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\begin{document}
%\frontmatter
\title{Ordinal Analysis II}
\subtitle{A course at the Nesin Matematics Village}
\author{David Pierce}
\date{February 3--9, 2020\\
Last edited February 14, 2020}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\.Istanbul\\
\url{dpierce@msgsu.edu.tr}\\
\url{mat.msgsu.edu.tr/~dpierce/}\\
\url{polytropy.com}}
\maketitle
\tableofcontents
\addchap{Preface}
Here are notes in English from
Ordinal Analysis II,
given at the Nesin Matematik K\"oy\"u in \c Sirince,
February 3--9, 2020.
English is the language I wrote on the blackboards
of the Ni\c sanyan K\"ut\"uphanesi,
while speaking mostly in Turkish.
One student proposed that I speak in English,
but another student objected.
At the beginning of the first day,
when there were around fifty students,
I wrote the sequence of ordinals---%
\eqref{eqn:on} on page \pageref{eqn:on}---%
across all four boards in the library.
On the last day, there were seven students.
I had given Ordinal Analysis I in the previous week,
in the Arf Dersli\u gi.
The two courses together together covered
much of what I had done
in Aksiyomatik K\"umeler Kuram\i\
in the previous semester at Mimar Sinan.
\begin{compactdesc}
\item[Ordinal Analysis I] introduced the Zermelo--Fraenkel axioms
as they were needed to develop and prove the desired properties
of the class $\on$ of ordinals.
\item[Ordinal Analysis II]
assumed those properties
(listed on page \pageref{on-props})
and developed the arithmetic of the ordinals.
\end{compactdesc}
The latter course
is summarized on page \pageref{sum}.
I did not typeset new notes for the first course.
I did not assume that students in the second course
had taken the first.
In all but one or two cases they had not.
One student from the Mimar-Sinan course
did attend the entirety of Ordinal Analysis II.
I started that course,
thinking that I would
progressively introduce concrete examples of well-ordered sets:
\begin{compactenum}[1)]
\item
$\upomega$,
closed under addition,
multiplication, and exponentiation,
but containing no limits;
\item
$\upomega\sqcup\upomega$,
isomorphic to $\upomega+\upomega$
and containing a single limit;
\item
$\upomega\times\upomega$,
isomorphic to $\upomega\cdot\upomega$,
closed under addition,
and containing infinitely many limits,
which themselves have no limit;
\item
the set of operations on $\upomega$
that have finite support: this set is
isomorphic to $\upomega^{\upomega}$,
is closed under addition and multiplication,
and has limits of limits;
\item
the set of finite rooted trees: this set is
isomorphic to $\upepsilon_0$
and closed under all of the operations.
\end{compactenum}
In the event, I did not isolate the last two examples,
but I interpreted arbitrary powers $\alpha^{\beta}$
as sets of functions,
and after $\upomega$ I considered only $\on$ itself
as being closed under all of the operations.
I often left proofs as exercises.
Had there been time,
I would have had students give their proofs at the board:
I do this at Mimar Sinan,
and I have done it in other courses at the Math Village.
This time I wanted to cover all of my Mimar-Sinan course
in two-thirds of the time.
Some students asked for references.
These would include:
\begin{compactitem}
\item
the notes (in Turkish) from my Mimar-Sinan course;
\item
any set-theory textbook,
particularly Levy, \emph{Basic Set Theory}
(Dover, 2002), which has perhaps been my own main reference;
\item
Ali Nesin's A\c c\i k Ders text,
\emph{Aksiyomatik K\"umeler Kuram\i\ (D\"onem 1),}
on the website of the T\"urkiye Bilimler Akademisi.
I have recommended this text to students at Mimar Sinan
and used as a source for Turkish terminology.
\end{compactitem}
\addchap{My Personal Journey}
I was not prepared for the question of references.
Since I started learning real mathematics from one Donald J. Brown
in my last two years (1981--3) at St.\ Albans School
in Washington, D.C.,
I have understood the main text of a mathematics course
to be the teacher's own lectures.
Mr Brown had us copy down exactly
what he wrote on the blackboards of his classroom,
and he inspected our notebooks
(though perhaps mainly to see that we had reinforced the holes
in our sheets of paper so that they would not easily tear
from our three-ring binders).
Mr Brown did have us buy
supplementary books,
to be used particularly as sources of exercises.%%%%%
\footnote{For the record,
the books
were the following,
roughly in order of use.
\begin{compactitem}
\item
Kutepov and Rubanov,
\emph{Problem Book: Algebra and Elementary Functions}
(Moscow: Mir, 1978);
\item
Dorofeev, Potapov, and Rozov,
\emph{Elementary Mathematics:
Selected Topics and Problem Solving}
(Moscow: Mir, 1973);
\item
Spivak, \emph{Calculus,}
Second Edition (Berkeley: Publish or Perish, 1980);
\item
Salas and Hille,
\emph{Calculus: One and Several Variables, Part I,}
Third Edition (New York: Wiley, 1978);
\item
Petit Bois, \emph{Tables of Indefinite Integrals}
(New York: Dover, 1961);
\item
Apostol, \emph{Mathematical Analysis,} Second Edition
(Reading, Massachusetts: Addison-Wesley, 1974).
\end{compactitem}
I am able to list these books,
because I have kept them.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
One of the books was Spivak's \emph{Calculus.}
I read this,
while understanding that Spivak's approach to the subject
was not ``official'' when it differed from our own
(as it did for example
in the development of the real numbers as a complete ordered field,
and later in the definitions of the trigonometric functions).
I might have preferred Spivak's or another approach;
If I ever taught my own course.
I could do things my way.
This is what I have often done,
starting when I was assigned to join two colleagues
in teaching Fundamentals of Mathematics at METU in Ankara
in the fall of 2001-2.
One of those colleagues (who was from Germany)
proposed that he and I write the text of the course.
Part of my own share of the writing
became the text that Ay\c se Berkman used
when teaching Set Theory at METU in the fall of 2003-4.
I revised and rewrote that text
when teaching the set-theory course for myself
in 2006-7, 2008-9, and 2010-1.
When I moved to Istanbul,
I prepared texts in Turkish
for the course Aksiyomatik K\"umeler Kuram\i,
which I have given so far in
2012-3, 2013-4, 2015-6, 2017-8, and 2019-20.
Without going back and reviewing all of those texts
(though they are all on my webpage at Mimar Sinan),
I think the progression has been as follows.
Initially I tried to write down what a \emph{teacher} should know,
or at least what \emph I wanted to know.
I pared down down what I wrote
as I learned the material and the students.
I have come to emphasize ordinal arithmetic,
through computation with Cantor normal forms,
because, as I understand it,
the Turkish national university entrance examinations
are largely computational,
at least as far as mathematics is concerned.
It has also come to seem worthwhile to me
to develop an analogy between the ``well-order'' of the ordinal numbers
and the complete dense ordering of the real numbers.
I am not a set theorist in the sense of publishing new research in the subject,
and I do not expect my students to become set theorists
(though one of them has).
Students are more likely to work with analysis than set theory.
The former had been my own introduction to modern mathematical rigor,
through Mr Brown's two-year course of honors precalculus and calculus.
All students of mathematics learn such techniques
as differentiation and integration,
which are justified by the aforementioned rigor.
Students may forget the techniques,
but they ought to retain an awareness of the possibility of justifying them.
The standards of justification are universal,
precisely because they are the individual property of each of us.
\addchap{Summary}\label{sum}
\begin{compactdesc}
\item[Monday]
Paradoxes of Russell and Burali-Forti.
$\upomega$ and $\upomega\sqcup\upomega$
as well-ordered sets.
Addition on $\upomega$ is defined by recursion;
its properties are proved by induction.
Addition of a natural number to an element of $\upomega\sqcup\upomega$
in two ways, only one being continuous.
\item[Tuesday]
Continuity of functions,
especially increasing functions,
from ordered sets to ordered sets.
\item[Wednesday]
The desired properties of $\R$
lead to Dedekind's construction;
of $\on$,
to von Neumann's.
Transfinite recursion.
$\alpha+\beta\cong\alpha\sqcup\beta$.
\item[Thursday]
History of our subject.
Properties of addition and multiplication of ordinals
proved by transfinite induction.
An increasing operation on $\on$ is continuous
if and only if its value at every limit is the supremum
of its values at the preceding ordinals.
$\alpha\cdot\beta\cong\alpha\times\beta$.
Cantor normal forms of elements of $\upomega\times\upomega$
and $\upomega^{\upomega}$.
\item[Friday]
Ordinal exponentiation.
Cantor normal forms of arbitrary ordinals
and the generalization to arbitrary bases.
$\alpha^{\beta}$ is isomorphic to the set of functions
from $\beta$ to $\alpha$ with finite support.
\item[Saturday]
Computation with Cantor normal forms.
Each of $\alpha+\beta$, $\alpha\cdot\beta$, and $\alpha^{\beta}$
is equipollent with the greater of $\alpha$ and $\beta$,
if these are infinite.
\end{compactdesc}
\chapter{Monday}
The \emph{ordinal numbers} or \textbf{ordinals}
compose this \emph{linear order}:%%%%%
\footnote{The definition of linear ordering
is spelled out on page \pageref{lo}.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{equation}\label{eqn:on}
\left.
\begin{gathered}
0,1,2,\dots;\upomega,\upomega+1,\upomega+2,\dots;
\upomega+\upomega=\upomega\cdot2,\\
\upomega\cdot2+1,\dots;\upomega\cdot3,\upomega\cdot3+1,\dots;
\upomega\cdot\upomega=\upomega^2,\\
\upomega^2+1,\dots;\upomega^2+\upomega,\upomega^2+\upomega+1,\dots;
\upomega^3,\dots;
\upomega^{\upomega},\\
\upomega^{\upomega}+1,\dots,\upomega^{\upomega^{\upomega}},\dots;
\upepsilon_0,\upepsilon_0+1,\dots;\upepsilon_1,\dots;
\upomega_1,\dots
\end{gathered}
\right\}
\end{equation}
\begin{compactitem}
\item
$\upomega$ is \emph{omega}, ``large o,''
the minuscule case of the last letter of the Greek alphabet;
we shall understand
\begin{equation*}
\upomega=\{0,1,2,\dots\},
\end{equation*}
the set of \textbf{natural numbers.}
(The set $\{1,2,3,\dots\}$ of \emph{counting numbers}
can be denoted by $\N$.)
\item
$\upepsilon_0$ will be the first solution; $\upepsilon_1$, the next;
to the equation
\begin{equation*}
\upomega^x=x.
\end{equation*}
\item
$\upomega_1$ will be the least ordinal $\alpha$ such that the set
\begin{equation*}
\{x\colon x<\alpha\}
\end{equation*}
is \textbf{uncountable,}
meaning there is no bijection with $\upomega$ or a subset of it.%%%%%
\footnote{People asked about the $\epsilon_x$ and $\upomega_1$,
so I talked about them.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{compactitem}
Moreover, the ordinals have the following properties:
\begin{compactenum}
%\item
% There is a first or least one.
\item
Every $\alpha$ of them has a \textbf{successor,} $\alpha'$ or $\alpha+1$,
which is the first of those that are greater than it.
\item
Every set $A$ of them that has no greatest element
still has a \textbf{supremum}
or least upper bound, $\sup(A)$.
\newcounter{onenum}
\setcounter{onenum}{\value{enumi}}
\end{compactenum}
\begin{theorem}[Burali-Forti Paradox]\label{thm:BF}
The class of ordinals is not a set.
\end{theorem}
\begin{proof}
Since every ordinal has a successor,
the class of ordinals has no supremum,
so it cannot be a set.
\end{proof}
The class of ordinals is ``too big'' to be a set.
Another example is given by the following.
\begin{theorem}[Russell Paradox]
The class $\{x\colon x\notin x\}$ is not a set.
\end{theorem}
A technical property of the ordinals is the following.
\begin{compactenum}
\setcounter{enumi}{\value{onenum}}
\item
They are \textbf{left-narrow:}%%%%%
\footnote{In class,
I did not use this term,
which is from Levy, I.1(vii), p.\ 33.}
%%%%%%%%%%%%%%%%%%%%%%%%%
for each $\alpha$ of them,
the class $\{x\colon x<\alpha\}$
or $\pred(\alpha)$ is a set.
\end{compactenum}
\begin{theorem}
The class of ordinals is \textbf{well-ordered:}
it is left-narrow, and
every non\-empty set $A$ of them has a least element, $\min(A)$.
\end{theorem}
\begin{proof}
Let $B$ be a nonempty set of ordinals, and let
\begin{equation*}
A=\bigcap_{x\in B}\pred(x)=\{x\colon\Forall y(y\in B\lto y0$.
Therefore, as we shall see,
the operation then is not \emph{continuous} at $\upomega$.
If we define
\begin{equation*}
x\mapsto k+ x
\end{equation*}
on $\upomega\sqcup\upomega$ by
\begin{equation*}
k+(\upomega+n)=\upomega+n,
\end{equation*}
then the operation is still strictly increasing.
Also,
\begin{equation*}
\sup\{k+x\colon x<\upomega\}=\upomega=k+\upomega,
\end{equation*}
so the operation \emph{is} continuous at $\upomega$.
\chapter{Tuesday}
To define continuity,
suppose $f$ is a function from a linear order $A$
to a linear order $B$, and $c\in A$.
For example, $A$ and $B$ could be the set of real numbers.
Then $f$ is \textbf{continuous} at $c$,
provided that for all $\epsilon_1$ and $\epsilon_2$ in $B$ such that
\begin{equation*}
\epsilon_10$.
Now we can define
\begin{equation}\label{eqn:ww}
(\upomega\cdot n+k)+(\upomega\cdot m+\ell)
=
\begin{cases}
\upomega\cdot n+(k+\ell),&\text{ if }m=0,\\
\upomega\cdot(n+m)+\ell,&\text{ if }m>0.
\end{cases}
\end{equation}
\begin{theorem}
For each $\alpha$ in $\upomega\times\upomega$,
the operation
\begin{equation*}
x\mapsto\alpha+x
\end{equation*}
is strictly increasing and continuous;
the operation
\begin{equation*}
x\mapsto x+\alpha
\end{equation*}
is increasing, though not strictly,
and is not continuous unless $\alpha=0$.
\end{theorem}
We could also define
\begin{equation*}
(\upomega\cdot n+k)\oplus(\upomega\cdot m+\ell)
=\upomega\cdot(n+m)+(k+\ell).
\end{equation*}
Then $\oplus$ is commutative and associative,
but continuous in neither argument.
\chapter{Wednesday}
What \emph{is} an ordinal number?
Well, what is an real number?
To do real analysis,
all one need know is that the real numbers together compose a
\emph{complete ordered field,} $\R$.
That this is \textbf{complete} means every nonempty subset with an upper bound
has a least upper bound, or supremum.
Every ordered field includes the ordered field $\Q$
of rational numbers.
In an ordered field,
if $\Q$ has an upper bound $a$,
then $a-1$ is also an upper bound;
therefore there is no least upper bound.
Ordered fields in which $\Q$ is bounded do exist.
For example, if
\begin{gather*}
\Q[X]=\{\text{polynomials in $X$ over $\Q$}\},\\
\Q(X)=\left\{\frac fg\colon f\in\Q[X]\land g\in\Q[X]\setminus\{0\}\right\},
\end{gather*}
this field can be ordered by the rule
\begin{equation*}
\frac{a_0+a_1X+\dots+a_mX^m}{b_0+b_1X+\dots+b_nX^n}>0
\liff \frac{a_m}{b_n}>0.
\end{equation*}
Hence for all $a$ in $\Q$, $X-a>0$, and so $X>a$.
Thus $X$ is an upper bound of $\Q$.
An ordered field in which $\Q$ has no upper bound
is called \textbf{Archimedean.}
Being complete,
$\R$ must be Archimedean.
Because of this, $\Q$ is \textbf{dense} in $\R$:
between any two real numbers lies a rational number.
Therefore the function $f$ on $\R$ given by
\begin{equation*}
f(\alpha)=\{x\in\Q\colon x<\alpha\}
\end{equation*}
is injective.
Moreover, its range is the set of subsets $A$ of $\Q$ such that
\begin{gather*}
\emptyset\pincluded A\pincluded\Q,\\
\Forall x\Forall y(x\beta.
\end{gather*}
\item
every nonempty subset $A$ of $\on$ has a least element, $\min(A)$.
\item
the class $\pred(\alpha)$ or $\{x\colon x<\alpha\}$
of predecessors of a given ordinal $\alpha$ is a set.%%%%%
\footnote{It follows now that every nonempty \emph{class}
of ordinals has a least element,
but I did not spell this out.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{compactenum}
\item
$\on$ is non-empty, so it has a least element, $0$.
\item
$\on$ has no greatest element, so
every element $\alpha$ of $\on$ has a successor, $\alpha'$,
namely $\min\{x\colon x>\alpha\}$.
\item
Every subset $A$ of $\on$ has an upper bound
and therefore a supremum, $\sup(A)$,
namely
\begin{equation*}
\min\{x\colon\Forall y(y\in A\lto y\leq x)\}.
\end{equation*}
\item
$\on$ contains limits,
namely elements that are neither $0$ nor successors;
the least limit is $\upomega$.
\end{compactenum}
Can we prove that there are more limits than $\upomega$?
As we did from $\upomega$ to $\upomega\sqcup\upomega$,
we recursively define
$x\mapsto\upomega+x$ from the subset $\pred(\upomega)$ of $\on$
to $\on$ itself by
\begin{align*}
\upomega+0&=\upomega,&\upomega+x'&=(\upomega+x)'.
\end{align*}
Then $\sup\{\upomega+x\colon x<\upomega\}$
is the next limit after $\upomega$.
It exists by the \textbf{Replacement Axiom:}
the image of a set under a function is still a set.
For any $\alpha$ in $\on$, we define $x\mapsto\alpha+x$
by \emph{transfinite recursion}:
\begin{compactenum}
\item
$\alpha+0=\alpha$.
\item
$\alpha+\beta'=(\alpha+\beta)'$.
\item
$\lim(\gamma)\lto\alpha+\gamma=\sup\{\alpha+x\colon x<\gamma\}$.
\end{compactenum}
We can do this by the following.
\begin{theorem}[Transfinite Recursion]\label{thm:tr}
If $\alpha\in\on$ and $\bm F$ is an operation on $\on$,
then a unique operation $\bm G$ on $\on$
exists such that
\begin{compactenum}[1)]
\item
$\bm G(0)=\alpha$,
\item
$\Forall x(\bm G(x')=\bm F(\bm G(x)))$,
\item
$\Forall x(\lim(x)\lto\bm G(x)=\sup(\bm G[\pred(x)])$.
\end{compactenum}
\end{theorem}
\begin{proof}
By transfinite induction,
there is at most one such $\bm G$.
For the same reason,
for each $\beta$ in $\on$, there is at most one function $g_{\beta}$
from $\pred(\beta')$ to $\on$ with the properties of $\bm G$,
and also
\begin{equation*}
\delta\leq\gamma\leq\beta\lto g_{\beta}(\delta)=g_{\gamma}(\delta).
\end{equation*}
Again by transfinite induction,
there is at least one such $g_{\beta}$ for each $\beta$;
for we can let
\begin{gather*}
g_0=\{(0,\alpha)\},\\
g_{\beta'}=g_{\beta}\cup\{(\beta',\bm F(g_{\beta}(\beta)))\},\\
\lim(\gamma)\lto g_{\gamma}
=\bigcup\{g_x\colon x<\gamma\}
\cup\{(\gamma,\sup\{g_x(x)\colon x<\gamma\})\}.
\end{gather*}
Then $\bm G$ can be $x\mapsto g_x(x)$.
\end{proof}
By a modification of the proof,
there is $\bm G$ such that
\begin{compactenum}[1)]
\item
$\bm G(0)=\emptyset$,
\item
$\Forall x\bm G(x')=\bm G(x)\cup\{\bm G(x)\}$,
\item
$\Forall x(\lim(x)\lto\bm G(x)=\bm G[\pred(x)])$.
\end{compactenum}
In this case,
\begin{equation*}
\Forall x(\bm G(x)=\bm G[\pred(x)]).
\end{equation*}
Then $\bm G[\on]$ consists of sets that are
\begin{compactitem}
\item
well-ordered by $\in$,
\item
\emph{transitive,}
\end{compactitem}
where a set $A$ is \textbf{transitive} if
\begin{equation*}
\Forall x\Forall y(y\in x\land x\in A\lto y\in A),
\end{equation*}
that is,
\begin{equation*}
\Forall x(x\in A\lto x\included A).
\end{equation*}
One can show that \emph{every} transitive set
that is well-ordered by $\in$ belongs to the range of $\bm G$.
Thus we can define $\on$ as this range.
Now each $\alpha$ in $\on$ is $\pred(\alpha)$;
also,
\begin{equation*}
\alpha'=\alpha\cup\{\alpha\}.
\end{equation*}
Hence
\begin{align*}
0&=\emptyset,&1=0'&=\{0\},&2=1'&=\{0,1\},
\end{align*}
and so on.
Also,
if $A$ is a set of ordinals, then
\begin{equation*}
\sup(A)=\bigcup A=\{x\colon\Exists y(y\in A\land x\in y)\}.
\end{equation*}
\begin{theorem}
For all $\alpha$ and $\beta$ in $\on$,
\begin{equation*}
\alpha+\beta\cong\alpha\sqcup\beta,
\end{equation*}
where $\alpha\sqcup\beta$ is well-ordered by the rule,
\begin{equation*}
(a,b)<(c,d)\liff b1$,
by an analogue of Theorem \ref{thm:a+x}.
\begin{theorem}
For all $\alpha$, $\beta$, and $\gamma$ in $\on$,
\begin{gather*}
\alpha\cdot(\beta\cdot\gamma)=(\alpha\cdot\beta)\cdot\gamma,\\
\alpha\cdot(\beta+\gamma)=\alpha\cdot\beta+\alpha\cdot\gamma.
\end{gather*}
\end{theorem}
\begin{proof}
Induction on $\gamma$ in either case.
\end{proof}
Multiplication is not commutative, because for example
for all $n$ in $\upomega$, if $n>1$, then
\begin{equation*}
n\cdot\upomega=\sup_{x<\upomega}(n\cdot x)=\upomega=\upomega\cdot1
<\upomega\cdot n.
\end{equation*}
\begin{theorem}
For all $\alpha$ and $\beta$ in $\on$,
\begin{equation*}
\alpha\cdot\beta\cong\alpha\times\beta,
\end{equation*}
where $\alpha\times\beta$ is well-ordered by the rule,
\begin{equation*}
(a,b)<(c,d)\liff bn_0>n_1>\dots>n_m,\\
0n_0$.
\chapter{Friday}
The following are now strictly increasing and continuous:
\begin{compactitem}
\item
$x\mapsto\alpha+x$ for all $\alpha$,
\item
$x\mapsto\alpha\cdot x$ when $\alpha>0$,
\item
$x\mapsto\alpha^x$ when $\alpha>1$.
\end{compactitem}
Also
\begin{gather*}
\alpha^0=1,\\
\alpha^1=\alpha^{0'}=\alpha^0\cdot\alpha=1\cdot\alpha=\alpha,\\
\alpha^2=\alpha^{1'}=\alpha^1\cdot\alpha=\alpha\cdot\alpha.
\end{gather*}
\begin{theorem}
For all $\alpha$ in $\on$,
\begin{compactenum}
\item
$0^{\alpha}=0$ if $\alpha>0$.
\item
$1^{\alpha}=1$.
\end{compactenum}
\end{theorem}
\begin{proof}
Induction.
\end{proof}
\begin{theorem}
For all $\alpha$, $\beta$, and $\gamma$ in $\on$,
\begin{gather*}
\alpha^{\beta+\gamma}=\alpha^{\beta}\cdot\alpha^{\gamma},\\
\alpha^{\beta\cdot\gamma}=(\alpha^{\beta})^{\gamma}.
\end{gather*}
\end{theorem}
\begin{proof}
Induction on $\gamma$.
\end{proof}
Yesterday we could assign to each element $\gamma$ of $\upomega^2$
a Cantor normal form $\upomega\cdot n+m$ as follows.
\begin{compactenum}
\item
If $\gamma<\upomega$, we are done.
Suppose now
\begin{equation*}
\upomega\leq\gamma<\upomega^2
=\upomega\cdot\upomega.
\end{equation*}
\item
We let
\begin{equation*}
n=\sup\{x\colon\upomega\cdot x\leq\gamma\},
\end{equation*}
a nonzero element of $\upomega$, and then
\begin{equation*}
\upomega\cdot n\leq\gamma<\upomega\cdot n'=\upomega\cdot n+\upomega.
\end{equation*}
\item
We let
\begin{equation*}
m=\sup\{x\colon\upomega\cdot n+x\leq\gamma\},
\end{equation*}
an element of $\upomega$, and then
\begin{equation*}
\gamma=\upomega\cdot n+m.
\end{equation*}
\end{compactenum}
To find the Cantor normal form of an arbitrary ordinal,
we need to know that it is bounded by a power of $\upomega$.
For this, we use the following.
\begin{theorem}\label{thm:xa}
For all $\alpha$ in $\on$, $x\mapsto x\cdot\alpha$ is increasing.
\end{theorem}
\begin{proof}
Induction on $\alpha$.
\end{proof}
\begin{lemma}\label{lem:a}
If $\alpha\geq2$, then for all $\gamma$ in $\on$,
\begin{equation*}
\gamma\leq\alpha^{\gamma}.
\end{equation*}
\end{lemma}
\begin{proof}
Induction.
\begin{compactenum}
\item
$0<1=\alpha^0$.
\item
Suppose
$\beta<\alpha^{\beta}$ for some $\beta$.
\begin{compactenum}
\item
If $\beta=0$, then $\beta'=1<\alpha=\alpha^{\beta'}$.
\item
If $\beta>0$, then by Theorem \ref{thm:xa},
\begin{equation*}
\beta'=\beta+1\leq\beta+\beta=\beta\cdot2
\leq\beta\cdot\alpha
\leq\alpha^{\beta}\cdot\alpha
=\alpha^{\beta'}.
\end{equation*}
\end{compactenum}
\item
Suppose $\beta$ is a limit and the claim is true when $\gamma<\beta$.
Then
\begin{equation*}
\beta=\sup_{x<\beta}x\leq\sup_{x<\beta}(\alpha^x)=\alpha^{\beta}.\qedhere
\end{equation*}
\end{compactenum}
\end{proof}
We can now find the Cantor normal form
of an arbitrary nonzero $\gamma$ in $\on$ as follows.
More generally, instead of $\upomega$ as a base,
we can use an arbitrary $\alpha$, as long as $\alpha>1$.
\begin{compactenum}
\item
We know from Lemma \ref{lem:a} that
\begin{equation*}
\gamma<\alpha^{\gamma'},
\end{equation*}
and therefore $\gamma'$ is an upper bound
of $\{x\colon\alpha^x\leq\gamma\}$,
which is also nonempty, because it contains $0$.
We let
\begin{equation*}
\beta_1=\sup\{x\colon\alpha^x\leq\gamma\},
\end{equation*}
and then, by Theorem \ref{thm:norm},
\begin{multline*}
\alpha^{\beta_1}
=\alpha^{\sup\{x\colon\alpha^x\leq\gamma\}}\\
=\sup\{\alpha^x\colon\alpha^x\leq\gamma\}
\leq\gamma
<\alpha^{\beta_1{}'}
=\alpha^{\beta_1}\cdot\alpha.
\end{multline*}
\item
We let
\begin{equation*}
\alpha_1=\sup\{x\colon\alpha^{\beta_1}\cdot x\leq\gamma\},
\end{equation*}
a nonzero element of $\alpha$, and then,
again by Theorem \ref{thm:norm},
\begin{multline*}
\alpha^{\beta_1}\cdot \alpha_1
=\alpha^{\beta_1}\cdot\sup\{x\colon\alpha^{\beta_1}\cdot x\leq\gamma\}\\
=\sup\{\alpha^{\beta_1}\cdot x\colon\alpha^{\beta_1}\cdot x\leq\gamma\}
\leq\gamma\\
<\alpha^{\beta_1}\cdot \alpha_1{}'
=\alpha^{\beta_1}\cdot \alpha_1+\alpha^{\beta_1}.
\end{multline*}
\item
We let
\begin{equation*}
\delta=\sup\{x\colon\alpha^{\beta_1}\cdot \alpha_1+x\leq\gamma\},
\end{equation*}
an element of $\alpha^{\beta_1}$,
and then,
again by Theorem \ref{thm:norm},
\begin{multline*}
\alpha^{\beta_1}\cdot \alpha_1+\delta
=\alpha^{\beta_1}\cdot \alpha_1+\sup\{x\colon\alpha^{\beta_1}\cdot \alpha_1+x\leq\gamma\}\\
=\sup\{\alpha^{\beta_1}\cdot \alpha_1+x\colon\alpha^{\beta_1}\cdot \alpha_1+x\leq\gamma\}
\leq\gamma\\
<\alpha^{\beta_1}\cdot \alpha_1+\delta'
=(\alpha^{\beta_1}\cdot \alpha_1+\delta)'.
\end{multline*}
Thus
\begin{align*}
\gamma&=\alpha^{\beta_1}\cdot \alpha_1+\delta,&
\delta&<\alpha^{\beta_1}.
\end{align*}
\end{compactenum}
If $\delta>0$,
then we apply to $\delta$ the same process that we did to $\gamma$.
Ultimately we obtain, for some $m$ in $\upomega$, the expansion
\begin{equation}\label{eqn:cnf}
\gamma=\alpha^{\beta_1}\cdot\alpha_1+\alpha^{\beta_2}\cdot\alpha_2
+\dots+\alpha^{\beta_m}\cdot\alpha_m,
\end{equation}
where
\begin{align*}
\beta_1&>\beta_2>\dots>\beta_m,&
\bigwedge_{1\leq j\leq m}\alpha>\alpha_j&>0.
\end{align*}
The expansion has finitely many terms, since $\on$ is well-ordered,
so the strictly decreasing sequence of $\beta_j$ must end.
If $m=0$, the expansion is $0$.
We have shown
\begin{align*}
\alpha+\beta&\cong\alpha\sqcup\beta,&
\alpha\cdot\beta&\cong\alpha\times\beta.
\end{align*}
Now let
\begin{equation*}
{}^{\beta}\alpha=\{\text{functions from $\beta$ to $\alpha$}\}.
\end{equation*}
Then we have established a function $f\mapsto f_{\gamma}$
from $\alpha^{\beta}$ to ${}^{\beta}\alpha$, where,
if $\gamma$ is as in \eqref{eqn:cnf}, with $\beta>\beta_1$,
then
\begin{equation*}
f_{\gamma}(x)=
\begin{cases}
\alpha_j,&\text{ if }x=\beta_j,\\
0,&\text{ otherwise}.
\end{cases}
\end{equation*}
The function is injective,
briefly because all of the functions mentioned at the beginning of today
are strictly increasing.
The range of the function is
\begin{equation*}
\bigl\{f\in{}^{\beta}\alpha\colon
\{x\in\beta\colon f(x)\neq0\}\text{ is finite}\bigr\},
\end{equation*}
and we can well-order this so that it is isomorphic to $\alpha^{\beta}$.
\chapter{Saturday}
We develop some rules for computing with Cantor normal forms.
We have already effectively proved the following.
\begin{lemma}\label{lem:sol}
If $\alpha\leq\beta$, then the equation
\begin{equation*}
\alpha+x=\beta
\end{equation*}
is solved by $\sup\{x\colon\alpha+x\leq\beta\}$.
\end{lemma}
\begin{proof}
Letting
$\gamma=\sup\{x\colon\alpha+x\leq\beta\}$,
We compute
\begin{multline*}
\alpha+\gamma
=\alpha+\sup\{x\colon\alpha+x\leq\beta\}\\
=\sup\{\alpha+x\colon\alpha+x\leq\beta\}
\leq\beta
<\alpha+\gamma'
=(\alpha+\gamma)',
\end{multline*}
so $\alpha+\gamma=\beta$.
\end{proof}
\begin{lemma}\label{lem:n+wa}
If $\alpha>0$ and $n<\upomega$,
then $n+\upomega^{\alpha}=\upomega^{\alpha}$.
\end{lemma}
\begin{proof}
\begin{asparaenum}
\item
We know the claim is true when $\alpha=1$.
\item
Hence, when $\beta\geq1$,
\begin{equation*}
n+\upomega^{\beta'}\leq\upomega^{\beta}+\upomega^{\beta+1}
=\upomega^{\beta}\cdot(n+\upomega)
=\upomega^{\beta'}.
\end{equation*}
(This did not require an inductive hypothesis.)
\item
If $\beta$ is a limit and the claim is true when $\alpha<\beta$,
then (in the standard way)
\begin{equation*}
n+\upomega^{\beta}
=n+\sup_{x<\beta}\upomega^x
=\sup_{x<\beta}(n+\upomega^x)
=\sup_{x<\beta}\upomega^x
=\upomega^{\beta}.\qedhere
\end{equation*}
\end{asparaenum}
\end{proof}
\begin{theorem}\label{thm:a+b=b}
If $\alpha<\beta$, then
\begin{equation*}
\upomega^{\alpha}+\upomega^{\beta}=\upomega^{\beta}.
\end{equation*}
\end{theorem}
\begin{proof}
By Lemma \ref{lem:sol}, for some $\gamma$,
$\beta=\alpha+\gamma$,
and then
\begin{equation*}
\upomega^{\alpha}+\upomega^{\beta}
=\upomega^{\alpha}+\upomega^{\alpha+\gamma}
=\upomega^{\alpha}\cdot(1+\upomega^{\gamma})
=\upomega^{\alpha}\cdot\upomega^{\gamma}=\upomega^{\beta}.
\qedhere
\end{equation*}
\end{proof}
We have seen that, every $\alpha$ has a Cantor normal form,
given by
\begin{equation*}
\alpha=\upomega^{\alpha_1}\cdot a_1
+\dots+\upomega^{\alpha_m}\cdot a_m,
\end{equation*}
where
\begin{equation*}
\alpha_1>\dots>\alpha_m,\qquad
\bigwedge_{1\leq j\leq m}\upomega>a_j>0.
\end{equation*}
If $\alpha>0$, so that $m\geq1$, let us say
\begin{align*}
\deg(\alpha)&=\alpha_1,&\lc(\alpha)&=a_1,
\end{align*}
the \textbf{degree} and \textbf{leading coefficient} of $\alpha$.
Thus, for every positive ordinal $\alpha$, for some ordinal $\beta$,
\begin{align*}
\alpha&=\upomega^{\deg(\alpha)}\cdot\lc(\alpha)+\beta,&
\deg(\beta)<\deg(\alpha).
\end{align*}
By Theorem \ref{thm:a+b=b},
if $\deg(\alpha)<\deg(\beta)$, then
\begin{equation*}
\alpha+\beta=\beta.
\end{equation*}
Now we can compute sums of Cantor normal forms. Thus for example
\begin{multline*}
(\upomega^{\upomega+5}\cdot2+\upomega^3\cdot5+1)
+(\upomega^{\omega}+\upomega^2\cdot6)\\
=\upomega^{\upomega+5}\cdot2+\upomega^{\omega}+\upomega^2\cdot6.
\end{multline*}
\begin{lemma}\label{lem:nwa}
If $\alpha>0$ and $n\in\N$,
then $n\cdot\upomega^{\alpha}=\upomega^{\alpha}$.
\end{lemma}
\begin{proof}
For some $\beta$, $\alpha=1+\beta$.
We know $n\cdot\upomega=\upomega$, and then
\begin{equation*}
n\cdot\upomega^{\alpha}
=n\cdot\upomega\cdot\upomega^{\beta}
=\upomega\cdot\upomega^{\beta}
=\upomega^{\alpha}.\qedhere
\end{equation*}
\end{proof}
\begin{theorem}
Let $\beta>\deg(\alpha)$ and $n\in\N$.
\begin{compactenum}
\item
If $k\in\N$, then
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot k
=\upomega^{\beta}\cdot n\cdot k+\alpha.
\end{equation*}
\item
If $\gamma\geq1$, then
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)\cdot\upomega^{\gamma}
=\upomega^{\beta+\gamma}.
\end{equation*}
\end{compactenum}
\end{theorem}
\begin{proof}
\begin{asparaenum}
\item
We use (finite) induction.
\begin{compactenum}
\item
The claim is trivially true when $k=1$.
\item
If it is true when $k=\ell$, then
\begin{align*}
(\upomega^{\beta}\cdot n+\alpha)\cdot(\ell+1)
&=(\upomega^{\beta}\cdot n+\alpha)\cdot\ell
+\upomega^{\beta}\cdot n+\alpha\\
&=\upomega^{\beta}\cdot n\cdot\ell+\alpha
+\upomega^{\beta}\cdot n+\alpha\\
&=\upomega^{\beta}\cdot n\cdot\ell
+\upomega^{\beta}\cdot n+\alpha\\
&=\upomega^{\beta}\cdot n\cdot(\ell+1)+\alpha.
\end{align*}
\end{compactenum}
\item
As in the proof of Lemma \ref{lem:nwa},
we need only prove the claim when $\gamma=1$.
We have
\begin{multline*}
\upomega^{\beta+1}
=\sup_{x<\upomega}(\upomega^{\beta}\cdot x)
\leq\sup_{x<\upomega}((\upomega^{\beta}\cdot n+\alpha)\cdot x)\\
\leq\sup_{x<\upomega}(\upomega^{\beta}\cdot(n+1)\cdot x)
= \upomega^{\beta+1},
\end{multline*}
while also $\sup_{x<\upomega}((\upomega^{\beta}\cdot n+\alpha)\cdot x)
=(\upomega^{\beta}\cdot n+\alpha)\cdot\upomega$.\qedhere
\end{asparaenum}
\end{proof}
Now we can compute products of Cantor normal forms;
for, we have (under the obvious conditions)
\begin{multline*}
(\upomega^{\beta}\cdot n+\alpha)\cdot(\upomega^{\gamma_1}\cdot d_1+\dots
+\upomega^{\gamma_m}\cdot d_m+d_{m+1})\\
=\upomega^{\beta+\gamma_1}\cdot d_1+\dots
+\upomega^{\beta+\gamma_m}\cdot d_m
+\upomega^{\beta}\cdot n\cdot d_{m+1}
+\alpha,
\end{multline*}
the last two terms being deleted when $d_{m+1}=0$.
\begin{theorem}
If $k\in\N$, $n\in\upomega$, and $\upomega\leq\alpha$, then
\begin{gather*}
k^{\upomega^{n+1}}
=k^{\upomega^{1+n}}
=(k^{\upomega})^{\upomega^n}
=\upomega^{\upomega^n},\\
k^{\upomega^{\alpha}}
=k^{\upomega^{1+\alpha}}
=\upomega^{\upomega^{\alpha}}.
\end{gather*}
\end{theorem}
\begin{lemma}
$\alpha$ is a non-successor if and only if, for some $\beta$,
$\alpha=\upomega\cdot\beta$.
\end{lemma}
\begin{proof}
$0$ is not a successor and is $\upomega\cdot0$.
Suppose $\alpha>0$.
In Cantor normal form,
\begin{equation*}
\alpha=\upomega^{\alpha_1}\cdot a_1+\dots+\upomega^{\alpha_m}\cdot a_m+a_{m+1},
\end{equation*}
where $\alpha_m>0$ and $a_{m+1}\in\upomega$.
By Lemma \ref{lem:sol}, for some $\beta_i$,
\begin{equation*}
\alpha
=\upomega\cdot(\upomega^{\beta_1}\cdot a_1+\dots+\upomega^{\beta_m}\cdot a_m)
+a_{m+1},
\end{equation*}
which is a limit if and only if $a_{m+1}>0$.
\end{proof}
\begin{theorem}
If $\gamma$ is a limit, then
\begin{equation*}
(\upomega^{\beta}\cdot n+\alpha)^{\gamma}=\upomega^{\beta\cdot\gamma}.
\end{equation*}
\end{theorem}
We end with \textbf{cardinalities.}
If there is, from $A$ to $B$,
\begin{compactitem}
\item
an injection, $A\preccurlyeq B$;
\item
a bijection, $A\approx B$;
\end{compactitem}
is written.
\begin{theorem}[Schr\"oder--Bernstein]
If
\begin{align*}
f&\colon A\xrightarrow{\preccurlyeq}B,&
g&\colon B\xrightarrow{\preccurlyeq}A,
\end{align*}
then
\begin{equation*}
A\approx B.
\end{equation*}
\end{theorem}
\begin{proof}[Zermelo's proof.]
Under the hypothesis,
\begin{align*}
A&\approx g\circ f[A],&
g\circ f[A]&\included g[B]\included A,&
g[B]&\approx B.
\end{align*}
Thus, assuming
\begin{align*}
A&\included B\included C,&
f&\colon C\xrightarrow{\approx}A,
\end{align*}
we find a bijection $g$ from $B$ onto $A$.
Let
\begin{equation*}
\mathscr E=\{X\included B\colon(B\setminus A)\cup f[X]\included X\}.
\end{equation*}
Then
\begin{align*}
B&\in\mathscr E,&
X\in\mathscr E&\lto(B\setminus A)\cup F[X]\in\mathscr E.
\end{align*}
Let
\begin{equation*}
D=\bigcap\mathscr E=\{t\colon\Forall X(X\in\mathscr E\lto t\in X)\}.
\end{equation*}
Then
\begin{align*}
D&\in\mathscr E,&D&=(A\setminus B)\cup f[D].\qedhere
\end{align*}
\end{proof}
\begin{theorem}
If $\alpha\cdot\beta\neq0$ and $\max(\alpha,\beta)\geq\upomega$, then
\begin{align*}
\alpha+\beta&\approx\max(\alpha,\beta),&
\alpha\cdot\beta&\approx\max(\alpha,\beta).
\end{align*}
\end{theorem}
\begin{proof}
Since
\begin{gather*}
\alpha+\beta\approx\alpha\sqcup\beta\approx\beta\sqcup\alpha\approx\beta+\alpha,\\
\alpha\cdot\beta\approx\alpha\times\beta\approx\beta\times\alpha\approx\beta\cdot\alpha,
\end{gather*}
and also by Lemmas \ref{lem:n+wa} and \ref{lem:nwa}
we may assume $\upomega\leq\alpha\leq\beta$. Then
\begin{equation*}
\beta
\leq\alpha+\beta
\leq\beta+\beta
=\beta\cdot2
<\beta\cdot\alpha
\leq\beta\cdot\beta
=\beta^2.
\end{equation*}
Thus it is enough to show
\begin{equation*}
\beta\approx\beta^2.
\end{equation*}
This is true when $\beta=\upomega$,
since for example if
\begin{equation*}
t_n=\frac{n\cdot(n+1)}2,
\end{equation*}
then $(k,m)\mapsto t_{k+m}+k$ is a bijection from $\upomega\times\upomega$
to $\upomega$.
Then by finite induction, for all $n$ in $\N$,
\begin{equation*}
\upomega^n\approx\upomega.
\end{equation*}
Letting $\gamma=\deg(\beta)$,
we know for some $\delta$, where $\deg(\delta)<\gamma$,
\begin{equation*}
\beta=\upomega^{\gamma}\cdot\lc(\beta)+\delta
\approx\delta+\lc(\beta)\cdot\upomega^{\gamma}
=\upomega^{\gamma}.
\end{equation*}
We may assume now $\gamma\geq\upomega$. Since for any $\alpha$ and $\beta$
\begin{equation}\label{eqn:a^b}
\alpha^{\beta}\approx\{f\in{}^{\beta}\alpha\colon\card{\supp(f)}<\upomega\},
\end{equation}
where by definition
\begin{equation*}
\card{\supp(f)}<\upomega
\liff
\{x\in\beta\colon f(x)\neq0\}\text{ is finite,}
\end{equation*}
we have generally
\begin{equation*}
\alpha\approx\gamma\land\beta\approx\delta
\lto\alpha^{\beta}\approx\gamma^{\delta}.
\end{equation*}
In our case,
\begin{equation*}
\beta^2\approx(\upomega^{\gamma})^2
\approx\upomega^{\upomega^{\deg(\gamma)}\cdot2}
\approx\upomega^{2\cdot\upomega^{\deg(\gamma)}}
=\upomega^{\upomega^{\deg(\gamma)}}
\approx\upomega^{\gamma}
\approx\beta.\qedhere
\end{equation*}
\end{proof}
\begin{corollary}
For all infinite $\alpha$, for all $n$ in $\N$,
\begin{equation*}
\alpha^n\approx\alpha.
\end{equation*}
\end{corollary}
\begin{theorem}
If $\min(\alpha,\beta)\leq2$ and $\upomega\leq\max(\alpha,\beta)$, then
\begin{equation*}
\alpha^{\beta}\approx\max(\alpha,\beta).
\end{equation*}
\end{theorem}
\begin{proof}
From Lemma \ref{lem:a} we know
\begin{equation*}
\max(\alpha,\beta)\leq\alpha^{\beta}.
\end{equation*}
From \eqref{eqn:a^b},
an element of $\alpha^{\beta}$ uniquely determines
an ordered pair consisting of, for some $n$ in $\upomega$,
\begin{compactitem}
\item
a subset $\{\beta_i\colon i< n\}$ of $\beta$, where
$\beta_0<\dots<\beta_{n-1}$;
\item
a function $(\alpha_i\colon i