cC$ is a subclass of $\dom{\cF}$, then the class 
\begin{equation}\label{eqn:image}
  \cls y{\Exists x(x\in\cC\land \cF(x)=y)}
\end{equation}
is the \defn{image}{} of $\cC$ under
  $\cF$ and can be denoted by either of
\begin{align*}
  &\cF\setimb{\cC},&
&\cls{\cF(x)}{x\in\cC}.
\end{align*}
Then $\ran{\cF}=\cF\setimb{\dom{\cF}}$.
\item
The class 
\begin{equation}\label{eqn:pre-image}
\cls x{x\in\dom{\cF}\land\cF(x)\in\cC}
\end{equation}
is the \defn{pre-image}{} of $\cC$ under
  $\cF$ and can be denoted by
\begin{equation*}
  \cF\inv\setimb{\cC}.
\end{equation*}
Pre\-{}-images of all classes exist, regardless of whether the
function $\cF$ itself
has an inverse (\parref{par:inv}).
In particular, $\dom{\cF}=\cF\inv\setimb{\cC}$ whenever
$\ran{\cF}\included\cC$.
\end{enumerate}
Note the great difference in form between~\eqref{eqn:image}
and~\eqref{eqn:pre-image}.  The difference is reflected in the
following. 
\end{parag}

\begin{theorem}\label{thm:functions}
  Suppose $\cF$ is a function.  Then
\begin{align}
\cF\setimb{\cC\cup\cD}&=\cF\setimb{\cC}\cup\cF\setimb{\cD},\\
\cF\setimb{\cC\cap\cD}&\included\cF\setimb{\cC}\cap\cF\setimb{\cD},\\
\cF\setimb{\dom{\cF}\setminus\cC}&\includes\ran{\cF}\setminus\cF\setimb{\cC}
\end{align}
for all subclasses $\cC$ and $\cD$ of $\dom{\cF}$, and
\begin{align}
\cF\inv\setimb{\cC\cup\cD}&=\cF\inv\setimb{\cC}\cup\cF\inv\setimb{\cD},\\
\cF\inv\setimb{\cC\cap\cD}&=\cF\inv\setimb{\cC}\cap\cF\inv\setimb{\cD},\\
\cF\inv\setimb{\ran{\cF}\setminus\cC}&=\dom{\cF}\setminus\cF\inv\setimb{\cC}
\end{align}
for all classes $\cC$ and $\cD$.  Moreover,
the following statements are equivalent:
\begin{enumerate}
  \item
$\cF$ is injective;
\item
$\cC\cup\cD\included\dom{\cF}\lto\cF\setimb{\cC\cap\cD}
  =\cF\setimb{\cC}\cap\cF\setimb{\cD}$ 
for all classes $\cC$ and $\cD$;
\item
$\cC\included\dom{\cF}\lto\cF\setimb{\dom{\cF}\setminus\cC}=
  \ran{\cF}\setminus\cF\setimb{\cC}$ for all classes $\cC$; 
\item
$a\cup b\included\dom{\cF}\lto\cF\setimb{a\cap b}=\cF\setimb
  a\cap\cF\setimb b$ for all sets $a$ and $b$;  
\item
$a\included\dom{\cF}\lto\cF\setimb{\dom{\cF}\setminus
  a}=\ran{\cF}\setminus\cF\setimb a$ for all 
  sets $a$.
\end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:functions}.
\end{proof}
\begin{parag}
If $\dom{\cF}$ is a set,
  then $\cF\inv\setimb{\cE}$ is a set, by Separation
  (\parref{ax:comprehension}).  Likewise, if 
  $\ran{\cF}$ is 
a set, and $\cE\included\dom{\cF}$, then $\cF\setimb{\cE}$ is a set.
  Now, possibly  
$\ran{\cF}$ is a set, while $\dom{\cF}$ is a proper class:
consider a constant function.  However, when we noted in
\parref{par:proper} that some classes are too big to be
  sets, we also suggested that, if a class is \emph{not} too big
  to be a set, then it \emph{is} a set.
Presumably the range of a function is not bigger than its domain,
  since the range contains no more than one element for each element
  of the domain; so we postulate the following.
\end{parag}

\begin{axiomscheme}[Replacement]\label{ax:replacement}
\dindexsub{axiom}{Replacement Scheme}%
\dindex{Replacement Scheme}%
\dindexsub{scheme}{Replacement S---}
%\axiomname{Replacement-Scheme: \[a\included\dom{\cF}\lto\Exists xx=\cF\setimb{a}.\]} 
The image of a set under a function is a set:  
\begin{equation*}
  a\included\dom{\cF}\lto\Exists xx=\cF\setimb{a}
\end{equation*}
for all functions $\cF$, so that the class
$\cls{\cF(x)}{x\in a}$ is a set whenever $a$ is a subset of
$\dom{\cF}$. 
\end{axiomscheme}

\begin{theorem}\label{thm:dom-inj-set}\mbox{}
  \begin{enumerate}
\item
A function whose domain is a set is a set.
    \item
  The domain of an injection into a set is a set.
\item
Suppose $f\colon a\to b$.  Then $x\mapsto f\setimb x$ is a
(well\-{}-defined) function
from $\pow a$ to $\pow b$, and $y\mapsto f\inv\setimb y$ is a
well\-{}-defined function from $\pow b$ to~$\pow a$.  Moreover,
\begin{gather*}
f\setimb{\bigcup c}=\bigcup\cls{f\setimb x}{x\in c},\\
f\setimb{\bigcap c}\included\bigcap\cls{f\setimb x}{x\in c}
\end{gather*}
 for all subsets
    $c$ of $\pow a$; and
\begin{gather*}
f\inv\setimb{\bigcup c}=\bigcup\cls{f\inv\setimb x}{x\in c},\\
f\inv\setimb{\bigcap c}=\bigcap\cls{f\inv\setimb x}{x\in c}\\
\end{gather*}
for all subsets $c$ of $\pow b$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:dom-inj-set}.
\end{proof}

\begin{parag}
Now functions whose domains are sets may be denoted by constants like
$f$ and $g$.  The class of functions from $a$ to $\cC$ is denoted by
\begin{equation*}
  \mapset a{\cC}.
\end{equation*}
So $\mapset a{\cC}\included a\times\cC$.  In particular, if we let
$0=\emptyset$, $1=\{0\}$, and $2=\{0,1\}$, then
\begin{equation*}
  f\colon\mapset a2\bijection\pow a,
\end{equation*}
where $f(g)=g\inv\setimb{\{1\}}$.
\end{parag}

\section*{Exercises}
%\addcontentsline{toc}{section}{Exercises}

%\begin{enumerate}
%  \renewcommand{\theenumi}{\arabic{enumi}}

\begin{xca}\label{ex:a=b->f(a)<->f(b)}
  Prove Theorem~\ref{thm:a=b->f(a)<->f(b)}.
\end{xca}

\begin{xca}
Prove carefully that $\{a\}=\{b\}\iff a=b$.
\end{xca}

\begin{xca}
  In~\parref{par:subclass}, the sentence~\eqref{eqn:ext-classes} is
  said to be a validity.  But then~\eqref{eqn:ext} is just a special
  case.  Why then do we need the Extension Axiom?
\end{xca}

\begin{xca}\label{ex:cop-bigcop}
Show~\eqref{eqn:cup-bigcup} and~\eqref{eqn:cap-bigcap}.
\end{xca}

\begin{xca}
  Find three proper classes.
\end{xca}

\begin{xca}\label{ex:proper}
Find:
  \begin{enumerate}
  \item
  a proper class that is not a proper subclass of any class;
\item
  a class with a proper subclass that is not a proper class.
  \end{enumerate}
\end{xca}

\begin{xca}\label{ex:UC-set}
Show that, if $\cC\included\pow a$, then $\bigcup\cC$ is a
set. 
\end{xca}

\begin{xca}\label{ex:cap}
Show that the intersection of a non\-{}-empty class is a set
(\parref{par:comprehension}). 
\end{xca}

\begin{xca}
  Determine whether $\cC=\bigcup\pow{\cC}$.
\end{xca}

\begin{xca}
  Give three examples of sets.
\end{xca}

\begin{xca}
Using only $\emptyset$, $\universe$, and the braces $\{$ and $\}$,
  compute 
  \begin{enumerate}
  \item
  $\bigcap\{\emptyset\}$ and $\bigcup\{\emptyset\}$,
\item
  $\bigcap\universe$ and $\bigcup\universe$.
  \end{enumerate}
\end{xca}

\begin{xca}\label{ex:u-pair}
Show that the class determined by $\phi(x,y)$ in the sense of
\parref{par:u-pair} is the same as the class determined by
$\phi(x,y)\lor\phi(y,x)$. 
\end{xca}

\begin{xca}\label{ex:op}
Show the definition of ordered pair in \parref{par:op}
causes~\eqref{eqn:op-iff} to be true.
\end{xca}

\begin{xca}\label{ex:pair-alt}
Show that the same is true if $(a,b)$ is defined as
$\{\{\emptyset,\{a\}\},\{\{b\}\}\}$. 
\end{xca}

\begin{xca}\label{ex:pair-pow}
Show that $\cC\times\cD\included\pow{\pow{\cC\cup\cD}}$. 
\end{xca}

\begin{xca}\label{ex:comp}
Verify
\eqref{eqn:dom} and~\eqref{eqn:ran}.
\end{xca}

\begin{xca}
  Suppose $\cE$ is a reflexive binary relation such that
  $a\mathrel{\cE}c\land b\mathrel{\cE}c\lto a\mathrel{\cE}b$.  Show
  that $\cE$ is an equivalence-relation.
\end{xca}

\begin{xca}\label{ex:=-equiv}
Prove that $\Delta_{\universe}$ is an equivalence-relation using only
the definition of equality in \parref{par:=def} (and not for example
the Extension Axiom).  
\end{xca}

\begin{xca}
  Show that an irreflexive transitive relation is a strict ordering.
\end{xca}

\begin{xca}\label{ex:conv-order}
Show that the converse of an ordering is an ordering.
\end{xca}

\begin{xca}\label{ex:least-unique}
Show that, if an ordered class has a least element, it is unique.
\end{xca}

\begin{xca}
Find a set with minimal elements, but no least element, with respect
to some ordering.
\end{xca}

\begin{xca}
  Is there an ordered class with a proper initial segment that is not
  a section?
\end{xca}

\begin{xca}\label{ex:sym}
Prove the claim in \parref{par:aRRa} that the
relation $\cR$ is symmetric if and only if $\Forall xx\cR=\cR x$.
\end{xca}

\begin{xca}\label{ex:aRaS}
  If $\cR$ and $\cS$ are binary relations, prove that they are equal
  if and only if $a\cR=a\cS$ for all $a$.
\end{xca}

\begin{xca}\label{ex:functional}
Prove that a binary relation $\cF$ is functional if and only if
 $\conv{\cF}/\cF=\Delta_{\ran{\cF}}$.
\end{xca}

\begin{xca}\label{ex:inj}
If $\cR$ is a binary relation, prove that
$\cR/\conv{\cR}\included\Delta_{\universe}$ if and only if
$\cR/\conv{\cR}=\Delta_{\dom{\cR}}$. 
\end{xca}

\begin{xca}\label{ex:conv-f-inv}
Show that the converse of an injective function is a function.
\end{xca}

\begin{xca}\label{ex:f-inj-bij}
Prove Theorem~\ref{thm:f-inj-bij}.
\end{xca}

\begin{xca}
Suppose $\cF\colon\cC\surjection\cD$.  Is there a
function $\cG$ from $\cD$ to $\cC$ such that $\cF\circ\cG=\id{\cD}$?
\end{xca}

\begin{xca}\label{ex:functions}
Prove Theorem~\ref{thm:functions}.
\end{xca}

\begin{xca}
Write out the Replacement Scheme (\parref{ax:replacement}) using only
the symbols of \parref{par:1}, and using an \emph{arbitrary} binary
formula $\phi(x,y)$ instead of $\cF$ (so your sentence will have
to express the condition that $\phi(x,y)$ defines a functional
relation). 
\end{xca}

\begin{xca}\label{ex:sep-from-rep}
  Prove the Separation Scheme from the Replacement Scheme.
\end{xca}

\begin{xca}\label{ex:dom-inj-set}
Prove Theorem~\ref{thm:dom-inj-set}.
\end{xca}

\chapter{Size and order}\label{ch:size-order}

%\setcounter{equation}{9}

\section{Cardinality}

\begin{parag}\label{par:equip}
Two
  classes $\cC$ and $\cD$ have the 
\defnplain{same cardinality}{}%
\dindexsub{same}{--- cardinality}%
\dindexsub{cardinal}{---ity}
  (or the 
\defnplain{same size}{}%
\dindexsub{same}{--- size}%
\dindex{size}) 
if there is a bijection between them.  In that case, we may
  write
  \begin{equation}\label{eqn:C-eq-D}
    \cC\equip \cD,
  \end{equation}
and we may also say that $\cC$ and $\cD$ are
\defnplain{equipollent}.%
\dindex{equipollent classes, equipollence of classes}
So we have a relation between classes, called 
\defnplain{equipollence}, which is an
equivalence-relation in the sense of~\parref{par:=inc}.  
Note however that the expression in~\eqref{eqn:C-eq-D} does \emph{not}
generally stand for a sentence of the logic of sets
(Exercise~\ref{ex:C-eq-D}). 
If a
\emph{set} is equipollent with a class, then that class is also a set,
by Replacement~(\parref{ax:replacement}).  We shall
generally be concerned only with equipollence of sets.
If $a\equip b$,
then we may write also
\begin{equation}\label{eqn:|a|=|b|}
  \card a=\card b;
\end{equation}
this \emph{does} stand for a sentence of the logic of sets
(Exercise~\ref{ex:a-eq-b}). 
Here, $\card a$ is the 
\defnplain{cardinality}{}%
\dindexsub{cardinal}{---ity} of $a$:
for now, we can understand this as the
equivalence-class $\cls x{x\equip a}$.  (See Exercise~\ref{ex:equip}.)
However, we shall ultimately (see~\parref{par:card}) be able to define
$\card a$ as a certain
\emph{set} that is equipollent with $a$.  In any case, we may write
\begin{equation*}
  \cC\nequip\cD
\end{equation*}
if $\cC$ and $\cD$ are not equipollent.
\end{parag}

\begin{parag}\label{par:injects}
  A class $\cD$ is \defn{larger}{} than or \defn{bigger}{}
  than, or
  has \defnplain{greater cardinality}{}
  than, a class $\cC$, if there is an
  injection from $\cC$ 
  into $\cD$, but no bijection; then
  $\cC$ is \defn{smaller}{} than, or has
  \defn{lesser cardinality}{} than, 
  the class $\cD$, and we may write
  \begin{equation*}
    \cC\pinjects \cD.
  \end{equation*}
If there is an injection from $\cC$ to $\cD$, then we write
\begin{equation*}
  \cC\injects\cD.
\end{equation*}
Therefore
\begin{equation*}
  \cC\injects\cD\text{\quad if and only if\quad
  }\cC\pinjects\cD\text{\quad or\quad }\cC\equip\cD. 
\end{equation*}
If $\cC\injects\cD$, and $\cD$ is a set, then so is
$\cC$, by Theorem~\ref{thm:dom-inj-set}. 
If $a\injects b$, then we may write also
\begin{equation*}
  \card a\leq\card b.
\end{equation*}
The relations $\injects$ and $\pinjects$ are transitive; the
former is reflexive, but the latter is irreflexive
(Exercise~\ref{ex:<-trans}).
Thus $\pinjects$ induces a strict ordering on cardinalities.
  By the
following theorem, $\injects$ is anti-symmetric on cardinalities; so
it induces a reflexive ordering on cardinalities.
\end{parag}

\begin{theorem}[Schroeder--Bernstein\footnote{The theorem is also
      called the Cantor--Bernstein Theorem,%
\index{theorem!Cantor--Bernstein Th---}%
\index{Cantor!---{}--Bernstein Theorem}
 as for example by Levy
  \cite[III.2.8, p.~85]{MR1924429}, who nonetheless observes that Dedekind
  gave the first proof in 1887.  The proof given here is due to
  Zermelo \cite[p.~208]{Zermelo-invest}.}]\label{thm:sch-b}%
\dindex{Schroeder--Bernstein Theorem}%
\dindexsub{theorem}{Schroeder--Bernstein Th---}
  $a\injects b\land b\injects a\lto a\equip b$.
\end{theorem}

\begin{proof}
  Suppose $f\colon a\injection b$ and $g\colon b\injection a$.  
We want to find a bijection $h$ from $a$ to $b$.  Since we have only
$f$ and $g$ to work with, we shall look for a subset $c$ of $b$ such
that we can define $h$ by
\begin{equation}\label{eqn:sch-b:h}
  h(x)=
  \begin{cases}
 g\inv(x),&\text{ if }x\in g\setimb c;\\
    f(x),&\text{ if }x\in a\setminus g\setimb c.
  \end{cases}
\end{equation}
To ensure that $h$ is injective, since $f$ and $g\inv$ are already
injective, we need only have 
$g\inv\setimb{g\setimb c}\cap f\setimb{a\setminus g\setimb
  c}=\emptyset$, 
that is,
\begin{equation}\label{eqn:ccapf}
  c\cap(f\setimb a\setminus(f\circ g)\setimb c)=\emptyset.
\end{equation}
To ensure that $h$ is surjective onto $b$, we need only have
$g\inv\setimb{g\setimb c}\cup f\setimb{a\setminus g\setimb
  c}=b$, 
that is,
\begin{equation}\label{eqn:ccupf}
  c\cup(f\setimb a\setminus(f\circ g)\setimb c)=b.
\end{equation}
These two conditions on $c$, given in~\eqref{eqn:ccapf}
and~\eqref{eqn:ccupf}, are equivalent to the conditions 
\begin{align*}
  c&\included b,&
b\setminus c&=f\setimb a\setminus(f\circ g)\setimb c
\end{align*}
and hence to
\begin{align}\label{eqn:sb}
  c&\included b,&
(b\setminus f\setimb a)\cup(f\circ g)\setimb c=c.
\end{align}
Now let $\cD$ be
  the class
  \begin{equation*}
    \cls x{x\included b\land(b\setminus f\setimb{a})\cup(f\circ
    g)\setimb{x}\included x}.
  \end{equation*}
Then $\cD$ contains $b$.  In particular, $\cD$ is non-empty,
so (by \parref{par:comprehension}) its intersection is a set.  If
$c=\bigcap\cD$, then~\eqref{eqn:sb} holds (Exercise~\ref{ex:sb}).
\end{proof}

\begin{parag}
  The Schroeder--Bernstein Theorem is often proved in the following
  way.  Assuming $f\colon a\injection b$ and $g\colon b\injection a$,
  we make the
  following definitions:
  \begin{align*}
    a_1&=g\setimb b,&
    a_2&=g\setimb{b_1},&
    a_3&=g\setimb{b_2},&
    a_4&=g\setimb{b_3},\\
    b_1&=f\setimb a,&
    b_2&=f\setimb{a_1},&
    b_3&=f\setimb{a_2},&
    b_4&=f\setimb{a_3},
  \end{align*}
and so on.  (We are not ready to be precise about what \Eng{and so on}
means here; this is why the proof above does not follow the present
lines.)  Then $a\includes a_1\includes a_2\includes\dotsb,$ and
$b\includes b_1\includes b_2\includes\dotsb$.  Also, $f$ determines a
bijection from $a\setminus a_1$ to $b_1\setminus b_2$, from
$a_2\setminus a_3$ to $b_3\setminus b_4$, and so on, while $g\inv$
determines a bijection from $a_1\setminus a_2$ to $b\setminus b_1$, from
$a_3\setminus a_4$ to $b_2\setminus b_3$, and so on.  Then there
is a bijection $h$ from $a$ to $b$ that agrees with these, and agrees
with $f$ at the elements of $a$ that are not yet accounted for.  In
fact, this will be the same $h$ found in the proof above.
\end{parag}

\begin{theorem}[Cantor]\label{thm:pow}%
\dindexsub{Cantor}{--- Theorem}%
\dindexsub{theorem}{Cantor Th---}
The power class of a set is larger than the set itself:
\begin{equation*}
  a\pinjects\pow a.
\end{equation*}
\end{theorem}

\begin{proof}
We have $x\mapsto\{x\}\colon a\injection\pow a$, so $a\injects\pow a$.
  Suppose $f\colon a\injection\pow a$.
  Let $b$ be the set $\set{x\in a}{x\notin f(x)}$.  Then
  \begin{align}\label{eqn:cantor1}
    c\in b&\lto c\in b\setminus f(c),\\\label{eqn:cantor2}
c\in a\setminus b&\lto c\in f(c)\setminus b
  \end{align}
(Exercise~\ref{ex:cantor}).
Thus, if $c\in a$, then $f(c)\neq b$.  So $b\notin\ran f$.  Therefore,
there is no bijection from $a$ to $\pow a$; so $a\pinjects\pow a$.
\end{proof}

\begin{parag}
  Note well how the preceding proof requires $b$ to be a set
  (Exercise~\ref{ex:why-set}).  If $a$
  were a proper class, then the proof would fail.  In particular, we
  cannot yet address the cardinality of $\pow{\pow a}$, since we have
  not established that $\pow a$ is a set.  This is what
  the following axiom is for.
\end{parag}

\begin{axiom}[Power Set]\label{ax:power-set} 
\dindexsub{axiom}{Power Set A---}%
\dindexsub{power set}{P--- Axiom}
%\axiomname{Power-Set: \[\Exists xx=\pow a.\]}
  The power class of a set is a set: that is,
  \begin{equation*}
    \Exists xx=\pow a.
  \end{equation*}
\end{axiom}

\begin{parag}
  As foretold in \parref{par:infinitary}, we may now refer to the
  power class of a set as its \defn{power set}.
  We can form 
\techplain{chain}{s:}%
\tindex{chain (ordered set)}
  \begin{equation*}
    a\pinjects\pow a\pinjects\pow{\pow a}\pinjects\pow{\pow{\pow
    a}}\pinjects\dotsb. 
  \end{equation*}
In particular, letting $a$ be the empty set, we have
\begin{equation*}
  \emptyset\pinjects
\{\emptyset\}\pinjects
\{\emptyset,\{\emptyset\}\}\pinjects
\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\emptyset,\{\emptyset\}\}\}
\pinjects\dotsb; 
\end{equation*}
but this is hardly surprising.  Cantor's Theorem becomes
remarkable when we have \techplain{infinite}{}%
\tindexsub{infinite}{--- set}%
\tindexsub{set}{infinite ---} sets.  
\end{parag}

\section{Ordinary induction}

\begin{parag}\label{par:inf}
A class is
  \defnplain{infinite}{}%
\dindexsub{infinite}{--- class}
 if it is equipollent with a proper subclass of
  itself.  Let us first observe that there
  \emph{is} an infinite class: 
\end{parag}

\begin{theorem}\label{thm:Vinf}
  The universal class~$\universe$ is
  infinite.
\end{theorem}

\begin{proof}
  By the Power Set Axiom
  (\parref{ax:power-set}), we
  have a function $x\mapsto\pow x$ from $\universe$ to itself.  This
  function is
  injective, since, if $a\neq b$, then we may assume that $a\setminus
  b$ is non-empty, so that $a\setminus b\in\pow a\setminus\pow b$ and
  hence $\pow a\neq\pow b$ (Exercise~\ref{ex:a-bneq0}).
  Also, every power set contains $\emptyset$, but $\emptyset$ contains
  nothing, so $\emptyset$ is not a power set; thus, $x\mapsto\pow x$
  is not surjective onto $\universe$.  So~$\universe$ is infinite, by
  definition.
\end{proof}

\begin{parag}\label{par:Vnotset}
  By the Russell Paradox (\parref{thm:R}), $\universe$
  has a subclass that is not a set; so $\universe$ itself is not a
  set, by Separation (\parref{ax:comprehension}).  Is
  there an infinite \emph{set?}  We cannot now \emph{prove} that there
  is an infinite set, or that there is not.
  In \parref{ax:infinity}, we shall \emph{assume} that there is an
  infinite set; but even without this assumption, we
  shall be able to find an infinite \emph{class} that we can define as
  the class of \tech{natural number}{s.}%
\tindexsub{number}{natural ---}
\end{parag}

\begin{parag}
Indeed, suppose $\cC$ is an infinite class.  This means there is an
injection $\cF$
from~$\cC$ into itself, and there is an element $i$ of $\cC$ that is
not in $\cF\setimb{\cC}$.  It appears we can form the list
\begin{equation}\label{eqn:list}
  i,\ \cF(i),\ \cF(\cF(i)),\ \cF(\cF(\cF(i))),\ \dots
\end{equation}
whose entries are distinct elements of $\cC$.  This list would appear
to have the
essential properties of the natural numbers.  However, it is not yet
clear that the members of the list compose a \emph{subclass} of $\cC$.
\end{parag}

\begin{parag}\label{par:iterative}
Suppose all we know is that $\cF$ is a function, and $\cC$ is a
subclass of $\dom{\cF}$ 
such that $\cF\setimb{\cC}\included\cC$.  Then $\cC$ is
\defnplain{closed under}{}%
\dindex{closed under an operation}%
\dindex{under: closed --- an operation}%
\dindex{operation}{closed under an ---}
$\cF$.  Suppose also $i\in\cC$.  Then we can still form a list
as in~\eqref{eqn:list}, although possibly not all of the entries are
distinct.  Indeed, in the simplest example,
$\cC=\{\emptyset\}$, and $\cF$ is the identity on this set, and $i$ is
the unique element $\emptyset$ of the set.  Then the
list~\eqref{eqn:list} is just $\emptyset$, $\emptyset$, $\emptyset$,
$\emptyset$, \dots
In general, there would appear to be the three kinds of
possibilities depicted in Figure~\ref{fig:induction}, namely 
\begin{inparaenum}
  \item
no entry in~\eqref{eqn:list} is repeated,
\item
some entries are repeated, but not $i$, or
\item
every entry is repeated.
\end{inparaenum}
\begin{figure}[t]
\mbox{}\hfill
  \begin{pspicture}(0,6)
\psset{linewidth=0.5mm, arrowsize=2mm 2}
\psset{arrows=->}
    \psline(0,6)(0,0)
\psdots(0,6)(0,5)(0,4)(0,3)(0,2)(0,1)
\uput[l](0,6){$i$}
\uput[l](0,5){$\cF(i)$}
\uput[l](0,4){$\cF(\cF(i))$}
  \end{pspicture}
\hfill
\begin{pspicture}(1,6)
\psset{linewidth=0.5mm, arrowsize=2mm 2}
    \psline(0,6)(0,2)
    \psarc(1,2){1}{180}{450}
\psset{arrows=->}
\psline(1,3)(0.2,3)
\psdots(0,6)(0,5)(0,4)%(0,3)(0,2)(0,1)  
\end{pspicture}
\hfill
\begin{pspicture}(2,6)
\psset{linewidth=0.5mm, arrowsize=2mm 2}
    \psline(0,6)(0,5)
    \psarc(1,5){1}{180}{450}
\psset{arrows=->}
\psline(1,6)(0.2,6)
\psdots(0,6)(0,5)%(0,4)(0,3)(0,2)(0,1)    
\end{pspicture}
\hfill\mbox{}
  \caption{Some iterative structures}\label{fig:induction}
\end{figure}
In any case, let us refer to a class $\cC$, with element $i$, and
closed under~$\cF$, as an \defn{iterative structure}.%
\dindexsub{structure}{iterative ---}  I choose this term%
\footnote{Stoll \cite{MR83e:04002} uses the term 
\techplain{unary system},%
\tindexsub{unary}{--- system}%
\tindexsub{system}{unary ---}
though he assumes $\cC$ is a set.} 
 simply
because we can start with $i$ and apply $\cF$ over and over, while
staying within $\cC$.  (Structures in
  general will be defined in \sref{sect:structures}.)  We may denote
  this iterative structure by
\begin{equation*}
  (\cC,i,\cF).
\end{equation*}
It is not clear whether this symbolism denotes a class; but nor does it
  really matter.  An iterative structure is simply a class `equipped with'
  a distinquished element and a function under which the class is
  closed; we write these three things together, to indicate that we
  are thinking of them together.  For example, from
  $\{\emptyset\}$ we obtained the iterative structure
  $(\{\emptyset\},\emptyset,\id{\{\emptyset\}})$; in the proof of
  Theorem~\ref{thm:Vinf}, we used the iterative structure
  $(\universe,\emptyset,x\mapsto\pow x)$.  In general, if
  $(\cC,i,\cF)$ is an iterative structure, then we may call $i$ the
  \defnplain{initial element}{}
\dindexsub{initial}{--- element}
 of the structure, and $\cF$ the operation
  of \defn{succession}; then $\cF(a)$ is the \defn{successor}{} of
  $a$, whenever $a\in\cC$.
\end{parag}

\begin{parag}
  Given a set $i$ belonging to the domain of a function $\cF$, we may
  consider the \defn{family}{} of classes that contain $i$ and are
  closed under $\cF$.  (Compare~\parref{par:=classes}.)
  This family (like all families of classes) is ordered 
  by inclusion (\parref{par:=inc}).  The notions of \defn{least}{} and
  \defn{minimal}{} member, defined in~\parref{par:minimum}, can be adapted
  to this ordered family.  A \emph{minimal} member $\cC$ of
  the family can be said to admit \defnplain{proof by (ordinary)
  induction};% 
\dindexsub{proof}{--- by (ordinary) induction}%
\dindexsub{ordinary}{proof by --- induction}%
\dindex{induction} 
more precisely, the corresponding iterative structure $(\cC,i,\cF)$
  admits induction.
This means the following.  Suppose that $\cD\included\cC$ and that we can
  establish two claims, namely 
  \begin{enumerate}
    \item
$i\in\cD$ (the 
\defnplain{base}{}%
\dindexsub{base}{--- of an induction}
 of the induction); and 
\item
if $a\in\cD$ (the \defnplain{inductive hypothesis}{}%
\dindexsub{inductive}{--- hypothesis}), then $\cF(a)\in\cD$.
\end{enumerate}
Then $\cD$ belongs to the family of which $\cC$ is a minimal member, and
so we have proved, by induction, that~$\cD=\cC$.  A basic lemma proved
by induction is the following.
\end{parag}

\begin{lemma}\label{lem:1-or-successor}
  Suppose the iterative structure $(\cC,i,\cF)$ admits induction.
  Then 
  \begin{equation*}
  \cC=\{i\}\cup\cF\setimb{\cC}.
  \end{equation*}
\end{lemma}

\begin{proof}
  The subclass $\{i\}\cup\cF\setimb{\cC}$ of $\cC$ contains $i$,
  and if it contains an element $a$ of $\cC$, then it contains
  $\cF(a)$; therefore it \emph{is} $\cC$.
\end{proof}

\begin{parag}\label{par:rec-set}
For $\cC$ to admit proof by induction, it is sufficient that $\cC$ be
the \emph{least}
member of the family of classes that contain the set~$i$ and are
closed under the function~$\cF$. 
In this case, the iterative structure $(\cC,i,\cF)$ can be said to be 
\defnplain{recursive};%
\dindexsub{recursive}{--- structure}%
\dindexsub{structure}{recursive ---}
and $\cC$ is
\defnplain{recursively defined},%
\dindexsub{recursive}{---ly defined structure}%
\dindexsub{structure}{recursively defined ---}
or defined by \defnplain{(ordinary) recursion}.%
\dindexsub{ordinary}{--- recursion}%
\dindexsub{recursion}{ordinary ---}%
\dindex{definition by recursion, recursive ---}
This means that
\begin{enumerate}
  \item
$\cC$ is a class $\cD$ such that
    \begin{enumerate}
      \item
$i\in\cD$, and
\item
$a\in\cD\lto\cF(a)\in\cD$; and
    \end{enumerate}
\item
whenever $\cD$ is such a class, then $\cC\included\cD$.
\end{enumerate}
Then $\cC$ can be
denoted suggestively by
\begin{equation*}
  \{i,\cF(i),\cF(\cF(i)),\dots\}.
\end{equation*}
Note well that possibly
$\cF\restriction\cC$ is not injective, and possibly
$i\in\cF\setimb{\cC}$.  (However, 
these possibilities seem to be mutually exclusive.)
%(Exercise~\ref{ex:rec-counter}).
\end{parag}

\begin{lemma}\label{lem:rec=ind}
An iterative structure is recursive if and only if it admits
induction.
\end{lemma}

\begin{proof}
The `only-if' part was implicit in \parref{par:rec-set}:
\emph{Least} elements
  of a class (with respect to some ordering) are minimal, and
  therefore recursive structures admit induction.
Now suppose $(\cC,i,\cF)$ is \emph{not} recursive.  Then there is a
  class $\cD$ containing $i$ and closed under $\cF$ of which $\cC$ is
  not a subclass.  Then $\cC\cap\cD$ is a proper subclass of $\cC$
  that contains $i$ and is closed under $\cF$ (by
  Theorem~\ref{thm:functions}).  Hence $(\cC,i,\cF)$ does not admit
  induction.
\end{proof}

\section{Ordinary recursion}

\begin{parag}\label{par:rec}
Suppose $(\cC,i,\cF)$ is a recursive structure, and
$(\cD,j,\cG)$ is another iterative structure (not necessarily
recursive).  There \emph{may be} a function $\cH$ from $\cC$ to $\cD$
satisfying two conditions:
\begin{enumerate}
  \item
$\cH(i)=j$,
\item
$a\in\cC\lto \cH(\cF(a))=\cG(\cH(a))$, that is,
  $\cH\circ\cF=\cG\circ\cH$ on $\cC$.
\end{enumerate}
These conditions can be depicted as in
Figure~\ref{fig:commutative-diagram}.
\begin{figure}[t]
\begin{equation*}
  \begin{CD}
    \{i\}@>{\id{\{i\}}}>>\cC @>{\cF}>> \cC\\
@V{\cH\restriction{\{i\}}}VV @VV{\cH}V @VV{\cH}V\\
\{j\} @>>{\id{\{j\}}}>\cD @>>{\cG}> \cD
  \end{CD}
\end{equation*}
\caption[A commutative diagram]{Here is a 
\techplain{commutative diagram}{}% 
\tindexsub{commutative}{--- diagram}%
\tindex{diagram, commutative} 
showing the definition of a function~$\cH$ by ordinary recursion.
Paths through the diagram that follow the arrows
  represent compositions of functions.  If two paths start from the
  same point and end at the same point, then the corresponding
  compositions of functions are equal: this is why the diagram is
  called \Eng{commutative.}}\label{fig:commutative-diagram}
\end{figure}
The first rule says what $\cH(i)$ is; the second says how to
obtain $\cH(\cF(a))$ from $\cH(a)$.  
If such a function $\cH$ does exist, then, since $(\cC,i,\cF)$ admits
induction by Lemma~\ref{lem:rec=ind}, we can use induction to prove
that $\cH$ is uniquely 
determined by these rules.
We may then say that $\cH$ is \defnplain{recursively defined}{}%
\dindexsub{recursive}{---ly defined function}%
\dindexsub{function}{recursively defined ---}%
\dindex{definition by recursion, recursive ---}
 by the
given rules; also, $\cH$ is a \defn{homomorphism}{} from $(\cC,i,\cF)$
to $(\cD,j,\cG)$.  See Corollary~\ref{cor:rec-uni} below.
\end{parag}

\begin{parag}\label{par:rec2}
  Note first another possible kind of recursive definition.
 If $(\cC,i,\cF)$ is recursive, 
$\cE\included\cD$, and
  $\cG:\cD\to\cD$, then perhaps there is a subclass $\cR$
  of $\cC\times\cD$ such that (in the notation of \parref{par:aRRa})
  \begin{enumerate}
    \item
$i\cR=\cE$,
\item
$a\in\cC\lto\cF(a)\cR=\cG\setimb{a\cR}$.
  \end{enumerate}
(Note that $\cF(a)\cR$ means $b\cR$, where $b=\cF(a)$.)
Then $\cR$ too is uniquely determined by these rules, so it too is
\defnplain{recursively defined},%
\dindexsub{recursive}{---ly defined relation}%
\dindexsub{relation}{recursively defined ---}%
\dindex{definition by recursion, recursive ---}
by the following.
\end{parag}

\begin{theorem}\label{thm:rec-uni}
  Suppose   $(\cC,i,\cF)$ is recursive,
$\cE\included\cD$, and  $\cG:\cD\to\cD$.  Then there is at most one
  relation $\cR$ as in \parref{par:rec2}.  
\end{theorem}

\begin{proof}
  Suppose $\cR_0$ and $\cR_1$ are two such relations.  Let
  \begin{equation*}
    \cC_1=\cls x{x\in\cC\land x\cR_0=x\cR_1}.
  \end{equation*}
Since $i\cR_0=\cE=i\cR_1$, we have $i\in\cC_1$.  Suppose $a\in\cC_1$,
so that $a\in\cC$ and $a\cR_0=a\cR_1$.  Then
\begin{equation*}
  \cF(a)\cR_0=\cG\setimb{a\cR_0}=\cG\setimb{a\cR_1}=\cF(a)\cR_1,
\end{equation*}
so $\cF(a)\in\cC_1$.  By (Lemma~\ref{lem:rec=ind} and) induction,
$\cC_1=\cC$. Since $\dom{\cR_0}\included\cC$ and
$\dom{\cR_1}\included\cC$, we conclude (by Exercise~\ref{ex:aRaS})
that $\cR_0=\cR_1$. 
\end{proof}

\begin{corollary}\label{cor:rec-uni}
  Suppose $(\cC,i,\cF)$ is recursive, and $(\cD,j,\cG)$ is iterative.
  Then there is at most one homomorphism from
  $(\cC,i,\cF)$ to $(\cD,j,\cG)$.
\end{corollary}

\begin{proof}
  Supposing such a homomorphism $\cH$ exists, it is a relation, namely
  a subclass 
  $\cR$ of $\cC\times\cD$.  Let $\cE=\{\cH(i)\}$.  Then
    \begin{enumerate}
      \item
$i\cR=\{\cH(i)\}=\cE$;
\item
$a\in\cC\lto\cF(a)\cR=
  \{\cH(\cF(a))\}=\{\cG(\cH(a))\}=\cG\setimb{\{\cH(a)\}}=\cG\setimb{a\cR}$.  
    \end{enumerate}
By the theorem, $\cR$ is unique, so $\cH$ is unique.
\end{proof}

\section{Countably infinite classes}

\begin{parag}\label{par:c-inf}
  Suppose $\cC$ is infinite.  By~\parref{par:inf}, this means:
  \begin{enumerate}
    \item
  there is an element $i$ of $\cC$, and
\item
 there is a function $\cF$
  under which $\cC$ is closed,
such that
    \item
$i\notin\cF\setimb{\cC}$, and
\item
$\cF\rest{\cC}$ is injective.
  \end{enumerate}
Let us refer then to the iterative structure $(\cC,i,\cF)$ as an
\defn{infinitary structure}.%
\dindexsub{structure}{infinitary ---}
Suppose also
\begin{enumerate}
  \setcounter{enumi}{4}
\item
the infinitary structure $(\cC,i,\cF)$ admits induction (or is
recursive). 
\end{enumerate}
Then $\cC$ is called 
\defn{countably infinite}.%
%\dindexsub{countable}{countably infinite}%
\dindexsub{infinite}{countably --- class}
I propose to refer to $(\cC,i,\cF)$ as an 
\defn{arithmetic structure}.%
\dindexsub{structure}{arithmetic structure}%
\footnote{Stoll \cite{MR83e:04002} uses the term 
\tech{integral system},%
\tindexsub{system}{integral ---}
 assuming $\cC$ is a set.} 
The five numbered 
conditions here are sometimes referred to as the 
\defn{Peano axioms}.%
\dindexsub{axiom}{Peano ---s}%
\footnote{In \S~1 of Chapter~1 of the \emph{Foundations of
  Analysis}~\cite{MR12:397m}, Landau formulates conditions like these 
  as Axioms~1--5 (respectively) for the \tech{natural number}{s;} he
  refers to these axioms as Peano's axioms in his `Preface for the Teacher'. 
However, for Peano~\cite{Peano}\nocite{MR1890980}, the five
  conditions are numbered 1, 6, 8, 7, and 9 in a list of \emph{nine}
  axioms for the positive integers.  (The other four of these axioms
  are purely logical.)  A year earlier, Dedekind~\cite[II.VI, (71),
  p.~67]{MR0159773} 
  defines the natural numbers as the elements of a set $N$ on which a
  function $\phi$ is defined, and which has an element $1$, such that
  \begin{enumerate}
\item
$\phi\setimb{N}\included N$;
\item
$N$ is the intersection of the class of all sets that contain $1$ and
  are closed under $\phi$ [however, Dedekind does not have the
  distinction between sets and classes];
\item
$1\notin\phi\setimb N$;
\item
$\phi$ is injective.
\end{enumerate}
}
Does any class $\cC$ meet these conditions?  The following gives a
sufficient condition.
\end{parag}

\begin{theorem}\label{thm:countably-infinite-subset}
  Every infinite \emph{set} has a countably infinite subset.
\end{theorem}

\begin{proof}
  Say $a$ is infinite.  Then there is an element $i$ of $a$ and an
  injection $f$ from $a$ into $a\setminus\{i\}$.  Let $\cC$ be the
  class
  \begin{equation*}
    \cls x{x\included a\land i\in x\land f\setimb x\included x};
  \end{equation*}
this is the class of subsets of $a$ that contain $i$ and are closed
under $f$.  Then $\bigcap\cC$ also belongs to $\cC$
(Exercise~\ref{ex:nc-in-c}) and is a set $a_0$.  Then $(a_0,i,f)$ is
recursive, so $a_0$ is countably infinite.
\end{proof}

\begin{parag}
  If we have two sets $a$ and $b$, then surely we can form a set that
  \emph{includes} $a$ and \emph{contains} $b$; that is, surely the
  class $a\cup\{b\}$ is a set.  The Pairing Axiom is a special case of
  this observation (Exercise~\ref{ex:pair-from-augmentation}); the Binary
  Union Axiom (\parref{ax:binary-union}) will be a generalization. 
\end{parag}

\begin{axiom}[Augmentation]\label{ax:augmentation}
\dindexsub{axiom}{Augmentation A---}%
\dindex{Augmentation Axiom}
%\axiomname{Weak Union: \[\Exists xx=a\cup\{b\}.\]}
  The union of a set and a singleton is a set:
  \begin{equation*}
    \Exists xx=a\cup\{b\}.
  \end{equation*}
\end{axiom}


\begin{theorem}[Recursion]\label{thm:rec}%
\dindexsub{recursion}{R--- Theorem}%
\dindexsub{theorem}{Recursion Theorem}
  Suppose $(\cC,i,\cF)$ is arithmetic, 
  $\cE\included\cD$, and
$\cG\colon\cD\to\cD$.
  Then there is (uniquely, by Theorem~\ref{thm:rec-uni}) one
  recursively defined relation
  $\cR$ as in~\parref{par:rec2}.
\end{theorem}

\begin{proof}
We want $\cR$ to be a relation from $\cC$ to $\cD$ such that
\begin{gather}\label{eqn:rec-thm-1}
 i\mathrel{\cR}a\iff a\in\cE,\\\label{eqn:rec-thm-2}
a\mathrel{\cR}b\lto\cF(a)\mathrel{\cR}\cG(b),\\\label{eqn:rec-thm-3}
\cF(a)\mathrel{\cR}c\lto\Exists x(a\mathrel{\cR}x\land\cG(x)=c).
\end{gather}
To this end, we define $\cR$ as the union of the class comprising
every subset of $\cC\times\cD$
whose elements can be listed as
\begin{equation*}
  (i,d),\ (\cF(i),\cG(d)),\ (\cF(\cF(i)),\ \cG(\cG(d))),\ \dots
\end{equation*}
for some $d$ in $\cE$.  To be precise,
  let $\cB$ be the class of sets $a$ whose every element is either
  $(i,d)$ for some $d$ in $\cE$ or else $(\cF(b),\cG(c))$ for some
  $(b,c)$ in $a\cap(\cC\times\cD)$.  That is, $\cB$ is the
  class
  \begin{equation*}
    \cls x{\Forall y(y\in x\lto\phi(y)\lor\psi(x,y))},
  \end{equation*}
where $\phi(y)$ is the formula
\begin{equation*}
  \Exists z(z\in\cE\land y=(i,z)),
\end{equation*}
and $\psi(x,y)$ is the formula
\begin{equation*}
  \Exists u\Exists v\bigl((u,v)\in x\cap(\cC\times\cD)\land
  y=(\cF(u),\cG(v))\bigr). 
\end{equation*}
Then $\cB\included\pow{\cC\times\cD}$, by induction.  
Let $\cR=\bigcup\cB$.
If $i\mathrel{\cR}a$, then $(i,a)\in b$ for some $b$ in $\cB$; but then
$\lnot\psi(b,(i,a))$ (since $i\not\in\cF\setimb{\cC}$, since
$(\cC,i,\cF)$ is arithmetic); so $\phi((i,a))$, and therefore
$a\in\cE$.
Conversely, if $a\in\cE$, then $\phi((i,a))$, so $\{(i,a)\}\in\cB$, so
$i\mathrel{\cR} a$.  This establishes~\eqref{eqn:rec-thm-1}.

Suppose $a\mathrel{\cR}b$.  Then $(a,b)\in c$ for some $c$ in $\cB$, and
then $\psi(c,(\cF(a),\cG(b)))$, so $c\cup\{(\cF(a),\cG(b))\}\in\cB$
(here we use the Augmentation Axiom).
Therefore $\cF(a)\mathrel{\cR}\cG(b)$.  Thus,~\eqref{eqn:rec-thm-2}.

Suppose finally $\cF(a)\mathrel{\cR}c$ for some $c$.  Then
$(\cF(a),c)\in d$ for some $d$ in $\cB$, so $\psi(d,(\cF(a),c))$
(again since $i\notin\cF\setimb{\cC}$).  Hence
$(\cF(a),c)=(\cF(e),\cG(k))$ for some $(e,k)$ in $d$.  Since $\cF$ is
injective on $\cC$, we have $a=e$, so $(a,k)\in d$ and
$a\mathrel{\cR}k$.  Thus,~\eqref{eqn:rec-thm-3}, and $\cR$ is as
desired. 
\end{proof}

\begin{corollary}[Recursion]\label{cor:rec}
  Suppose $(\cC,i,\cF)$ is an arithmetic, and $(\cD,j,\cG)$
  an iterative, structure.  Then there is (uniquely, by
  Corollary~\ref{cor:rec-uni}) a homomorphism from $\cC$ to $\cD$.
\end{corollary}

\begin{proof}
  Exercise~\ref{ex:cor-rec}.
\end{proof}

\begin{parag}\label{par:isom1}
  An 
\defnplain{isomorphism}{}%
\index{function|seealso{isomorphism}}
\dindexsub{isomorphism}{iterative structures}
of iterative structures is a bijective
  homomorphism.  If an isomorphism does exist, then its inverse is
  also an isomorphism (Exercise~\ref{ex:isom1}), and
the two structures are
  \defnplain{isomorphic}.%
\end{parag}

\begin{theorem}\label{thm:cnt-equip}
All countably infinite classes are equipollent; indeed, all arithmetic
structures are isomorphic.
\end{theorem}

\begin{proof}
Suppose $(\cC,i,\cF)$ and $(\cD,j,\cG)$ are arithmetic structures.
By the corollary (\ref{cor:rec}) to the Recursion Theorem,
  there are a homomorphism $\cH$ from
 $(\cC,i,\cF)$ to $(\cD,j,\cG)$ and a homomorphism $\cK$ from
 $(\cD,j,\cG)$ to $(\cC,i,\cF)$.
 Then also
\begin{enumerate}
  \item
$(\cK\circ \cH)(i)=i$;
\item
$(\cK\circ \cH)\circ \cF=\cF\circ(\cK\circ \cH)$ since,
  by~\eqref{eqn:comp-f-assoc}, we can compute
$(\cK\circ \cH)\circ \cF=\cK\circ(\cH\circ \cF)=\cK\circ(\cG\circ
  \cH)=(\cK\circ \cG)\circ 
  \cH=(\cF\circ \cK)\circ \cH=\cF\circ(\cK\circ \cH)$.
\end{enumerate}
Thus $\cK\circ \cH$ is a homomorphism from $(\cC,i,\cF)$ to itself.
But so is $\id{\cC}$.  Therefore
(\parref{cor:rec-uni}) $\cK\circ \cH=\id{\cC}$.  Likewise $\cH\circ
\cK=\id{\cD}$. 
Therefore, by Theorem~\ref{thm:f-inj-bij}, $\cH$ is a bijection from
$\cC$ to~$\cD$. 
\end{proof}

\section{Are there countably infinite classes?}

\begin{parag}\label{par:post-inf}
There are infinite classes, by Theorem~\ref{thm:Vinf}.  
  We have defined countably infinite classes (\parref{par:c-inf}),
  and we have shown that all of them have the
  same cardinality (\parref{thm:cnt-equip}).  
We have shown that, if there
  is an infinite \emph{set,}
  then there is a countably infinite set
  (\parref{thm:countably-infinite-subset}).  We have \emph{not} shown
  that a countably infinite \emph{class} exists, much less a set.  
\end{parag}


\begin{parag}
If we believe that there are
  infinite sets, then we can \emph{postulate} their existence as an
  axiom: the
\techplain{Axiom of Infinity}.%
\tindexsub{axiom}{A--- of Infinity}%
\tindex{Infinity, Axiom of}
But this is a delicate matter, since we know from the Russell Paradox
  that some infinite classes \emph{cannot} be sets.  Again, by
  Theorem~\ref{thm:countably-infinite-subset}, if there is an infinite
  set at all, then there must be a \emph{countably} infinite set.  So
  safest form of the Axiom of Infinity is that there is a
  countably infinite set.  Then we can obtain larger sets by the
  Power Set Axiom and Cantor's Theorem.
\end{parag}

\begin{parag}
  However, all of our axioms so far, except for Extension, are
  assertions that certain classes are sets.  One way to write the
  Axiom of Infinity in the same style is as follows.  Pick an
  infinitary structure
  $(\cC,i,\cF)$.  For example, we might pick
  $(\universe,\emptyset,x\mapsto\pow x)$.  Let $\cD$ be the class
  \begin{equation*}
    \cls x{x\included\cC\land i\in\cC\land\cF\setimb x\included x}
  \end{equation*}
(as in the proof of
  Theorem~\ref{thm:countably-infinite-subset}).  All elements of $\cD$ 
  are infinite sets.  If there \emph{are} elements of $\cD$,
then $\bigcap\cD$ is a countably infinite set (and an element of
  $\cD$).  On the other hand, if $\cD$ is empty,
  then $\bigcap\cD=\universe$ (Exercise~\ref{ex:cop}), which is not a
  set (\parref{par:Vnotset}).  So, by postulating that $\bigcap\cD$
  is a set, we would ensure that $\cD$ is not empty, so that there
  would be
  infinite sets.  However, before asserting that $\bigcap\cD$ is a
  set, we would not have \emph{exhibited} an element of~$\cD$.
  The question 
  arises:  Can we exhibit a countably infinite \emph{class} without
  assuming that there are infinite sets?
\end{parag}

\begin{parag}\label{par:aF}
We can try to define a countably infinite class as in the proof of the
Recursion Theorem (\ref{thm:rec}).
Suppose
$(\cC,i,\cF)$ is an infinitary structure, and let
\begin{equation}\label{eqn:D}
\cD=  \cls x{\Forall y(y\in x\lto y=i\lor\Exists z(z\in
    x\cap\cC\land\cF(z)=y))}.
\end{equation}
Then $\cD\included\pow{\cC}$, and $\{i\}\in \cD$; also, if $b\in
\cD$, and $a\in b$, then $b\cup\{\cF(a)\}\in \cD$.
Thus, $i\in\bigcup \cD$; and if $a\in\bigcup \cD$, then
$\cF(a)\in\bigcup \cD$.  So
$(\bigcup\cD,i,\cF)$ is an infinitary structure.  However, it is
not clearly arithmetic.  For example, suppose $\cF$
is the injective function $x\mapsto\{x\}$ on $\universe$, and $i$ is
$\emptyset$.  If there is a set $a$ that is equal to the set $\{a\}$,
then $\cF(a)=a$, so that $\{a\}\in \cD$, and then $a\in\bigcup
\cD$.  So $\bigcup\cD$ properly includes $\bigcup\cD\setminus\{a\}$,
although the latter class still contains $i$ and is closed under
$\cF$.  So, in this case, $(\bigcup\cD,i,\cF)$
is \emph{not}
arithmetic, and we do not know whether $\bigcup\cD$ is countably
infinite.  We shall ultimately rule out the possibility that $a=\{a\}$
with the Foundation Axiom (\ref{ax:foundation}).  Meanwhile, we can
proceed as follows.
\end{parag}

\section{Well-ordered classes}\label{sect:wo}

\begin{parag}\label{par:strong}
Suppose $(\cC,<)$ is a strict total order.
\begin{enumerate}
  \item
$(\cC,<)$ 
\defnplain{admits (proof by) strong}{} 
or 
\defnplain{transfinite induction}{}%
\dindexsub{proof}{--- by strong or transfinite induction}%
\dindexsub{strong}{--- induction}%
\dindexsub{transfinite}{--- induction}%
\dindexsub{induction}{strong or transfinite ---} 
if
    \begin{equation*}
      \Forall x(x\in\cC\land\pred x\included\cC_0\lto x\in\cC_0)
\lto\cC_0=\cC
    \end{equation*}
whenever $\cC_0\included\cC$;
\item
$(\cC,<)$ 
\defnplain{admits (definition by) strong}{}
 or 
\defnplain{transfinite recursion}{}%
\dindexsub{strong}{--- recursion}%
\dindexsub{transfinite}{--- recursion}%
\dindexsub{recursion}{strong or transfinite ---} 
if
  \begin{inparaenum}
    \item
all sections are sets, and
\item
for every class $\cD$ and every
  function $\cF$ from $\pow{\cD}$ to $\cD$, there is a unique function
  $\cG$ from $\cC$ to $\cD$ such that
  \begin{equation}\label{eqn:strong-rec}
    \cG(a)=\cF(\cG\setimb{\pred a}).
  \end{equation}
  \end{inparaenum}
\end{enumerate}
\end{parag}

\begin{theorem}\label{thm:o-eq}
  Suppose $(\cC,<)$ is a strict total order whose every section
  is a set.  The following conditions are equivalent:
  \begin{enumerate}
    \item\label{item:wo}
$(\cC,<)$ is well-ordered;
    \item\label{item:si}
$(\cC,<)$ admits induction;
    \item\label{item:sr}
$(\cC,<)$ admits recursion.
  \end{enumerate}
\end{theorem}

\begin{proof}
There are three implications to prove.
\begin{asparaenum}[1.]
  \item
Suppose~\eqref{item:wo}.
If $\cC_0\pincluded\cC$, then $\cC\setminus\cC_0$ is non-empty, so it
has a least element $a$; then $\pred a\included\cC_0$, but
$a\notin\cC_0$.  The contrapositive of this implication
  is~\eqref{item:si}.

\item
Suppose next~\eqref{item:si}, and $\cD$ is a class, and
$\cF\colon\pow{\cD}\to\cD$.
We show by induction that, for all
$a$ in $\cC$, there is a unique function $g_a$ with domain $\pred
a\cup\{a\}$ such that
\begin{equation*}
  g_a(c)=\cF(g_a\setimb{\pred c})
\end{equation*}
whenever $c\leq a$.  Suppose the claim holds whenever $a<b$.
If $a<d<b$, then 
$g_d\restriction{(\pred a\cup\{a\})}$ has the defining property of $g_a$,
so it is equal to $g_a$; in particular, $g_d(a)=g_a(a)$.  Therefore we
can define $g_b$ by
\begin{equation}\label{eqn:gb}
  g_b(x)=
  \begin{cases}
    g_x(x),&\text{ if }x<b;\\
\cF(\{g_y(y)\colon y<b\}),&\text{ if }x=b.
  \end{cases}
\end{equation}
Moreover, as
before, any $g_b$ as desired must agree with $g_a$ on $\pred
a\cup\{a\}$ when 
$a<b$, and then $g_b(b)$ must be as in~\eqref{eqn:gb}.  By induction,
we have a function $g_a$ as desired for all $a$ in $\cC$.  Then we
have~\eqref{eqn:strong-rec} for all $c$ in $\cC$ if and only if $\cG$ is
$x\mapsto g_x(x)$.
\item
Suppose finally~\eqref{item:wo} fails.  Then some nonempty subclass
$\cC_0$ of
$\cC$ has no least element.  Then the subclass $\cls
x{x\in\cC\land\Exists y(y\in\cC_0\land y\leq x)}$ of $\cC$ also has no
least element.  Call this subclass $\cC_1$.
Let $\cF\colon\pow{\{0,1\}}\to\{0,1\}$, where $\cF(a)=1$ if and only
if $1\in a$.  If $e\in\{0,1\}$, let
$\cG_e$ be the function from $\cC$ into $\{0,1\}$ given by
\begin{equation*}
  \cG_e(x)=
  \begin{cases}
    0,&\text{ if }x\in\cC\setminus\cC_1;\\
    e,&\text{ if }x\in\cC_1.
  \end{cases}
\end{equation*}
Then $\cG_e(a)=\cF(\cG_e\setimb{\pred a})$, so~\eqref{item:sr}
fails. \hfill\qed 
\end{asparaenum}
%The circle of implications is now closed.
\renewcommand{\qed}{}
\end{proof}

\begin{theorem}\label{thm:s-rec-simp}
  A strict total order $(\cC,<)$ whose every section is a set admits
  strong recursion if and only if, for every class $\cD$, if
  $\cF\colon\cE\to\cD$, where
\begin{equation}\label{eqn:tf-rec}
\cE=\cls x{\Exists
  y(y\in\cC\land x\in\mapset{\pred y}{\cD})},
\end{equation}
(so $\cF(g)\in\cD$ whenever $g$ is a function from a section of
$\cD$ into $\cD$,)
then there is a unique function $\cG$ from $\cC$ into $\cD$ such that
\begin{equation}\label{eqn:G(a)=F}
  \cG(a)=\cF(\cG\rest{\pred a}).
\end{equation}
\end{theorem}

\begin{proof}
Suppose first $(\cC,<)$ admits strong recursion, $\cD$ is a class,
$\cE$ is as in~\eqref{eqn:tf-rec}, and
$\cF\colon\cE\to\cD$.  We have $\cE\included\pow{\cC\times\cD}$.  Let
$\cK$ be a function from $\pow{\cC\times\cD}$ to $\cC\times\cD$ such
that, if $f\in\cE$, then $\cK(f)=(a,\cF(f))$, where $\dom f=\pred a$.
Then there is a unique function $\cH$ from $\cC$ to $\cC\times\cD$
such that
\begin{equation*}
  \cH(a)=\cK(\cH\setimb{\pred a})=\cK(\set{\cH(x)}{x<a}).
\end{equation*}
By  Theorem~\ref{thm:o-eq}, we can use induction to show
$\pi_0\circ\cH=\id{\cC}$, where $\pi_0$ is $(x,y)\mapsto x$ (as
in~\parref{par:f}).  Indeed,
if $\pi_0(\cH(b))=b$ whenever $b<a$, then
$\cH\setimb{\pred a}$ is a function from $\pred a$ to $\cD$, and in
particular $\cH\setimb{\pred a}=(\pi_1\circ\cH)\restriction{\pred
  a}$.  So
$\cH\setimb{\pred a}\in\cE$, and $\cH(a)=(a,\cF(\cH\setimb{\pred a}))$;
in particular, $\pi_0(\cH(a))=a$.  Then
\begin{equation*}%\label{eqn:p1H}
(\pi_1\circ\cH)(a)=\cF((\pi_1\circ\cH)\restriction{\pred a}).
\end{equation*}
Thus the function $\pi_1\circ\cH$ is a
function $\cG$ as in~\eqref{eqn:G(a)=F}; it is unique, 
since $\cH$ can be recovered from $\pi_1\circ\cH$ as
$x\mapsto(x,\pi_1\circ\cH(x))$.

Now suppose conversely that $\cG$ exists uniquely as in~\eqref{eqn:G(a)=F}
whenever $\cF\colon\cE\to\cD$, and now $\cK\colon\pow{\cD}\to\cD$.
Then we may let $\cF$ be $x\mapsto\cK(\ran x)$ and obtain $\cG$
uniquely from $\cC$ to $\cD$ so that $\cG(a)=\cF(\cG\restriction{\pred
  a})=\cK(\cG\setimb{\pred a})$.  
\end{proof}

\begin{parag}\label{par:succ}
  Suppose $(\cC,<)$ is a non-empty well-ordered class.  Then $\cC$ has
  a least element, which we may call $0$ (or more precisely
  $0^{(\cC,<)}$).  If $\cD$ is a proper initial segment of $\cC$, then
  $\cD$ is the section $\pred a$, where $a=\min(\cC\setminus\cD)$.
Every element $a$ of 
  $\cC$ that is not a greatest element has a \defn{successor}, 
which we may denote by 
  \begin{equation*}
    \scr a
  \end{equation*}
(or more precisely $a^{+^{(\cC,<)}}$);
  it is  $\min(\cC\setminus(\pred a\cup\{a\}))$.
Immediately, 
\begin{equation*}
  \pred{\scr a}=\pred a\cup\{a\}.
\end{equation*}
  An element of $\cC$ is a \defnplain{successor}, if it is the
  successor of some element.  An element of $\cC$ that is neither a
  successor nor
  $0$ is a \defn{limit}.  
So $a$ is a limit if and only if
  \begin{enumerate}
    \item
$a\neq0$ and
\item
$b<a\lto\scr b<a$.
  \end{enumerate}
By distinguishing between $0$, successors, and limits, we obtain
alternative formulations of induction and recursion:
\end{parag}

\begin{theorem}\label{thm:si}
  Suppose $(\cC,<)$ is a well-ordered class, and $\cD$ is a subclass
  such that:
  \begin{enumerate}
\item
$0\in\cD$ (that is, $0^{(\cC,<)}\in\cD$), if $\cC\neq\emptyset$;
    \item
$a\in\cD\lto\scr a\in\cD$, if $a$ is not the greatest element of $\cC$;
\item
if $a$ is a limit of $\cC$ and $\pred a\included\cD$, then $a\in\cD$.
  \end{enumerate}
Then $\cD=\cC$.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:si}.
\end{proof}

\begin{theorem}\label{thm:rec-alt}
  Suppose $(\cC,<)$ is a well-ordered class, $\cD$ is a class, $b\in\cD$,
  $\cF\colon\cD\to\cD$, and $\cG\colon\pow{\cD}\to\cD$.
Then there is a unique function $\cH$ from
  $\cC$ to $\cD$ such that
  \begin{enumerate}
\item
$\cH(0)=b$, if $\cC\neq\emptyset$;
\item
$\cH(\scr a)=\cF(\cH(a))$, if $a$ is not the greatest element of $\cC$;
\item
$\cH(d)=\cG(\cH\setimb{\pred d})$,
if $d$ is a limit.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:rec-alt}, using Theorem~\ref{thm:s-rec-simp}.
\end{proof}

\begin{parag}\label{par:wo-embedding}
  An 
\defnplain{embedding}{}%
\dindexsub{embedding}{ordered classes}%
 of an ordered class $(\cC,\cR)$ in another one,
  $(\cD,\cS)$, is an injection $\cF$ of the \emph{class} $\cC$ in
  $\cD$ (\parref{par:inv}) such that
  $a\mathrel{\cR}b\iff a\mathrel{\cS}b$.  A surjective embedding is an
  \defnplain{isomorphism}.%
\dindexsub{isomorphism}{ordered classes}
\end{parag}

\begin{theorem}\label{thm:wo}
  Of any two well-ordered classes, one is uniquely isomorphic to a
  unique initial segment of the other.
\end{theorem}

\begin{proof}
Let $(\cC,\cR)$ and $(\cD,\cS)$ be well-ordered classes.  There is at
most one way that $(\cC,\cR)$ can be isomorphic to an initial segment
of $(\cD,\cS)$.  Indeed, let $\cF$ and $\cG$ be two such
isomorphisms.  If they agree \emph{on} $\pred a$, then they must agree
\emph{at} $a$ (Exercise~\ref{ex:wo-isom}).  By induction, the
functions agree on $\cC$.   

Now let $\cC^*$ comprise the elements $a$ of $\cC$ for which there is
a function $f_a$ such that, for all $b$ in $\pred a\cup\{a\}$, the
restriction $f_a\restriction{(\pred b\cup\{b\})}$ is an isomorphism
from $(\pred b\cup\{b\},\cR)$ to 
an initial segment of $(\cD,\cS)$.  If $a\in\cC^*$, and
$b\mathrel{\cR}a$, then the function $f_a\restriction{(\pred
  b\cup\{b\})}$ shows $b\in\cC^*$.  Thus $\cC^*$ is an initial segment
of $\cC$.  Conversely, if $a$ and $b$ are in $\cC^*$, and
$b\mathrel{\cR}a$, then the uniqueness result above shows
$f_b=f_a\restriction{(\pred a\cup\{a\})}$.  Therefore the union
$\bigcup\cls{f_x}{x\in\cC^*}$ is the function $x\mapsto f_x(x)$ on
$\cC^*$, and this function is an isomorphism from $(\cC^*,\cR)$ to an
initial segment $(\cD^*,\cS)$ of $(\cD,\cS)$.

Let this function be called $\cF$, and suppose $\pred a\included\cC^*$
and $\cF\setimb{\pred a}=\pred c$ for some $a$ in $\cC$ and $c$ in
$\cD$.  Then the function $\cF\restriction{\pred a}\cup\{(a,c)\}$
shows that $a\in\cC^*$.  We can conclude that either $\cC^*=\cC$ or
else $\cD^*=\cD$.  In the former case, $\cF$ is an isomorphism from
$(\cC,\cR)$ to an initial segment of $(\cD,\cS)$; in the latter case,
$\conv{\cF}$ is an isomorphism from $(\cD,\cS)$ to an initial segment
of $(\cC,\cR)$.
\end{proof}

\section{Construction of countably infinite classes}

\begin{lemma}\label{lem:inf-bin-rel}
  Suppose $(\cC,i,\cF)$ is an infinitary structure, and $\cA$ is a
  subclass of $\cC$ whose every element is either $i$ or $\cF(b)$ for
  some $b$ in $\cA$.  Then there is at most one binary relation $<$ on
  $\cA$ such that
  \begin{enumerate}
  \item\label{item:a<F(a)}
$a<\cF(a)$ whenever $a\in\cC$ and $\cF(a)\in\cA$;
\item\label{item:Awo}
$\cA$ is well-ordered by $<$.
  \end{enumerate}
Moreover, if $<$ is such a relation and $a\in\cA$, then $\cF(a)$ is the least
element of $\cls x{x\in\cA\land a<x}$, if this class is non-empty.
\end{lemma}

\begin{proof}
  Suppose $<$ is a strict ordering of $\cA$ such
  that~\eqref{item:a<F(a)} holds.  Suppose there is some $a$ in $\cA$
  such that $\cF(a)$ is not the least element of $\cls x{x\in\cA\land
    a<x}$, although this class is non-empty.  If there is no
  \emph{least} such $a$, then $\cA$ is not well-ordered by~$<$.  So
  suppose $a$ is least.  If there is no least element of $\cls
  x{x\in\cA\land a<x}$, then $\cA$ is not well-ordered by $<$.  So
  suppose $b$ is the least element of the class.  Then
  \begin{equation*}
    a<b<\cF(a).
  \end{equation*}
  If $b=i$, then $\cA$
  has no least element, so $\cA$ is not well-ordered by $<$.  So
  suppose $b\neq i$.  Then $b=\cF(c)$ for some $c$ in $\cA$, and then
  $c<b$, but $c\neq a$.  Also $c\not<a$, by the
  leastness of $a$; and $a\not<c$, by the leastness of $b$.  Therefore
  $<$ is not a total ordering of $\cA$.

Now suppose $\cR$ is a binary relation $<$ on $\cA$
satisfying~\eqref{item:a<F(a)} and~\eqref{item:Awo}.  We have shown
$\cF(a)$ is the least element of $\cls x{x\in\cA\land a<x}$, assuming
this is nonempty.  Suppose $\cS$ is a binary relation on $\cA$,
distinct from $\cR$, but such that
\begin{equation*}
  \cls{(c,\cF(x))}{x\in\cC\land \cF(x)\in\cA}\included\cS.
\end{equation*}
Let $a$ be least such that $a\cR\neq a\cS$, and let $b$ be the least
element of $a\cR\symdiff a\cS$, always with respect to $\cR$.  There
are two cases.
\begin{asparaenum}[1.]
\item
Suppose $a\mathrel{\cR}b$, that is, $a<b$, but
$\lnot(a\mathrel{\cS}b)$.  Then $b\neq 
i$, so $b=\cF(c)$ for some $c$ in $\cA$.  
Therefore $c<b$ and $c\mathrel{\cS}b$, so $a\neq c$ and hence $a<c$,
that is, $a\mathrel{\cR}c$.
By leastness of $b$, we have $a\mathrel{\cS}c$, so $\cS$ is not
transitive. 
\item
Suppose $\lnot(a\mathrel{\cR}b)$, but $a\mathrel{\cS}b$.
Then $b\leq a$.  If $b=a$, then $\cS$ is not irreflexive.  If $b<a$,
then $b\cR=b\cS$, so $b\mathrel{\cS}a$, so $\cS$ is not an
ordering. 
\end{asparaenum}

 This establishes uniqueness of $<$.
\end{proof}

\begin{axiom}[Binary union]\label{ax:binary-union}
\dindexsub{axiom}{Binary Union A---}%
\dindexsub{binary}{B--- Union Axiom}%
\dindexsub{union}{Binary U--- Axiom}
  The union of two sets is a set:
  \begin{equation*}
    \Exists xx=a\cup b.
  \end{equation*}
\end{axiom}

\begin{theorem}\label{thm:rel-set}
  A relation from a set to another is a set.
\end{theorem}

\begin{proof}
  $a\times b\included\pow{\pow{a\cup b}}$ by~\parref{par:cartesian}.
\end{proof}

\begin{theorem}\label{thm:inf-bin-rel}
  Let $(\cC,i,\cF)$ be an infinitary structure.  Then $\cC$ has a
  subclass $\cD$ such that
  \begin{enumerate}
  \item
 $(\cD,i,\cF)$ is an arithmetic structure;
\item
$\cD$ is well-ordered by some unique $<$ such that $a<\cF(a)$ for all
  $a$ in $\cD$.  
  \end{enumerate}
\end{theorem}

\begin{proof}
  Let $\cE$ comprise all subsets $a$ of $\cC$ such that
  \begin{enumerate}
  \item 
every element of $a$ is either $i$ or $\cF(b)$ for some $b$ in $a$,
and
\item
$a$ is well-ordered by some $<$ such that $b<\cF(b)$ whenever
  $\cF(b)\in a$.
  \end{enumerate}
Here $<$ is a set by Theorem~\ref{thm:rel-set}; by
Lemma~\ref{lem:inf-bin-rel}, it is
unique.  We have $\{i\}\in\cE$.  Suppose $a\in\cE$, and $r$ is the
associated ordering $<$.  If $b\in a$, but $\cF(b)\notin a$, then
$a\cup\{\cF(b)\}$ is well-ordered by $r\cup\set{(x,\cF(b))}{x\in a}$,
and this shows $a\cup\{\cF(b)\}\in\cE$.  So $\bigcup\cE$ contains $i$
and is closed under $\cF$; hence $(\bigcup\cE,i,\cF)$ is an infinitary
structure. 

We shall show that $\bigcup\cE$ is the desired
class $\cD$.  To do so, suppose $a$ and $c$ are elements of
$\cE$.  By Theorem~\ref{thm:wo}, we may
assume that there is an isomorphism $g$ from $(a,<)$ onto an initial
segment of $(c,<)$.  
Then $g(i)=i$, and if $g(b)=b$, and $\cF(b)\in a$, then
$g(\cF(b))=\cF(b)$ by Lemma~\ref{lem:inf-bin-rel}.  Since $(a,<)$ has
no limits, we have $g=\id a$ by induction, by Theorem~\ref{thm:si}.
That is, $(a,<)$ is an initial segment of 
$(c,<)$.  If we now write $r_a$ for the ordering of $a$, then
$\bigcup\cE$ is well-ordered by $\bigcup\cls{r_x}{x\in\cE}$;
this ordering is unique by~\ref{lem:inf-bin-rel}.
Since every element of $\bigcup\cE$ is either $i$ or $\cF(b)$ for some
$b$ in $\bigcup\cE$, we can conclude by Theorem~\ref{thm:si} that
$\bigcup\cE$ is the desired class $\cD$.
\end{proof}

\begin{parag}
In Theorem~\ref{thm:inf-bin-rel}, the unique ordering $<$ may be
called the ordering 
\defnplain{associated}{}%
\index{associated|see{order}}%
\dindexsub{order}{---ing associated with an arithmetic structure}
with the arithmetic structure.
  Since an infinitary structure does exist by Theorem~\ref{thm:Vinf},
  we can use Theorem~\ref{thm:inf-bin-rel} to obtain a particular
  arithmetic structure, say $(\cC,i,\cF)$.
By Theorem~\ref{thm:s-rec-simp}, there is a
  function $\cG$ on $\cC$ such that
  \begin{equation*}
    \cG(a)=\cG\setimb{\pred a}.
  \end{equation*}
We denote $\ran{\cG}$ by
  \begin{equation*}
    \vnn.
  \end{equation*}
\end{parag}

\begin{theorem}\label{thm:omega}
The definition of $\vnn$ is independent of the choice of infinitary
structure $(\cC,i,\cF)$, and
  $(\vnn,\emptyset,x\mapsto x\cup\{x\})$ is an arithmetic structure,
  whose associated ordering is both membership ($\in$) and proper
  inclusion ($\pincluded$).
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:omega}.
\end{proof}

\begin{parag}\label{par:s(n)}
The elements of $\vnn$ are the \defn{natural number}{s.}  If $n$ is
one of them, we denote
the \defn{successor}{} $n\cup\{n\}$ of $n$ by
\begin{equation*}
  \vscr n.
\end{equation*}
There are standard 
\defn{numeral}s for denoting some natural numbers:
\begin{align*}
0&=\emptyset,&2&=\vscr 1,&4&=\vscr 3,&6&=\vscr 5,&8&=\vscr 7,\\
1&=\vscr 0,&3&=\vscr 2,&5&=\vscr 4,&7&=\vscr 6,&9&=\vscr 8.
\end{align*}
In general, a natural number $n$ is the set that can be denoted by
\begin{equation*}
  \{0,\dots,n-1\}.
\end{equation*}
\end{parag}



\section*{Exercises}

\begin{xca}\label{ex:C-eq-D}
Show that~\eqref{eqn:C-eq-D} does not generally stand for a sentence of
the logic of sets.  
\end{xca}

\begin{xca}\label{ex:a-eq-b}
  Prove that~\eqref{eqn:|a|=|b|} stands for a sentence of
the logic of sets.  
\end{xca}

\begin{xca}\label{ex:equip}
Show that equipollence of sets is an equivalence-relation, and hence that it
is justifiable to define the cardinality of a set $a$ as $\cls
x{x\equip a}$ (as in \parref{par:equip}).
\end{xca}

\begin{xca}\label{ex:<-trans}
Show that $\injects$ is transitive and reflexive, but $\pinjects$ is
transitive and irreflexive.
\end{xca}

\begin{xca}\label{ex:sb}
Complete the proof of the Schroeder--Bernstein Theorem.
\end{xca}

\begin{xca}
Is the Schroeder--Bernstein Theorem true for classes in general?
\end{xca}

\begin{xca}\label{ex:cantor}
Establish the implications~\eqref{eqn:cantor1} and~\eqref{eqn:cantor2}
in the proof of Cantor's Theorem.
\end{xca}

\begin{xca}\label{ex:why-set}
Why does the proof of Cantor's Theorem (\parref{thm:pow}) require
$\set{x\in a}{x\notin f(x)}$ to be a set?
\end{xca}

\begin{xca}\label{ex:a-bneq0}
  If $a\neq b$, so that one of $a\setminus b$ and $b\setminus a$ is
  nonempty.  If $a\setminus b\neq\emptyset$, show that $a\setminus
  b\in\pow a\setminus \pow b$ and hence $\pow a\neq\pow b$.
\end{xca}

\begin{xca}
  Find a set $a$ of minimal cardinality such that $(a,i,f)$ is
  recursive for some $i$ in $a$ and $f$ in $\mapset aa$.
\end{xca}


\begin{xca}\label{ex:rec-counter}
Find an example of a structure $(\cC,i,\cF)$ admitting induction and an
iterative structure
$(\cD,j,\cG)$ such that there is \emph{no} function $\cH$ from $\cC$
to $\cD$ such that $\cH(i)=j$ and $\cH\circ\cF=\cG\circ\cH$.
\end{xca}

\begin{xca}\label{ex:nc-in-c}
Prove the claim in the proof of Theorem~\ref{thm:countably-infinite-subset}.
\end{xca}

\begin{xca}\label{ex:pair-from-augmentation}
Derive the Pairing Axiom from the Augmentation Axiom and the
Separation Scheme.
\end{xca}

\begin{xca}\label{ex:cor-rec}
Concerning the corollary (\ref{cor:rec}) to the Recursion Theorem:
\begin{enumerate}
\item 
prove it by means of the Theorem;
\item
prove it directly by considering the class of sets $a$ whose every
element is either $(i,j)$ or else $(\cF(b),\cG(c))$ for some element
$(b,c)$ of $a$.
\end{enumerate}
\end{xca}

\begin{xca}\label{ex:isom1}
  Prove that the inverse of an isomorphism of iterative structures is
  an isomorphism (\parref{par:isom1}) and that being isomorphic is an
  equivalence-relation. 
\end{xca}

\begin{xca}
Show that every subset of an ordered set that admits strong induction
admits strong induction.
\end{xca}

\begin{xca}\label{ex:si}
Prove Theorem~\ref{thm:si}.
\end{xca}

\begin{xca}\label{ex:rec-alt}
Prove Theorem~\ref{thm:rec-alt}.
\end{xca}

\begin{xca}\label{ex:isom}
Suppose $(\cA,\cR)$ and $(\cB,\cS)$ are strict total orders,
$\cH\colon\cA\to\cB$, and
$c\mathrel{\cR}d\lto\cH(c)\mathrel{\cS}\cH(d)$.   Show that $\cH$ is
an isomorphism from $(\cA,\cR)$ to $(\cH\setimb{\cA},\cS)$.
\end{xca}

\begin{xca}\label{ex:wo-isom}
Let $\cF$ and $\cG$ be isomorphisms
from a well-ordered class $(\cC,\cR)$ to initial segments of a
well-ordered class $(\cD,\cS)$.  Assuming that $\cF$ and $\cG$ agree
on $\pred a$ for some $a$ in $\cC$, show that $\cF(a)=\cG(a)$.
\end{xca}

\begin{xca}\label{ex:omega}
  Prove Theorem~\ref{thm:omega}.
\end{xca}

\chapter{The natural numbers}\label{ch:N}

%\setcounter{equation}{9}

\section{Structures}\label{sect:structures}

\begin{parag}
If $n\in\vnn$, then the class $\mapset n{\cC}$ may also be denoted by 
\begin{equation*}
  \cC^n,
\end{equation*}
although some cases of this notation will
get a different interpretation in \parref{par:card-exp}.
An element $f$ of $\cC^n$ can be written also as $x\mapsto f_x$.  Then
we have
\begin{equation}\label{eqn:C0}
\cC^0
=\{0\}=1,
\end{equation}
as well as
\begin{align}\label{eqn:C1}
x\mapsto x_0
&\colon\cC^1\bijection\cC,\\\label{eqn:C2}
  x\mapsto(x_0,x_1)
&\colon\cC^2\bijection\cC\times\cC.
\end{align}
Note in particular that~\eqref{eqn:C0} still holds if $\cC=\emptyset$,
while $\emptyset^n=\emptyset$ if $n>0$.
A notation like $\cC\times\cC\times\cC$ is ambiguous; it
could be $(\cC\times\cC)\times\cC$ or $\cC\times(\cC\times\cC)$; but the
distinction is usually unimportant, since the two classes are in
bijection with each other and with ${\cC}^3$ (Exercise~\ref{ex:3}).
\end{parag}

\begin{parag}
An arbitrary element of ${\cC}^n$ can now be written as one of
\begin{align*}
  &(a_i\colon i\in n),&
  &(a_0,\dots,a_{n-1}),&
  &\vec a.
\end{align*}
If the free variables of a formula $\phi$ are among the variables $x_i$,
where $i\in n$, then we can write $\phi$ as $\phi(x_0,\dots,x_{n-1})$
or $\phi(\vec x)$.  Such a formula is called 
\defnplain{$n$-ary},%
\dindexsub{formula}{n-ary ---@$n$-ary ---}%
\dindexsub{n-ary@$n$-ary}{--- formula}
and the result
of replacing each free occurrence of
$x_i$ with $a_i$, for each $i$ in $n$, is
$\phi(a_0,\dots,a_{n-1})$ or $\phi(\vec a)$.
In general, a subclass of ${\cC}^n$ can be called an
\defnplain{$n$-ary relation on}{}%
\dindexsub{relation}{n-ary ---@$n$-ary ---}%
\dindex{n-ary relation@$n$-ary relation} 
$\cC$: such a relation is
\begin{equation*}
  \cls{\vec x\in\cC^n}{\phi(\vec x)}
\end{equation*}
for some $n$-ary formula $\phi$.
  By~\eqref{eqn:C1}, a
$1$-ary relation on $\cC$ can be considered as a
subclass of $\cC$; by~\eqref{eqn:C2}, a 
$2$-ary relation can be considered as a binary
relation.  A function from ${\cC}^n$  
into $\cC$ is an 
\defnplain{$n$-ary operation on}{}%
\dindex{n-ary operation@$n$-ary operation}%
\dindexsub{operation}{n-ary ---@$n$-ary ---} 
$\cC$.  By~\eqref{eqn:C0} and~\eqref{eqn:C1}, a $0$-ary operation can
be considered as an element of $\cC$.  A singulary
($1$-ary) operation $\cF$ is sometimes written as
\begin{equation*}
  x\mapsto x^{\cF}
\end{equation*}
(as in the case of the operation $x\mapsto\vscr x$ of succession on
$\vnn$) instead of $x\mapsto\cF(x)$;
a binary ($2$-ary) operation $\cG$ is often
written as
\begin{equation*}
  (x,y)\mapsto x\mathop{\cG}y
\end{equation*}
instead of $(x,y)\mapsto\cG(x,y)$.
\end{parag}

\begin{parag}\label{par:structure}
A \defn{structure}{} is a class equipped with some relations and
operations on it.  The class is then the 
\defn{universe}{}
 of the
structure.  Examples include:
\begin{asparaenum}
\item
orders (\parref{par:ordering}), and in particular well-ordered sets
(\parref{par:least}); 
  \item
iterative structures (\parref{par:iterative}), and in particular
recursive structures (\parref{par:rec-set}), infinitary structures,
and arithmetic structures (\parref{par:c-inf});
\item
$(\universe,\in)$;
\item
$(\pow{a},\cap,\cup,x\mapsto x\comp,\emptyset,a)$.
\end{asparaenum}
We can understand a structure formally as a class
(Exercise~\ref{ex:str}), but it is not necessary to do so.
When we speak in general terms, we may denote a structure by the
Fraktur form of the letter denoting its universe.  (See
Appendix~\ref{app:German}.)  So the structure
$(\cC,\dots)$ may be denoted by $\str C$, and $(a,\dots)$ by $\str a$.
\end{parag}

\begin{parag}\label{par:sig}
  A structure $\str C$ has a 
\defn{signature},
  which comprises: 
  \begin{enumerate}
\item
    an 
\defnplain{$n$-ary function-symbol}{}%
\dindex{n-ary function-symbol@$n$-ary function-symbol}%
\dindexsub{function}{n-ary ---{}-symbol@$n$-ary ---{}-symbol}%
\dindexsub{symbol}{function-{}---}
    for each $n$-ary operation of
    $\str C$;
\item
an 
\defnplain{$n$-ary predicate}{}%
\dindex{n-ary predicate@$n$-ary predicate}%
\dindexsub{predicate}{n-ary ---@$n$-ary ---}
for each $n$-ary relation of $\str C$. 
  \end{enumerate}
If $s$ is one of these symbols, then the corresponding operation
  or relation can be denoted by
  \begin{equation*}
    s^{\str C}.
  \end{equation*}
We may need this notation when another structure $\str D$ has the same
signature as $\str C$.  Then $s^{\str C}$ and $s^{\str D}$ are
different operations, although they are represented by the same symbol
$s$. 
\end{parag}

\begin{parag}\label{par:hom}
Suppose $\str C$ and $\str D$ have a common signature, and
$\cH\colon\cC\to\cD$.  For each $n$ in $\vnn$, we can understand $\cH$
also as the function $\vec x\mapsto(\cH(x_i)\colon i\in n)$ from
$\cC^n$ to $\cD^n$.
If 
\begin{enumerate}
  \item
$\cH(F^{\str C}(\vec a))=F^{\str D}(\cH(\vec a))$ for all $\vec
    a$ in $\cC^n$ when
 $n\in\vnn$ and $F$ is
    an $n$-ary function-symbol of the signature, and
\item
$\vec a\in R^{\str C}\lto\cH(\vec a)\in
  R^{\str D}$ for all $\vec a$ in $\cC^n$ when $n\in\vnn$ and $R$ is
  an $n$-ary predicate of the 
  signature, 
\end{enumerate}
then $\cH$ is a \defn{homomorphism}{} from $\str C$ to $\str D$, and we
may write
\begin{equation*}
  \cH\colon\str C\to\str D.
\end{equation*}
The homomorphisms defined in \parref{par:rec} are a special case.
\end{parag}

\begin{parag}\label{par:embed}
  Again suppose $\str C$ and $\str D$ have a common signature.  A
  function $\cH$ from $\cC$ into $\cD$ is an \defn{embedding}{} of $\str
  C$ in $\str D$ if $\cH$ is an injective homomorphism and also
  \begin{equation*}
(\cH(a_i)\colon i\in n)\in
  R^{\str D}\lto
  (a_i\colon i\in n)\in R^{\str C}
\end{equation*}
when $R$ is an $n$-ary predicate of the signature.  This is a special
case of the embeddings defined in~\P\P~\ref{par:inv}
and~\ref{par:wo-embedding}. 
In the general cas, if $\cC\included\cD$, and $\id{\cC}$ is an
embedding, then $\str C$ is a
\defn{substructure}{}%
\dindexsub{structure}{sub---}
 of $\cD$, and we may write
\begin{equation*}
  \str C\included\str D.
\end{equation*}
Suppose in
 particular that $\str D$ is an iterative structure.  Then $\str D$
 admits induction if and only if, whenever $\str C$ is an iterative
 structure, and $\cH$ is an embedding of $\str C$ in $\str D$, then
 $\cH$ is surjective onto $\cD$ (Exercise~\ref{ex:embed}).
\end{parag}

\begin{parag}\label{par:N}
If $\cH$ is a bijection from $\cC$ to $\cD$, and $\cH$ is a
homomorphism from $\str C$ to $\str D$, and $\cH\inv$ is a
homomorphism from $\str D$ to $\str C$,
then $\cH$ is an 
\defnplain{isomorphism}{}%
\dindexsub{isomorphism}{arbitrary structures}
 from $\str C$ to $\str D$.
\begin{comment}



, and we
may write
\begin{equation*}
  \cH\colon\str C\overset{\cong}{\longrightarrow}\str D
\end{equation*}
or simply
\begin{equation*}
  \str C\cong\str D.
\end{equation*}



\end{comment}
By Theorem~\ref{thm:cnt-equip}, all arithmetic
  structures are isomorphic.  We know of an arithmetic structure,
  namely $(\vnn,\emptyset,x\mapsto\vscr x)$.  However, throughout this
  chapter, let
  \begin{equation*}
    (\N,0,x\mapsto\scr x)
  \end{equation*}
denote an arbitrary arithmetic structure; also, let $\scr0$ be denoted
by
\begin{equation*}
  1.
\end{equation*}
  We shall think of $\N$ as
the class of natural numbers, when the understanding of natural
numbers as ordinals in the sense of \parref{par:ord} is unimportant.
To simplify writing, we may
use $\N$ and~$\vnn$ 
by themselves to refer to structures of which they are universes.
\end{parag}

\section{Addition on recursive structures}

  \begin{thmdefn}\label{thm:rec+}
On $\N$, there is a unique binary operation, called 
\defnplain{addition}{}%
\index{addition|see{operation}}
\dindexsub{operation}{addition}
and denoted by $+$, such that
\begin{enumerate}
  \item
$m+0=m$,
\item
$m+\scr n=\scr{(m+n)}$.
\end{enumerate}
  \end{thmdefn}

  \begin{proof}
    Exercise~\ref{ex:+}.
  \end{proof}

\begin{parag}
It is sometimes suggested that the validity of the definition of
addition given in the theorem
is an easy consequence of induction.
The intended argument seems to run as follows: 
  \begin{quote}
We define $x+0$ as $x$, for all $x$ in $\N$.  If $x+n$ has been
defined for all $x$ in $\N$, then we let
$x+\scr n=\scr{(x+n)}$.  Therefore, by induction, $x+y$ is defined for all
$x$ and $y$ in $\N$.
  \end{quote}
Indeed, Peano~\cite{Peano} seems to have had in mind such an
argument.  Landau~\cite{MR12:397m} reports having accepted such an
argument until he was forced to recognize its wrongness.  But
Dedekind~\cite[II.IX (130)]{MR0159773} has had the truth all
along, even though he refers to definition by recursion as definition
by induction.  Proof by induction is not a method of definition; it is
a method of proving that two classes are equal.
The proposed `proof' here of the definition of addition would have to
start out in the following fashion: 
\begin{quote}
  Let $\cC$ be the subclass of $\N$ comprising all $y$ for which
  the operation $x\mapsto x+y$ is defined on $\N$ so that $\Forall
  xx+0=x$ and $\Forall x\Forall zx+\scr z=\scr{(x+z)}$.
\end{quote}
But this definition of $\cC$ makes no sense.  Nonetheless, it
\emph{is} possible to define addition on every recursive structure
$(\cA,i,\cS)$, as we shall now see.
\end{parag}

\begin{lemma}\label{lem:+}
    if $\cF\colon\cB\to\cC$ and $\cG\colon\cC\to\cC$, then there
    is a unique function $\cH$ from $\cB\times\N$ to $\cC$ such that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=\cF(a)$,
\item
$a\mathop{\cH}\scr n=\cG(a\mathop{\cH}n)$.
    \end{enumerate}
Informally,
\begin{equation*}
  a\mathop{\cH}n={\underbrace{\cG\circ\dotsb\circ\cG}_n}\circ\cF(a).
\end{equation*}
\end{lemma}

  \begin{proof}
Let the operation $(x,y)\mapsto(x,\cG(y))$ on $\cB\times\cC$ be denoted
    by $\cK$.
    By the Recursion Theorem (\ref{thm:rec}), there is a unique
    subclass $\cR$ of $\N\times(\cB\times\cC)$ such that
    \begin{enumerate}
      \item
$0\cR=\cF$,
\item
$n\in\N\lto\scr n\cR=\cK\setimb{n\cR}=
  \cls{(x,\cG(y))}{n\mathrel{\cR}(x,y)}$.  
    \end{enumerate}
By induction, if $n\in\N$, then $n\cR$ is a function from $\cB$ to
$\cC$.  Indeed, $0\cR$ is such a function (namely $\cF$), and if
$n\cR$ is such a function, then $\scr n\cR=\cG\circ(n\cR)$.
We can now define the binary function $\cH$ as 
$(x,y)\mapsto y\cR(x)$.
Then
\begin{enumerate}
  \item
$a\mathop{\cH}0=0\cR(a)=\cF(a)$,
\item
$a\mathop{\cH}\scr n=\cG\circ(n\cR)(a)=\cG(a\mathop{\cH}n)$.
\end{enumerate}
So $\cH$ is as desired.  To see that $\cH$ is unique, note that $\cR$
determines $\cH$, and conversely.  Indeed,
\begin{equation*}
\cH=\cls{((x,y),z)}{y\mathrel{\cR}(x,z)},\qquad
  \cR=\cls{(y,(x,z))}{x\mathop{\cH}y=z}.  
\end{equation*}
Since $\cR$ uniquely satisfies the given conditions, so does~$\cH$. 
  \end{proof}

  \begin{thmdefn}\label{thm:addition}
Suppose $(\cA,i,\cS)$ is recursive.  Then there is a unique binary
operation of 
\defnplain{addition}{}%
\dindexsub{operation}{addition} on $\cA$ given by
\begin{enumerate}
  \item
$a+i=a$,
\item
$a+\cS(b)=\cS(a+b)$.
\end{enumerate}
  \end{thmdefn}

  \begin{proof}
    By Lemma~\ref{lem:+}, there is a unique function $\cH$ from
    $\cA\times\N$ to $\cA$ such that 
    \begin{enumerate}
      \item
$a\mathop{\cH}0=a$,
\item
$a\mathop{\cH}\scr n=\cS(a\mathop{\cH}n)$.
    \end{enumerate}
So $\cH$ is recursively defined in its second argument.  We shall show
that $\cH$ is also recursively definable in its first argument, that
is,
\begin{enumerate}
\item\label{item:iHn}
$i\mathop{\cH}n=\cF(n)$,
\item\label{item:+n}
$\cS(a)\mathop{\cH}n=\cS(a\mathop{\cH}n)$,
\end{enumerate}
where $\cF$ is just the function $x\mapsto i\mathop{\cH}x$.
We use induction.  We have~\eqref{item:+n} when $n=0$, since
$\cS(a)\mathop{\cH}0=\cS(a)=\cS(a\mathop{\cH}0)$. 
Suppose we have~\eqref{item:+n} when $n=k$.  Then
\begin{align*}
  \cS(a)\mathop{\cH}\scr k
&=\cS(\cS(a)\mathop{\cH}k)&&\text{[by definition of $\cH$]}\\
&=\cS(\cS(a\mathop{\cH}k))&&\text{[by inductive hypothesis]}\\
&=\cS(a\mathop{\cH}\scr k).&&\text{[by definition of $\cH$]}
\end{align*}
Thus~\eqref{item:+n} holds when $n=\scr k$.  This completes the
induction.

Now~\eqref{item:iHn} and~\eqref{item:+n} hold.  In particular, the
function $x\mapsto x\mathop{\cH}n$ on $\cA$ is uniquely determined by
$\cF(n)$, by Theorem~\ref{thm:rec-uni}.  That is,
\begin{equation*}
\cF(m)=\cF(n)\iff\Forall xx\mathop{\cH}m=x\mathop{\cH}n.  
\end{equation*}
Moreover, $\cF\colon\N\surjection\cA$, by induction, since
$\cF(0)=i\mathop{\cH}0=i$, while if $\cF(n)=a$, then 
\begin{equation*}
\cF(\scr n)=i\mathop{\cH}{\scr
  n}=\cS(i\mathop{\cH}n)=\cS(\cF(n))=\cS(a).
\end{equation*}
Now we can define addition on $\cA$ by
\begin{equation*}
  a+b=c\iff\Exists x(\cF(x)=b\land a\mathop{\cH}x=c).
\end{equation*}
Since $\cF(0)=i$, we have $a+i=a\mathop{\cH}0=a$.  
Also, if $b=\cF(n)$, so that $\cS(b)=\cF(\scr n)$, then
\begin{equation*}
  a+\cS(b)
=a\mathop{\cH}\scr n
=\cS(a\mathop{\cH}n)
=\cS(a+b).
\end{equation*}
Thus $+$ is as desired; it is unique by Theorem~\ref{thm:rec-uni}.
  \end{proof}

\begin{lemma}\label{lem:+2}
Suppose $(\cA,i,\cS)$ is recursive.
For all $a$ and $b$ in $\cA$,
  \begin{enumerate}
    \item
$i+a=a$,
\item\label{item:cSb+a}
$\cS(b)+a=\cS(b+a)$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  By definition of $+$, we have $i+i=i$.  Suppose $i+b=b$.  Then also
  by definition of $+$, we have $i+\cS(b)=\cS(i+b)=\cS(b)$.  By
  induction, $\Forall xi+x=x$.

Let $\cA_0=\cls x{\Forall y\cS(y)+x=\cS(y+x)}$.  Since
$\cS(b)+i=\cS(b)=\cS(b+i)$, we have $i\in\cA_0$.  Suppose
$a\in\cA_0$, so that $\cS(b)+a=\cS(b+a)$.  Then
$\cS(b)+\cS(a)=\cS(\cS(b)+a)=\cS(\cS(b+a))=\cS(b+\cS(a))$, so
$\cS(a)\in\cA_0$. 
\end{proof}

\begin{parag}
  In Lemma~\ref{lem:+2}, we cannot establish~\eqref{item:cSb+a}
  by induction on $b$.
\end{parag}

\begin{parag}
If we know that $\cA$ is a \emph{set,} so that operations on $\cA$ are
sets, then we can prove
Theorem~\ref{thm:addition} as follows.  
For each $a$ in $\cA$, by Corollary~\ref{cor:rec-uni}, there is at
\emph{most} one operation 
$x\mapsto x+a$ on $\cA$ such that
\begin{enumerate}
  \item
$i+a=a$,
\item
$\Forall x\cS(x)+a=\cS(x+a)$.
\end{enumerate}
This much does not require $\cA$ to be a set; but the next observation
does. 
By induction, for each $a$ in $\cA$, there is at \emph{least} one such
operation $x\mapsto x+a$; for we can let $x\mapsto x+0$ be $\id{\cA}$, and if
$x\mapsto x+b$ has been defined, then let $x+\cS(b)=\cS(x+b)$.
This gives addition as desired.
But again, if we do not know that $\cA$ is a set, then it seems we have to
follow a more roundabout route, as above.
\end{parag}

\begin{theorem}\label{thm:+}
Suppose $(\cA,i,\cS)$ is recursive.
For all $a$, $b$, and $c$ in $\cA$,
  \begin{enumerate}
    \item
$\cS(a)=a+\cS(i)$;
in particular, $\scr n=n+1$ in $\N$;
\item
$a+b=b+a$, that is, $+$ is 
\defnplain{commutative};%
\index{commutative|seealso{operation}}
\index{binary|seealso{operation}}%
\dindexsub{operation}{commutative!addition is ---}
\item
$(a+b)+c=a+(b+c)$, that is, $+$ is 
\defnplain{associative}.%
\index{associative|see{operation}}%
\index{binary|seealso{operation}}%
\dindexsub{operation}{associative!addition is ---}
  \end{enumerate}
Moreover, if $\cS$ is injective, then
\begin{enumerate}
\setcounter{enumi}3
\item\label{item:canc}
$a+c=b+c\lto a=b$, that is, $+$ is
\defnplain{cancellative}.%
\index{cancellative|see{operation}}%
\dindexsub{operation}{cancellative: addition is ---} 
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:+thm}.
\end{proof}

\section{Multiplication on recursive structures}

  \begin{thmdefn}\label{thm:multiplication}
On $\N$, there is a unique binary operation, called
\defnplain{multiplication}{}%
\index{multiplication|see{operation}}
\dindexsub{operation}{multiplication}
and denoted by $\cdot$ or by juxtaposition ($(x,y)\mapsto xy$), 
such that
\begin{enumerate}
  \item
$m\cdot0=0$,
\item
$m\cdot(n+1)=m\cdot n+m$.
\end{enumerate}
  \end{thmdefn}

  \begin{proof}
    Exercise~\ref{ex:.}.
  \end{proof}

\begin{lemma}\label{lem:.}
If $\cF\colon\cB\to\cC$ and
    $\cG\colon\cC\times\cB\to\cC$, then there 
    is a unique function $\cH$ from $\cB\times\N$ to $\cC$ such that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=\cF(a)$,
\item
$a\mathop{\cH}(n+1)=(a\mathop{\cH}n)\mathop{\cG}a$.
    \end{enumerate}
\end{lemma}

  \begin{proof}
Let the operation $(x,y)\mapsto(x,y\mathop{\cG}x)$ on $\cB\times\cC$
    be denoted by $\cK$.
    By the Recursion Theorem,~\ref{thm:rec}, there is a unique subclass
    $\cR$ of $\N\times(\cB\times\cC)$ such that 
    \begin{enumerate}
      \item
$0\cR=\cF$,
\item
$n\in\N\lto(n+1)\cR=\cK\setimb{n\cR}=
  \cls{(x,y\mathop{\cG}x))}{n\mathrel{\cR}(x,y)}$.  
    \end{enumerate}
By induction, if $n\in\N$, then $n\cR$ is a function from $\cB$ to
$\cC$.  Indeed, $0\cR$ is such a function (namely $\cF$), and if
$n\cR$ is such a function, then $(n+1)\cR$ is
$x\mapsto n\cR(x)\mathop{\cG}x$.  We can
now define the binary function $\cH$ as
$(x,y)\mapsto y\cR(x)$ on $\cA\times\N$.  Then
\begin{enumerate}
  \item
$a\mathop{\cH}0=0\cR(a)=\cF(a)$,
\item
$a\mathop{\cH}(n+1)
=(n+1)\cR(a)
=n\cR(a)\mathop{\cG}a=(a\mathop{\cH}n)\mathop{\cG}a$. 
\end{enumerate}
So $\cH$ is as desired; its uniqueness follows from that of $\cR$, as
in the proof of Lemma~\ref{lem:+}.
  \end{proof}

  \begin{thmdefn}\label{thm:.}
Suppose $(\cA,i,\cS)$ is recursive.  Then there is a unique binary
operation of 
\defnplain{multiplication}{}%
\dindexsub{operation}{multiplication} on $\cA$ given by
\begin{enumerate}
  \item
$a\cdot0=0$,
\item
$a\cdot\cS(b)=a\cdot b+b$.
\end{enumerate}
  \end{thmdefn}

  \begin{proof}
    We follow the pattern of the proof of Theorem~\ref{thm:addition}.
    By Lemma~\ref{lem:.}, there is a unique function
    $\cH$ from $\cA\times\N$ into $\cA$ such
    that
    \begin{enumerate}
      \item
$a\mathop{\cH}0=i$,
\item
$a\mathop{\cH}(n+1)=a\mathop{\cH}n+a$.
    \end{enumerate}
By induction, $i\mathop{\cH}n=i$ for all $n$ in $\N$; indeed, this is
given when $n=0$, and if it holds when $n=m$, then $i\mathop{\cH}\scr
m=i\mathop{\cH}m+i=i\mathop{\cH}m=i$.  Let $\cF$ be the unique
homomorphism from $(\N,0,{}\scr{})$ into $(\cA,i,\cS)$; by induction,
it is surjective.  The equation
\begin{equation}\label{eqn:.n}
  \cS(a)\mathop{\cH}n=a\mathop{\cH}n+\cF(n)
\end{equation}
holds when $n=0$, since
$\cS(a)\mathop{\cH}0=i=i+i=a\mathop{\cH}0+\cF(0)$.
Suppose~\eqref{eqn:.n} holds for some $n$ in $\N$.  Then
\begin{align*}
  \cS(a)\mathop{\cH}(n+1)
&=\cS(a)\mathop{\cH}n+\cS(a)&&\text{[by definition of $\cH$]}\\
&=(a\mathop{\cH}n+\cF(n))+\cS(a)&&\text{[by inductive hypothesis]}\\
&=a\mathop{\cH}n+(\cF(n)+\cS(a))&&\text{[by associativity of $+$]}\\
&=a\mathop{\cH}n+\cS(\cF(n)+a)&&\text{[by definition of $+$]}\\
&=a\mathop{\cH}n+(\cS(\cF(n))+a)&&\text{[by Lemma~\ref{lem:+2}]}\\
&=a\mathop{\cH}n+(\cF(n+1)+a)&&\text{[because $\cF$ is a homomorphism]}\\
&=a\mathop{\cH}n+(a+\cF(n+1))&&\text{[by commutativity of $+$]}\\
&=(a\mathop{\cH}n+a)+\cF(n+1)&&\text{[by associativity of $+$]}\\
&=a\mathop{\cH}(n+1)+\cF(n+1).&&\text{[by definition of $\cH$]}
\end{align*}
So~\eqref{eqn:.n} holds for all $n$ in $\N$.  Therefore each of the
operations $x\mapsto x\mathop{\cH}n$ is
recursively defined by
\begin{enumerate}
  \item
$i\mathop{\cH}n=i$,
\item
$\cS(a)\mathop{\cH}n=a\mathop{\cH}n+\cF(n)$.
\end{enumerate}
In particular,
\begin{equation*}
\cF(m)=\cF(n)\iff\Forall xx\mathop{\cH}m=x\mathop{\cH}n.
\end{equation*}
Now we can define multiplication on $\cA$ by
\begin{equation*}
  a\cdot b=c\iff\Exists x(\cF(x)=b\land a\mathop{\cH}x=c).
\end{equation*}
Then $a\cdot i=a\mathop{\cH}0=i$.  Also, if $b=\cF(n)$, so that
$\cS(b)=\cF(n+1)$, then
\begin{equation*}
  a\cdot\cS(b)
=a\mathop{\cH}(n+1)
=a\mathop{\cH}n+a
=a\cdot b+a.
\end{equation*}
Thus $\cdot$ is as desired; it is unique by Theorem~\ref{thm:rec-uni}.
  \end{proof}

\begin{lemma}\label{lem:mult}
Suppose $(\cA,i,\cS)$ is recursive.
For all $a$ and $b$ in $\cA$,
  \begin{enumerate}
    \item
$i\cdot a=i$,
\item
$\cS(b)\cdot a=b\cdot a+a$.
  \end{enumerate}
\end{lemma}

\begin{proof}
  Exercise~\ref{ex:mult-lem}.
\end{proof}

\begin{theorem}\label{thm:mult}
Suppose $(\cA,i,\cS)$ is recursive.
For all $a$, $b$, and $c$ in $\cA$,
  \begin{enumerate}
    \item
$\cS(i)\cdot a=a$;
\item
$a\cdot b=b\cdot a$, that is, multiplication is
  \defnplain{commutative};%
\dindexsub{operation}{commutative!multiplication is ---}%
\item
$(a+b)\cdot c=a\cdot c+b\cdot c$, that is, multiplication
  \defnplain{distributes}{}%
\dindex{distributes, multiplication, over addition}
 over addition;
\item
$(a\cdot b)\cdot c=a\cdot (b\cdot c)$, that is, multiplication is
  \defnplain{associative}.%
\dindexsub{operation}{associative!multiplication is ---}
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:mult-thm}.
\end{proof}

\section{Exponentiation}

\begin{thmdefn}\label{thm:exp}
On $\N$, there is a unique binary operation, called
\defn{exponentiation}{}%
\dindexsub{operation}{exponentiation}
and denoted by $(x,y)\mapsto x^y$,
such that
\begin{enumerate}
\item
$m^0=1$,
\item
$m^{n+1}=m^n\cdot m$.
\end{enumerate}
More generally, if $(\cA,i,\cS)$ is recursive,
  there is a unique function from $\cA\times\N$ into $\cA$, also called
  exponentiation and denoted the same way,
such that, for all $a$ in $\cA$ and $m$ in~$\N$,
\begin{enumerate}
  \item
$a^0=\cS(i)$,
\item
$a^{m+1}=a^m\cdot a$.
\end{enumerate}
\end{thmdefn}

\begin{proof}
  Exercise~\ref{ex:exp0}.
\end{proof}

\begin{comment}


\begin{proof}
By the Recursion Theorem (\parref{thm:rec}), for each $a$ in $\cA$,
there is a unique function $\cF_a$ from $\N$ into $\cA$ such that
\begin{enumerate}
  \item
$\cF_a(0)=\cS(i)$,
\item
$\cF_a(\scr b)=\cF_a(b)\cdot a$.
\end{enumerate}
By the \emph{proof} of the Recursion Theorem, there is a binary
formula $\phi(x,y)$ such that $\cF_a=\cls y{\phi(a,y)}$.  Hence we can
define $(x,y)\mapsto x^y$ as $\cls{(x,y)}{x\in\cA\land\phi(x,y)}$.
Since the functions $\cF_a$ are unique, so is $(x,y)\mapsto x^y$.
\end{proof}



\end{comment}


\begin{parag}
  An \defn{endomorphism}{} of a structure is a homomorphism from the
  structure into itself.  The endomorphisms of $\str C$ compose a
  family (\parref{par:=classes}),\index{family} which we may denote by
  \begin{equation*}
    \End{\str C};
  \end{equation*}
this is closed under composition, so we have a structure
$(\End{\str C},\circ)$. 
\end{parag}

\begin{theorem}\label{thm:exp-prop}
Suppose $(\cA,i,\cS)$ is a recursive structure.  For all $a$ in $\cA$,
and $m$ and $n$ in $\N$,
  \begin{enumerate}
    \item
$a^{m+n}=a^m\cdot a^n$, that is, $x\mapsto a^x$ is a homomorphism from
      $(\N,+)$ into $(\cA,\cdot)$;
\item
$(a\cdot b)^m=a^m\cdot b^m$, that is, $x\mapsto x^m$ is an endomorphism
  of $(\cA,\cdot)$;
\item
$(a^m)^n=a^{m\cdot n}$, that is, $x\mapsto(y\mapsto y^x)$ is a
  homomorphism from $(\N,\cdot)$ into $(\End{\cA,\cdot},\circ)$. 
  \end{enumerate}
\end{theorem}

\begin{proof}
Exercise~\ref{ex:exp}.
\end{proof}

\section{The ordering of the natural numbers}

\begin{parag}
  In the ordering of $\N$ guaranteed by Theorem~\ref{thm:inf-bin-rel},
  we have
  \begin{equation*}
    \pred{\scr n}=\pred n\cup\{n\}.
  \end{equation*}
This formula suggests that the ordering can be obtained by a kind of
recursion. 
\end{parag}

\begin{theorem}[Recursion with Parameter]\label{thm:rec-p}%
\index{recursion!R--- Theorem with Parameter}%
\index{theorem!Recursion Th--- with Parameter}%
\index{parameter!Recursion Theorem with P---}
  Suppose $i\in\cC$ and $\cF\colon\N\times\cC\to\cC$.  Then there is a
  unique function $\cG$ from $\N$ into $\cC$ such that
  \begin{enumerate}
    \item
$\cG(0)=i$,
\item
$\cG(a+1)=a\mathop{\cF}\cG(a)$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Let $\cF_1$ be the operation $(x,y)\mapsto(x+1,x\mathop{\cF}y)$ on
  $\N\times\cC$.  Then $(\N\times\cC,(0,i),\cF_1)$ is an iterative
  structure, so by the corollary (\parref{cor:rec}) to the Recursion
  Theorem, there is a unique homomorphism
  $\cH$ to this structure from $(\N,0,x\mapsto\scr x)$.
Let $\pi_0$ and $\pi_1$ be as in~\parref{par:f}.
Now define $\cG$ as 
$\pi_1\circ\cH$, so that $\cG(0)=i$.  By induction, we can prove
\begin{equation*}
 \pi_0(\cH(a))=a.
\end{equation*}
Indeed,
the claim is true when $a=0$.  Suppose it is true when $a=b$; that is,
assume
\begin{equation}\label{eqn:rec-ih}
  \cH(b)=(b,\cG(b)).  
\end{equation}
Then 
\begin{equation*}
  \cH(b+1)=\cF_1(\cH(b))
=\cF_1(b,\cG(b))
=(b+1,b\mathop{\cF}\cG(b)).
\end{equation*}
Hence $\pi_0(\cH(b+1))=b+1$.  This computation establishes the
claim.  In particular, the inductive hypothesis~\eqref{eqn:rec-ih} has
been proved correct.  Hence the computation also establishes that $\cG(\scr
b)=\pi_1(\cH(b+1))=b\mathop{\cF}\cG(b)$.  So $\cG$ exists as desired.

The uniqueness of $\cG$ follows from the uniqueness of $\cH$, since
each of these functions is a function of the other:
$\cG=\pi_1\circ\cH$, and also $\cH$ is $x\mapsto(x,\cG(x))$.
\end{proof}

\begin{theorem}\label{thm:<}
In $\N$, for all $a$, $b$, and $c$,
\begin{enumerate}
\item
$0\leq a$,
\item
$a<b\iff a+c<b+c$,
\item
$a<b\iff a\cdot(c+1)<b\cdot(c+1)$,
\item\label{item:m-canc}
$a=b\iff a\cdot(c+1)=b\cdot(c+1)$,
  \item\label{item:leq}
$a\leq b\iff\Exists xa+x=b$.
\end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:<}.
\end{proof}

\begin{parag}
  Theorem~\ref{thm:rec-p} allows to make some standard definitions:
  \begin{asparaenum}
    \item
The operation $x\mapsto x!$ on $\N$ is defined recursively by
\begin{align*}
0!&=1,&
(n+1)!&=n!\cdot(n+1).
\end{align*}
\item
Suppose $\cF\colon\N\to\cC$, where $\cC$ is the universe of a structure
equipped with addition and multiplication.  Then the sum
$\sum_{k=0}^n\cF(k)$ and the product $\prod_{k=0}^n\cF(k)$ are defined
recursively as follows:
  \begin{align*}
    &\begin{gathered}
      \sum_{k=0}^0\cF(k)=\cF(0),\\
\sum_{k=0}^{n+1}\cF(k)=\sum_{k=0}^n\cF(k)+\cF(n+1),
     \end{gathered}&
&\begin{gathered}
\prod_{k=0}^0\cF(k)=\cF(0),\\
\prod_{k=0}^{n+1}\cF(k)=\prod_{k=0}^n\cF(k)\cdot\cF(n+1).  
 \end{gathered}
  \end{align*}
  \end{asparaenum}
\end{parag}

\section{The integers and the rational numbers}

\begin{parag}\label{par:a-b}
  In $\N$, if $a\leq b$, then (Theorem~\ref{thm:<}) there is a
  solution to the equation
  \begin{equation*}
    a+x=b;
  \end{equation*}
this solution is unique (Theorem~\ref{thm:+}~\eqref{item:canc}) and
can be denoted by
\begin{equation*}
  b-a.
\end{equation*}
If $a\leq b$ and $c\leq d$, then $b-a=d-c\iff a+d=b+c$; if $a+d=b+c$
and $a\leq b$, then $c\leq d$ (Exercise~\ref{ex:a-b}).
\end{parag}

\begin{thmdefn}\label{thmdefn:Z}
  On $\N\times\N$, let $\cE$ be the binary relation given by
  \begin{equation*}
    (a,b)\mathrel{\cE}(c,d)\iff a+d=b+c.
  \end{equation*}
Then $\cE$ is an equivalence-relation.  The family
$(\N\times\N)/\cE$ is denoted by
\begin{equation*}
  \Z;
\end{equation*}
its elements are the \defn{integer}{s.}
  The class
\begin{equation*}
  \cls{(x,y)}{(x,y)\in\N\times\N\land(x=0\lor y=0)}
\end{equation*}
contains exactly one representative for each $\cE$-class, and nothing
else; so we can treat $\Z$ as this class. 
Let the $\cE$-class $(a,b)\cE$
be denoted by
\begin{equation*}
  b-a.
\end{equation*}
Then addition and \defn{additive inversion}{} and multiplication of
$\cE$-classes, and a total ordering of them, can be defined by 
the following rules:
\begin{align*}
  (b-a)+(d-c)&=(b+d)-(a+c),\\
-(b-a)&=a-b,\\
  (b-a)\cdot(d-c)&=(b\cdot d+a\cdot c)-(b\cdot c+a\cdot d),\\
b-a<d-c&\iff b+c<a+d.
\end{align*}
In $\Z$, let $0-0$ be denoted by $0$, and let $1-0$ be denoted by
$1$.  Then $(\Z,+,-,\cdot,0,1,<)$ is an 
\defnplain{ordered ring}:%
\dindexsub{order}{---ed ring}%
\dindex{ring}
\begin{enumerate}
  \item
$+$ and $\cdot$ are commutative and associative;
\item
$\cdot$ distributes over $+$;
\item
$a+0=a$;
\item
$a\cdot 1=a$;
\item
$<$ is a total ordering;
\item
$0<a\land 0<b\lto0<a+b\land0<a\cdot b$.
\end{enumerate}
There is an embedding of $(\N,+,\cdot,0,1,<)$ in $(\Z,+,\cdot,0,1,<)$
by the rule taking $a$ to $a-0$.
\end{thmdefn}

\begin{proof}
  Exercise~\ref{ex:Z}.  To validate the definitions of $+$ and $\cdot$
  in $\Z$, one must show that, if $(a_0,b_0)\mathrel{\cE}(a_1,b_1)$ and
$(c_0,d_0)\mathrel{\cE}(c_1,d_1)$, then
\begin{align}\notag
    (a_0+c_0,b_0+d_0)&\mathrel{\cE}(a_1+c_1,b_1+d_1),\\\label{eqn:Z-m}
(a_0\cdot c_0+b_0\cdot d_0,b_0\cdot c_0+a_0\cdot
  d_0)&\mathrel{\cE}(a_1\cdot c_1+b_1\cdot d_1,b_1\cdot c_1+a_1\cdot
  d_1),\\\notag
b_0+c_0<a_0+d_0&\iff b_1+c_1<a_1+d_1. 
\end{align}
For~\eqref{eqn:Z-m}, consider  $(a_1\cdot c_0+b_1\cdot
d_0,b_1\cdot c_0+a_1\cdot d_0)$.
\end{proof}

\begin{parag}\label{par:b/a}
  In $\Z$, if $a\neq0$, if there is a solution to the equation
  \begin{equation*}
    a\cdot x=b,
  \end{equation*}
then the solution is unique by
Theorem~\ref{thm:<}~\eqref{item:m-canc} and
can be denoted by
\begin{equation*}
  \frac ba
\end{equation*}
or by $b/a$; also $a$ is a \defn{divisor}{} of $b$.
If $b/a$ and $d/c$ exist in $\Z$, then $b/a=d/c\iff a\cdot
  d=b\cdot c$ (Exercise~\ref{ex:b/a}).
\end{parag}

\begin{thmdefn}\label{thmdefn:Q}
  On $(\Z\setminus\{0\})\times\Z$, let $\cE$ be the binary relation
  given by 
  \begin{equation*}
    (a,b)\mathrel{\cE}(c,d)\iff a\cdot d=b\cdot c.
  \end{equation*}
Then $\cE$ is an equivalence-relation.  The family
$((\Z\setminus\{0\})\times\Z)/\cE$ is denoted by
\begin{equation*}
  \Q;
\end{equation*}
its elements are the 
\defnplain{rational number}{s.}%
\dindexsub{number}{rational ---}
Let the $\cE$-class $(a,b)\cE$
be denoted by
\begin{equation*}
  \frac ba.
\end{equation*}
Then addition, additive inversion, multiplication, and multiplicative
inversion of
$\cE$-classes, and a total ordering of them, can be defined by
the following rules:
\begin{align*}
  \frac ba+\frac dc&=\frac{a\cdot d+b\cdot c}{ac},\\
-\frac ba&=\frac{-b}a,\\
\left(\frac ba\right)\inv&=\frac ab\quad\text{ (if $b\neq0$)},\\
  \frac ba\cdot\frac dc&=\frac{b\cdot d}{a\cdot c},\\
\frac ba<\frac dc&\iff a\cdot b\cdot c\cdot c<a\cdot a\cdot c\cdot d.
\end{align*}
In $\Q$, let $0-0$ be denoted by $0$, and let $1-0$ be denoted by
$1$.  Then $(\Q,+,-,\cdot,0,1,<)$ is an 
\defnplain{ordered field}:%
\index{field|see{ordered field}}
\dindexsub{order}{---ed field} 
it is an ordered ring, and
\begin{equation*}
  \frac ab\cdot\frac ba=1
\end{equation*}
if $a\neq0$ and $b\neq0$.
Then $(\Z,+,\cdot,0,1,<)$ embeds in $(\Q,+,\cdot,0,1,<)$
by the rule taking $a$ to $a/1$.
\end{thmdefn}

\begin{proof}
  Exercise~\ref{ex:Q}.
\end{proof}

\section*{Exercises}

\begin{xca}\label{ex:3}
Write down bijections from $(\cC\times\cC)\times\cC$ to
$\cC\times(\cC\times\cC)$ and ${\cC}^3$. 
\end{xca}

\begin{xca}\label{ex:str}
Show that an ordered pair $(\cC,\cD)$ of \emph{classes} can be defined
as the class $(\cC\times\{0\})\cup(\cD\times\{1\})$: that is, show
that an equivalence like~\eqref{eqn:op-iff} holds in this case.
Define an ordered triple $(\cC,\cD,\cE)$ of classes.
\end{xca}

\begin{xca}\label{ex:embed}
Prove that an iterative structure admits induction if and only if every
embedding into it is a surjection onto the universe
(\parref{par:embed}).
  \end{xca}

\begin{xca}\label{ex:+}
  Prove Theorem~\ref{thm:rec+} by means of the corollary
  (\parref{cor:rec}) to the Recursion Theorem. 
\end{xca}

\begin{xca}\label{ex:+thm}
Prove Theorem~\ref{thm:+}.
\end{xca}

\begin{xca}
Find a recursive structure on which addition is not cancellative.
\end{xca}

\begin{xca}\label{ex:.}
  Prove Theorem~\ref{thm:multiplication}.
\end{xca}

\begin{xca}\label{ex:mult-lem}
Prove~Lemma~\ref{lem:mult}.
\end{xca}

\begin{xca}\label{ex:mult-thm}
Prove~Theorem~\ref{thm:mult}.
\end{xca}

\begin{xca}
Find a recursive structure $(\cA,i,\cS)$ where $\cA$ has elements $a$,
$b$, and $c$, all different from $i$, such that $a\cdot c=b\cdot c$,
but $a\neq b$.
\end{xca}

\begin{xca}\label{ex:exp0}
Prove Theorem~\ref{thm:exp}.
\end{xca}

\begin{xca}\label{ex:exp}
Prove Theorem~\ref{thm:exp-prop}.
\end{xca}

\begin{xca}
Find a recursive structure in which exponentiation cannot be defined
 as a binary operation.
\end{xca}

\begin{xca}
Show that there is a unique binary operation $(x,y)\mapsto\binom xy$
on $\N$ such that $\binom a0=1$ and $\binom 0{b+1}=0$ and
$\binom{a+1}{b+1}=\binom ab+\binom a{b+1}$.  Can such an
operation be defined on any structures that admit induction, but are
\emph{not} 
arithmetic? 
\end{xca}

\begin{xca}\label{ex:<}
Prove Theorem~\ref{thm:<}.
\end{xca}

\begin{xca}
Prove $1+a\cdot b\leq (1+a)^b$ in $\N$.
\end{xca}

\begin{xca}
Prove $3<a\lto a^2<2^a$ in $\N$.
\end{xca}

\begin{xca}\label{ex:a-b}
Prove the claims in \parref{par:a-b}.
\end{xca}

\begin{xca}\label{ex:Z}
Prove Theorem~\ref{thmdefn:Z}.
$a\neq0$. 
\end{xca}

\begin{xca}\label{ex:b/a}
Prove the claims in \parref{par:b/a}.
\end{xca}

\begin{xca}\label{ex:Q}
Prove Theorem~\ref{thmdefn:Q}.
\end{xca}

\chapter{Ordinality}

%\setcounter{equation}{9}

\section{Ordinals}

\begin{parag}
  A class is called 
\defnplain{transitive}{}%
\dindexsub{transitive}{--- class}%
\dindexsub{class}{transitive ---}
 if it \emph{includes}
  each of its elements.  That is, the class~$\cC$ is transitive if
  (and only if)
  $a\in\cC\lto a\included\cC$, or equivalently 
  \begin{equation*}
    b\in a\land a\in\cC\lto b\in\cC.
  \end{equation*}
Compare with~\parref{par:composite}:
transitivity of \emph{classes} must be distinguished from transitivity
of \emph{relations} on classes. 
Consider for example the relation $\in$ of membership:
\begin{enumerate}
  \item
 $\Bigl\{\emptyset,     \{\emptyset\}     ,\bigl\{\{\emptyset\}\bigr\}\Bigr\}$
    is a transitive 
    set, but membership is
    not transitive on this set.
\item
On the set
$\biggl\{\{\emptyset\},\bigl\{\{\emptyset\}\bigr\},\Bigl\{\{\emptyset\},\bigl\{\{\emptyset\}\bigr\}\Bigr\}\biggr\}$,
membership is transitive, but the set itself is not
transitive (it does not include its member $\{\emptyset\}$).
\end{enumerate}
Extending an operation on $\vnn$ (\parref{par:s(n)}) to all of
$\universe$, we say that 
the \defnplain{(set-theoretic) successor}{}%
\dindexsub{successor}{set-theoretic ---}%
\dindexsub{set}{---{}-theoretic successor} of a set $a$ is $\vscr a$,
where
\begin{equation*}
  \vscr a=a\cup\{a\}.
\end{equation*}
\end{parag}

\begin{lemma}\label{lem:trans}
  Every transitive set includes the set-theoretic
successor of each of its elements.  The set-theoretic successor of
every transitive set is transitive.
\end{lemma}

\begin{proof}
Suppose $a$ is transitive.  If $b\in a$, then
$\{b\}\included a$, but also $b\included a$ by transitivity of $a$, so
that $\vscr b\included a$.  If $c\in \vscr a$, then either $c=a$ or
$c\in a$; in either case, $c\included \vscr a$; thus $\vscr a$ is
transitive. 
\end{proof}

\begin{parag}\label{par:ord}
A set is an 
\defnplain{ordinal (number)}{}%
\dindex{ordinal}%
\dindexsub{number}{ordinal ---}
 if it is transitive and also
well-ordered by membership.  The class of ordinals is denoted by
\begin{equation*}
  \on.
\end{equation*}
We shall generally denote arbitrary elements of this class by letters
from the beginning of the Greek alphabet: $\alpha$,~$\beta$,
$\gamma$,~$\delta$, and so on.
\end{parag}

\begin{theorem}
  $\on$ contains $0$ and is closed under set-theoretic succession.  In
  particular, $\vnn\included\on$.
\end{theorem}

\begin{proof}
Vacuously, $0$ is transitive and well-ordered by membership.  Suppose
$\alpha\in\on$.  Then $\vscr{\alpha}$ is transitive by
Lemma~\ref{lem:trans}.  Also, since $\alpha$ is well-ordered by
membership, so is $\vscr{\alpha}$; indeed, $\alpha$ is the
$\in$-greatest member.
\end{proof}

\begin{theorem}\label{thm:on}
  $\on$ is transitive and well-ordered by membership.  With respect to
  this ordering, the least element is $0$, and succession is
  set-theoretic succession $(\scr{\alpha}=\vscr{\alpha})$.  On $\on$
  and on any ordinal, membership is proper inclusion.
\end{theorem}

\begin{proof}
For transitivity of $\on$, 
  supposing $\alpha\in\on$, we show $\alpha\included\on$.  To do this,
  say $\beta\in\alpha$; we show $\beta\in\on$.  But
  $\beta\included\alpha$, since $\alpha$ 
  is transitive; therefore $\beta$, like $\alpha$, is well-ordered by 
  membership.  To show $\beta$ is transitive, suppose
  $\gamma\in\beta$.  Then $\gamma\in\alpha$, so
  $\gamma\included\alpha$.  Suppose $\delta\in\gamma$; then
  $\delta\in\alpha$.  Since membership is transitive on $\alpha$, from
  $\delta\in\gamma$ and $\gamma\in\beta$, we
  can conclude $\delta\in\beta$.  Thus $\gamma\included\beta$.
  Therefore $\beta$ is transitive, so it is an ordinal.  Thus $\on$ is
  transitive.

%Now we show $\on$ is well-ordered by membership.  

Since the
members of $\on$ are transitive, membership is a transitive relation on $\on$.
Indeed, if $\on$ contains $\alpha$, $\beta$, and $\gamma$, and
$\alpha\in\beta$ and $\beta\in\gamma$, then $\beta\included\gamma$, so
$\alpha\in\gamma$.  Since membership is irreflexive on an ordinal, it
is so on $\on$.  That is, if $\alpha$ is in $\on$, and
$\beta\in\alpha$, then $\beta\notin\beta$, so $\beta\neq\alpha$.  In
particular, $\alpha\notin\alpha$.
Thus membership strictly orders $\on$.

Since $0$ has no members, it is the least member of $\on$.  
%Also, every ordinal that contains $\alpha$ must also \emph{include} $\alpha$
Also, every element of $\vscr{\alpha}$ is
either $\alpha$ itself or an element of $\alpha$; so $\vscr{\alpha}$
is minimal among ordinals that contain $\alpha$.  

%In particular, if $\alpha$ is in an ordinal $\beta$, but is not the
%greatest member of this ordinal, then $\vscr{\alpha}$ is the
%successor of $\alpha$ with respect to membership on $\beta$.


Suppose $\alpha$ and $\beta$
are arbitrary elements of $\on$.
By Theorem~\ref{thm:wo}, we may
assume there is an isomorphism $f$ from $\alpha$ to an
initial segment of $\beta$. 
By induction then, $\alpha$ must \emph{be} an initial segment of
$\beta$.  Indeed, say $\gamma\in\alpha$, so $\pred{\gamma}=\gamma$.
If $f\restriction{\gamma}=\id{\gamma}$, then $f(\gamma)=\gamma$.  So
either $\alpha=\beta$ or else $\alpha$ is a section $\pred{\delta}$ of
$\beta$; but in the latter case, $\alpha=\delta$, so $\alpha\in\beta$.
So membership is a
strict total ordering of $\on$.  

Every section of $\on$ is an element
of $\on$;
in particular, it is a set.  If $d$ is a subset of $\on$ with an
element $\alpha$, then either $\alpha$ is the least element of
$d$, or else the least element of $d$ is the least element of
$d\cap\alpha$.  So $\on$ is well-ordered by $\in$.

Finally, on $\on$ and its members, membership implies
\emph{inclusion;} and this inclusion is proper because membership is
irreflexive \emph{on} $\on$.  Conversely, if the ordinal $\beta$
properly includes the ordinal $\alpha$,
then $\alpha$ is a section of $\beta$, by transitivity of the latter;
hence $\alpha\in\beta$ as just shown.
\end{proof}

\begin{corollary}[Burali-Forti
    Paradox]\label{cor:bf-paradox}\dindex{Burali-Forti Paradox}
  $\on$ is not a set, so it is not an ordinal.
\end{corollary}

\begin{proof}
  Exercise~\ref{ex:bf-paradox}.
\end{proof}

\begin{parag}\label{par:omega-as-class}
If $\vnn$ is a proper class, then it is just $\on$
(Exercise~\ref{ex:omega=on}); if it is a set, then it is a limit in
$\on$ (\parref{par:succ}).  In any case, $\vnn$ is the class of
ordinals that neither \emph{are} limits nor \emph{contain} limits
(Exercise~\ref{ex:omega-limit}). 
\end{parag}

\section{Order-types}

\begin{parag}\label{par:on<}
  By definition (\parref{par:ord}), every ordinal $\alpha$ is
  well-ordered by membership.  We may then understand $\alpha$ to
  denote the structure $(\alpha,\in)$.  Likewise, since $\on$ is also
  well-ordered by membership (\parref{thm:on}), we may take $\on$
  to denote the
  structure $(\on,\in)$.  Finally, since for
  ordinals, $\alpha\in\beta\iff\alpha\pincluded\beta$
  (\parref{thm:on}), we may use $\in$ and $\pincluded$
  interchangeably in $\on$; we may also use $<$ for either of them,
  and we may use $\leq$ in place of $\included$.  But let us continue
  to use $\vscr{\alpha}$ (rather than $\scr{\alpha}$) to denote the
  successor
  $\alpha\cup\{\alpha\}$ of the ordinal $\alpha$.
\end{parag}

\begin{parag}
  Suppose $(a,<)$ is an order, and $b\included a$, and $c\in a$.  Then
  $c$ is an 
\defn{upper bound}{}%
\dindexsub{bound}{upper ---} for $b$ (with respect to $<$) if $d\in
  b\lto d\leq c$; and $c$ is a 
\defnplain{strict upper bound}{}%
\dindexsub{strict}{--- upper bound}
if $d\in
  b\lto d<c$.  If $b$ has a \emph{least} upper bound, then this is
unique and is the 
\defn{supremum}{} of $b$; it is denoted by
  \begin{equation*}
    \supr b.
  \end{equation*}
\end{parag}

\begin{axiom}[Union]\label{ax:union} 
\dindexsub{axiom}{Union A---}%
\dindexsub{union}{U--- Axiom}
%\axiomname{Union: \[\Exists xx=\bigcup a.\]}
  The union of a set is a set:
  \begin{equation*}
    \Exists xx=\bigcup b.
  \end{equation*}
\end{axiom}

\begin{theorem}\label{thm:sup}
  The union of a set of ordinals is an ordinal, which is the supremum
  of the set:
  \begin{equation*}
   b\pincluded\on\lto\bigcup b=\supr b.
  \end{equation*}
\end{theorem}

\begin{proof}
  Let $b$ be a set of ordinals.  Ordinals are sets of ordinals, by
  transitivity of $\on$ (\parref{thm:on}), so $\bigcup b$ is a
  subclass of $\on$. 
  Hence $\bigcup b$ is well-ordered by proper inclusion.  
Also $\bigcup b$ is a set by the Union Axiom.
Let $\beta\in\bigcup b$.  Then some
  element $\alpha$ of $b$ contains $\beta$.  But then $\alpha$ is
  transitive, so $\beta\included\alpha\included\bigcup b$.  Therefore
  $\bigcup b$ is an ordinal.

Finally, if $\alpha\in b$, then $\alpha\included\bigcup b$; so
$\bigcup b$ is an upper bound of $b$.  If $\beta<\bigcup b$, then
$\beta$ belongs to an element of $b$; that is, $\beta$ is less than
that element, so $\beta$ is not an upper bound of $b$.
\end{proof}

\begin{theorem}\label{thm:least}
  If $b$ is a set of ordinals, then $\bigcup\set{\vscr x}{x\in b}$ is
  the strict upper bound of $b$.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:least}.
\end{proof}

\begin{thmdefn}\label{thm:ot}
  Every well-ordered set $\str b$ is uniquely isomorphic to a unique
  ordinal, denoted by
  \begin{equation*}
    \ord{\str b}
  \end{equation*}
and called
  the 
\defnplain{order-type}{}%
\dindexsub{order}{---{}-type}
 or the 
\defn{ordinality}{} of $\str b$.
  Every well-ordered
  proper class is uniquely isomorphic to $\on$.
\end{thmdefn}

\begin{proof}
Theorem~\ref{thm:wo} and its proof show that every well-ordered set is
isomorphic to an initial segment of $\on$, since the latter is closed
under succession.  The initial segment must then be proper, since it
is a set.  Therefore it is an ordinal, as in the proof of
Theorem~\ref{thm:on}. 

Similarly, of two well-ordered proper classes, one is isomorphic to an
initial segment of the other.  But \emph{proper} initial segments of
well-ordered classes must be sets; therefore all well-ordered proper
classes are isomorphic.
\end{proof}

\section{Ordinal addition}

\begin{parag}\label{par:lex}
  Suppose $(\cC,<)$ and $(\cD,<)$ are total orders.  The
  \defnplain{(right) lexicographic ordering}{}%
\dindex{lexicographic ordering}%
\dindexsub{order}{lexicographic ---ing}
  of $\cC\times\cD$ is given by 
  \begin{equation*}
    (a,b)<(c,d)\iff b<d\lor (b=d\land a<c).
  \end{equation*}
This is a \emph{total} ordering of $\cC\times\cD$
(Exercise~\ref{ex:lex-tot}; see also Figure~\ref{fig:lex}). 
  If $(\cC,<)$ and $(\cD,<)$ are \emph{well-ordered,} then the
  lexicographic order well-orders $\cC\times\cD$: Indeed,
  if $\cE\included\cC\times\cD$, then its least element is $(a,b)$,
  where $b$ is the least element of
$\cls y{\Exists x(x,y)\in\cE}$, and $a$ is the least element of $\cls
  x{(x,b)\in\cC}$.
\end{parag}

\begin{figure}[t]
  \begin{center}
        \begin{pspicture}(0,-9.2)(12,0)
\psset{arrows=->}
\rput(1,-1){\rnode{00}{$(0,0)$}}
\rput(3,-1){\rnode{01}{$(0,1)$}}
\rput(1,-3){\rnode{10}{$(1,0)$}}
\rput(3,-3){\rnode{11}{$(1,1)$}}
\rput(5,-1){\rnode{02}{$(0,2)$}}
\rput(5,-3){\rnode{12}{$(1,2)$}}
\rput(1,-5){\rnode{20}{$(2,0)$}}
\rput(3,-5){\rnode{21}{$(2,1)$}}
\rput(5,-5){\rnode{22}{$(2,2)$}}
\rput(7,-1){\rnode{03}{$(0,3)$}}
\rput(7,-3){\rnode{13}{$(1,3)$}}
\rput(7,-5){\rnode{23}{$(2,3)$}}
\rput(1,-7){\rnode{30}{$(3,0)$}}
\rput(3,-7){\rnode{31}{$(3,1)$}}
\rput(5,-7){\rnode{32}{$(3,2)$}}
\rput(7,-7){\rnode{33}{$(3,3)$}}
\rput(9,-1){\rnode{04}{$(0,4)$}}
\rput(9,-3){\rnode{14}{$(1,4)$}}
\rput(9,-5){\rnode{24}{$(2,4)$}}
\rput(9,-7){\rnode{34}{$(3,4)$}}
\rput(1,-9){\rnode{40}{$(4,0)$}}
\rput(3,-9){\rnode{41}{$(4,1)$}}
\rput(5,-9){\rnode{42}{$(4,2)$}}
\rput(7,-9){\rnode{43}{$(4,3)$}}
\rput(9,-9){\rnode{44}{$(4,4)$}}
\rput(11,-1){\rnode{05}{$(0,5)$}}
\rput(11,-3){\rnode{15}{$(1,5)$}}
\rput(11,-5){\rnode{25}{$(2,5)$}}
\rput(11,-7){\rnode{35}{$(3,5)$}}
\rput(11,-9){\rnode{45}{$(4,5)$}}

\ncline{00}{10}
\ncline{10}{20}
\ncline{20}{30}
\ncline{30}{40}
\ncline{40}{01}
\ncline{01}{11}
\ncline{11}{21}
\ncline{21}{31}
\ncline{31}{41}
\ncline{41}{02}
\ncline{02}{12}
\ncline{12}{22}
\ncline{22}{32}
\ncline{32}{42}
\ncline{42}{03}
\ncline{03}{13}
\ncline{13}{23}
\ncline{23}{33}
\ncline{33}{43}
\ncline{43}{04}
\ncline{04}{14}
\ncline{14}{24}
\ncline{24}{34}
\ncline{34}{44}
\ncline{44}{05}
\ncline{05}{15}
\ncline{15}{25}
\ncline{25}{35}
\ncline{35}{45}
\ncline{45}{55}
    \end{pspicture}
  \end{center}
\caption{The lexicographic ordering of $5\times 6$}\label{fig:lex}
\end{figure}

\begin{parag}\label{par:ord-sum}
  The 
\defnplain{(ordinal) sum}{}%
\dindexsub{ordinal}{--- sum}%
\dindexsub{sum}{ordinal ---}
 of two
  ordinals $\alpha$ and $\beta$ is the
  ordinality of a well-ordered set that is like $\alpha$
  \emph{followed by} $\beta$.  Suppose there are embeddings $f$ and
  $g$ of $\alpha$ and $\beta$ respectively in a common well-ordered
  set such that $f(\gamma)<g(\delta)$ whenever $\gamma\in\alpha$ and
  $\delta\in\beta$; then we can take the sum of $\alpha$ and $\beta$
  to be the ordinality of $f\setimb{\alpha}\cup g\setimb{\beta}$.  To
  be precise, we define
  \begin{equation*}
    \alpha+\beta=\ord{(\alpha\times\{0\})\cup(\beta\times\{1\})},
  \end{equation*}
where $(\alpha\times\{0\})\cup(\beta\times\{1\})$ has the lexicographic
ordering of $(\alpha\cup\beta)\times2$.  But we must confirm that this
definition agrees with the definition in \parref{thm:addition} when
the ordinals are natural numbers.
\end{parag}

\begin{theorem}\label{thm:ord-add}
For all ordinals $\alpha$, $\beta$, and $\gamma$, 
\begin{enumerate}
  \item
  $0+\alpha=\alpha+0=\alpha,$
  \item
  $\vscr\alpha=\alpha+1,$
  \item
  $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma),$
\item\label{item:c+a<c+b}
$\alpha<\beta\lto\gamma+\alpha<\gamma+\beta,$
\item\label{item:a+x=b}
if $\alpha<\beta$, then $\alpha+x=\beta$ has a unique ordinal solution.
\end{enumerate}
\end{theorem}

\begin{proof}
Exercise~\ref{ex:ord-add}.  In \eqref{item:a+x=b}, the solution is
$\ord{\alpha\setminus\beta}$; it is unique by \eqref{item:c+a<c+b}.
\end{proof}

\begin{theorem}\label{thm:ord+rec}
  Ordinal addition can be defined recursively by the method given by
  Theorem~\ref{thm:rec-alt}: 
  \begin{enumerate}
\item
    $\alpha+0=\alpha$,
\item
$\alpha+\vscr{\beta}=\vscr{(\alpha+\beta)}$,
\item\label{item:+normal}
$\alpha+\beta=\bigcup\set{\alpha+x}{x\in\beta}$ when $\beta$ is a limit.
  \end{enumerate}
In particular, addition on $\vnn$ is as defined in
\parref{thm:rec+}. 
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:ord+rec}.
\end{proof}

\begin{lemma}\label{lem:ord<}
  Suppose $\str a$ and $\str b$ are well-ordered sets, and
  $f\colon\str a\to\str b$.  Then $\ord{\str a}\leq\ord{\str b}$.
\end{lemma}

\begin{proof}
  The homomorphism $f$ is an embedding, which induces an embedding $g$
  of  $\ord{\str a}$ in $\ord{\str b}$.  We shall show by induction
  that
 $\alpha<\ord{\str a}\lto\alpha\leq g(\alpha)$.  Suppose this is so
  when $\alpha<\beta$.  Say also $\beta<\ord{\str a}$.  Then
  $\alpha\leq g(\alpha)< g(\beta)$, so $\alpha<g(\beta)$.  Briefly,
  $\alpha<\beta\lto\alpha<g(\beta)$; so $\beta\leq g(\beta)$.  This
  completes the induction.  Since $g(\alpha)<\ord{\str b}$ whenever
  $\alpha<\ord{\str a}$, we conclude $\alpha<\ord{\str
  a}\lto\alpha<\ord{\str b}$, and hence $\ord{\str a}\leq\ord{\str b}$. 
\end{proof}

\begin{theorem}\label{thm:ord<}
  $\alpha\leq\beta\lto\alpha+\gamma\leq\beta+\gamma$.
\end{theorem}

\begin{proof}
  If $\alpha\leq\beta$, then
  $(\alpha\times\{0\})\cup(\gamma\times\{1\})\included 
   (\beta\times\{0\})\cup(\gamma\times\{1\})$; this inclusion is an
  embedding of the well-ordered sets, so
  $\alpha+\gamma\leq\beta+\gamma$ by Lemma~\ref{lem:ord<}.
\end{proof}

\begin{axiom}[Infinity]\label{ax:infinity}
\dindexsub{axiom}{A--- of Infinity}%
\dindex{Infinity, Axiom of}
%\axiomname{Infinity: \[\Exists xx=\vnn.\]}
The class of natural numbers is a set:
\begin{equation*}
  \Exists xx=\vnn.
\end{equation*}
\end{axiom}

\begin{theorem}\label{thm:+not-comm}
  $n<\vnn\lto n+\vnn=\vnn$.
\end{theorem}

\begin{proof}
  Define $f$ from $\vnn$ into $(n\times\{0\})\cup(\vnn\times\{1\})$ by 
  \begin{equation*}
    f(x)=
    \begin{cases}
      (x,0),& \text{ if }x<n;\\
      (y,1),& \text{ if }x=n+y.
    \end{cases}
  \end{equation*}
Then $f$ is an isomorphism.
\end{proof}

\begin{parag}\label{par:is+}
  By Theorem~\ref{thm:ord-add}~\eqref{item:c+a<c+b}, along with the
  Axiom of Infinity, we have the following initial segment of $\on$:
  \begin{equation*}
    \{0,1,2,\dots;\vnn,\vnn+1,\vnn+2,\dots;
  \vnn+\vnn,\vnn+\vnn+1,\dots;\vnn+\vnn+\vnn,\dots\}.
  \end{equation*}
Here the ordinals following the semicolons (;) are limits.  By
Theorem~\ref{thm:+not-comm}, addition involving infinite ordinals is not
commutative; also the ordering in Theorem~\ref{thm:ord<} cannot be
made strict: we have $0<1$, but $0+\vnn=\vnn=1+\vnn$.
\end{parag}

\section{Ordinal multiplication}

\begin{parag}\label{par:ord-prod}
  The 
\defnplain{(ordinal) product}{}%
\dindexsub{ordinal}{--- product}%
\dindex{product} 
of two ordinals is the 
  ordinality of their product with the lexicographic ordering
  (\parref{par:lex}):
  \begin{equation*}
    \alpha\cdot\beta=\ord{\alpha\times\beta}.
  \end{equation*}
We must confirm that this agrees with
Definition~\ref{thm:multiplication} on $\vnn$. 
\end{parag}

\begin{theorem}\label{thm:ord-mult}
  For all ordinals $\alpha$, $\beta$, and $\gamma$,
  \begin{enumerate}
    \item
$1\cdot\alpha=\alpha\cdot 1=\alpha$;
\item
$(\alpha\cdot\beta)\cdot\gamma=\alpha\cdot(\beta\cdot\gamma)$;
\item\label{item:left}
$\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma$;
\item\label{item:mult-strict}
$\alpha<\beta\land 0<\gamma\lto\gamma\cdot\alpha<\gamma\cdot\beta$;
\item\label{item:mult-loose}
$\alpha\leq\beta\lto\alpha\cdot\gamma\leq\beta\cdot\gamma$;
\item\label{item:xy-sol}
the system $\alpha\cdot x+y=\beta\land y<\alpha$ has a unique ordinal
solution if $0<\alpha$.  
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:ord-mult}.  
For \eqref{item:xy-sol}, show
  $\beta\leq\alpha\cdot\beta$.  If $\beta<\alpha\cdot\beta$, then
  $\beta$ is isomorphic to a section $\pred{(\gamma,\delta)}$ of
  $\alpha\times\beta$.  But
  $\pred{(\gamma,\delta)}=
  (\alpha\times\delta)\cup(\gamma\times\{\delta\})$; so
  $\alpha\cdot\delta+\gamma=\beta$ and $\gamma<\alpha$.  Now show
  uniqueness.  
\end{proof}

\begin{theorem}\label{thm:ord.rec}
  Ordinal multiplication can be defined recursively by the method
  given by Theorem~\ref{thm:rec-alt}:  
  \begin{enumerate}
    \item
$\alpha\cdot0=0$,
\item
$\alpha\cdot\vscr{\beta}=\alpha\cdot\beta+\alpha$,
\item\label{item:.normal}
$\alpha\cdot\beta=\bigcup\set{\alpha\cdot x}{x\in\beta}$ when $\beta$
  is a limit.
  \end{enumerate}
In particular, multiplication on $\vnn$ is as in
\parref{thm:multiplication}. 
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:ord.rec}.
\end{proof}

\begin{theorem}\label{thm:nw=w}
  $0<n<\vnn\lto n\cdot\vnn=\vnn$.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:nw=w}.  (Write out an isomorphism from
  $n\times\vnn$ to $\vnn$.)
\end{proof}

\begin{parag}
We can now extend the initial segment of $\on$ in \parref{par:is+}:
\begin{equation*}
  \{0,1,\dots;\vnn,\vnn+1,\dots;
  \vnn\cdot2,\dots;\vnn\cdot3,\dots;
  \vnn\cdot\vnn,\dots;\vnn\cdot\vnn\cdot\vnn,\dots\}.
\end{equation*}
By Theorem~\ref{thm:ord-mult}~\eqref{item:left} and
Theorem~\ref{thm:nw=w}, ordinal multiplication is not commutative, nor
does it distribute from the right over addition:
\begin{gather*}
(1+1)\cdot\vnn=2\cdot\vnn=\vnn<\vnn+\vnn
=1\cdot\vnn+1\cdot\vnn;
%=\vnn\cdot 1+\vnn\cdot 1=\vnn\cdot2;
\end{gather*}
also the ordering in
Theorem~\ref{thm:ord-mult}~\eqref{item:mult-loose} cannot be made strict.
\end{parag}

\section{Ordinal exponentiation}

\begin{parag}\label{par:ord-exp}
The binary operation of 
\defnplain{(ordinal) exponentiation}{}%
\dindexsub{ordinal}{--- exponentiation}%
\dindexsub{exponentiation}{ordinal ---}
on $\on$, denoted by
\begin{equation*}
(x,y)\mapsto x^y,
\end{equation*}
is such that
$0^0=1$ and $0^\beta=0$ when $\beta\neq0$, and, when $0<\alpha$, is
defined by the method given by Theorem~\ref{thm:rec-alt}:
\begin{enumerate}
  \item
  $\alpha^0=1$,
  \item
  $\alpha^{\beta+1}=\alpha^\beta\cdot\alpha$,
  \item
  $\alpha^{\beta}=\bigcup\set{\alpha^x}{x\in\beta}$ when
  $\beta$ is a limit. 
\end{enumerate}
In particular, on $\vnn$, this operation is as given in
\parref{thm:exp}.  
\end{parag}

\begin{theorem}\label{thm:1abc}
  If $1<\alpha$ and $\beta<\gamma$, then $\alpha^{\beta}<\alpha^{\gamma}$.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:1abc}.
\end{proof}

\begin{parag}\label{par:normal}
  An endomorphism $\cF$ of $\on$ is called 
\defn{normal}{} if
  \begin{equation*}
    \cF(\alpha)=\supr{\cF\setimb\alpha},
  \end{equation*}
that is, $\cF(\alpha)=\bigcup\cF\setimb{\alpha}$,
whenever $\alpha$ is a limit.  
\end{parag}

\begin{lemma}
The following operations on $\on$ are normal:
  \begin{enumerate}
  \item
$x\mapsto\alpha+x$;
\item
$x\mapsto\alpha\cdot x$, when $0<\alpha$;
\item
$x\mapsto\alpha^x$, when $1<\alpha$.
  \end{enumerate}
\end{lemma}

\begin{proof}
By Theorems~\ref{thm:ord-add}~\eqref{item:c+a<c+b},
\ref{thm:ord-mult}~\eqref{item:mult-strict}, and~\ref{thm:1abc},
the operations are endomorphisms.
Hence they are normal by
Theorems~\ref{thm:ord+rec}~\eqref{item:+normal}
and~\ref{thm:ord.rec}~\eqref{item:.normal} and~\parref{par:ord-exp}.  
\end{proof}

\begin{lemma}\label{lem:normal}
If $\cF$ is normal and $c\pincluded\on$, then
\begin{equation*}
  \cF(\supr c)=\supr{\cF\setimb c}.
\end{equation*}
\end{lemma}

\begin{proof}
  Let $\alpha=\supr c$.  If $\alpha\in c$, then $\alpha$ is the
  greatest element of $c$,
  so $\supr{\cF\setimb c}=\cF(\alpha)$ since $\cF$ preserves order.
  Suppose $\alpha\notin c$. 
Then $c\included\alpha$, so $\supr{\cF\setimb
  c}\leq\supr{\cF\setimb{\alpha}}$.  Also, if
  $\beta<\alpha$, then $\beta<\gamma<\alpha$ for some $\gamma$ in
  $c$, so $\supr{\cF\setimb\alpha}\leq\bigcup\cF\setimb c$.
Therefore $\supr{\cF\setimb\alpha}=\bigcup\cF\setimb c$.
But $\alpha$ must be a limit, so $\supr{\cF\setimb\alpha}=\cF(\alpha)$
  by normality of~$\cF$.
\end{proof}

\begin{theorem}\label{thm:ord-exp-prop}
  For all ordinals $\alpha$, $\beta$, and $\gamma$,
  \begin{enumerate}
\item
$1^{\alpha}=1$,
\item
$\alpha^1=\alpha$,
\item\label{item:a^(b+c)}
$\alpha^{\beta+\gamma}=\alpha^\beta\cdot\alpha^\gamma$,
\item\label{item:a^(bc)}
$\alpha^{\beta\cdot\gamma}=(\alpha^\beta)^\gamma$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:ord-exp-prop}.
  For~\eqref{item:a^(b+c)}, by
  Lemma~\ref{lem:normal}, if $1<\alpha$ and $\gamma$ is a limit, one
  has by induction
  \begin{align*}
\alpha^\beta\cdot\alpha^\gamma
&=\alpha^\beta\cdot\supr{\set{\alpha^x}{x\in\gamma}}\\
&=\supr{\set{\alpha^\beta\cdot\alpha^x}{x\in\gamma}}\\
&=\supr{\set{\alpha^{\beta+x}}{x\in\gamma}}\\
&=\alpha^{\supr{\set{\beta+x}{x\in\gamma}}}\\
&=\alpha^{\beta+\supr{\set{x}{x\in\gamma}}}
=\alpha^{\beta+\gamma}.
  \end{align*}
Likewise for \eqref{item:a^(bc)}.
\end{proof}

\begin{parag}
We now have the following initial segment of $\on$:
\begin{multline*}
  \{0,1,\dots;\vnn,\vnn+1,\dots,\vnn\cdot 2,\dots;\vnn^2,\vnn^2+1,\dots;
  \vnn^2+\vnn,\dots;\\
\vnn^2\cdot2,\dots;
\vnn^3,\dots;\vnn^\vnn,\dots;
  \vnn^{\vnn\cdot2},\dots;\vnn^{\vnn^2},\dots;\vnn^{\vnn^\vnn},
  \dots;\vnn^{\vnn^{\vnn^\vnn}},\dots\}.
\end{multline*}
This set is closed under the operations that we have defined so far.
The supremum of the set is a solution of the equation
\begin{equation}\label{eqn:epsilon}
  \vnn^x=x
\end{equation}
(Exercise~\ref{ex:epsilon});
in fact it is the least solution and so is denoted by
\begin{equation*}
  \upepsilon_0.
\end{equation*}
\end{parag}

\section{Base omega}

\begin{theorem}
  $\alpha\leq\vnn^{\alpha}$.
\end{theorem}

\begin{proof}
  The claim holds by definition of exponentiation (\parref{par:ord-exp})
  when $\alpha=0$.  Suppose it holds when $\alpha=\beta$.  Then
  \begin{equation*}
    \vnn^{\beta+1}
   =\vnn^{\beta}\cdot\vnn^1
   =\vnn^{\beta}\cdot\vnn
   >\vnn^{\beta}\cdot 2
   =\vnn^{\beta}+\vnn^{\beta}
\geq\vnn^{\beta}+\vnn^0
   =\vnn^{\beta}+1\geq\beta+1.
  \end{equation*}
Finally, if $\beta$ is a limit, and the claim holds when
$\alpha<\beta$, then it holds when $\alpha=\beta$ by normality.
\end{proof}

\begin{theorem}
  Suppose $\alpha>0$.  There are, uniquely, a natural number $n$, 
an $(n+1)$-tuple $(\beta_0,\dots,\beta_n)$ of ordinals, 
and 
an $(n+1)$-tuple $(c_0,\dots,c_n)$ of non-zero natural numbers, 
such that
  \begin{align*}
    &\beta_0>\dotsb>\beta_n,&
    &\alpha=\vnn^{\beta_0}\cdot c_0+\dotsb+\vnn^{\beta_n}\cdot c_n.
  \end{align*}
\end{theorem}

\begin{proof}
Uniqueness follows from observing
\begin{equation*}
  \vnn^{\beta_0}\leq
\vnn^{\beta_0}\cdot c_0+\dotsb+\vnn^{\beta_n}\cdot c_n<
\vnn^{\beta_0+1}.
\end{equation*}
For existence, we proceed recursively.
Assuming
 $(\beta_0,\dots,\beta_{k-1})$ and $(c_0,\dots,c_{k-1})$ have been
  defined, we consider the inequality
  \begin{equation*}
    \sum_{i<k}\vnn^{\beta_i}\cdot c_i+\vnn^x\leq \alpha.
  \end{equation*}
If this has no solution, then $n=k-1$, and we are done.
If there is a solution, let $\beta_k$ be the supremum of the set of
solutions; then this is a solution by normality.  Now let $c_k$ be the
maximal solution of
  \begin{equation*}
    \sum_{i<k}\vnn^{\beta_i}\cdot c_i+\vnn^{\beta_k}\cdot x\leq \alpha.
  \end{equation*}
By induction, we have $\beta_0>\beta_1>\dotsb$, so the construction
must terminate in finitely many steps.
\end{proof}

\begin{lemma}
  If $\alpha>0$, then $1+\vnn^{\alpha}=\vnn^{\alpha}$.
\end{lemma}

\begin{proof}
  Use induction.  The claim holds when $\alpha=1$ by
  Theorem~\ref{thm:+not-comm}.  Suppose it holds when $\alpha=\beta$.
  Then
  \begin{align*}
 \vnn^{\beta+1}
&\leq 1+\vnn^{\beta+1}\leq \vnn^{\beta}+\vnn^{\beta+1}
          &&\text{[Theorem~\ref{thm:ord<}]}\\
&=     \vnn^{\beta}\cdot
1+\vnn^{\beta}\cdot\vnn&&\text{[Theorems~\ref{thm:ord-mult}
  and~\ref{thm:ord-exp-prop}]}\\
&=     \vnn^{\beta}\cdot
(1+\vnn)=\vnn^{\beta}\cdot\vnn=\vnn^{\beta+1},&&
  \end{align*}
so the claim holds when $\alpha=\beta+1$.  The claim holds at limits
by normality of exponentiation.
\end{proof}

\begin{theorem}\label{thm:b<a}
  If $\beta<\alpha$, and $n\in\vnn$, then
  $\vnn^{\beta}\cdot n+\vnn^{\alpha}=\vnn^{\alpha}$. 
\end{theorem}

\begin{proof}
Exercise~\ref{ex:b<a}.
\end{proof}

\begin{theorem}
  If $\beta+\alpha=\alpha$, then $\beta\leq\alpha$ and
  \begin{equation*}
    (\alpha+\beta)\cdot\gamma=
    \begin{cases}
      \alpha\cdot\gamma+\beta,&\text{ if $\gamma$ is a successor};\\
      \alpha\cdot\gamma,&\text{ if $\gamma$ is a limit or $0$}.
    \end{cases}
  \end{equation*}
\end{theorem}

\begin{proof}
For the first claim, note that if $\alpha<\beta$, then
$\alpha<\beta\leq\beta+\alpha$. 
  For the second claim, use induction.  The claim holds when
  $\gamma=0$ by definition of 
  multiplication.  Suppose it holds when $\gamma=\delta$; then
  \begin{equation*}
    (\alpha+\beta)\cdot(\delta+1)
=   (\alpha+\beta)\cdot\delta+\alpha+\beta
=    \alpha\cdot\delta+\alpha+\beta
=    \alpha\cdot(\delta+1)+\beta,
  \end{equation*}
so the claim holds when $\gamma=\delta+1$.  Finally, since
\begin{equation*}
\alpha\cdot\delta+\beta
\leq\alpha\cdot\delta+\alpha
=\alpha\cdot(\delta+1),
\end{equation*}
by normality the claim holds at limits.
\end{proof}


\begin{parag}
In short, ordinals can be written in
\defnplain{base $\vnn$}{},%
\dindexsub{base}{--- $\vnn$} 
and when they are so written, 
we can add and multiply them.  For example,
  \begin{gather*}
  (\vnn^{\vnn+1}\cdot 3+\vnn^6\cdot 4+1)\cdot(\vnn^{\vnn^2}\cdot 2+3)\\
=  (\vnn^{\vnn+1}\cdot 3+\vnn^6\cdot 4+1)\cdot(\vnn^{\vnn^2}\cdot 2)+
  (\vnn^{\vnn+1}\cdot 3+\vnn^6\cdot 4+1)\cdot3\\
=  (\vnn^{\vnn+1}\cdot 3)\cdot(\vnn^{\vnn^2}\cdot 2)+
  (\vnn^{\vnn+1}\cdot 3)\cdot 3+\vnn^6\cdot 4+1\\
=  \vnn^{\vnn+1}\cdot(3\cdot\vnn^{\vnn^2})\cdot 2+
  \vnn^{\vnn+1}\cdot(3\cdot 3)+\vnn^6\cdot 4+1\\
=  (\vnn^{\vnn+1}\cdot\vnn^{\vnn^2})\cdot 2+
  \vnn^{\vnn+1}\cdot9+\vnn^6\cdot 4+1\\
=  (\vnn^{\vnn+1+\vnn^2})\cdot 2+
  \vnn^{\vnn+1}\cdot9+\vnn^6\cdot 4+1\\
=  (\vnn^{\vnn^2})\cdot 2+
  \vnn^{\vnn+1}\cdot9+\vnn^6\cdot 4+1.
\end{gather*}
\end{parag}


\section*{Exercises}

\begin{xca}%\label{ex:a'}
Prove that
\begin{equation}%\label{eqn:a'}
\vscr a=\bigcap\cls x{a\in x\land a\included x}.
\end{equation}

\end{xca}

\begin{xca}\label{ex:o-wf}
Show that, if $a=\{a\}$, then $a$ is not an ordinal.
\end{xca}

\begin{xca}\label{ex:bf-paradox}
Prove the Burali-Forti Paradox (\parref{cor:bf-paradox}).
\end{xca}

\begin{xca}\label{ex:omega=on}
Show that, if $\vnn$ is a proper class, then $\vnn=\on$. 
\end{xca}

\begin{xca}\label{ex:omega-limit}
  Show that $\vnn$ is the class of ordinals that neither are limits
  nor contain limits.
\end{xca}

\begin{xca}
Prove that an ordinal $\alpha$ is $0$ or a limit if and only if
$\alpha=\bigcup\alpha$. 
\end{xca}

\begin{xca}\label{ex:least}
Prove Theorem~\ref{thm:least}.
\end{xca}

\begin{xca}\label{ex:lex-tot}
Prove that the lexicographic ordering is total (\parref{par:lex}).
\end{xca}

\begin{xca}\label{ex:ord-add}
Prove Theorem~\ref{thm:ord-add}.
\end{xca}

\begin{xca}\label{ex:ord+rec}
Prove Theorem~\ref{thm:ord+rec}.
\end{xca}

\begin{xca}\label{ex:ord-mult}
Prove Theorem~\ref{thm:ord-mult}.
\end{xca}

\begin{xca}\label{ex:ord.rec}
Prove Theorem~\ref{thm:ord.rec}.
\end{xca}

\begin{xca}\label{ex:nw=w}
Prove Theorem~\ref{thm:nw=w}.
\end{xca}

\begin{xca}\label{ex:1abc}
  Prove Lemma~\ref{thm:1abc}.
\end{xca}
\begin{xca}
Prove $1<\alpha\lto\beta\leq\alpha^\beta$.
\end{xca}

\begin{xca}\label{ex:ord-exp-prop}
Prove Theorem~\ref{thm:ord-exp-prop}.
\end{xca}

\begin{xca}\label{ex:epsilon}
  Defining $(a_n\colon n\in\vnn)$ by $a_0=0$ and
  $a_{n+1}=\vnn^{a_n}$, show that $\sup\{a_n\colon n\in\vnn\}$ is
  closed under the operations of addition, multiplication, and
  exponentiation, and is the least solution of~\eqref{eqn:epsilon}.
\end{xca}

\begin{xca}\label{ex:b<a}
  Prove Theorem~\ref{thm:b<a}.
\end{xca}

\begin{xca}
  Compute the following in base $\vnn$:
  \begin{enumerate}
    \item
$(\vnn+1)+(\vnn+2)$;
\item
$(\vnn^{\vnn^\vnn}+\vnn^2\cdot 3+4)+(\vnn^{\vnn^2}+\vnn^\vnn\cdot
  5+1)$;
\item
$(\vnn+1)\cdot(\vnn+ 2)$;
\item
$(\vnn^2\cdot 3+2)\cdot(\vnn^\vnn+5)$
  \end{enumerate}
\end{xca}
\begin{xca}
  Compute the following in base $\vnn$:
  \begin{enumerate}
\item
    $(\vnn+n)^2$;
\item
    $(\vnn+n)^3$;
\item
    $(\vnn+n)^m$;
\item
    $(\vnn+n)^\vnn$;
\item
    $(\vnn+n)^{\vnn+m}$.
  \end{enumerate}
\end{xca}


\chapter{Cardinality}\label{ch:cardinality}

%\setcounter{equation}{9}

\section{Finite sets}

\begin{parag}\label{par:inductive}

A set $a$ is \defn{finite}{}%
\dindexsub{set}{finite ---} 
if it belongs to every subset of $\pow a$ that
\begin{enumerate}
\item
contains $\emptyset$, and
\item
if closed under the operation $x\mapsto x\cup\{b\}$, for every $b$ in
$a$. 
\end{enumerate}
(If one wants a word, one may refer to such a subset of $\pow a$ as
\defnplain{inductive}.%
\dindexsub{inductive}{--- set}%
\dindexsub{set}{inductive ---}) 
We shall investigate the relation between finite sets and infinite
sets as defined in \parref{par:inf}.
\end{parag}

\begin{lemma}\label{lem:fin}
$\emptyset$ is finite, and 
  if $a$ is finite, then so is $a\cup\{b\}$.
\end{lemma}

\begin{proof}
Since $\emptyset$ belongs to every subset of $\pow{\emptyset}$ that
contains $\emptyset$, it is finite.

Suppose $a$ finite, and $c$ is subset of $\pow{a\cup\{b\}}$ that
contains $\emptyset$ and is closed under $x\mapsto x\cup\{d\}$ for
every $d$ in $a\cup\{b\}$.  Then $c\cap\pow a$ contains $\emptyset$
and is closed under $x\mapsto x\cup\{d\}$ whenever $d\in a$.
Therefore $a\in c\cap\pow a$, since $a$ is finite; hence $a\in c$, so
$a\cup\{b\}\in c$.  Therefore $a\cup\{b\}$ is finite.
\end{proof}

\begin{theorem}[Induction]\label{thm:set-ind}
\dindexsub{induction}{I--- Theorem}%
\dindexsub{theorem}{Induction Th---}
A class of finite sets that contains $\emptyset$ and is closed under
each operation $x\mapsto x\cup\{b\}$  contains all finite sets.
\end{theorem}

\begin{proof}
Let $\cC$ be such a class, and let $a$ be a finite set.  Then
$a\in\cC\cap\pow a$ (since this set is inductive in the sense
of~\parref{par:inductive}). 
\end{proof}

\begin{lemma}\label{lem:fin-im}
  The image of a finite set is finite.
\end{lemma}

\begin{proof}
  Let $\cC$ comprise those finite sets whose images are all finite.
  Trivially, $\emptyset\in \cC$.  Suppose $a\in\cC$ and
  $a\cup\{b\}\included\dom{\cF}$.  Then $\cF\setimb a$ is finite, and
  $\cF\setimb{a\cup\{b\}}=\cF\setimb a\cup\{\cF(b)\}$, so this is also
  finite by Lemma~\ref{lem:fin}.  Thus $a\cup\{b\}\in\cC$.  By induction
  (\parref{thm:set-ind}), $\cC$ contains all finite sets.
\end{proof}

\begin{theorem}\label{thm:fin-char}
  A set is finite if and only if it is equipollent with a natural
  number. 
\end{theorem}

\begin{proof}
  All natural numbers are finite, by induction:
  \begin{enumerate}
    \item
$\emptyset$ is finite;
\item
if $n$ is finite, then so is $\vscr n$, that is, $n\cup\{n\}$, by
Lemma~\ref{lem:fin}. 
  \end{enumerate}
Hence all sets equipollent with natural numbers are finite, by
Lemma~\ref{lem:fin-im}.  
Conversely, every finite set is equipollent with a natural number, by
induction (\parref{thm:set-ind}):
\begin{enumerate}
  \item
$\emptyset\equip\emptyset$;
\item
if $a\equip n$, then $a\cup\{b\}\equip n\cup\{n\}$ (assuming $b\notin
a$). \hfill\qed
\end{enumerate}
\renewcommand{\qed}{}
\end{proof}

\begin{lemma}\label{lem:finite}
  Suppose $a$ and $b$ are in $\vnn$, and $f\colon\vscr
  a\injection\vscr b$.  Then $g\colon a\injection b$, where 
  \begin{equation*}
    g(x)=
    \begin{cases}
      f(x),&\text{ if }f(x)\neq b;\\
      f(a),&\text{ if }f(x)=b.
    \end{cases}
  \end{equation*}
Also, $f\colon\vscr a\bijection\vscr b\iff g\colon a\bijection b$.
\end{lemma}

\begin{proof}
If $f$ is a bijection, then
\begin{align*}
  \ran g
&=\set{f(x)}{x\in a\land f(x)\neq b}\cup\set{f(a)}{x\in a\land
    f(x)=b}\\
&=f\setimb{a\setminus\{f\inv(b)\}}\cup
  f\setimb{\{a\}\setminus\{f\inv(b)\}}\\
&=f\setimb{a'\setminus\{f\inv(b)\}}\\
&=f\setimb{a'}\setminus\{b\}\\
&=b,
\end{align*}
so $g$ is surjective.  If $g$ is a bijection, then
\begin{equation*}
  \ran f=\ran g\cup f\setimb{f\inv\setimb{\{b\}}}=b\cup\{b\}=\vscr b,
\end{equation*}
so $f$ is surjective.
For the  injectivity of $g$, assuming the injectivity of $f$, there
  are two cases.  If $f(a)=b$, then $g$ is simply $f\rest{a}$, so it
  preserves the injectivity of $f$.  Suppose $f(c)=b$, where
  $c<b$.  Then $g(c)=f(a)$, but $a\notin\dom g$, so $g(d)=f(d)\neq
  f(a)$ when $d\in a\setminus\{c\}$.  Again $g$ is injective.  
\end{proof}

\begin{theorem}\label{thm:nat-fin}
  No finite set is infinite.
\end{theorem}

\begin{proof}
By Theorem~\ref{thm:fin-char}, it is enough to show that no natural
  number is infinite.
  We show that every injection from a natural number into itself is
  surjective.  This is trivially true for $0$, and if it is true for
  $a$, then it is true for $\vscr a$, by Lemma~\ref{lem:finite}: if
  $f\colon\vscr a\injection\vscr a$, then there is an injection
  from $a$ into itself, but this is then a surjection too by
  inductive hypothesis, so $f$ is surjective. 
\end{proof}

\begin{theorem}
  Equipollent natural numbers are equal.
\end{theorem}

\begin{proof}
  We show
  \begin{equation*}
    a\in\vnn\land b\in\vnn\land a+b\equip a\lto b=0.
  \end{equation*}
This is trivially true when $a=0$.  Suppose it is true when $a=c$.
Say $\vscr a+b\equip\vscr a$.  Since $\vscr a+b=\vscr{(a+b)}$, by
Lemma~\ref{lem:finite} there is a bijection from $a+b$ to $a$.  By the
inductive hypothesis then, $b=0$.
\end{proof}

\section{Cardinals}


\begin{parag}\label{par:card}
If a set $a$ can be well-ordered, then $a$ is equipollent with some
  ordinal number (Theorem~\ref{thm:ot}).  In this case, 
  we can define the 
\defn{cardinality}{} of $a$ as the \emph{least}
  ordinal that is equipollent with $a$.  
If $a$ cannot be well-ordered, then we still have to understand its
\defn{cardinality}{} as the class $\cls x{x\equip a}$, as in
\parref{par:equip}.
In either case, as suggested in
  \parref{par:equip}, the cardinality of $a$ is denoted by
  \begin{equation*}
    \card a.
  \end{equation*}
We shall ultimately (with \parref{ax:choice}) decide that there are
\emph{no} sets that cannot be well-ordered; but not everything that
can be said about cardinalities requires this assumption.  So, for now,
we have two kinds of cardinalities:
\begin{enumerate}
  \item
those that are ordinal numbers: these can be called 
\defnplain{cardinal number}s
 or just 
\defnplain{cardinal}{s;}%
\dindexsub{cardinal}{--- (number)}
\dindexsub{number}{cardinal ---} 
\item
the other cardinalities: the classes $\cls x{x\equip a}$: where $a$
cannot be well-ordered.
\end{enumerate}
The cardinals compose the class denoted by
  \begin{equation*}
    \cn;
  \end{equation*}
this is a subclass of $\on$.  Then $\cn$ inherits the ordering of
$\on$,
usually denoted by $<$ as suggested in \parref{par:on<}; and this
ordering coincides with $\pinjects$ (\parref{par:injects};
Exercise~\ref{ex:cn<}). 
\end{parag}

\begin{lemma}[Hartogs]\dindex{Hartogs Theorem}%
\dindexsub{theorem}{Hartogs Th---}
  For every set, there is an ordinal that does not embed in it.
\end{lemma}

\begin{proof}
  If $a$ is a set, let $b$ be the set of well-ordered sets $(c,<)$
  such that $c\included a$.  If $\ord{c,<}=\beta$, and
  $\gamma<\beta$, then $\ord{d,<}=\gamma$ for some section $d$ of $c$.
  This shows that $\set{\ord{\str c}}{\str c\in b}$ is a transitive
  subset of $\on$; so it is an ordinal~$\alpha$.  If
  $f:\beta\injection a$, then $f$ determines an element of $b$ whose
  ordinality is $\beta$; so $\beta\in\alpha$.  Since
  $\alpha\notin\alpha$, there is no injection of $\alpha$ in $a$.
\end{proof}

\begin{parag}
  As a consequence of the Hartogs Lemma, for every cardinal $\kappa$,
  there is an ordinal $\alpha$ such that $\kappa<\alpha$, but
  $\kappa\nequip\alpha$.  Therefore $\kappa<\card{\alpha}$.  Thus
  $\kappa$ has a 
\defn{successor},
  \begin{equation*}
    \scr{\kappa};
  \end{equation*}
it is the \emph{least} of the cardinals that are greater than $\kappa$.
\end{parag}

\begin{lemma}
  The union of a set of cardinals is a cardinal.
\end{lemma}

\begin{proof}
  Let $a$ be a set of cardinals.  The union $\bigcup a$ is an ordinal,
  namely $\supr a$,
  by Theorem~\ref{thm:sup}.  Hence, if $\kappa$ is a cardinal
  less than
  $\supr a$, then $\kappa<\lambda$ for some $\lambda$ in $a$, and
  therefore $\kappa\neq\card{\supr a}$.  Therefore $\supr a$ must be a
  cardinal (namely its own cardinality).
\end{proof}

\begin{parag}
The function
\begin{equation*}
  x\mapsto\aleph_x
\end{equation*}
from $\on$ into $\cn$ is given (as in Theorem~\ref{thm:rec-alt}) by
\begin{enumerate}
  \item
$\aleph_0=\vnn$,
\item
$\aleph_{\vscr{\alpha}}=\scr{(\aleph_\alpha)}$,
\item
$\aleph_{\alpha}=\bigcup\set{\aleph_x}{x\in\alpha}$ if $\alpha$ is a limit.
\end{enumerate}
Here $\aleph$ is \emph{aleph,} the first letter of the Hebrew
alphabet, as mentioned in \parref{par:aleph}.
If $0<\alpha$, then $\aleph_{\alpha}$ is called
\defn{uncountable}.
\end{parag}

\begin{theorem}\label{thm:aleph}
  The function $x\mapsto\aleph_x$ is an isomorphism between $\on$ and
  the class of infinite cardinals.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:aleph}.
\end{proof}

\section{Cardinal addition and multiplication}

\begin{parag}
The 
\defnplain{(cardinal) sum}{}%
\dindexsub{cardinal}{--- sum}%
\dindexsub{sum}{cardinal ---}
 and
\defnplain{(cardinal) product}{}%
\dindexsub{cardinal}{--- product}%
\dindexsub{product}{cardinal ---}
  of two
  cardinals $\kappa$ and $\lambda$ are defined by
  \begin{align*}
    \kappa+\lambda&=\card{(\kappa\times\{0\})\cup(\lambda\times\{1\})},\\
    \kappa\cdot\lambda&=\card{\kappa\times\lambda}.
  \end{align*}
If $\kappa$ and $\lambda$ are merely cardinalities, not ordinals, then
  \begin{align}\label{eqn:card+}
    \kappa+\lambda&=\card{(a\times\{0\})\cup(b\times\{1\})},\\\label{eqn:card.}
    \kappa\cdot\lambda&=\card{a\times b},
  \end{align}
where $a\in\kappa$ and $b\in\lambda$ (Exercise~\ref{ex:card+.}).
One must distinguish the cardinal operations from the ordinal operations of
  \P\P~\ref{par:ord-sum} and~\ref{par:ord-prod}.  However, the
  cardinal operations, when involving infinite \emph{cardinals} (and not
  merely cardinalities), will turn
  out to be very simple.
\end{parag}

\begin{theorem}\label{thm:card-arith}
If $\kappa$, $\lambda$, and $\mu$ are cardinalities, then
\begin{enumerate}
  \item
$\kappa+\lambda=\lambda+\kappa$,
\item
$\kappa+0=\kappa$,
\item
$(\kappa+\lambda)+\mu=\kappa+(\lambda+\mu)$,
  \item
$\kappa\cdot\lambda=\lambda\cdot\kappa$,
\item
$\kappa\cdot1=\kappa$,
\item
$(\kappa\cdot\lambda)\cdot\mu=\kappa\cdot(\lambda\cdot\mu)$,
\item
$\kappa\cdot(\lambda+\mu)=\kappa\cdot\lambda+\kappa\cdot\mu$,
\item
$\kappa\leq\lambda\lto\kappa+\mu\leq\lambda+\mu$,
\item
$\kappa\leq\lambda\lto\kappa\cdot\mu\leq\lambda\cdot\mu$.
\end{enumerate}
  The cardinal operations agree with the ordinal operations on
  $\vnn$.  
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:card-arith}.
\end{proof}

\begin{lemma}\label{lem:onon}
The class $\on$ of ordinals becomes isomorphic to
  $\on\times\on$ when the latter is ordered by $<$, where
\begin{multline*}
  (\alpha,\beta)<(\gamma,\delta)\iff
  \max(\alpha,\beta)<\max(\gamma,\delta)\lor{}\\ 
{}\lor\Bigl(\max(\alpha,\beta)=\max(\gamma,\delta)\land
  \bigl(\alpha<\gamma\lor(\alpha=\gamma\land\beta<\delta)\bigr)\Bigr).  
\end{multline*}
(Here $\max(\alpha,\beta)$ is the maximal element of
$\{\alpha,\beta\}$.  See Figure~\ref{fig:onon}.) 
\end{lemma}

\begin{figure}[t]
  \begin{center}
    \begin{pspicture}(0,-9.2)(12,0)
\psset{arrows=->}
\rput(1,-1){\rnode{00}{$(0,0)$}}
\rput(3,-1){\rnode{01}{$(0,1)$}}
\ncline{00}{01}
\rput(1,-3){\rnode{10}{$(1,0)$}}
\ncline{01}{10}
\rput(3,-3){\rnode{11}{$(1,1)$}}
\ncline{10}{11}
\rput(5,-1){\rnode{02}{$(0,2)$}}
\ncline{11}{02}
\rput(5,-3){\rnode{12}{$(1,2)$}}
\ncline{02}{12}
\rput(1,-5){\rnode{20}{$(2,0)$}}
\ncline{12}{20}
\rput(3,-5){\rnode{21}{$(2,1)$}}
\ncline{20}{21}
\rput(5,-5){\rnode{22}{$(2,2)$}}
\ncline{21}{22}
\rput(7,-1){\rnode{03}{$(0,3)$}}
\ncline{22}{03}
\rput(7,-3){\rnode{13}{$(1,3)$}}
\ncline{03}{13}
\rput(7,-5){\rnode{23}{$(2,3)$}}
\ncline{13}{23}
\rput(1,-7){\rnode{30}{$(3,0)$}}
\ncline{23}{30}
\rput(3,-7){\rnode{31}{$(3,1)$}}
\ncline{30}{31}
\rput(5,-7){\rnode{32}{$(3,2)$}}
\ncline{31}{32}
\rput(7,-7){\rnode{33}{$(3,3)$}}
\ncline{32}{33}
\rput(9,-1){\rnode{04}{$(0,4)$}}
\ncline{33}{04}
\rput(9,-3){\rnode{14}{$(1,4)$}}
\ncline{04}{14}
\rput(9,-5){\rnode{24}{$(2,4)$}}
\ncline{14}{24}
\rput(9,-7){\rnode{34}{$(3,4)$}}
\ncline{24}{34}
\rput(1,-9){\rnode{40}{$(4,0)$}}
\ncline{34}{40}
\rput(3,-9){\rnode{41}{$(4,1)$}}
\ncline{40}{41}
\rput(5,-9){\rnode{42}{$(4,2)$}}
\ncline{41}{42}
\rput(7,-9){\rnode{43}{$(4,3)$}}
\ncline{42}{43}
\rput(9,-9){\rnode{44}{$(4,4)$}}
\ncline{43}{44}
\rput(11,-1){\rnode{05}{$(0,5)$}}
\ncline{44}{05}
\rput(11,-3){\rnode{15}{$(1,5)$}}
\ncline{05}{15}
\rput(11,-5){\rnode{25}{$(2,5)$}}
\ncline{15}{25}
\rput(11,-7){\rnode{35}{$(3,5)$}}
\ncline{25}{35}
\rput(11,-9){\rnode{45}{$(4,5)$}}
\ncline{35}{45}
    \end{pspicture}
  \end{center}
\caption{$\on\times\on$, well-ordered}\label{fig:onon}
\end{figure}

\begin{proof}
  Exercise~\ref{ex:onon}.  After showing that $<$ is a total ordering,
  one can show that the least element of a non-empty subset $a$ of
  $\on\times\on$ is $(\beta,\gamma)$, where 
  \begin{align*}
\gamma&=\min\set y{(\beta,y)\in a},\\
\beta&=\min\set x{\Exists y(x,y)\in a\land\max(x,y)=\alpha)},\\
\alpha&=\min\set{\max(x,y)}{(x,y)\in a}.
  \end{align*}
Now show that every proper initial segment is a set. 
\end{proof}

\begin{theorem}\label{thm:card-easy}
If $\kappa$ and $\lambda$ are cardinals,
 $0<\kappa\leq\lambda$, and $\lambda$ is infinite, then
  \begin{equation*}
    \kappa+\lambda=\kappa\cdot\lambda=\lambda.
  \end{equation*}
In particular,
\begin{equation*}
  \aleph_\alpha+\aleph_\beta=
  \aleph_\alpha\cdot\aleph_\beta=
  \aleph_{\max(\alpha,\beta)}.
\end{equation*}
\end{theorem}

\begin{proof}
We shall show 
\begin{equation}\label{eqn:lll}
\lambda\cdot\lambda=\lambda.
\end{equation}
Then we can complete the argument by observing
\begin{gather*}
  \lambda\leq\kappa+\lambda\leq\lambda+\lambda=\lambda\cdot
  2\leq\lambda\cdot\lambda,\\ 
\lambda\leq\kappa\cdot\lambda\leq\lambda\cdot\lambda.
\end{gather*}
We establish~\eqref{eqn:lll} by induction on the infinite cardinals.
Suppose the equation holds 
whenever $\vnn\leq\lambda<\mu$, for some cardinal $\mu$.  Let $\cF$ be
the isomorphism from 
$\on\times\on$ onto $\on$ guaranteed by Lemma~\ref{lem:onon}.  As
$\alpha\times\alpha$ is always a section (namely, $\pred{0,\alpha}$)
of $\on\times\on$, so
$\cF\setimb{\alpha\times\alpha}$ must be a section of $\on$: that is,
$\cF\setimb{\alpha\times\alpha}$ is an ordinal.  Suppose
$\cF\setimb{\mu\times\mu}=\alpha$.
Then
\begin{equation*}
  \mu\leq\mu\cdot\mu=\card{\mu\times\mu}=\card{\alpha}\leq\alpha.
\end{equation*}
So $\mu\leq\alpha$.
Suppose $\nu$ is an infinite cardinal and $\nu<\alpha$.  Then
$\nu=\cF(\beta,\gamma)$, where $(\beta,\gamma)\in\mu\times\mu$.  Since
$\mu$ is a limit ordinal (Exercise~\ref{ex:card-lim}), the successor
$\delta$ of $\max(\beta,\gamma)$ is also less than $\mu$.  Hence
\begin{gather*}
\nu\in\cF\setimb{\delta\times\delta},\\
\nu\included\cF\setimb{\delta\times\delta},\\
\nu\leq\card{\delta\times\delta}=\card\delta\cdot\card\delta
\end{gather*}
(Exercise~\ref{ex:nnn}).  By inductive hypothesis,
$\card\delta\cdot\card\delta=\card\delta$, so
$\nu\leq\card\delta<\delta<\mu$.  In short, $\nu<\alpha\lto\nu<\mu$.
Therefore $\alpha\leq\mu$; but since $\mu\leq\alpha$, we have
$\mu=\alpha=\mu\cdot\mu$. 
\end{proof}

\section{Exponentiation}

\begin{parag}\label{par:card-exp}
\defnplain{Cardinal exponentiation}{}%
\dindexsub{cardinal}{--- exponentiation}%
\dindexsub{exponentiation}{cardinal ---}
 is as easy to define as cardinal
  addition and multiplication:  If $\kappa$ and $\lambda$
  are cardinals, then
  \begin{equation*}
    \kappa^\lambda=\card{\mapset{\lambda}{\kappa}};
  \end{equation*}
if $\kappa$ and $\lambda$ are merely cardinalities, containing
representatives $a$ and $b$ respectively, then
\begin{equation*}
  \kappa^\lambda=\card{\mapset ab}.
\end{equation*}
\end{parag}

\begin{theorem}\label{thm:card-exp}
  For all cardinalities $\kappa$, $\lambda$, $\mu$ and $\nu$,
  \begin{enumerate}
    \item
$\kappa^0=1$,
\item
$0<\lambda\lto0^\lambda=0$,
\item
$1^\lambda=1$,
\item
$\kappa^1=\kappa$,
\item
$\kappa^{\lambda+\mu}=\kappa^\lambda\cdot\kappa^\mu$,
\item
$\kappa^{\lambda\cdot\mu}=(\kappa^\lambda)^\mu$,
\item
$\kappa\leq\mu\land\lambda\leq\nu\lto\kappa^\lambda\leq\mu^\nu$.
  \end{enumerate}
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:card-exp}.
\end{proof}

\begin{theorem}\label{thm:k<2^k}
For all sets $a$,
\begin{equation}\label{eqn:pow-card}
    \card{\pow a}=2^{\card a}; 
\end{equation}
hence for all
  cardinalities $\kappa$,
  \begin{equation}\label{eqn:k<2^k}
    \kappa<2^\kappa.
  \end{equation}
\end{theorem}

\begin{proof}
There is a bijection between $\mapset a2$ and $\pow a$  that takes the
function $f$ to the set $\set{x\in a}{f(x)=1}$.  This
establishes~\eqref{eqn:pow-card}.  Then~\eqref{eqn:k<2^k} follows by
Cantor's Theorem (\parref{thm:pow}). 
\end{proof}

\begin{corollary}
  If $\kappa$ and $\lambda$ are cardinals such that
  $2\leq\kappa\leq2^\lambda$ and $\lambda$ is infinite, then
  \begin{equation*}
    \kappa^\lambda=2^\lambda.
  \end{equation*}
\end{corollary}

\begin{proof}
  $2^\lambda\leq\kappa^\lambda\leq\lambda^\lambda\leq(2^\lambda)^\lambda
  =2^{\lambda\cdot\lambda}=2^\lambda$ by Theorem~\ref{thm:card-easy}.
\end{proof}

\begin{parag}
The function
\begin{equation*}
  x\mapsto\beth_x
\end{equation*}
from $\on$ into $\cn$ is given recursively (as in
Theorem~\ref{thm:rec-alt}) by 
\begin{enumerate}
  \item
$\beth_0=\vnn$,
\item
$\beth_{\vscr{\alpha}}=2^{\beth_{\alpha}}$,
\item
$\beth_{\alpha}=\bigcup\set{\beth_x}{x\in\alpha}$ if $\alpha$ is a limit.
\end{enumerate}
Here $\beth$ is \emph{beth,} the second letter of the Hebrew
alphabet.
\end{parag}

\begin{theorem}
The function $x\mapsto\beth_x$ is an embedding of $\on$ in $\cn$, and
\begin{equation}\label{eqn:aleph-beth}
  \aleph_\alpha\leq\beth_\alpha.
\end{equation}
\end{theorem}

\begin{proof}
Theorem~\ref{thm:k<2^k} and induction.
\end{proof}

\begin{parag}\label{par:GCH}
The letter $\beth$ is \emph{beth,} the second letter of the Hebrew
alphabet.  The 
\defnplain{Continuum Hypothesis}{}%
\dindex{continuum, (Generalized) C--- Hypothesis}, or $\gch$,
is that $\aleph_1=\beth_1$;
the 
\defnplain{Generalized Continuum Hypothesis}{}%
is that 
$\aleph_\alpha=\beth_\alpha$ for all ordinals $\alpha$.  Here the
\defnplain{continuum}{}
 is the set of 
\techplain{real number}{s,}%
\tindexsub{number}{real ---}
defined in \sref{sect:real}.  We shall see in
Chapter~\ref{ch:models} that we \emph{can} make these hypotheses
without contradicting our other axioms about
sets.  We shall not see what is also the case, that these hypotheses
are not \emph{implied} by our axioms \cite{MR0232676}.
\end{parag}

\section{The Axiom of Choice}

\begin{parag}
Suppose $a$ is not finite.  If $n\in\vnn$ and $f\colon n\injection a$,
then $f$ is not surjective; so $a\setminus f\setimb n$ has an element
$b$, and then $f\cup\{(n,b)\}\colon n+1\injection a$.  Since,
trivially, $0\colon 0\injection a$, we have by induction that every
natural number embeds in $a$.  However, this by itself does not allow
us to conclude that $\vnn$ embeds in $a$.
\end{parag}

\begin{parag}
  A \defnplain{choice-function}{}%
\dindexsub{choice}{---{}-function}%
\dindexsub{function}{choice-{}---} for a set is a function that assigns, to
  each non-empty subset of the set, an element of that subset.
  Suppose $f$ is a
  choice-function for $a$, so that
  \begin{equation*}
    b\in\pow a\setminus\{\emptyset\}\lto f(b)\in b.
  \end{equation*}
If also $a$ is non-finite, then there is an embedding $g$ of $\vnn$ in
$a$ defined recursively by
\begin{equation*}
  g(n)=f(a\setminus g\setimb n).
\end{equation*}
Thus every non-finite set with a choice-function is infinite.
\end{parag}

\begin{theorem}\label{thm:ac=wo}
  A set has a choice-function if and only if the set can be
  well-ordered.
\end{theorem}

\begin{proof}
  Suppose $a$ has the choice-function $f$.  Assume also $f(\emptyset)$
  is defined, but is not in $a$.  Then there is a function $\cG$ on
  $\on$ defined recursively by
  \begin{equation*}
    \cG(\alpha)=f(a\setminus\cG\setimb{\alpha}).
  \end{equation*}
Then $\cG\inv{\setimb a}$ is an initial segment of $\on$, and
$\cG\rest{\cG\inv\setimb a}$ is a bijection onto $a$.  So, by means of
$\cG$, the ordering of $\cG\inv\setimb a$ induces a well-ordering of
$a$.

Now suppose conversely that $a$ is well-ordered.  Then there is a
choice\-{}-function for $a$ that assigns to each non-empty subset of $a$
its least element.
\end{proof}

\begin{theorem}\label{thm:zorn-lem}
  Assume $a$ has a choice-function.  If $(a,<)$ is an order such that
  every totally ordered subset of $a$ has an upper bound in $a$,
  then $a$ has a maximal element.
\end{theorem}

\begin{proof}
Let $f$ be the operation on $\pow a$ taking each element $b$ to the
set (possibly empty) of strict upper bounds of $b$.  So $f$ is
  order-reversing, in the sense that
  \begin{equation*}
    c\included b\lto f(c)\includes f(b).
  \end{equation*}
  Let $g$ be a choice-function for $a$, and assume
  $g(\emptyset)\notin a$.  Then define $\cH$
  on $\on$ by
  \begin{equation*}
    \cH(\alpha)=g(f(a\cap\cH\setimb{\alpha})).
  \end{equation*}
In particular, if $\cH\setimb{\alpha}\pincluded a$ and has a strict
upper bound, then $\cH(a)$ is such an upper bound.
So $\cH$ embeds $\cH\inv\setimb a$ in $a$, and
$a\cap\ran{\cH}$ is totally ordered, but it has no strict upper bound
in $a$.  By hypothesis though, it has an upper bound; this is a
maximal element of $a$.
\end{proof}

\begin{theorem}\label{thm:zorn}
For a set $a$, let $b$ be the set of choice-functions for subsets of
$a$.  Then $b$ is ordered by inclusion, and every totally ordered
subset $c$ of $b$ is bounded above by $\bigcup c$.  If $b$ has a
maximal element $f$, then $f$ is a choice-function for $a$.
\end{theorem}

\begin{proof}
Suppose $f\in b$, but is a choice-function only for a proper subset
$d$ of $a$.   Then $a\setminus d$ has an element $e$, and $f\pincluded
g$, where $g$ is the choice-function for $d\cup\{e\}$ that agrees with
$f$ on $\pow d\setminus\{\emptyset\}$, but takes $x$ to $e$ if $e\in
x$.  Thus $f$ is not a maximal element of $b$.
\end{proof} 

\begin{axiom}[Choice]\label{ax:choice}
\dindexsub{axiom}{A--- of Choice}%
\dindexsub{choice}{Axiom of Ch---}
%\axiomname{Choice: \[\text{Every set has a choice-function.}\]}
  Every set has a choice-function.
\end{axiom}

\begin{parag}
The Axiom of Choice, or $\ac$, is a completely new kind of axiom, since it
  asserts the existence of sets that we do not already have as
  classes.  However, as with the Generalized Continuum Hypothesis
  (\parref{par:GCH}), so we shall see in Chapter~\ref{ch:models} that we
  \emph{can} assume the Axiom of Choice without contradicting our
  other axioms.  The Axiom of Choice is convenient for mathematics in
  that it allows many theorems to be proved.  Often it does this
  through (the conclusion of)
  Theorem~\ref{thm:zorn-lem}, which is often known as Zorn's Lemma (though it
  appears to be due to Hausdorff).  Then Theorem~\ref{thm:zorn} shows that
  assuming the Axiom of Choice is \emph{equivalent} to assuming Zorn's
  Lemma (this equivalence apparently \emph{is} due to Zorn).  By
  Theorem~\ref{thm:ac=wo} now, every set has a cardinality in the
  sense of \parref{par:card}; and this
  too is equivalent to the Axiom of Choice.
\end{parag}

\section{Computations}

\begin{parag}
  From the definitions of ordinal and cardinal addition and
  multiplication, we have
  \begin{align*}
    \card{\alpha+\beta}&=\card\alpha+\card\beta,\\
    \card{\alpha\cdot\beta}&=\card\alpha\cdot\card\beta.
  \end{align*}
Here the ordinal operations are on the left; the cardinal, on the
right.  So these equations illustrate a convention:  If an operation
is applied to ordinals that are not necessarily cardinals, then the
operation is the ordinal operation; if the ordinals \emph{are} known
to be cardinals, then the operation is the cardinal operation (even
though the ordinal operation would also make sense).  Following
strictly the notation of \parref{par:sig}, we would write for example
\begin{equation*}
\card{\alpha+^{\on}\beta}=\card\alpha+^{\cn}\card\beta.
\end{equation*}
\end{parag}

\begin{parag}\label{par:rlo}
  The ordinal power $\alpha^{\beta}$ is not defined as the order-type
  of $(\mapset{\beta}{\alpha},<)$ for some~$<$.  Indeed, when
  $1<\alpha$ and $\vnn\leq\beta$, there is no obvious relation that
  well-orders $\mapset{\beta}{\alpha}$.  Thinking that
  \begin{equation*}
    \mapset{\beta}{\alpha}=\overbrace{\alpha\times\alpha\times\dotsb}^{\beta},
  \end{equation*}
we can define the left lexicographic ordering $<_{\mathrm{\ell}}$ on
$\mapset{\beta}{\alpha}$, where, if
$\gamma=\min\set{x\in\beta}{f(x)\neq g(x)}$, and
$f(\gamma)<g(\gamma)$, then $f<_{\mathrm{\ell}}g$.  But then
$<_{\mathrm{\ell}}$ does not well-order $\mapset{\beta}{\alpha}$,
since $\set{f_n}{n\in\vnn}$ has no least element when
\begin{equation*}
  f_n(x)=
  \begin{cases}
    1,&\text{ if }x=n;\\
    0,&\text{ otherwise}.
  \end{cases}
\end{equation*}
There is also a right lexicographic ordering $<_{\mathrm r}$, where,
if $f(\delta)\leq g(\delta)$ when $\gamma<\delta<\beta$, and
$f(\gamma)<g(\gamma)$, then $f<_{\mathrm r}g$.  But this ordering of
$\mapset{\beta}{\alpha}$ is not generally total
(Exercise~\ref{ex:rlo}). 
\end{parag}

\begin{parag}
  If $f\in\mapset{\beta}{\alpha}$, let the \defn{support}{} of $f$ be
  the set $\set{x\in\beta}{f(x)\neq0}$; this can be denoted by
  \begin{equation*}
    \supp f.
  \end{equation*}
\end{parag}

\begin{theorem}\label{thm:ord-exp2}
  $\mapset{\beta}{\alpha}=\ord{\set{x\in\mapset{\beta}{\alpha}}{\card{\supp
        x}<\vnn}, <_{\mathrm r}}$.
\end{theorem}

\begin{proof}
  Exercise~\ref{ex:ord-exp2}.  See Figure~\ref{fig:a^b}.
\end{proof}


\begin{figure}[t]
\begin{gather*}
  (0,0,0,0,\dots)\\
(1,0,0,0,\dots)\\
(2,0,0,0,\dots)\\
\text{\makebox[2.2cm]{\dotfill}}\\
(0,1,0,0,\dots)\\
(1,1,0,0,\dots)\\
(2,1,0,0,\dots)\\
\text{\makebox[2.2cm]{\dotfill}}\\
(0,0,1,0,\dots)\\
(1,0,1,0,\dots)\\
\text{\makebox[2.2cm]{\dotfill}}\\
(0,1,1,0,\dots)\\
\text{\makebox[2.2cm]{\dotfill}}
\end{gather*}
\caption{The ordering of $\set{x\in\mapset\beta\alpha}{\card{\supp
      x}<\vnn}$.}\label{fig:a^b}
\end{figure}

\begin{lemma}\label{lem:inf-sum}
  If $b$ is a non-empty set, and $x\mapsto a_x$ is a function on $b$
  such that each set $a_c$ is non-empty, but $c\neq d\lto a_c\cap
  a_d=\emptyset$, 
  and one of 
  $b$ and $\sup\set{\card{a_x}}{x\in b}$ is infinite, then
  \begin{equation*}
    \operatorname{card}\bigl(\bigcup\set{a_x}{x\in b}\bigr)=\card
    b\cdot\sup\set{\card{a_x}}{x\in b}. 
  \end{equation*}
\end{lemma}

\begin{proof}
Let $\kappa=\card{\bigcup\set{a_x}{x\in b}}$ and
  $\lambda=\sup\set{\card{a_x}}{x\in b}$.
  For each $c$ in $b$, there is a bijection $f_c$ from $a_c$ onto
  $\card{a_c}$.  Now let $f$ be the function from
  $\bigcup\set{a_x}{x\in b}$ into $b\times\lambda$ such that
  $f(d)=(c,f_c(d))$ when $d\in a_c$.  Since $f$ is injective, we have
  $\kappa=\card{\ran f}\leq\card b\cdot\lambda$.  Also $\ran f$
  includes $b\times
  1$, so $\card b\leq\kappa$.  If $\alpha<\lambda$, then
  $\alpha<\card{a_c}$ for some $c$ in $b$, but then
  $\{c\}\times\alpha\included\ran f$, so $\alpha\leq\kappa$.  Thus
  $\lambda\leq\kappa$ (since $\lambda$ is a limit ordinal).  By
  Theorem~\ref{thm:card-easy}, we are done.
\end{proof}

\begin{parag}
  The preceding proof used the Axiom of Choice, since a bijection
  $f_c$ from $a_c$ onto $\card{a_c}$ was chosen for each $c$ in $b$.
  There may be many such bijections.  To define $f$, we need to
  know more than that such functions $f_c$ exist; we must be able to
  specify one of them for each $c$ in $b$.
  So let $g$ be a choice-function for the set $e$ of bijections
  between elements of $\set{a_x}{x\in b}$ and their cardinals.  (See
  Exercise~\ref{ex:choice}.)  Then we may let $f_c=g(\set{x\in e}{\dom
  x=a_c})$.
\end{parag}

\begin{lemma}\label{lem:fin-subsets}
If $a$ is an infinite set, then
\begin{equation*}
  \card{\set{x\in\pow a}{\card x<\aleph_0}}=\card a.
\end{equation*}
\end{lemma}

\begin{proof}
  Exercise~\ref{ex:fin-subsets}.  Use Lemma~\ref{lem:inf-sum}.
\end{proof}

\begin{theorem}
  If $1<\alpha$, and $0<\beta$, and one of $\alpha$ and $\beta$ is
  infinite, then
  \begin{equation*}
    \card{\alpha^\beta}=\card\alpha\cdot\card\beta.
  \end{equation*}
\end{theorem}

\begin{proof}
Let  $b$ be the set of finite subsets of $\beta$.  Then
\begin{equation*}
  \card{\alpha^\beta}
=\card{\set{x\in\mapset{\beta}{\alpha}}{\card{\supp x}<\aleph_0}}
=\card{\bigcup\set{\mapset
  x{(\alpha\setminus\{0\})}}{x\in b}}.
\end{equation*}
If $\alpha$ is infinite and $b$ is finite, then $\card{\mapset
  b{(\alpha\setminus\{0\})}}=\card{\alpha}$ (Exercise~\ref{ex:k^n}).
  If $\beta$ is infinite, then $\card b=\card\beta$ by
  Lemma~\ref{lem:fin-subsets}.  In either case, we are done by
  Lemma~\ref{lem:inf-sum}. 
\end{proof}

\section{The real numbers}\label{sect:real}

\begin{parag}
  A \defn{cut}{} of a total order $(\cC,<)$ is a non-empty proper
  initial segment that does not contain its supremum (if it has a
  supremum).  The cuts of $\on$ are precisely the limit ordinals, and
  every cut of $\on$ does have a supremum, namely itself.  Some cuts
  of $\Q$ have suprema, others don't: $\pred{a,\Q,<}$ has the supremum
  $a$, but $\set{x\in\Q}{x^2<2\lor x<0}$ has no supremum in $\Q$.
  If we picture $\Q$ as comprising points on a horizontal line,
  with a cut on the left and its complement on the right, then the two
  possibilities are as in Figure~\ref{fig:cuts}. (If $a$ is a cut of
  $\Q$, then the ordered pair $(a,\Q\setminus a)$ is a cut in the
  original sense of Dedekind~\cite[p.~13]{MR0159773}.)
\end{parag}

\begin{figure}[t]
  \begin{center}
    \begin{pspicture}(0,-0.2)(12,1.2)
\psset{linewidth=0.8mm}
      \psline(0,1)(5.6,1)
      \psline(0,0)(5.6,0)
      \psline(6.4,1)(12,1)
      \psline(6.4,0)(12,0)
\psdots(6.4,1)
\psdots[dotstyle=o](5.6,1)(5.6,0)(6.4,0)
    \end{pspicture}
  \end{center}
\caption{Cuts}\label{fig:cuts}
\end{figure}

\begin{parag}
  Let $\R$ be the set of cuts of $\Q$; these are the so-called
  \defnplain{real number}{s.}%
\dindexsub{number}{real ---}  
Then $\R$ is totally ordered 
  by inclusion, and
  \begin{equation*}
    x\mapsto\pred x\colon(\Q,<)\to(\R,\pincluded).
  \end{equation*}
Now define the ring-operations on $\R$ so that $x\mapsto\pred x$ is an
embedding of ordered fields.  First,
\begin{align*}
  a+b&=\set{x+y}{x\in a\land y\in b},\\
-a&=\set{-x}{x\in\Q\setminus a\land\Exists y(y\in\Q\setminus a\land y<x)}.
\end{align*}
Multiplication is defined by cases.  First, if $0\in a\cap b$ (so that
$a$ and $b$ are \emph{positive} real numbers), then
\begin{equation*}
  a\cdot b=\set{x\cdot y}{x\in a\land y\in b}.
\end{equation*}
If $a=0^{\R}$ or $b=0^{\R}$ (that is, 
if $a=\pred 0$ or $b=\pred 0$), then $a\cdot b=0$.  Finally,
\begin{equation*}
  a\cdot b=
  \begin{cases}
    -(-a\cdot b),&\text{ if }0\in b\setminus a;\\
-(a\cdot-b),&\text{ if }0\in a\setminus b;\\
-a\cdot-b,&\text{ if }0\notin a\cup b.
  \end{cases}
\end{equation*}
Then $x\mapsto\pred x$ is an embedding of ordered fields
(Exercise~\ref{ex:Q-in-R}). 
\end{parag}

\begin{parag}
  If $a$ is a cut of $\R$, then $\bigcup a$ is a cut of $\Q$ and is
  the supremum of $a$ (Exercise~\ref{ex:complete}).  Then every subset
  of $\R$ with an upper bound 
  has a supremum: this makes $\R$ a 
\defn{complete ordered field}.%
\dindexsub{order}{complete ---ed field}
\end{parag}

\begin{parag}
We have a surjection of $\N\times\N$ onto $\Z$, so (Exercise~\ref{ex:ac-surj})
$\card{\Z}=\aleph_0$.  We have a surjection of
$\Z\times(\Z\setminus\{0\})$ onto $\Q$, so $\card{\Q}=\aleph_0$.  (See
also Exercise~\ref{ex:ZQcount}.)  Since $\R\included\pow{\Q}$, we have
\begin{equation}\label{eqn:R<=}
  \card{\R}\leq 2^{\aleph_0}.
\end{equation}
\end{parag}

\begin{parag}
  There is an embedding of $\mapset\vnn2$ into $\R$, so that
  $2^{\aleph_0}\leq\card{\R}$; because of~\eqref{eqn:R<=}, by the
  Schroeder--Bernstein Theorem (\parref{thm:sch-b}), we then have
  \begin{equation*}
    \card{\R}=2^{\aleph_0}.
  \end{equation*}
To define the embedding, replace $\Q$ with its image in $\R$; that is,
treat $\Q$ as a subfield of $\R$.  Let $f\colon\mapset\vnn2\to\R$,
where
\begin{equation*}
  f(\sigma)=\sup\Bigl\{\sum_{k=0}^x\frac{2\cdot\sigma(k)}{3^{k+1}}\colon
  x\in\vnn\Bigr\}. 
\end{equation*}
Then $f$ is well-defined, since, by induction, 
\begin{equation*}
  \sum_{k=0}^n\frac{2\cdot\sigma(k)}{3^{k+1}}\leq 1-\frac 1{3^{n+1}}.
\end{equation*}
Also, $f$ is injective, since, if $\sigma\rest n=\tau\rest n$, but
$\sigma(n)=0<1=\tau(n)$, then
\begin{equation*}
  f(\sigma)\leq\sum_{k=0}^{n-1}\frac{2\cdot\sigma(k)}{3^{k+1}}+\frac1{3^n} 
< \sum_{k=0}^{n-1}\frac{2\cdot\sigma(k)}{3^{k+1}}+\frac2{3^n} 
\leq f(\tau).
\end{equation*}
Here $f\setimb{\mapset\omega2}$ is called the 
\defnplain{Cantor set};%
\dindexsub{Cantor}{--- set}%
\dindexsub{set}{Cantor ---} it is
the intersection of the sets depicted in Figure~\ref{fig:cantor}.
\end{parag}

\begin{figure}[t]
  \begin{center}
\begin{comment}
    \begin{pspicture}(0,-2.6)(12.15,1)
\uput[u](0,0    ){$0                   $}
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\caption{Towards the Cantor set}\label{fig:cantor}
\end{figure}

\section*{Exercises}

\begin{xca}\label{ex:ind-defn}
If $b$ is an inductive subset of $\pow a$, and $c\included a$, prove that
$b\cap\pow c$ is an inductive subset of $\pow c$.
\end{xca}

\begin{xca}\label{ex:cn<}
Prove that $\in$ coincides with $\pinjects$ on $\cn$.
\end{xca}

\begin{xca}\label{ex:aleph}
  Prove Theorem~\ref{thm:aleph}.
\end{xca}
\begin{xca}\label{ex:card+.}
Show that~\eqref{eqn:card+} and~\eqref{eqn:card.} are independent of
the choice of $a$ in $\kappa$ and $b$ in $\lambda$.
\end{xca}

\begin{xca}\label{ex:card-lim}
Prove that infinite cardinals are limit ordinals.
\end{xca}

\begin{xca}\label{ex:nnn}
Show that $\card{a\times b}=\card a\cdot\card b$.
\end{xca}

\begin{xca}\label{ex:card-arith}
Prove Theorem~\ref{thm:card-arith} without using
Theorem~\ref{thm:card-easy}. 
\end{xca}

\begin{xca}\label{ex:onon}
Prove Lemma~\ref{lem:onon}.
\end{xca}

\begin{xca}\label{ex:card-exp}
Prove Theorem~\ref{thm:card-exp}.
\end{xca}

\begin{xca}\label{ex:k^n}
If $\kappa$ is an infinite cardinal, and $1<n<\vnn$, prove
$\kappa^n=\kappa$. 
\end{xca}

\begin{xca}\label{ex:ac-surj}
Use the Axiom of Choice to show that, if $f\colon a\surjection b$,
then $b\injects a$. 
\end{xca}

\begin{xca}\label{ex:choice}
If $a$ is a set, why is there a \emph{set} of bijections between
elements of $a$ and their cardinals?
\end{xca}

\begin{xca}\label{ex:fin-subsets}
Prove Lemma~\ref{lem:fin-subsets}.
\end{xca}

\begin{xca}\label{ex:rlo}
  Prove that the right lexicographic ordering of
  $\mapset{\beta}{\alpha}$ defined in \parref{par:rlo} is not
  generally total.
\end{xca}

\begin{xca}\label{ex:ord-exp2}
  Prove Theorem~\ref{thm:ord-exp2}.
\end{xca}

\begin{xca}\label{ex:Q-in-R}
Show that $x\mapsto\pred x$ is an embedding of $\Q$ in $\R$ as ordered
fields. 
\end{xca}

\begin{xca}\label{ex:complete}
Show that the union of a cut of $\R$ is the supremum of the cut.
\end{xca}

\begin{xca}\label{ex:ZQcount}
Show that $\Z$ and $\Q$ are countable without using the Axiom of Choice.
\end{xca}

\chapter{Models}\label{ch:models}

%\setcounter{equation}{9}

\section{Well-founded sets}\label{sect:wf}

\begin{parag}\label{par:ill-founded}
  A binary relation $\cR$ is 
\defn{well-founded}{} on a class $\cC$ if
  \begin{enumerate}
  \item
$\cR a\cap\cC$ (that is, $\cls x{x\in\cC\land x\mathrel{\cR} a}$)
is a set whenever $a\in \cC$, and
\item
$b\cap\cR c$ (that is, $\set x{x\in b\land x\mathrel{\cR}c}$)
is empty for some $c$ in $b$,
  whenever $b\included\cC$ 
  and is nonempty.
  \end{enumerate}
So an \emph{ordering} of a class is well-founded on the class if and
only if each section of the class is a set and each non-empty subset
of the class has a minimal element.  In particular, a relation that
well-orders a class is well-founded on the class.  A class itself is
\defn{well-founded}{} if \emph{membership} is well-founded on the class.  If
$\cR$ is membership, then $\cR a\cap \cC$ is just $a\cap\cC$, a set;
and $b\cap\cR c$ is just $b\cap c$;
so $\cC$ is well-founded if and only if
\begin{equation*}
  a\included \cC\land a\neq\emptyset\lto \Exists x(x\in a\land a\cap
  x=\emptyset). 
\end{equation*}
A class that is not well-founded is \defn{ill-founded}.
\end{parag}

\begin{parag}
Suppose there are \emph{distinct} sets $a$ and $b$ such that $a=\{b\}$
and $b=\{a\}$.  Then $a$ and $b$ are well-founded, but not
transitive.  However, $a\cup\bigcup a=\{b\}\cup b=\{a,b\}$, which is
transitive, but ill-founded.
\end{parag}

\begin{thmdefn}\label{thm:tc}
  Given a set $a$, we recursively define the sequence $(\bigcup^na\colon
  n\in\vnn)$: 
  \begin{enumerate}
    \item
$\bigcup a=a$,
\item
$\bigcup^{n+1}a=\bigcup\bigcup^na$.
  \end{enumerate}
Now we define 
\begin{equation*}
\tc a =\sideset{}{^{\vnn}}\bigcup a
      =\bigcup\bigl\{\sideset{}{^x}\bigcup a\colon x\in\vnn\bigr\}; 
\end{equation*}
this is the 
\defnplain{transitive closure}{}%
\dindexsub{transitive}{--- closure}%
\dindex{closure, transitive} of $a$ in the sense that it
includes $a$ and is transitive and is included in every class that
includes $a$ and is transitive.
\end{thmdefn}

\begin{proof}
  Exercise~\ref{ex:tc}.
\end{proof}

\begin{theorem}
  If $\cC$ is non-empty and well-founded, then $\cC\cap a=\emptyset$
  for some $a$ in 
  $\cC$.
\end{theorem}

\begin{proof}
  Suppose $\cC$ is well-founded, and $b\in\cC$, but $\cC\cap
  b\neq\emptyset$.  Let $a=\tc
  b\cap\cC$.  Then $a\cap d=\emptyset$ for some $d$ in $a$.  But then
  $d\in\cC$, and $d\in\tc b$, so $d\included\tc b$, and therefore
  \begin{equation*}
    \cC\cap d=\cC\cap\tc b\cap d=a\cap d=\emptyset,
  \end{equation*}
as desired.
\end{proof}

\begin{parag}\label{par:wf-universe}
  The function $\rwf$ on $\on$ is defined recursively by
  \begin{enumerate}
    \item
$\rwf(0)=\emptyset$,
\item
$\rwf(\alpha+1)=\pow{\rwf(\alpha)}$,
\item
$\rwf(\alpha)=\bigcup\rwf\setimb\alpha$ if $\alpha$ is a limit.
  \end{enumerate}
Then we define
\begin{equation*}
  \wf=\bigcup\rwf\setimb{\on}.
\end{equation*}
\end{parag}

\begin{lemma}
  Each set $\rwf(\alpha)$ is transitive, so $\wf$ is transitive.  If
  $\beta<\alpha$, then $\rwf(\beta)\included\rwf(\alpha)$.
\end{lemma}

\begin{proof}
  Use induction.  Trivially, $\rwf(0)$ is transitive; the power set of
  a transitive set is transitive; the union of a set of transitive
  sets is transitive.  (See Exercise~\ref{ex:trans}.)  Therefore, by
  induction, each set $\rwf(\alpha)$ is transitive, so $\wf$ is
  transitive. 

Consequently, as $\rwf(\alpha)\in\rwf(\alpha+1)$, so
$\rwf(\alpha)\included\rwf(\alpha+1)$.  If $\beta<\alpha$, and
$\alpha$ is a limit, then immediately
$\rwf(\beta)\included\rwf(\alpha)$.  By induction,
$\beta<\alpha\lto\rwf(\beta)\included\rwf(\alpha)$. 
\end{proof}

\begin{parag}\label{par:rank}
  If $c\in\wf$, then the least ordinal $\alpha$ such that
  $c\in\rwf(\alpha)$ must be a successor, $\beta+1$.  Then $\beta$ is
  the \defn{rank}{} of $c$:
  \begin{equation*}
    \rank c=\min\set{x\in\on}{c\in\rwf(x+1)}
=\min\set{x\in\on}{c\included\rwf(x)}.
  \end{equation*}
\end{parag}

\begin{lemma}\label{lem:wf-wf}
\mbox{}
\begin{enumerate}
\item\label{item:sub-is-el}
Every subset of $\wf$ is an element of $\wf$.
\item\label{item:wf-wf}
  $\wf$ and its members are well-founded.
\end{enumerate}
\end{lemma}

\begin{proof}
Supposing $a\included\wf$, let $\beta=\sup\set{\rank x}{x\in a}$; then
$a\included\rwf(\beta+1)$, so $a\in\rwf(\beta+2)$.

Suppose $a\included\wf$, and let $b$ be an element of $a$ of minimal
rank.  If $c\in b$, then $\rank c<\rank b$
(Exercise~\ref{ex:rank}), so $c\notin a$.  Thus $b\cap a=\emptyset$.
So $\wf$ is well-founded; since it is transitive, its members are also
well-founded.
\end{proof}

\begin{theorem}
  $\wf$ comprises the sets whose transitive closures are
  well-founded.\footnote{It is such sets that are called
    \tech{well-founded}{} in \cite{MR0345816}.}  
\end{theorem}

\begin{proof}
Suppose $a\in\wf$.  Then $a\included\wf$, so $\tc a\included\wf$ by
Theorem~\ref{thm:tc}, and then $\tc a\in\wf$ by
Lemma~\ref{lem:wf-wf}~\eqref{item:sub-is-el},
so $\tc a$ is well-founded by Lemma~\ref{lem:wf-wf}~\eqref{item:wf-wf}.   

Now suppose $a\notin\wf$.  Then $a\nincluded\wf$, so $\tc
a\nincluded\wf$.  Let $b\in\tc
a\setminus\wf$.  Then $b\included\tc a$, but $b\nincluded\wf$.
Consequently 
\begin{equation*}
(\tc a\setminus\wf)\cap b=\tc
a\cap(b\setminus\wf)=b\setminus\wf\neq\emptyset.   
\end{equation*}
Thus $\tc a$ is
not well-founded.
\end{proof}

\section{Families of classes}\label{sect:e-wf}

\begin{axiom}[Foundation]\label{ax:foundation}
\dindexsub{axiom}{Foundation Scheme}%
\dindex{Foundation Scheme}%
\dindexsub{scheme}{Foundation S---}
%\axiomname{Foundation: \[\universe=\wf.\]}
All sets are well-founded:
\begin{equation*}
  a\neq0\lto\Exists y(y\in a\land y\cap a=\emptyset);
\end{equation*}
equivalently, $\universe=\wf$ (Exercise~\ref{ex:V=WF}).
\end{axiom}

\begin{parag}\label{par:tau}
The Foundation Axiom allows us to make the following definition, which
will allow us to understand the family $\cC/\cE$ of
\parref{par:=classes} as a certain class.
  Indeed, along with the class-operations of \P\P~\ref{par:infinitary}
  and~\ref{par:pair-class}, there 
  is an operation that assigns to the class $\cC$ a \emph{set}
  $\tau(\cC)$ such that
  \begin{enumerate}
    \item
$\tau(\cC)\included\cC$;
\item
$\cC\neq\emptyset\lto\tau(\cC)\neq\emptyset$.
  \end{enumerate}
Indeed, if $\cC\neq\emptyset$, let $\alpha=\min\cls x{\Exists
 y(y\in\cC\land\rank y=x)}$, and let
  \begin{equation*}
\tau(\cC)=\rwf(\alpha+1)\cap\cC.
  \end{equation*}
\end{parag}

\begin{theorem}\label{thm:e-wf}
  If $\cE$ is an equivalence-relation on $\cC$, then there is a
  function $\cF$ on $\cC$ such that
  \begin{equation*}
    \cF(a)=\cF(b)\iff a\mathrel{\cE}b.
  \end{equation*}
\end{theorem}

\begin{proof}
Let $\cF(a)=\tau(a\cE)=\bigl\{x\colon a\mathrel{\cE}x\land\rank x
  =\min\set{\rank y}{a\mathrel{\cE}y}\bigr\}$.
\end{proof}

\section{Consistency}

\begin{parag}\label{par:zf}
As axioms of set theory, Zermelo~\cite{Zermelo-invest} proposed
Extension, Pairing, Separation, Power Set, Union, Choice, and
Infinity.  Then Fraenkel (see~\cite[p.~50, n.~3]{MR0345816}), and
independently Skolem~\cite{Skolem-some-remarks}, proposed Replacement
(which makes Separation redundant by Exercise~\ref{ex:sep-from-rep});
Skolem and more definitely von Neumann~\cite{von-Neumann-ax} proposed
Foundation.  The list of all of these axioms, besides Choice, is
called $\zf$ for 
Zermelo and Fraenkel.  When Choice is added, the list is called
  $\zfc$.  It was mentioned in~\parref{par:incomp} that we cannot
prove the consistency of $\zfc$.  Nonetheless, there are
\emph{relative} consistency results, as for example that \emph{if}
$\zf$ is consistent, then so is $\zfc$.  We shall prove this and
similar results.
\end{parag}

\begin{parag}\label{par:model}
A class $\cM$ serves as a truth-assignment in the sense
of~\parref{par:ta}.  If $a$ and $b$ are in $\cM$, and $a\in b$, then
we write $\cM\models a\in b$; if $a\notin b$, then $\cM\nmodels a\in
b$.  Recursively, we have
\begin{enumerate}
\item
$\cM\models\lnot\sigma$ if and only if $\cM\nmodels\sigma$;
\item
$\cM\models\sigma\lto\tau$ if and only if $\cM\nmodels\sigma$ or
  $\cM\models\tau$;
\item
$\cM\models\Exists x\phi$ if and only if $\cM\models\phi(a)$ for some
  $a$ in $\cM$.
\end{enumerate}
Again, $\cM\models\sigma$ means $\sigma$ is
\defnplain{true in}{}%
\dindexsub{true}{--- in}
$\cM$.  If $\phi$ is an $n$-ary formula with constants from $\cM$,
then
\begin{equation*}
  \phi^{\cM}=\cls{\vec x}{\vec x\in\cM^n\land\cM\models\phi(\vec x)};
\end{equation*}
this is the class 
\defnplain{defined by}{}%
\dindexsub{define}{class ---d by a formula} $\phi$ in $\cM$.  If
\begin{equation*}
\phi^{\cM}=\cM^n\cap\phi^{\universe}, 
\end{equation*}
then $\phi$ is 
\defnplain{absolute for}{}%
\dindex{absolute: formula --- for a class}%
\dindexsub{formula}{absolute --- for a class} $\cM$.
\end{parag}

\begin{theorem}\label{thm:R(oo)}
  In $\rwf(\vnn)$, the axioms of $\zfc$ besides Infinity are true, but
  Infinity is false.
\end{theorem}

\begin{proof}
  Say $a$ and $b$ are in $\rwf(\vnn)$.  
  \begin{asparaenum}[1.]
    \item
Since $\{a\}$ and $\{b\}$ are
  in $\rwf(\vnn)$, \emph{equality} (that is, the
  formula $x=y$) is absolute for $\rwf(\vnn)$.  If $a\nincluded b$,
  then, since $\rwf(\vnn)$ is transitive, it contains an element of
  $a\setminus b$, so $\rwf(\vnn)\models a\nincluded b$.  This shows
  that \emph{inclusion} (the formula $x\included y$) is absolute for
  $\rwf(\vnn)$; therefore Extension is true in $\rwf(\vnn)$.  
\item
Since
  $\{a,b\}\in\rwf(\vnn)$, Pairing is true in $\rwf(\vnn)$.
\item
Every element of $\rwf(\vnn)$ is finite, and every finite subset of
$\rwf(\vnn)$ is an element.  This shows that Replacement (hence
Separation) is true in $\rwf(\vnn)$.
\item
For the same reason, Infinity is false in $\rwf(\vnn)$.
\item
We have $a\in\rwf(n+1)$ for some $n$ in $\vnn$, so
$a\included\rwf(n)$, hence $\pow a\included\rwf(n+1)$, so $\pow
a\in\rwf(n+2)$.  Thus Power Set is true in $\rwf(\vnn)$.  
\item
If $c\in a$,
then $c\in\rwf(n)$, so $c\included\rwf(n-1)$.  Hence $\bigcup
a\included\rwf(n-1)$, so $\bigcup a\in\rwf(n)$.  This shows Union is
true in $\rwf(\vnn)$.  
\item
If $c$ is an element of $a$ of minimal rank,
then $c\in\rwf(\vnn)$ (by transitivity) and $a\cap c=\emptyset$; thus
Foundation is true in $\rwf(\vnn)$.
\item
Finally, Choice is true in $\rwf(\vnn)$ because a choice-function for
$a$ is a subset of $\pow a\times a$, hence of $\pow{\pow{\pow a\cup
    a}}$ (\parref{par:cartesian}), hence of $\rwf(n+3)$, so it belongs
to $\rwf(n+4)$.  (See also Exercise~\ref{ex:ac-in-R(oo)}.)
\hfill\qed
  \end{asparaenum}
\renewcommand{\qed}{}
\end{proof}

\begin{theorem}\label{thm:R(oo2)}
  In $\rwf(\vnn\cdot 2)$, the axioms of $\zfc$, besides Replacement,
  are true, but Replacement is false.
\end{theorem}

\begin{proof}
  The proof of Theorem~\ref{thm:R(oo)} shows that all axioms but
  Separation, Replacement, and Infinity are true in $\rwf(\vnn\cdot
  2)$.  Moreover, Separation is true in $\rwf(\vnn\cdot 2)$ since, if
  $a\in\rwf(\vnn\cdot2)$ and $b\included a$, then
  $b\included\rwf(\alpha)$ for some $\alpha$ in $\vnn\cdot2$, and then
  $b\in\rwf(\alpha+1)$, so $b\in\rwf(\vnn\cdot2)$.  Infinity is true
  in $\rwf(\vnn\cdot2)$ since it contains $\vnn$.  Replacement is not
  true in $\rwf(\vnn\cdot2)$ since $\set{\rwf(\vnn+n)}{n\in\vnn}$ is
  the image of $\vnn$ under a function definable in
  $\rwf(\vnn\cdot2)$, but is not an element of 
  $\rwf(\vnn\cdot 2)$.
\end{proof}

\begin{theorem}\label{thm:wf}
  In $\wf$, the axioms of $\zfc$ are true.
\end{theorem}

\begin{proof}
By the proof of Theorem~\ref{thm:R(oo2)}, it remains to show that
Replacement is true in $\wf$.  If $a\in\wf$, and $\cF\colon a\to\wf$,
let $\beta=\sup\set{\rank{\cF(x)}}{x\in a}$; then $\cF\setimb
  a\in\rwf(\beta+2)$, so $\cF\setimb a\in\wf$.  
\end{proof}

\begin{parag}
  Theorems~\ref{thm:R(oo)} and~\ref{thm:R(oo2)} show that neither
  Infinity nor Replacement is deducible from the other axioms in
  $\zfc$, provided that these others are consistent.  Since the
  definition of $\wf$ and the proof of Theorem~\ref{thm:wf} do not
  assume Foundation, we can conclude that, if the $\zf$ axioms without
  Foundation are consistent, then $\zf$ is consistent, and likewise
  for $\zfc$.
\end{parag}

\section{Constructible sets}

\begin{parag}
If $a\included b$, a relation on $b$ is
\defnplain{definable  over}{}%
\dindexsub{define}{relation ---able over a set}
$a$ if the relation is $\phi^b$ (in the sense of~\parref{par:model})
for some formula $\phi$ with constants from $a$.  A
relation on $a$ is \defn{definable}, simply, if it is definable
over $a$.  By Theorem~\ref{thm:const} below, for all $n$ in $\vnn$,
the $n$-ary definable relations on a set $a$ compose a set, which we
shall denote by
\begin{equation*}
  \defr na.
\end{equation*}
Then $\defr na\included\pow{a^n}$.  However, if $a$ is infinite, then
$\card{\defr na}=\card a$ (Exercise~\ref{ex:def-count}); so not every
relation on $a$ is definable. 
\end{parag}

\begin{theorem}\label{thm:const}
  For all $n$ in $\vnn$, the $n$-ary relations on a set compose a set.
\end{theorem}

\begin{proof}
  Considering the recursive definition of formulas (\parref{par:1}),
  we obtain the definable relations as follows.  Given a set $a$, we
  let $b$ be the set of all sequences $(c_n\colon n\in\vnn)$, where
  $c_n\included a^n$, and
  \begin{enumerate}
  \item
$\set{\vec x\in a^n}{x_i\in x_j}\in c_n$ whenever $\{i,j\}\included
    n$;
\item
$\set{\vec x\in a^n}{x_i\in d}\in c_n$ whenever $i\in n$ and $d\in a$;
\item
$\set{\vec x\in a^n}{d\in x_i}\in c_n$ whenever $i\in n$ and $d\in a$;
\item
$c_n$ is closed under the operations $x\mapsto a^n\setminus x$ and
  $(x,y)\mapsto y\setminus x$ on $a^n$;
\item
if $f_n$ is the function $(x_0,\dots,x_n)\mapsto(x_0,\dots,x_{n-1})$
from $a^{n+1}$ to $a^n$, and $e\in c_{n+1}$, then $f_n\setimb e\in
c_n$. 
  \end{enumerate}
Let $b_k$ be the set of all $c_k$ such that $(c_n\colon n\in\vnn)\in
b$ for some $c_n$ where $n\neq k$.  Then the definable $k$-ary
relations on $a$ compose the set $\bigcap b_k$.
\end{proof}

\begin{parag}
  A function $x\mapsto\const(x)$ on $\on$ is defined recursively as follows:
  \begin{enumerate}
    \item
$\const(0)=\emptyset$,
\item
$\const(\alpha+1)=\defr 1{\const(\alpha)}$,
\item
$\const(\alpha)=\bigcup\const\setimb{\alpha}$ if $\alpha$ is a limit.
  \end{enumerate}
We now use $\cu$ to denote $\bigcup\cls{\const(x)}{x\in\on}$ (rather
than the function $x\mapsto\const(x)$).
The elements of $\cu$ are the
\defn{constructible set}{s.}
\end{parag}

\begin{lemma}%\label{lem:L(a)}
Each set $\const(\alpha)$ is transitive, so $\cu$ is transitive.
If $\beta<\alpha$, then $\const(\beta)\included\const(\alpha)$.
\end{lemma}

\begin{proof}
Suppose $a$ is transitive.  Then every element $b$ of $a$ is the
subset of $a$ defined by $x\in b$.  Thus $a\included\defr1a$.  Since
every element of $\defr1a$ is a subset of $a$, we conclude that
$\defr1a$ is transitive.  Since $\const(0)$ is trivially transitive,
by induction the claim follows.
\end{proof}

\begin{lemma}[Tarski--Vaught Test]
\index{theorem!Tarski--Vaught Test}
\index{Tarski--Vaught Test}
  Suppose $\cC\included\cD$ and $\phi$ is an $n$-ary formula with
  constants from $\cC$ such that, for every subformula of $\phi$ of
  the form $\Exists x\psi(x,\vec y)$, for all tuples $\vec b$ of
  elements of $\cC$, if $\cD\models\Exists x\psi(x,\vec b)$, then
  $\cD\models\psi(a,\vec b)$ for some $a$ in $\cC$.  Then
  \begin{equation}\label{eqn:absCD}
    \phi^{\cC}=\cC^n\cap\phi^{\cD}.
  \end{equation}
\end{lemma}

\begin{proof}
  We prove by induction that~\eqref{eqn:absCD} holds whenever $\phi$
  is a subformula of the original formula.  The claim holds easily
  when $\phi$ is atomic, and if it holds when $\phi$ is $\psi$ or
  $\rho$, then it holds when $\phi$ is $\lnot\chi$ or $\chi\lto\rho$.
  Suppose~\eqref{eqn:absCD} holds when $\phi$ is $\psi(x,\vec y)$, and
  $\vec b$ is from $\cC$.  The following statements are
  equivalent: 
  \begin{enumerate}
  \item 
$\cD\models\Exists x\psi(x,\vec b)$;
\item
$\cD\models\psi(a,\vec b)$ for some $a$ in $\cC$ (by assumption);
\item
$\cC\models\psi(a,\vec b)$ for some $a$ in $\cC$ (by inductive hypothesis);
\item
$\cC\models\Exists x\psi(x,\vec b)$.
  \end{enumerate}
Thus~\eqref{eqn:absCD} holds when $\phi$ is $\Exists x\psi(x,\vec
y)$.  Therefore it holds for all subformulas of the original formula. 
\end{proof}

\begin{lemma}\label{lem:Lbn}
  For every $n$ in $\vnn$, for every $n$-ary formula $\phi$ with
  constants from $\cu$, there is $\beta$ such that
  \begin{equation*}
\phi^{\const(\beta)}=    \const(\beta)^n\cap\phi^{\cu}.
  \end{equation*}
\end{lemma}

\begin{proof}
For every ordinal $\alpha$, let
  $\alpha^*$ be the least ordinal $\gamma$ greater than $\alpha$ such
  that, if $\vec b$ is from $\const(\alpha)$, and $\Exists
  x\psi(x,\vec y)$ is a subformula of $\psi$, and $\cu\models\Exists
  x\psi(x,\vec b)$, then $\cu\models\psi(a,\vec b)$ for some $a$ in
  $\const(\gamma)$.   Let $\alpha_0$ be such that the constants of
  $\phi$ are from $\const(\alpha_0)$, and let
  $\alpha_{k+1}=\alpha_k{}^*$.  By the Tarski--Vaught Test, it
  suffices to let
  $\beta=\sup\set{\alpha_n}{n\in\vnn}$, which is a limit.
  Indeed, if $\vec b$ is from $\const(\beta)$, then $\vec b$ is from
  $\const(\alpha_n)$ for some $n$, so if $\const\models\Exists
  x\psi(x,\vec b)$, where $\Exists x\psi(x,\vec y)$ is a subformula of
  $\phi$,
  then $\cu\models\psi(a,\vec b)$ for some $a$ in
  $\const(\alpha_{n+1})$ and hence in $\const(\beta)$. 
\end{proof}

\begin{parag}
  By analogy with~\parref{par:rank}, if $a\in\cu$, we define
  \begin{equation*}
    \rankl a=\min\set x{a\in\const(x+1)}.
  \end{equation*}
Note however that possibly $a\in\cu$, and $a\included\const(\beta)$,
but $a$ is not a \emph{definable} subset of $\const(\beta)$, so
$a\notin\const(\beta+1)$, and so $\beta<\rankl a$. 
\end{parag}

\begin{theorem}\label{thm:zf-l}
  $\zf$ is true in $\cu$.
\end{theorem}

\begin{proof}
  As in the proof of Theorem~\ref{thm:R(oo)}, Foundation is true in
  $\cu$; also equality and inclusion
  are absolute for $\cu$, and therefore Extension is true in $\cu$,
  and so is Pairing.

Suppose $a\in\cu$.  Let $\beta=\sup\set{\rankl x}{x\in\cu\land
  x\included a}$.  Then $\cu\cap\pow a\in\const(\beta+2)$.  Thus Power
Set is true in $\cu$.  

Say $\rankl a=\gamma$.  Then $a$ is a subset of $\const(\gamma)$
defined by a formula $\phi$.  Since $\const(\gamma)$ is transitive,
$\bigcup a$ is defined in $\const(\gamma)$ by $\Exists y(\phi(y)\land
x\in y)$.  Therefore $\bigcup a\in\const(\gamma+1)$.  Thus Union is
true in $\cu$.

If $b\in\const(\alpha)$, then $b\cup\{b\}\included\const(\alpha+1)$,
and so $\vscr b\in\const(\alpha+1)$.  Since $0\in\const(1)$, we have
$\vnn\included\const(\vnn)$.  Also $x\in\vnn$ is absolute for $\cu$.
Therefore $\vnn\in\const(\vnn+1)$, so Infinity is true in $\cu$.

Suppose $\phi(x,y)$ defines in $\cu$ a function $\cF$, and $a\in
\cu$.  Some $\const(\alpha)$ contains all constants in $\phi$ and
elements of $\cF\setimb a$.  Then
$\phi(a,y)^{\cu}\included\const(\alpha)$.  By Lemma~\ref{lem:Lbn},
there is $\beta$ such that
$\phi(a,y)^{\cu}=\phi(a,y)^{\const(\beta)}$.  Thus $\cF\setimb
a\in\const(\beta+1)$.  Therefore Replacement is true in $\cu$.
\end{proof}

\begin{theorem}
  The Axiom of Choice is
  true in $\cu$.  Therefore $\zfc$ is consistent (assuming $\zf$ is
  consistent). 
\end{theorem}

\begin{proof}
There is a binary formula $\phi$ such that $\cu$ is well-ordered by
$\phi^{\cu}$.  Indeed, because of the recursive construction of the
sets $\defr na$, there is a ternary formula $\psi$ such that, if
$r$ well-orders $\const(\alpha)$, then $\psi(r,x,y)^{\cu}$ well-orders
$\const(\alpha+1)$.  We define the ordering $<$ on $\cu$ recursively
as $\bigcup\set{r_x}{x\in\on}$ so that
$a<b$ if $\rankl a<\rankl b$ or if $\rankl
a=\rankl b$ and $\cu\models\psi(r_{\rankl a},a,b)$.  This argument
does not use the Axiom of Choice.
\end{proof}

\section{The Generalized Continuum Hypothesis}

\begin{theorem}[L\"owenheim--Skolem]%
\index{theorem!Lowenheim--Skolem Th---@L\"owenheim--Skolem Th---}%
\index{Lowenheim--Skolem Theorem@L\"owenheim--Skolem Theorem}
  For every set $a$ and every formula $\phi$ with constants from $a$,
  there is a set $b$ such that $a\included b$, and
  \begin{equation*}
    \card b\leq\card a+\aleph_0,
  \end{equation*}
and $\phi$ is absolute for $b$.
\end{theorem}

\begin{proof}
  Suppose $a\included c$.  Let $\alpha$ be the supremum of the set of
  all ordinals 
  \begin{equation*}
  \min\set{\rank z}{z\in\psi(x,\vec b)^{\universe}},
  \end{equation*}
  where $\Exists x\psi(x,\vec y)$ is a subformula of $\phi$, and $\vec
  b$ is from $c$, and $\Exists x\psi(x,\vec b)$ is true.  Fix a
  relation $r$ that well-orders 
  $\rwf(\alpha+1)$.  Define $c(r)$ to comprise each element of $c$ and
  the least element of each nonempty set $\rwf(\alpha+1)\cap\psi(x,\vec
  b)^{\universe}$, again where $\Exists x\psi(x,\vec y)$ is a
  subformula of $\phi$, and $\vec b$ is from $c$.  Note that
  $\card{c(r)}\leq\card c+\aleph_0$.

Now let $(a_n\colon n\in\vnn)$ be such that $a_0=a$ and
$a_{n+1}=a_n(r)$ for some choice of $r$ as above.  Let
$b=\bigcup\set{a_n}{n\in\vnn}$.  Then $b$ is as desired, by the
Tarski--Vaught Test.
\end{proof}

\begin{parag}
A relation $\cR$ on a class $\cC$ is
  \defnplain{extensional}{}%
\dindexsub{extension}{---al relation}
\dindexsub{relation}{extensional ---}
if the Extension Axiom is true in the structure $(\cC,\cR)$ in the
sense that
\begin{equation*}
  \cR a=\cR b\lto a=b.
\end{equation*}
\end{parag}

\begin{theorem}[Mostowski Collapse]
Let $\cR$ be a well-founded relation on $\cC$.  There is a unique
function $\cF$ on $\cC$ given by
\begin{equation*}
  \cF(a)=\cF\setimb{\cR a}.
\end{equation*}
Then $\cF\setimb{\cC}$ is transitive.  If $\cR$ is extensional,
then $\cF$ is an isomorphism from $(\cC,\cR)$ to
$(\cF\setimb{\cC},\in)$. 
\end{theorem}

\begin{proof}
  We follow the proof of Theorem~\ref{thm:o-eq}, though since $\cR$
  need not be transitive, we shall need
  the following definition.  if $a\in\cC$, then
  \begin{gather*}
    \begin{aligned}
    \cl[0]a&=\{a\},&
\cl[n+1]a&=\set x{\Exists y(y\in\cl[n]a\land x\mathrel{\cR}y)},
    \end{aligned}\\
\cl a=\bigcup\set{\cl[n]a}{n\in\vnn}.
  \end{gather*}
The structure $(\cC,\cR)$ admits induction in the sense that, if
$\cC_0\included\cC$ and, for all $a$ in $\cC$, we have $a\in\cC_0$
whenever $\cR a\included\cC_0$, then $\cC_0=\cC$.  Indeed, suppose
$\cC_0\pincluded\cC$, and $a\in\cC\setminus\cC_0$.  By definition
(\parref{par:ill-founded}), 
$(\cC\setminus \cC_0)\cap\cl a$ has an element $b$ such that
\begin{equation*}
(\cC\setminus\cC_0)\cap\cl a\cap \cR b=\emptyset.  
\end{equation*}
But $\cR
b\included\cl a$ (since $b\in\cl[n]a$ for some $n$, and then $\cR
b\included\cl[n+1]a$).  Hence $(\cC\setminus\cC_0)\cap\cR
b=\emptyset$, so $\cR b\included\cC_0$.

Now we can show by induction that, for all $a$ in $\cC$, there is a
unique function $f_a$ on $\cl a$ such that
\begin{equation*}
  f_a(c)=f_a\setimb{\cR c}.
\end{equation*}
Indeed, suppose the claim holds when $a\mathrel{\cR}b$.  If $a$ and
$d$ are in $\cR b$, then $f_a$ and $f_d$ must agree on $\cl a\cap \cl
d$ since, if $f_a$ and $g$ disagree on $\cl a\cap\cl d$, then by
well-foundedness this set
has an element $e$ such that $f_a$ and $g$ agree on $\cR e$, but
\begin{equation*}
g(e)\neq f_a(e)=f_a\setimb{\cR e}=g\setimb{\cR e}.  
\end{equation*}
Now we can
define $f_b$ on $\cl b$ so that, if $c\in\cl b\setminus\{b\}$, then
$f_b(c)=f_a(c)$, where $a$ is such that $a\mathrel{\cR}b$ and $c\in\cl
a$; and 
$f_b(b)=f_b\setimb{\cR b}$.  

The desired function $\cF$ is now $\bigcup\set{f_a}{a\in\cC}$.
Indeed, this is a function, since any two functions $f_a$ agree as
before on the intersection of their domains.  Likewise, $\cF$ itself
is unique.
Since $\cF(a)=\cF\setimb{\cR a}\included\cF\setimb{\cC}$, it
follows that $\cF\setimb{\cC}$ is transitive.

We have $a\mathrel{\cR}b\lto\cF(a)\in\cF(b)$.  If $\cF$ is injective,
then $\cF(a)\in\cF(b)\lto a\mathrel{\cR}b$, so $\cF$ is an
isomorphism.  Suppose $\cF$ is not injective.  Let $\cC_0$
comprise those $a$ in $\cC$ for which there is no distinct $b$ such
that $\cF(a)=\cF(b)$.  As in the proof that $(\cC,\cR)$ admits
induction, there is an element $a$ of $\cC\setminus\cC_0$ such that
$\cR a\included\cC_0$.  Then $\cF(a)=\cF(b)$ for some distinct $b$.
This means
\begin{equation*}
  \set{\cF(x)}{x\mathrel{\cR}a}=
  \set{\cF(y)}{y\mathrel{\cR}b}.
\end{equation*}
Since $\cR a\included\cC_0$, we conclude $\cR b=\cR a$.  Thus $\cR$ is
not extensional.
\end{proof}

\begin{theorem}
  The Generalized Continuum Hypothesis is true in $\cu$.  Thus $\gch$
  is consistent with $\zfc$ (assuming $\zf$ is consistent).
\end{theorem}

\begin{proof}
  Suppose $a\in\pow{\const(\alpha)}\cap\cu$, where $\alpha$ is
  infinite.  We shall show 
  \begin{equation*}
  a\in\const(\card{\alpha}\pl).   
  \end{equation*}
  Since
  $\card{\const(\beta)}=\card{\beta}$ (in $\universe$ and in $\cu$),
  it will follow that $\card{\pow{\kappa}}=\kappa\pl$ in $\cu$.

Apply the L\"owenheim--Skolem Theorem to the set
$\const(\alpha)\cup\{a\}$ and the conjunction of the Extension Axiom
with the formula $a\in\const(x)$.  We get a set $b$ such that
$\const(\alpha)\included b$, $a\in b$, $\card b=\card{\alpha}$,
$(b,\in)$ is extensional, and
\begin{equation*}
  b\models\Exists xa\in\const(x).
\end{equation*}
By the Mostowski Collapse Theorem, we may assume further that $b$ is
\emph{transitive.}  In particular, an element $\beta$ of $b$ such that
$b\models a\in\const(\beta)$ really is an ordinal.  Then
$a\in\const(\beta)$, but $\card{\beta}=\card{\alpha}$, so
$a\in\const(\card{\alpha}\pl)$. 
\end{proof}

\begin{parag}
  About a quarter century after G\"odel
\index{Goedel@G\"odel} 
proved that $\ac$ and $\gch$
  are consistent with $\zf$, Cohen (see~\cite{MR0232676}) proved the
  same of their negations.
\end{parag}

\section*{Exercises}

\begin{xca}
  Prove that class is ill-founded if and only if there is a sequence
  $(a_n\colon n\in\vnn)$ of elements of the class such that
  $a_{n+1}\in a_n$ in each case.  (One direction requires the Axiom of
  Choice.) 
\end{xca}

\begin{xca}\label{ex:tc}
  Prove Theorem~\ref{thm:tc}.
\end{xca}

\begin{xca}\label{ex:trans}
Show that both the power set of
  a transitive set and the union of a set of transitive
  sets are transitive.
\end{xca}

\begin{xca}
Show $\beta<\alpha\lto\rwf(\beta)\in\rwf(\alpha)$.
\end{xca}

\begin{xca}\label{ex:rank}
Assuming $a\in\wf$ and $b\in a$,
show that $b\in\wf$ and $\rank b<\rank a$.
\end{xca}

\begin{xca}\label{ex:wf-subsets}
Show that all subsets of $\wf$ are elements of $\wf$.
\end{xca}

\begin{xca}\label{ex:wf-ac}
Using the Axiom of Choice, assuming $a$ is transitive, but not a
subset of $\wf$, show that there is a function $f$ on $\vnn$ such that
$f(0)=a$ and $f(n+1)\in f(n)\setminus\wf$.
\end{xca}

\begin{xca}\label{ex:V=WF}
  Show that the two given formulations of the Foundation Axiom are
  equivalent. 
\end{xca}

\begin{xca}\label{ex:ac-in-R(oo)}
  Prove that the Axiom of Choice is true in $\rwf(\vnn)$ without using
  the Axiom of Choice.
\end{xca}

\begin{xca}\label{ex:def-count}
If $a$ is infinite, show that
$\card{\defr na}=\card a$.
\end{xca}

\appendix

\chapter{The Greek alphabet}\label{app:Greek}

  \begin{center}
      \begin{tabular}{c c c c}
capital&minuscule&transliteration&name\\\hline
\Gk{A}& \Gk{a} & a & alpha \\ 
\Gk{B}& \Gk{b} & b & beta   \\ 
\Gk{G}& \Gk{g} & g & gamma \\ 
\Gk{D}& \Gk{d} & d & delta \\ 
\Gk{E}& \Gk{e} & e & epsilon\\ 
%&\Gk{\stigma}& stigma\\
\Gk{Z}& \Gk{z} & z & zeta   \\ 
\Gk{H}& \Gk{h} & \^e & eta \\ 
 \Gk{J}& \Gk{j} & th & theta \\
%\hline 
 \Gk{I}& \Gk{i} & i & iota\\ 
\Gk{K}& \Gk{k} & k & kappa\\ 
 \Gk{L}& \Gk{l} & l & lambda\\ 
 \Gk{M}& \Gk{m} & m & mu \\
 \Gk{N}& \Gk{n} & n & nu \\ 
 \Gk{X}& \Gk{x} & x & xi \\ 
 \Gk{O}& \Gk{o} & o & omicron\\ 
 \Gk{P}& \Gk{p} & p & pi\\ 
%&\Gk{\qoppa}& koppa\\
%\hline
 \Gk{R}& \Gk{r} & r & rho\\ 
 \Gk{S}& \Gk{s, c} & s & sigma \\
 \Gk{T}& \Gk{t} & t & tau \\ 
 \Gk{U}& \Gk{u} & y, u & upsilon \\ 
 \Gk{F}& \Gk{f} & ph & phi\\ 
 \Gk{Q}& \Gk{q} & ch & chi\\ 
 \Gk{Y}& \Gk{y} & ps & psi\\ 
 \Gk{W}& \Gk{w} & \^o & omega\\
%&\Gk{\sampi}& sampi\\
\hline
  \end{tabular}
  \end{center}

The following remarks pertain to \emph{ancient} Greek.
The vowels are 
%\begin{center}
\Gk{a, e, h, i, o, u, w,} 
%\end{center}
where
\Gk h is a long \Gk e, and \Gk w is a long \Gk o; the other vowels
(\Gk{a, i, u}) can be long or short.  
Some vowels may be given tonal accents
(\Gk{'a, ~a, `a}).  
An initial vowel takes either a rough\-{}-breathing mark (as in
\Gk{<a}) or a 
smooth\-{}-breathing mark (\Gk{>a}): the former mark is transliterated
by a preceding
\lett h, and the latter can be ignored, as in
  \Gk{<uperbol'h} hyperbol\^e \emph{hyperbola,}
\Gk{>orjog'wnion} orthog\^onion \emph{rectangle.}
Likewise, \Gk{<r} is
transliterated as \lett{rh}, as in
  \Gk{<r'omboc} rhombos \emph{rhombus.}
A long vowel may have an iota subscript (\Gk{a|, h|, w|}), especially
in case-endings of nouns.
Of the two forms of minuscule sigma, the \Gk{c} appears at
the ends of words; elsewhere, \Gk{sv} appears, as in
\Gk{b'asic} basis \emph{base.}

\appendix\setcounter{chapter}1 % needed because of language-switching

\chapter{The German script}\label{app:German}

Writing in 1993, Wilfrid Hodges \cite[Ch.~1, p.~21]{MR94e:03002} observes
\begin{quotation}
  Until about a dozen years ago, most model theorists named structures
  in horrible Fraktur lettering.  Recent writers sometimes adopt a
  notation according to which all structures are named $M$, $M'$,
  $M^*$, $\bar M$, $M_0$, $M_i$ or occasionally $N$.  
%I hope I cause no offence by using a more freewheeling notation.
\end{quotation}
For Hodges, structures are $A$, $B$, $C$, and so forth; he refers to
their universes as
\defnplain{domain}s%
\index{domain!of a structure}
and denotes these by $\dom A$ and so forth.  
\begin{comment}
This practice is
convenient if one is using a typewriter (as in the preparation of
another of Hodges's books \cite{Hodges-Building}, from 1985).
In 2002, David Marker \cite{MR1924282} uses `calligraphic' letters for
structures, so that $M$ is the universe of~$\mathcal M$.
\end{comment}
I still prefer the Fraktur letters:
%In \AmS\ \LaTeX\ (by which these notes are typeset) these letters are:
\begin{align*}
&\begin{array}{ccccccccc}
\mathfrak A&\mathfrak B&\mathfrak C&\mathfrak D&\mathfrak E&\mathfrak F&\mathfrak G&\mathfrak H&\mathfrak I\\\mathfrak J&\mathfrak K&\mathfrak L&\mathfrak M&\mathfrak N&\mathfrak O&\mathfrak P&\mathfrak Q&\mathfrak R\\\mathfrak S&\mathfrak T&\mathfrak U&\mathfrak V&\mathfrak W&\mathfrak X&\mathfrak Y&\mathfrak Z&
  \end{array}
&
&\begin{array}{ccccccccc}
\mathfrak a&\mathfrak b&\mathfrak c&\mathfrak d&\mathfrak e&\mathfrak f&\mathfrak g&\mathfrak h&\mathfrak i\\\mathfrak j&\mathfrak k&\mathfrak l&\mathfrak m&\mathfrak n&\mathfrak o&\mathfrak p&\mathfrak q&\mathfrak r\\\mathfrak s&\mathfrak t&\mathfrak u&\mathfrak v&\mathfrak w&\mathfrak x&\mathfrak y&\mathfrak z&
  \end{array}
\end{align*}
A way to write these by hand is seen in a textbook
of German from 1931 \cite{German}:
\begin{center}
\includegraphics[width=417pt,height=292pt]{german-script-cropped.eps}
\end{center}

\backmatter



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\begin{figure}[t]%\label{back}
\vspace{3cm}
  \begin{center}
    \includegraphics{plantation}
  \end{center}
%\caption{METU forest.  Now I am trying to see how wide these lines
%  are; how wide are they?  Wide enough to cover more than one line?}
\end{figure}
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