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\refoot{Infinitesimal Analysis}
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\begin{document}
\title{Infinitesimal analysis}
\subtitle{Also known as\\
nonstandard analysis}
\author{David Pierce}
\date{September 1, 2015}
\publishers{Matematik B\"ol\"um\"u\\
Mimar Sinan G\"uzel Sanatlar \"Universitesi\\
\url{mat.msgsu.edu.tr/~dpierce/}}
\maketitle\thispagestyle{empty}
\begin{abstract}
These notes are based on my course,
officially called
Rudiments of Nonstandard Analysis,
given
August 10--22, 2015,
at the Nesin Matematik K\"oy\"u,
Kayser Da\u g\i\ Mevkii,
\c Sirince, Sel\c cuk, \.Izmir, Turkey.
I gave similar courses in earlier years.
Before the course of 2009,
I typed up notes on everything that I might possibly want to talk about,
if I had time.
I revised the notes before the 2010 course.
In 2014, after each lecture,
I typed up notes of what I actually \emph{had} talked about;
later I tried to polish them.
I used these notes in 2015,
but did not follow them carefully.
I created the present notes
as I did the 2014 notes,
though with (perhaps even more)
embellishments, omissions, and rearrangements.
\end{abstract}
\tableofcontents
\mysection{Preliminary discussion}
Throughout these notes,
the symbols $\N$ and $\upomega$
represent what I would call
the \emph{number-theorist's natural numbers}
and the \emph{set-theorist's natural numbers,}
respectively:
%More precisely,
\begin{align*}
\N&=\{1,2,3,\dots\},&
\upomega&=\{0,1,2,\dots\}.
\end{align*}
If the proof of a theorem is not supplied,
it is an exercise for the reader.
High-school algebra
%(as opposed to ``abstract algebra'')
is largely a study
of the properties of arbitrary \emph{ordered fields,}
such as the ordered field $\Q$ of \emph{rational numbers}
or the ordered field $\R$ of \emph{real numbers.}
Calculus is a study of $\R$ as a \emph{complete} ordered field.
Here \textbf{completeness} is the property whereby every nonempty set of numbers
with an upper bound has a \emph{least upper bound,} or \textbf{supremum.}
Suppose $f$ is a function from $\R$ to itself
(that is, $f\colon\R\to\R$),
and $a$ and $L$ are in $\R$.
We say $f$ has the \textbf{limit} $L$ at $a$,
and we may write either of
\begin{align}\label{eqn:lim}
\lim_{x\to a}f(x)&=L,&\lim_af&=L,
\end{align}
provided that $f(x)$ is close to $L$
whenever $x$ is close to, but not equal to, $a$.
This is vague.
By the ``standard'' definition,
\eqref{eqn:lim} means
we can make $f(x)$ \emph{arbitrarily} close to $L$
by making $x$ \emph{sufficiently} close to
(though not equal to) $a$.
Here ``arbitrarily close'' means ``within $\epsilon$,
for any positive $\epsilon$ given to us.''
Then ``sufficiently close'' means ``within $\delta$,
where $\delta$ is positive and depends on $\epsilon$.''
More precisely then,
the ``standard'' definition of \eqref{eqn:lim} is that,
for all positive $\epsilon$ in $\R$,%%%%%
\footnote{I do not say ``for all $\epsilon>0$,''
because I prefer to treat the expression $\epsilon>0$
always as a \emph{sentence} (``$\epsilon$ is greater than $0$''),
not as a noun phrase (``$\epsilon$ that are greater than $0$''
or ``$\epsilon$ greater than $0$'').}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for some positive $\delta$ in $\R$,%%%%%
\footnote{Alternatively, one may say,
``there exists a positive $\delta$ in $\R$ such that\dots''}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
for all $x$ in $\R$,
if $0<\abs{x-a}<\delta$, then $\abs{f(x)-L}<\epsilon$.
Even more symbolically, the definition is,
\begin{multline*}
\Forall{\epsilon}\Bigl(\epsilon>0\lto\Exists{\delta}\bigl(\delta>0\land\Forall x{}\\
\bigl(0<\abs{x-a}\land\abs{x-a}<\delta\lto\abs{f(x)-L}<\epsilon\bigr)\Bigr),
\end{multline*}
which can also be written as
\begin{multline}\label{eqn:e-d}
\Forall{\epsilon}\Exists{\delta}\Forall x\Bigl(\epsilon>0\lto\delta>0\land{}\\
\bigl(0<\abs{x-a}\land\abs{x-a}<\delta\lto\abs{f(x)-L}<\epsilon\bigr)\Bigr).
\end{multline}
These two expressions are equivalent sentences of \emph{first-order logic,}
in which the letters $\epsilon$, $\delta$, and $x$
are \emph{bound variables.}
One could use other letters as bound variables instead;
but $\epsilon$ and $\delta$ are almost always the letters used,
as an aid to the memory,
because the sentences are so logically complicated.
It is complicated to have \emph{three} blocks of quantifiers,
as shown at the head of \eqref{eqn:e-d};
most definitions in algebra require at most two,
as in the axiom ensuring existence of multiplicative inverses in fields,
\begin{equation*}
\Forall x\Exists y(x\neq0\lto xy=1).
\end{equation*}
The axiom of associativity of multiplication,
\begin{equation*}
\Forall x\Forall y\Forall zx(yz)=(xy)z,
\end{equation*}
has three quantifiers,
but they form one block:
they are all universal.
\textbf{Infinitesimal} or \textbf{``nonstandard'' analysis} provides a way to define limits
without using any such letters as $\epsilon$ and $\delta$.
By the ``nonstandard'' definition, $\lim_af=L$,
provided that, for all $x$,
if $x$ is \emph{infinitesimally close} to $a$,
without being equal to $a$,
then $f(x)$ is infinitesimally close to $L$.
Symbolically, the definition is
\begin{equation}\label{eqn:non}
\Forall x(x\simeq a\land x\neq a\lto f(x)\simeq L).
\end{equation}
This is evidently much simpler than \eqref{eqn:e-d},
although, as we shall see below,
there is a quantifier in the precise definition of
$a\simeq x$ and $f(x)\simeq L$.
\mysection{Non-Archimedean fields}
The adjective ``infinitesimal''
is obtained from the adjective ``infinite''
by adding a Latin ending,
which corresponds to the English ordinal ending \mbox{``-th''}
seen in ``fourth,'' ``fifth,'' ``sixth,'' and so on.
The English ordinals are used also to denote fractional parts:
one fourth, or the fourth part,
is one of four equal parts, or $1/4$.
Then an infinitesimal part is an ``infinitieth'' part, or $1/\infty$.
To be precise then, a number $a$ is \textbf{infinitesimal}
if, for all $n$ in $\N$,
\begin{equation*}
\abs a<\frac1n.
\end{equation*}
A number $b$ is \textbf{infinite} if,
for all $n$ in $\N$,
\begin{equation*}
n<\abs b.
\end{equation*}
Then $b$ is infinite if and only if $b\neq0$ and $1/b$ is infinitesimal.
The ``nonstandard'' definition \eqref{eqn:non}
can be spelled out as
\begin{multline*}
\Forall x\left(
\Forall n\Bigl(n\in\N\lto\abs{x-a}<\frac1n\Bigr)\land x\neq a\lto\right.\\
\left.\Forall n\Bigl(n\in\N\lto\abs{f(x)-L}<\frac1n\Bigr)\right)
\end{multline*}
and hence as
\begin{multline}\label{eqn:non-long}
\Forall x\Exists m\Forall n\left(\Bigl(m\in\N\land\abs{x-a}>\frac1m\Bigr)\lor{}\right.\\
\left.\Bigl(x\neq a\land\abs{f(x)-L}<\frac1n\Bigr)\right).
\end{multline}
Thus there are still three blocks of quantifiers.
But again, two of them can be hidden,
as in \eqref{eqn:non};
this cannot be done in \eqref{eqn:e-d}.
This makes the ``nonstandard'' definition of limit
simpler than the ``standard.''
However, there is a new complication
with the ``nonstandard'' definition.
The only infinitesimal in $\R$ is $0$,
and $\R$ contains no infinite numbers.
This follows from the theorem below.
Thus if the variable $x$ must always represent a real number,
then \eqref{eqn:non-long} is always true.
So we shall have to allow $x$ to be a ``nonstandard'' number,
in a sense that we shall develop.
An ordered field is \textbf{Archimedean} if,
for every element $c$ of the field,
for some $n$ in $\N$,
\begin{equation*}
\abs c\leq n.
\end{equation*}
That is, an ordered field is Archimedean \emph{precisely}
when none of its elements is infinite.
\begin{theorem}\label{thm:Arch}
$\R$ is Archimedean.
\end{theorem}
\begin{proof}
If $\R$ had an infinite element,
then it would have a positive infinite element,
and this would be an upper bound of $\N$.
Then $\N$ would have a supremum, say $d$.
In this case, $d-1$ would not be an upper bound of $\N$,
so $d-10\iff\frac ab>0,
\end{equation*}
where $f\in\R[X]$ and $g\in\R[X]\setminus\{0\}$,
and $f$ and $g$ have leading coefficients $a$ and $b$ respectively.
In the ordered field,
$X$ is positive and infinite.
\end{theorem}
\mysection{Ultrafilters}
We shall denote by
\begin{equation*}
\R^{\upomega}
\end{equation*}
the set of functions from $\upomega$ to $R$,
that is, the set of real-valued sequences on $\upomega$.
The same such sequence can be denoted by any of the expressions
\begin{align*}
&(a_0,a_1,a_2,\dots),&&(a_k\colon k\in\upomega),&&a.
\end{align*}
These sequences can be added, subtracted, and multiplied term by term:
\begin{align*}
a\pm b&=(a_k\pm b_k\colon k\in\upomega),&
a\cdot b=(a_k\cdot b_k\colon k\in\upomega).
\end{align*}
Then $\R^{\upomega}$ is a ring,
like $\R[X]$.
For each $n$ in $\upomega$,
there is an element $\chi^n$ of $\R^{\upomega}$, where
\begin{equation*}
\chi^n{}_m=
\begin{cases}
1,&\text{ if }n=m,\\
0,&\text{ if }n\neq m.
\end{cases}
\end{equation*}
Then the set $\{\chi^n\colon n\in\upomega\}$
generates a proper ideal of $\R^{\upomega}$.
Let this ideal be included in a maximal ideal $M$:
this exists by the \emph{Maximal Ideal Theorem}
(see page \pageref{MITh}).
We shall do infinitesimal analysis in the field
\begin{equation*}
\R^{\upomega}/M.
\end{equation*}
If $a\in\R^{\upomega}$, we define
\begin{equation*}
\supp(a)=\{k\in\upomega\colon a_k\neq0\};
\end{equation*}
this is the \textbf{support} of $a$.
Thus $\supp\colon\R^{\upomega}\to\pow{\upomega}$.
Like any function, the support function induces two additional functions,
as follows.
If $A\included\R^{\upomega}$
and $B\included\pow{\upomega}$,
we have
\begin{gather*}
\supp[A]=\{\supp(x)\colon x\in A\},\\
\supp\inv(B)=\{x\in\R^{\upomega}\colon\supp(x)\in B\}.
\end{gather*}
Then
\begin{align*}
A&\included\supp\inv(\supp[A]),&
B&=\supp[\supp\inv(B)].
\end{align*}
The ideal of $\R^{\upomega}$ generated by $\{\chi^n\colon n\in\upomega\}$
consists of the functions from $\upomega$ to $\R$
having finite support;
also, if this ideal is $I$,
then $\supp[I]$ is the set of all finite subsets of $\upomega$.
\begin{theorem}\label{thm:id}
A subset $I$ of $\R^{\upomega}$ is an ideal of $\R^{\upomega}$
if and only if
\begin{equation}\label{eqn:I}
\supp\inv(\supp[I])=I
%\{x\in\R^{\upomega}\colon\supp x\in\Supp I\}=I
\end{equation}
and,
for all $X$ and $Y$ in $\pow{\upomega}$,
\begin{compactenum}[(1)]
\item
$\emptyset\in\supp[I]$,
\item
$X,Y\in\supp[I]\implies X\cup Y\in\supp[I]$,
\item
$X\included Y\And Y\in\supp[I]\implies X\in\supp[I]$.
\end{compactenum}
If $I$ and $J$ are distinct ideals of $R^{\upomega}$,
then
\begin{equation*}
\supp[I]\neq\supp[J].
\end{equation*}
A subset $I$ of $\R^{\upomega}$ is a maximal ideal of $\R^{\upomega}$
if and only if \eqref{eqn:I} and,
for all $X$ and $Y$ in $\pow{\upomega}$,
\begin{compactenum}[(1)]
\item
$\emptyset\in\supp[I]$ and $\upomega\notin\supp[I]$,
\item
$X,Y\in\supp[I]\implies X\cup Y\in\supp[I]$,
\item
$X\notin\supp[I]\implies\upomega\setminus X\in\supp[I]$.
\end{compactenum}
A maximal ideal $M$ of $\R^{\upomega}$ is nonprincipal
if and only if $\supp[M]$
contains every finite subset of $\upomega$.
\end{theorem}
It may be observed that $\pow{\upomega}$ is a ring
when the \textbf{sum} of two elements is defined as their symmetric difference,
and their \textbf{product} as their intersection:
\begin{align*}
X+Y&=X\mathrel{\triangle}Y=(X\setminus Y)\cup(Y\setminus X),\\
X\cdot Y&=X\cap Y.
\end{align*}
In this case
\begin{align*}
0&=\emptyset,&
1&=\upomega,&
-X&=X.
\end{align*}
The ring $\pow{\upomega}$ is then a \textbf{Boolean ring}
because it satisfies
\begin{equation}\label{eqn:x^2=x}
\Forall xx^2=x.
\end{equation}
From this axiom (along with the ring axioms), it follows that
\begin{equation}\label{eqn:2x=0}
\Forall x2x=0,
\end{equation}
or $\Forall x-x=x$.
Indeed, from \eqref{eqn:x^2=x} we have
$2x=(2x)^2=4x^2=4x$,
and then \eqref{eqn:2x=0}.
Now part of the last theorem can be formulated as follows.
\begin{theorem}
Let $I\included\pow{\upomega}$.
\begin{compactenum}
\item
$I$ is an ideal of $\pow{\upomega}$
if and only if $\supp\inv[I]$ is an ideal of $\R^{\upomega}$.
\item
$I$ is a maximal ideal of $\pow{\upomega}$
if and only if $\supp\inv[I]$ is a maximal ideal of $\R^{\upomega}$.
\end{compactenum}
\end{theorem}
But it will be more useful to think of ideals of $\pow{\upomega}$
in the terms of Theorem \ref{thm:id},
rather than in the usual terms of ring theory.
In fact, it will be useful to replace the notion of an ideal
with a ``dual'' notion, as follows.
A subset $F$ of $\pow{\upomega}$ is a
\textbf{filter of} $\pow{\upomega}$ and a
\textbf{filter on} $\upomega$
if for all $X$ and $Y$ in $\pow{\upomega}$,
\begin{compactenum}[(1)]
\item
$\upomega\in F$,
\item
$X,Y\in F\implies X\cap Y\in F$,
\item
$X\in F\And X\included Y\implies Y\in F$.
\end{compactenum}
Thus $I$ is an ideal of $\pow{\upomega}$
if and only if $\{\upomega\setminus X\colon X\in I\}$
is a filter on $\upomega$.
A subset $\mathscr U$ of $\pow{\upomega}$ is an \textbf{ultrafilter on} $\upomega$
if for all $X$ and $Y$ in $\pow{\upomega}$,
\begin{compactenum}[(1)]
\item
$\upomega\in\mathscr U$ and $\emptyset\notin\mathscr U$,
\item
$X,Y\in\mathscr U\implies X\cap Y\in\mathscr U$,
\item
$X\notin\mathscr U\implies\upomega\setminus X\in\mathscr U$.
\end{compactenum}
Thus $M$ is a maximal ideal of $\pow{\upomega}$
if and only if $\{\upomega\setminus X\colon X\in M\}$
is an ultrafilter on $\upomega$;
moreover, in this case,
\begin{equation*}
\{\upomega\setminus X\colon X\in M\}=\pow{\upomega}\setminus M.
\end{equation*}
Suppose $\mathscr U$ is a \emph{nonprincipal}
ultrafilter on $\upomega$.
By Theorem \ref{thm:id},
$\mathscr U$ must contain every \textbf{cofinite} subset of $\upomega$,
that is, every subset whose complement is finite.
We may think of the elements of $\mathscr U$ as \textbf{large,}
and all other subsets of $\upomega$ as \textbf{small.}
Thus:
\begin{compactenum}[(1)]
\item
every set is large or small, but not both;
\item
the small sets are precisely the complements of the large sets;
\item
all finite sets are small, and so their complements are large;
\item
the union of two finite sets is small,
and the intersection of two large sets is large.
\end{compactenum}
However, some infinite sets will be small.
If $\upomega$ is a \emph{disjoint} union $A\sqcup B$,
then exactly one of $A$ and $B$ is large.
Hence if $\upomega=A_0\sqcup\dots\sqcup A_{n-1}$,
%and $A_i\cap A_j=\emptyset$ whenever $i1\land x>2\land x>3\land\cdots),&
&\Exists x\bigwedge_{n\in\N}x>n
\end{align*}
is true in $\R^{\upomega}/\mathscr U$, but false in $\R$.
Being infinite,
in the sense of involving the conjunction of infinitely many formulas,
this sentence is not first order.
\item
In proving Theorem \ref{thm:Arch} (that $\R$ has no infinite elements),
we showed that an ordered field with infinite elements cannot be complete.
In particular, the sentence
\begin{multline*}
\Forall X\Forall y\Exists z\Forall v\Exists u\\
\biggl(
y\in X\lto
\Bigl((v\in X\lto v\leq z)\land
\bigl((u\in X\land u>v)\lor z\leq v\bigr)\Bigr)\biggr),
\end{multline*}
%(``For all $X$, if $X$ has an element $y$,
%then for some $z$,
%$z$ is an upper bound of $X$,
%that is, $z$ is at least as great as every element $v$ of $X$,
%and moreover,
%every $v$ either is not an upper bound of $X$,
%because it is less than some element $u$ of $X$,
%or else $v$ is at least as great as $z$''),
which expresses completeness, is true in $\R$,
but false in $\R^{\upomega}/\mathscr U$.
The sentence is not first order,
because the variable $X$ ranges over sets as such,
not individuals.
\end{asparaenum}
We may denote the ordered field $\R^{\upomega}/\mathscr U$ by
\begin{equation*}
\stR.
\end{equation*}
We now establish the equivalence of
the standard and nonstandard definitions of limit.
\begin{theorem}\label{thm:f}
Suppose $f$ is a function from $\R$ to itself, and $a$ and $L$ are in $\R$.
There is a well-defined function $\st f$ from $\stR$ to itself
given by
\begin{equation*}
\st f[x]=[f(x_k)\colon k\in\upomega].
\end{equation*}
Then $\lim_af=L$ if and only if, for all $x$ in $\R^{\upomega}$,
\begin{equation*}
[x]\simeq a\And [x]\neq a\implies \st f[x]\simeq L.
\end{equation*}
\end{theorem}
\begin{proof}
For all $a$ and $b$ in $\R^{\upomega}$, we have
\begin{equation*}
\{k\in\upomega\colon a_k=b_k\}\included\{k\in\upomega\colon f(a_k)=f(b_k)\}.
\end{equation*}
If $[a]=[b]$, then the former set is large,
and therefore the latter set is large, which means
\begin{equation*}
[f(a_k)\colon k\in\upomega]=[f(b_k)\colon k\in\upomega].
\end{equation*}
Thus $\st f$ is well defined.
Suppose $\lim_af=L$, and $[x]\simeq a$, but $[x]\neq a$.
For all $n$ in $\N$,
for some positive $\delta$ in $\R$,
for all $k$ in $\upomega$,
if $0<\abs{x_k-a}<\delta$, then $\abs{f(x_k)-L}<1/n$.
Thus
\begin{equation}\label{eqn:xafL}
\{k\in\upomega\colon0<\abs{x_k-a}<\delta\}
\included
\{k\in\upomega\colon\abs{f(x_k)-L}<1/n\}.
\end{equation}
The former set is
\begin{equation*}
\{k\in\upomega\colon x_k\neq a\}\cap\{k\in\upomega\colon x_k-a<\delta\}
\cap
\{k\in\upomega\colon a-x_k<\delta\},
\end{equation*}
the intersection of three sets,
all of which are large,
since, respectively, $[x]\neq0$,
and $[x]-a<\delta$, and $a-[x]<\delta$.
So the intersection itself is large,
and therefore the latter set in \eqref{eqn:xafL}
%$\{k\in\upomega\colon\abs{f(x_k)-L}<1/n\}$
is large, which means
\begin{equation*}
\abs{\st f[x]-L}<\frac1n.
\end{equation*}
This being the case for all $n$ in $\N$,
$\st f[x]\simeq L$.
Now suppose $\lim_af\neq L$.
Then for some positive $\epsilon$ in $\R$,
for every $n$ in $\upomega$,
there is $x_n$ in $\R$ such that
$0<\abs{x_n-a}<1/(n+1)$,
but $\abs{f(x_n)-L}\geq\epsilon$.
Then $[x]\simeq a$, but $[x]\neq a$,
and $\abs{\st f[x]-L}\geq\epsilon$,
so $\st f[x]\not\simeq L$.
\end{proof}
The nonstandard definition makes proofs about limits easier,
at least once one has the following.
\begin{theorem}\label{thm:ideal}
In any ordered field,
the finite elements compose a sub-ring,
and the infinitesimal elements compose a maximal ideal of the ring of finite elements.
In fact the ideal is the only maximal ideal of the ring.
\end{theorem}
\begin{corollary}
In the ring of finite elements of any ordered field,
the relation $\simeq$ of being infinitesimally close
is an equivalence relation,
and
\begin{gather*}
a\simeq b\And c\simeq d\implies a+c\simeq b+d\And ac\simeq bd,\\
a\simeq b\And a\not\simeq0\implies b\not\simeq0\And\frac1a\simeq\frac1b.
\end{gather*}
\end{corollary}
The ``standard'' proof of the following
involves taking different deltas for different epsilons,
and then taking the maximum.
The ``nonstandard'' proof is more straightforward.
\begin{theorem}
Suppose $\lim_af=L$ and $\lim_af=M$.
Then
\begin{align*}
\lim_a(f+g)&=L+M,&
\lim_a(fg)&=LM.
\end{align*}
If $L\neq0$, then
\begin{equation*}
\lim_a\frac1f=\frac1L.
\end{equation*}
\end{theorem}
\begin{proof}
Suppose $[x]\simeq a$, but $[x]\neq a$.
By hypothesis and Theorem \ref{thm:f},
$\st f[x]\simeq L$ and $\st g[x]\simeq M$.
Then the corollary of Theorem \ref{thm:ideal}
yields the desired results.
\end{proof}
\mysection{Standard parts}
Every finite element $a$ of $\stR$
is infinitesimally close to a unique element $\stp a$ of $\R$,
namely
\begin{equation*}
\sup\{x\in\R\colon x0\lto0<\abs{x-a}<\delta\land\abs{f(x)-L}\geq\epsilon)
\end{equation*}
is true in $\R$, hence in $\stR$.
In particular, letting $\delta$ be a positive infinitesimal,
we obtain $x$ in $\stR$ such that $x\simeq a$, but $x\neq a$,
and $\st f(x)\not\simeq L$.
If we call $\stR$ a \textbf{(nonstandard) extension} of $\R$,
then $\st f$ as above is the corresponding extension of the function $f$.
Similarly,
being a subset of $\R$, the set $\N$ is a singulary relation on $\R$,
and so it has an extension $\stN$,
which is the set of all $a$ in $\stR$ such that the sentence $a\in\N$
is true in $\stR$.
(By the formal definition above,
the sentence $a\in\N$ would be written as $\N a$.)
This means
\begin{equation*}
\stN=\bigl\{[x]\colon x\in\R^{\upomega}\And\{k\in\upomega\colon x_i\in\N\}\in\mathscr U\bigr\}.
\end{equation*}
\begin{theorem}
$\N\pincluded\stN$,
and the finite elements of $\stN$
are precisely the elements of $\N$.
\end{theorem}
\begin{proof}
For every $n$ in $\N$, the sentence $n\in\N$ is true in $\R$, hence in $\stR$;
so $\N\included\stN$.
Since the sentence
\begin{equation*}
\Forall x\bigl(x\in\N\lto1\leq x\land(n\leq xx)$ is true in $\R$,
hence in $\stR$, so letting $x$ be a positive infinite element of $\stR$
gives us an infinite element of $\stN$;
this element is not in $\N$, so $\N\pincluded\stN$.
\end{proof}
If now $a$ is a sequence $(a_k\colon k\in\N)$ of elements of $\R$,
this means $a$ is the binary relation $\{(k,a_k)\colon k\in\N\}$ on $\R$,
so it has an extension $\st a$.
This extension must be a sequence $(\st a_n\colon n\in\stN)$,
where $\st a_n=a_n$ if $n\in\N$;
for the sentences
\begin{gather*}
\Forall x\Forall y(x\mathrel ay\lto x\in\N),\\
\Forall x\Forall y\Forall z(x\mathrel ay\land x\mathrel az\lto y=z)
\end{gather*}
are true in $\R$, hence in $\stR$.
\begin{theorem}
A real-valued sequence $a$ on $\N$ is bounded
if and only if, for each infinite $n$ in $\stN$,
$\st a_n$ is finite.
\end{theorem}
\begin{theorem}
For all real-valued sequences $a$ on $\N$,
for all $L$ in $\R$,
\begin{equation*}
\lim_{n\to\infty}a_n=L
\end{equation*}
if and only if, for all infinite $n$ in $\stN$,
\begin{equation*}
\st a_n\simeq L.
\end{equation*}
\end{theorem}
Recall that a real-valued sequence $a$ on $\N$
is a \textbf{Cauchy sequence} if for all positive $\epsilon$
in $\R$,
for some $\ell$ in $\N$,
\begin{equation*}
m>\ell\And n>\ell\implies\abs{a_m-a_n}<\epsilon.
\end{equation*}
Easily, every convergent sequence is a Cauchy sequence.
The converse is more difficult.
The following is the ``nonstandard'' version of the theorem:
the proof will use standard parts as defined above.
\begin{theorem}
A real-valued sequence $a$ on $\N$ is convergent
if and only if, for all infinite $m$ and $n$,
\begin{equation*}
a_m\simeq a_n.
\end{equation*}
\end{theorem}
\mysection{Mock second-order logic}
We may consider a structure
as having a universe partitioned into several subsets,
called \textbf{sorts.}
For example, a vector space
has a sort of vectors and a sort of scalars.
Correspondingly, in the logic of vector spaces,
there will be vector variables and scalar variables.
In a many-sorted structure,
operations and relations do not simply take
$n$ arguments for some $n$ in $\N$,
but the sort of each argument must be specified.
Thus, in a vector space,
two vectors can be added together,
and two scalars can be added together,
but not a vector and a scalar.
We can consider the natural numbers
as composing a two-sorted structure,
the sorts being $\N$ and $\pow{\N}$.
The variables for these sorts
are minuscule and capital letters, respectively.
In addition to the relation symbol $<$,
which takes two arguments from $\N$,
let the signature contain the membership relation $\in$,
taking an argument from $\N$ and an argument from $\pow{\N}$.
Then we can express the well-ordering property of $(\N,<)$
as a \emph{first-order} sentence in this signature:
\begin{equation}\label{eqn:wo}
\Forall X\forall y\Bigl(y\in X\lto
\Exists z\bigl(z\in X\land\Forall w(w\in X\lto z\leq w)\bigr)\Bigr).
\end{equation}
Denoting our two-sorted structure by $\N\amalg\pow{\N}$,
we can form the extension $\st(\N\amalg\pow{\N})$,
which can be understood as having the two sorts $\stN$ and $\st{\pow{\N}}$.
Then the same sentence \eqref{eqn:wo} is true in the extension.
However,
this does not mean that $\stN$ is well ordered.
This is because $\st{\pow{\N}}$ is not $\pow{\stN}$.
If $a\in\N^{\upomega}$ and $B\in\pow{\N}^{\upomega}$,
we have
\begin{equation*}
[a]\mathrel{\st{\in}}[B]\iff\{k\in\upomega\colon a_k\in B_k\}\in\mathscr U.
\end{equation*}
Thus we can identify $[B]$ with the subset
\begin{equation}\label{eqn:B}
\bigl\{[x]\in\stN\colon\{k\in\upomega\colon x_k\in B_k\}\in\mathscr U\bigr\}
\end{equation}
of $\stN$.
We can denote this subset of $\stN$ by
\begin{equation*}
\left.\prod_{k\in\upomega}B_k\right/\mathscr U;
\end{equation*}
however, it must be understood that if $[a]$ belongs to this set,
this does not mean $a\in\prod_{k\in\upomega}B_k$,
but only that $\{k\in\upomega\colon a_k\in B_k\}\in\mathscr U$,
as in \eqref{eqn:B}.
For example, if some $B_k$ is empty,
then $\prod_{k\in\upomega}B_k=\emptyset$,
but this does not imply that the set in \eqref{eqn:B} is empty.
The embedding $[X]\mapsto\left.\prod_{k\in\upomega}X_k\right/\mathscr U$
of $\st{\pow{\N}}$ in $\pow{\stN}$ is not surjective.
For example, $\N$ itself (considered as a subset of $\stN$)
is not in the image of $\st{\pow{\N}}$.
We can see this in two ways.
\begin{asparaenum}
\item
Every nonempty subset of $\st{\pow{\N}}$ has a least element,
because \eqref{eqn:wo} is true in $\st(\N\sqcup\pow{\N})$;
but $\stN\setminus\N$ has no least element.
\item
More directly,
suppose $B$ in $\pow{\N}^{\upomega}$ is such that
the element of $\pow{\stN}$ in \eqref{eqn:B} includes $\N$.
This means, for each $n$ in $\N$,
\begin{equation*}
\{k\in\upomega\colon n\in B_k\}\in\mathscr U.
\end{equation*}
In particular,
there are infinitely many $k$ in $\upomega$ such that $n\in B_k$.
Thus if, for all $k$ in $\upomega$, we define
\begin{equation*}
a_k=\max\{n\in B_k\colon n\leq k+1\},
\end{equation*}
then the sequence $(a_k\colon k\in\upomega)$ is unbounded,
and so $[a_k\colon k\in\upomega]$ is not in (the image of) $\N$,
though it is in $\left.\prod_{k\in\upomega}B_k\right/\mathscr U$.
\end{asparaenum}
We can introduce power sets as sorts
in order to define \emph{integration.}
By one standard definition (attributed to Riemann),
a real-valued function $f$
whose domain includes a closed interval $[a,b]$
is \textbf{integrable} on $[a,b]$ if, for some $I$ in $\R$,
for all positive $\epsilon$ in $\R$,
there is a positive $\delta$ in $\R$
such that,
for all $n$ in $\N$,
for all subsets $\{a_0,\dots,a_n\}$ of $[a,b]$,
for all subsets $\{\xi_0,\dots,\xi_{n-1}\}$ of $[a,b]$,
if
\begin{equation}\label{eqn:part}
\left.\qquad
\begin{gathered}
a=a_0<\dots}[r]\ar@{=>}[d]&\text{AC}\ar@{<=>}[r]&\text{MAX}\ar@{=>}[d]\\
\text{COM}\ar@{<=>}[rr] & &\text{PRI}}
\end{equation*}
\setbibpreamble{When I set out to learn nonstandard analysis for this course,
I used the book of its creator,
Abraham Robinson \cite{MR1373196}.
A professor had referred me to this book
when I was starting graduate school;
but I could not make sense of the book then.
There are textbooks of nonstandard analysis,
but I confess to not having sought them out.
I have consulted two \emph{calculus} textbooks for beginners
that use the nonstandard approach:
Keisler \cite{Keisler-calculus} and
Henle and Kleinberg \cite{MR1999278}.
Most books on model theory will say something about ultraproducts;
the ones that I have used the most are Hodges \cite{MR94e:03002},
followed by Chang and Keisler \cite{Chang--Keisler}
and Bell and Slomson \cite{MR0269486}.\\\mbox{}}
%\bibliographystyle{plain}
%\bibliography{../../../references}
\begin{thebibliography}{1}
\bibitem{MR0269486}
J.~L. Bell and A.~B. Slomson.
\newblock {\em Models and ultraproducts: {A}n introduction}.
\newblock North-Holland Publishing Co., Amsterdam, 1969.
\newblock Reissued by Dover, 2006.
\bibitem{Chang--Keisler}
C.~C. Chang and H.~J. Keisler.
\newblock {\em Model theory}.
\newblock North-Holland Publishing Co., Amsterdam, third edition, 1990.
\newblock First edition 1973.
\bibitem{MR1999278}
James~M. Henle and Eugene~M. Kleinberg.
\newblock {\em Infinitesimal calculus}.
\newblock Dover Publications Inc., Mineola, NY, 2003.
\newblock Reprint of the 1979 original [MIT Press, Cambridge, MA.
\bibitem{MR533327}
Wilfrid Hodges.
\newblock Krull implies {Z}orn.
\newblock {\em J. London Math. Soc. (2)}, 19(2):285--287, 1979.
\bibitem{MR94e:03002}
Wilfrid Hodges.
\newblock {\em Model theory}, volume~42 of {\em Encyclopedia of Mathematics and
its Applications}.
\newblock Cambridge University Press, Cambridge, 1993.
\bibitem{Keisler-calculus}
H.~Jerome Keisler.
\newblock {\em Elementary Calculus: {A}n Infinitesimal Approach}.
\newblock 2013.
\newblock First edition 1976, second edition 1986, both by Prindle, Weber \&\
Schmidt. Published online at \url{www.math.wisc.edu/~keisler/calc.html}.
\bibitem{MR1373196}
Abraham Robinson.
\newblock {\em Non-standard analysis}.
\newblock Princeton Landmarks in Mathematics. Princeton University Press,
Princeton, NJ, 1996.
\newblock Reprint of the second (1974) edition. With a foreword by Wilhelmus A.
J. Luxemburg. First edition 1965.
\end{thebibliography}
\end{document}