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\begin{document}

\title{Basic Analysis on Countable Sets}
\author{Jeffrey Lassahn}
\date{A THESIS\\
Presented to the Department of Mathematics\\
of the University of Oregon\\
June 8, 1994}
\maketitle

\tableofcontents
\endsection

%---------------------------------------------------------------------
\section*{Acknowledgements}
I would like to take this opportunity to thank some of the many people who
helped this thesis get written, including Professor Dick Koch,
 who suggested and
managed the whole project, Professors Peter Gilkey and Bill Kantor who
read it, thought about it and provided suggestions and commentary,
Professor Kenneth Ross for ${\bf Q}\times{\bf Q}$ and Alex West for the 
constant stream of advice and counterexamples.
\endsection
%-----------------------------------------------------------------------
\section{Introduction}
It is usual in an introduction to analysis to begin with a discussion of
the topology and completeness properties of the real numbers.  This set
is presented as the foundation for all further work, and ideas such as
continuity and convergence which are important to all of analysis depend
critically on the details its properties.  As a
result, while the real numbers are originally introduced as one set of numbers
among many, and while subjects such as algebra routinely explore
properties of sets such as the rational numbers, the algebraic numbers,
and other more specialized objects, analysis is usually done only on a
few specific sets; if work is being done in one dimension, functions are
assumed to be between sets of
 real numbers, while in two or more dimensions the choices
expand to allow ${\bf R}^n$ or the complex numbers or some combination of
these. Other possibilities are dismissed early on.  The notion that analysis
cannot be done on the rational numbers because they are not complete is 
generally presented, accepted and forgotten within the first few days of
the first introductory class taught on the subject.

The reasons for this are not unconvincing, of course.  Limits
and convergent sequences are some of the most powerful tools of analysis,
and their properties are much clearer in complete spaces.  Notions such
as compactness make no sense at all on sets like {\bf Q} except under
conditions so restrictive that their power as analytical tools is 
eliminated.  The whole modern notion of the real numbers was created
essentially so that the methods of analysis would have a space to work in.

Although the real numbers provide a convenient place to do much of analysis,
they come with their own collection of problems.  Their algebraic structure
is far from simple; while some sets like the integers or the rationals can
be described easily and completely by their algebraic properties, the
reals are difficult to characterize.  They are not the closure of any
finite set under any binary operations.  There cannot even be any notation
which can be used to write out an arbitrary real number, since such a
notation would map each number to a finite string from some alphabet and
the set of all such strings is countable.

The uncountability of the real numbers produces other problems as well.  
Even operations which seem trivial on finite or countable sets can become
the basis for deep, even inherently undecidable questions for uncountable sets.
Can all infinite subsets of ${\bf R}$ be put into one-to-one correspondence
with either ${\bf Z}$ or ${\bf R}$?  No set can be found which won't do this,
but is that the same as saying there are none?  This can not be proven.
Several important results in measure theory apply to objects which only
exist if the axiom of choice is assumed.  While analysis becomes definite
about what domains its functions should take it becomes ambiguous about
questions of what sort of logic is acceptable for deriving results.

I would like to propose that for some purposes the logical definiteness
 gained by
working in countable sets may outweigh the loss of the standard methods
from analysis, and that it is a reasonable and interesting problem to look 
for different methods which produce analogous results on countable spaces to
those given by standard analysis and topology on uncountable spaces.
In particular, completeness and connectedness are usually only properties of
uncountable spaces.  Certainly countable spaces like the integers with the
usual metric are complete, but in an uninteresting way.  There are also 
sets smaller than the real numbers which can be made connected, including
such things as the single point, and the indiscrete topology on any countable
set, but again these seem to lack some desirable structure.  It would be nice
to have analogous concepts which say useful things about countable metric 
spaces.

This thesis actually has two related goals. One is to develop some simple
analogies to standard analytic concepts which can be used on incomplete sets.
In particular, it is partially an attempt to give some description of how
completeness is actually related to other properties of a space, and to remove
this concept from ideas which can be developed without it.  This will among
other things give an examination of continuity which separates the notion
of total boundedness (which is the generalization of boundedness from finite
dimensional metric spaces to arbitrary metric spaces) from that of completeness
even though the two are usually intertwined in the hybrid idea of compactness.
It will also examine how much of basic calculus can be reconstructed without 
the intermediate value and extreme value theorems which come directly from the
 properties of completeness.

In addition to looking for extensions of analytic concepts to more general
domains, I am also attempting to give some sort of impression of what 
calculus might look like if the real numbers were never considered a valid
concept at all.  Many of the definitions and results given in the text have
logically equivalent forms which depend explicitly on the properties of the
real numbers.  For instance, the notion of strong continuity on a dense
 subset of ${\bf R}$, which I define in Chapter 2 in terms of distances
between sets,
can be shown to have the equivalent formulation: $f$ is strongly continuous
if and only if it is the restriction of a continuous function
 $F$ defined everywhere on
 ${\bf R}$.  This is not the definition I choose to work with, even though it
makes the demonstration of certain results trivial.  Of course strongly
continuous functions map bounded sets to bounded sets, since the domain of
the function can be extended to a compact set, which maps to another 
compact set because of a theorem from standard analysis.  This is a perfectly
reasonable proof of Theorem \ref{bounded}, but it is unenlightening.  In
general, of the proofs given the only ones which depend on the properties
of the real numbers are those such as Theorems \ref{completeness},
\ref{c-one} and \ref{riemann} which explicitly describe the relationship
between the concepts presented here and those of standard analysis.
In other cases, even when ${\bf R}$ appears explicitly in the theorem I
am often not depending on its completeness or cardinality.  In particular
some theorems use the terminology $X\subset{\bf R}$ to specify the domain
$X$ of some function.  Unless specific facts about $X$ are given, it can be 
assumed to be arbitrary, and the statement is simply a short way
 of specifying any
set such as ${\bf Q}$ which has the ordering and some arithmetic properties of
${\bf R}$.  

Either one of these goals extends into large enough areas of mathematics
that this thesis can only address a few preliminary points, and must ignore
many more.  In particular, the question of what can be done to topology
to produce something more consistent with this version of analysis is
almost totally ignored.  The chapters have also been separated by interludes
which are intended to informally present extensions to ideas in the
main text which are generally conjectural.  A major theme of the interludes
is that when the real numbers have been removed from consideration there are
many different possible sets which can take their place.  None of these have
all of the characteristics of ${\bf R}$ and each has its own character.  The
exploration and construction of countable sets which have different analytic
properties seems to me to be an interesting project on its own merits and these
interludes are my way of encouraging others to play with the possibilities.
 
\endsection
%----------------------------------------------------------------------
\interlude{Cardinality Paradoxes}
\subsection{Some Unusual Countable Sets}
In the introduction I made some fuss about difficulties arising from the
formal manipulation of uncountable sets.  There are many examples of problems
both logical and conceptual which these sets present, not the least of which
arise from the sheer difficulty of describing such objects.  As previously
mentioned, any notation consisting of finite strings of characters from a
finite alphabet can distinguish only a countable number of objects.  This 
includes essentially all reasonable forms of communication; even pictures
can be included as long as they are not expected to be examined at infinitely
fine detail.  There simply is no way to write down an arbitrary real number.

By similar reasoning, any objects which could reasonably be called
algorithms can only be part of a set which is at most countable.  There 
cannot be an algorithm which approximates each real number.  For that matter,
the number of proofs is countable, as is the number of possible conjectures;
there cannot be a theorem, or even a conjecture specific to each real number.

These facts suggest a picture of the real numbers as a set with a small
minority of its members available for examination and use, while the majority 
sit out of reach.  It is even possible to examine the set of numbers whose
properties are actually available for inspection.  Perhaps this can be 
described as the set of all expressions which when evaluated over the positive
integers produce convergent sequences, or the set of propositions which
can be evaluated for every rational number and for which the rationals where
they are true form a Dedekind cut.  Neither of these sets consists of all
of ${\bf R}$, since both the number of propositions and the number of
expressions are countable.  It is not clear what other properties they have,
or even if they are actually the same set.

For that matter, it is not immediately clear that such things are useful at
all.  Why bother defining a set by merely observing that the notation we
usually use only produces a subset of the objects we try to think about, if
in practice this observation has no consequences for the way calculation is
carried out?   A set with membership determined by a rule such as ``if
you can write it down it's in the set'' would seem to be so nebulous that
it couldn't possibly be useful.

\subsection{The Paradox of Algorithms}
There are, however, interesting results which can be discovered with the
aid of such sets.  For instance, consider the set of all decimal expansions
which can be produced by algorithms.  That is, suppose there is a procedure
which can be described by a finite set of instructions which when given a
positive integer $n$ returns after a finite number of calculations the
 $n$th digit of
some real number between 0 and 1.  The set of these algorithms must be
countable, since they can be listed by a procedure like alphabetization.

Suppose such a list exists and contains all possible algorithms from this set.
A real number can be approximated by a decimal expansion which has its
$n$th digit constructed from the $n$th digit of the number $d$ produced by the
$n$th algorithm in the list by making 
$$s_n=
\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
                          3&d\ne 3;\cr 2&d=3.\cr}}\right.
$$
The number which $s_n$ converges to will be different than that produced by
any algorithm in the list.  But isn't the procedure just described the
outline of an algorithm?  Shouldn't it already be on the list?  If it is
then the number produced by it would be the same as the number produced by some 
algorithm on the list, and if it is not then any listing of algorithms must
not contain all possible ones, so the set of these algorithms can not be
countable.  There appears to be a contradiction here.

One way to resolve this problem as it stands is to observe that what was
presented was not an algorithm of finite length.  In order to do the
calculation we must have access to every algorithm in the list, and if this
infinite list must be included to specify what the algorithm does, its
description is no longer finite.  Perhaps this requirement for infinite
length can be avoided by having the algorithm we are constructing generate
the list of algorithms internally.  Each algorithm is a string in some
alphabet of symbols, so our new algorithm can simply start producing 
strings in order of increasing length, examine each one to see if it is
the appropriate type of algorithm, then calculate based on this information.
By this method, it can give any number of digits in its decimal expansion 
after only finding a finite number of other algorithms and evaluating each one
for a single value of $n$, so it should still terminate, and it now has a
finite description (programmers of the language LISP might even consider 
implementing such an algorithm).  It would appear that our paradox has
returned.

One plausible conclusion to be made from this is that I am defining much too
loosely what counts as a computational operation.  Is it reasonable to expect
an algorithmic process to evaluate the computations of another algorithmic
process?  In the field of computer science, such algorithms are used all
the time.  The entire concept behind a programmable computer is that it is an
implementation of a formal algorithm which processes other algorithms.  The
more doubtful question is whether there is an algorithmic method for 
recognizing other algorithms.  If there is no strictly formal analysis of
text strings which can determine whether they will behave in some predictable
way when executed as algorithms, then the proposed method cannot be
guaranteed to complete its calculation successfully and the paradox is resolved.

This conclusion corresponds nicely to a result from the theory of computer
science that states that no algorithm can exist which can evaluate other
algorithms and reliably determine whether they terminate.  It is usually
demonstrated by entirely different methods that are proposed here, but it is
perhaps surprising to see how such a result can arise from questions about
cardinality.

\subsection{Suggestions of an Incompleteness Theorem}
Another line of argument with similar form uses the fact that the number of
real numbers is larger than the number of theorems about them.  Suppose
that there is a formal system which allows proofs of the correctness of
mathematical calculation (choose your favorite one, like the axioms of
set theory combined with the predicate calculus).  It is possible to 
assemble a list of pairs of strings in this formal system, the first of which
is an expression which returns a decimal digit for each positive integer, and
the second of which is a proof that the first actually does this.  These can
be found by a strictly formal examination of all strings; simply check that the 
second string is a series of statements which can be derived from previous
statements and the first string by the transformation rules of the system,
and that the concluding statement is of the form 
$<\mbox{first string}>\in\{1\dots 9\}\mbox{\ for all\ }n\in{\bf N}$
however this is best expressed in the particular formal system.  

If the formal system is powerful enough that it can represent the logic of 
Cantor's diagonal proof, and the typographical processes used in its rules
of inference can be represented as operations on sets in the system, then
a construction which evaluates all strings for the above properties and 
composes the expressions it finds with the function 
$$f(x)=
\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
                          3&x\ne 3;\cr 2&x=3.\cr}}\right.
$$
would constitute an expression which returns a decimal digit for each positive
integer, but this fact has no proof in the formal system.

Again, results of this sort have been proven by arguments much different from 
the ones briefly sketched here, and actually showing that this argument leads
to valid conclusions requires much more careful work, but these examples
are suggestive of a method of reasoning which uses the countability of 
constructable subsets of uncountable sets to help discover surprising results.
 
\endsection

%-----------------------------------------------------------------
\section{Continuity}

\subsection{Continuity and Strong Continuity}

The concept of a continuous function is useful in many mathematical contexts.
This has led to a large number of characterizations of continuity,
 all of which are 
equivalent, but each emphasizes different features of the functions which
are of interest in different circumstances.
 
\begin{definition}
For a function $f:X\to Y$ between metric spaces $X$ and $Y$ the following are equivalent and any can be taken as the definition for continuity of the function $f$.
\begin{verse}
\gap
(a) For any $A\subset X$, $f(\cls(A))\subset \cls(f(A))$.

\gap
(b) For any open set $B\subset Y$, $f^{-1}(B)$ is open in $X$.

\gap
(c) For any closed set $B\subset Y$, $f^{-1}(B)$ is closed in $X$.

\gap
(d) For any $x\in X$ and $\epsilon>0$, there is some $\delta>0$
such that $d(x,z)<\delta$ implies $d(f(x),f(z))<\epsilon$.

\gap
(e) For any sequence $a_n$ which converges to $a\in X$,
the sequence $f(a_n)$ converges to $f(a)$.
\end{verse}
\end{definition}

These characterizations were designed to correspond to the intuitive idea of
a continuous function on a complete metric space such as the real numbers,
but give somewhat questionable results on metric spaces which are not complete.

\begin{example}
\label{step}
A function $f:{\bf Q}\to {\bf Q}$ is defined as follows:
$$
f(x)=
\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
                          1&x^2<2;\cr 0&x^2>2.\cr}}\right.
$$
This function is continuous on ${\bf Q}$.
\end{example}

A definition of continuity which allows this sort of function to be continuous is not going to be particularly helpful for doing work on ${\bf Q}$.
In general, these definitions allow continuous functions to have steplike
changes in value anywhere that their domains are not complete.  Since many
interesting countable subsets of ${\bf R}^n$ including ${\bf Q}$ are not
complete {\it anywhere}, any attempt to extend notions of analysis to
countable sets must begin with a rethinking of continuity.

One property which continuous functions should have is that pieces of their domains which touch should have images which touch.  This notion can be formalized.  $A$ is called {\bf disconnected} from $B$ if
$\cls(A)\cap B= \emptyset$ and $A\cap \cls(B)=\emptyset$.

Two sets are {\bf connected} to each other if they are not disconnected.  This
does not assume anything about the individual sets; in particular, two sets
can be connected to each other even when each one can be divided into parts 
which are disconnected.

This gives a new equivalent characterization for continuity.
\begin{theorem}
\label{connected}
A function between metric spaces $f:X\to Y$ is continuous
 if and only if for any $A\subset X$ 
connected to $B\subset X$, $f(A)$ is connected to $f(B)$ in $Y$.
\end{theorem}
\proof 
Choose any $A\subset X$ and $B\subset X$ such that $A$ is connected to $B$.
This means either $\cls(A) \cap B \ne \emptyset$ or 
$A \cap \cls(B) \ne \emptyset$. Without loss of generality I can assume
$\cls(A) \cap B \ne \emptyset$.  If $f$ is continuous, 
$f(\cls(A)) \subset \cls(f(A))$, so in particular 
$f(\cls(A) \cap B) \subset \cls(f(A))$.  This means that $\cls(f(A))$ contains 
some element of $f(B)$, so $f(A)$ is connected to $f(B)$.


Conversely, if $f$ is not continuous, there must exist some $A\subset X$
such that $f(\cls(A)) \not\subset \cls(f(A))$. There must be some point 
$b\in\cls(A)$, $f(b)\not\in \cls(f(A))$.  $A$ is connected to $b$, since 
$b\in\cls(A)$, but $f(A)$ is disconnected from $f(b)$, because 
$f(b)\not\in \cls(f(A))$ by hypothesis and $\cls(f(b))=f(b)\not\in f(A)$.
 \endproof

Of course, since this characterization of continuity is equivalent to the
others, it will allow the same functions (such as Example \ref{step})
 to be continuous.
It does, however, make clearer why such functions are continuous. In particular, the function in Example \ref{step} is continuous around $\sqrt 2$
even though its image is the disconnected sets $\{0\}$ and $\{1\}$ because
the set $(-\infty ,\sqrt 2\,]$ is disconnected from $[\sqrt 2,\infty )$ in
${\bf Q}$.  Since countable metric spaces tend to be disconnected everywhere,
they tend to impose only weak conditions on their continuous functions.
Perhaps if connectedness in this characterization of continuity is replaced
with a slightly weaker notion, a natural analogue to continuity will appear
which is more useful on countable spaces.

\begin{definition}
Two subsets $A$ and $B$ of a metric space are said to be {\bf close}
 if and only if there are nonempty totally bounded sets 
$\alpha \subset A$ and $\beta\subset B$ 
such that $d(\alpha,\beta)=0$.
\end{definition} 

The requirement that the sets $\alpha$ and $\beta$ be totally bounded
prevents sets from being close if their distance only goes to zero at
infinity.  For example the sets $\{(x,0)\}$ and $\{(x,y):y=1/x\}$ in
${\bf R}^2$ are not close, even though the distance between them is zero.

A criterion similar to continuity can now be constructed.

\begin{definition}
A function between metric spaces $f:X\to Y$ is {\bf strongly continuous}
if and only if for any two sets $A\subset X$ and $B\subset X$ which are close,
$f(A)$ and $f(B)$ are close in $Y$.
\end{definition}

This new notion of strong continuity has some striking similarities to 
continuity of the usual sort.  In particular, the two are equivalent on
complete metric spaces, such as ${\bf R}^n$.  On spaces which aren't
complete, such as ${\bf Q}^n$ or open subsets of ${\bf R}^n$, strong
continuity is a more restrictive condition than continuity;
some properties which continuous functions preserve on complete spaces
are preserved on all metric spaces by strong continuity.  Under strong
continuity completeness can often be thought of as an incidental property
of a space instead of being inextricably intertwined with many other
characteristics of the space.  This conceptual separation is in large
part a result of the replacement of connectedness, which is strongly
affected by completeness, by closeness, which is not.  Unfortunately,
closeness and connectedness are not equivalent even in the case where
spaces are assumed to be complete, (for example $[-1,0)$ is close to $(0,1]$
in ${\bf R}$,) but there is a particularly simple
characterization for closeness in complete spaces.

\begin{theorem}
\label{closures}
Two subsets of a complete metric space $X$ are close if and only if their
 closures intersect.
\end{theorem}

\proof
If $A$ and $B$ are close in $X$, they have totally bounded subsets
$\alpha$ and $\beta$ such that $d(\alpha,\beta)=0$.  This means there 
are sequences $a_n\in\alpha$ and $b_n\in\beta$ such that 
$\lim_{n\to\infty}d(a_n,b_n)=0$.
Since compact is equivalent to complete and totally bounded, 
$\cls(\alpha)$ is compact. This means
the sequence $a_n$ has a convergent subsequence, $a_k$.  $\cls(\beta)$
is also compact, so the subsequence $b_k$ has a convergent subsubsequence,
$b_l$.  Let $a=\lim_{l\to\infty} a_l$ and $b=\lim_{l\to\infty} b_l$.
For any $\epsilon>0$, $l$ can be made large enough so that
 $d(a_l,a)<\epsilon$, $d(b_l,b)<\epsilon$, and $d(a_l,b_l)<\epsilon$ 
simultaneously.  This implies $d(a,b)<3\epsilon$, for all $\epsilon>0$,
so it must be true that $a=b$.  But $a\in\cls(A)$ and $b\in\cls(B)$, so
$\cls(A)\cap\cls(B)\ne \emptyset$.

Conversely, if there is some $x\in\cls(A)\cap\cls(B)$, then there are 
sequences $a_n\in A$ and $b_n\in B$ so that
 $\lim_{n\to\infty}a_n=x=lim_{n\to\infty}b_n$.
All convergent sequences are totally bounded,
 so $\alpha=\{a_n\}$ and $\beta=\{b_n\}$
are totally bounded subsets of $A$ and $B$ and 
$d(\alpha,\beta)\le d(\alpha,x)+d(x,\beta)=0$, so $A$ and $B$ must be close.
\endproof
 
Even if closeness and connectedness are not equivalent on complete spaces,
they are comparable in some weaker sense.  Connectedness is preserved by
continuous functions, and closeness is preserved by strongly continuous
functions.  In complete spaces, the two types of continuity are equivalent,
so any function that preserves connectedness preserves closeness.

\begin{theorem}
\label{completeness}
If $X$ and $Y$ are complete metric spaces, a function $f:X \to Y$
 is strongly continuous if and only if it is continuous.
\end{theorem}

\proof
Choose any two sets $A$ and $B$ in $X$ which are close. 
The closures of $A$ and $B$ must intersect.
This means there must be some $x\in\cls(A)\cap\cls(B)$.
If $f$ is continuous, $f(x)\in\cls(f(A))$ and $f(x)\in\cls(f(B))$.
Thus $\cls(f(A))\subset Y$ and $\cls(f(B))\subset Y$
intersect, so $f$ must be strongly continuous.



Suppose $f:X \to Y$ is not continuous.  There is some
$A\subset X$ such that $f(\cls(A))\not\subset\cls(f(A))$.  A point
$b$ can be chosen so that $b\in\cls(A)$, but $f(b)\not\in\cls(f(A))$.
$\cls(b)\cap\cls(A)=b$, so $b$ is close to $A$, but
 $Cls(f(b))\cap\cls(f(A))=\emptyset$ so $f(b)$ is not close to $f(A)$ and
$f$ cannot be strongly continuous.
\endproof

\subsection{Properties of Strong Continuity}

In cases where strong continuity is not equivalent to continuity, strongly
continuous functions may still preserve interesting properties of their
domains. In fact, as the name suggests, strong continuity implies continuity
on any metric space, so every property of a metric space which is preserved
by continuous functions is also preserved by strongly continuous functions.

\begin{theorem}
\label{strong}
If $f:X\to Y$ is strongly continuous, it is continuous.
\end{theorem}
\proof
Suppose $f$ is not continuous.  There must be some $A\subset X$ so that
$f(\cls(A))\not\subset \cls(f(A))$.   There must be some $a_0\in\cls(A)$
such that $f(a_0)\not\in\cls(f(A))$.  Since the closure of a set consists
of all points with distance zero from it, $d(f(a_0),f(A))>0$, but
$d(a_0,A)=0$.  This means that $a_0$ is close to $A$, but $f(a_0)$ is not
close to $f(A)$, so $f$ is not strongly continuous.
\endproof

Strongly continuous functions must, therefore, map connected sets to connected
sets, compact sets to compact sets, and so forth.  It should not be surprising
that strongly continuous functions also preserve properties which continuous
functions in general do not.  Closeness is one obvious example.  The images
of close sets under continuous functions are in general not close unless the
domain of the function is complete.  Compactness, which in metric spaces is
simply the combination of completeness and total boundedness, is preserved
by all continuous functions, but strongly continuous functions preserves 
total boundedness as a separate property. 

\begin{theorem}
\label{bounded}
If $A$ is totally bounded and $f:A\to B$ is strongly continuous, then $f(A)$ is
totally bounded.
\end{theorem}

\proof
Suppose $f(A)$ contains a sequence $b_n$ and there is an $\epsilon>0$ so that
 $d(b_i,b_j)>\epsilon$ for all $i$ and $j$, and $f$ is strongly continuous. 
There must be some sequence $a_n\in f^{-1}(b_n)$ and some $\epsilon^\prime>0$
so that $d(a_i,a_j)>\epsilon^\prime$ for all $i$ and $j$.  Otherwise,
there would be some two subsets of $a_n$ which were close, while all pairs of
subsets of $b_n$ would be separated by at least $\epsilon$ and not be close.

But a space cannot have a sequence $a_n$ where $d(a_i,a_j)>\epsilon^\prime$
 if it can be covered by finitely many balls of radius $\epsilon^\prime/2$
or less, because each of these
balls can contain at most one member of such a sequence, and there are
infinitely many members of the sequence.
Thus, if the sequence $b_n$ exists, $A$ cannot be covered by a finite number
of balls with radius less than $\epsilon^\prime/2$, so $A$ cannot be totally
bounded.

If $f(A)$ is not totally bounded, it must have a sequence $b_n$ such that
$d(b_i,b_j)>\epsilon$ for all $i$ and $j$.  This sequence can be constructed
by taking $\epsilon$ small enough that $f(A)$ cannot be covered by balls of 
radius $\epsilon$, then choosing an arbitrary $b_1\in f(A)$.
The ball of radius $\epsilon$ around $b_1$ does not cover $f(A)$,
 so a $b_2$ can be chosen outside of it.  If there is a partial sequence
$b_1\ldots b_n$ such that $d(b_i,b_j)>\epsilon$ for all $i,j\le n$, then the
union of balls of radius $\epsilon$ centered on the members of $b_1\ldots b_n$
must not cover all of $f(A)$, so a point outside of this union can be chosen as
$b_{n+1}$.  This point must have distance at least $\epsilon$ from all of the
$b_1\ldots b_n$, so there is now a larger partial sequence $b_1\ldots b_{n+1}$
such that $d(b_i,b_j)>\epsilon$ for all $i,j\le n+1$.  By induction, there
must be an entire sequence with this property.

Since any $B$ which is not totally bounded allows the creation of a sequence
$b_n$, which in turn forces $A$ not to be totally bounded, the image of a 
totally bounded set under $f$ must be totally bounded.
\endproof

Completeness, however, is not preserved independently of total boundedness.  For example, there are strongly continuous functions $f:{\bf R}\to (0,1)$.

I would now like to develop some tools for working with strongly continuous 
functions, including some basic properties of sums and products and such,
as well as the behavior of sequences and their images. It will be useful
to first examine set products.  Unlike continuity which depends only on the 
topology imposed on a space, strong continuity depends on the metric, so the
question of what metrics should be used for product spaces becomes important.
There are several obvious and useful metrics to place on the product of metric
spaces.  If $X$ is the set product of metric spaces $X_1\dots X_n$, with
each $X_i$ having its own metric $d_i$, the metric $d_X$ on $X$ could be
chosen as the Euclidian one $d_X(p,q)=\sqrt{\sum d_i(p_i,q_i)^2}$, or as
$d_X(p,q)=\max_{all\ i}(d_i(p_i,q_i))$, or $d_X(p,q)=\sum d_i(p_i,q_i)$,
all of which are useful under different circumstances.  Because of the 
inequality
$$\max_{all\ i}(d_i(p_i,q_i))\le (\sum d_i(p_i,q_i)^k)^{1/k}\le \cdots 
 \le \sum d_i(p_i,q_i) \le
 n\mdot\max_{all\ i}(d_i(p_i,q_i))$$
where $i=1,\dots, n$,
if any of these distance functions makes two sets close the others must
 as well. Two metrics which specify the same sets as close must specify
the same functions as strongly continuous, so results about
strong continuity proved for
one of these metrics hold for each of the others. I will assume unless
stated otherwise that product spaces have one of these metrics.  Another
fact which is used in the following proof is that if two sequences have
a distance which tends to zero in the limit, their images under a strongly
continuous function must as well, since not only must those two sequences
have images which are close, but all of their subsequences must as well.

\begin{theorem}
\label{products}
If $f:A\to C$ and $g:B\to D$ are strongly continuous, the function
$h:A\times B \to C\times D$ defined by $h(x,y)=(f(x),g(y))$ is strongly continuous.
\end{theorem}

\proof
Choose two close sets $X\subset A\times B$ and $Y\subset A\times B$.
There must be some sequences $x_n=(a_{x_n},b_{x_n})\in X$ and
$y_n=(a_{y_n},b_{y_n})\in Y$ such that $x_n$ and $y_n$ are both
totally bounded and
 $$\lim_{n\to\infty}\max(d(a_{x_n},a_{y_n}),d(b_{x_n},b_{y_n}))=0.$$
$$\mbox{This means that}\quad
\lim_{n\to\infty}d(a_{x_n},a_{y_n})=0\quad \mbox{and} \quad
\lim_{n\to\infty}d(b_{x_n},b_{y_n})=0.$$
  Since $f$ and $g$ are strongly continuous,
 $$\lim_{n\to\infty}d(f(a_{x_n}),f(a_{y_n}))=0
 \quad \mbox{and} \quad
\lim_{n\to\infty}d(g(b_{x_n}),g(b_{y_n}))=0,
 \quad \mbox{so}$$
$$\lim_{n\to\infty}\max(d(f(a_{x_n}),f(a_{y_n})),d(g(b_{x_n}),g(b_{y_n})))=0.$$
But $(f(a_{x_n}),g(b_{x_n}))=h(x_n)$ and $(f(a_{y_n}),g(b_{y_n}))=h(y_n)$
so $h(x_n)$ and $h(y_n)$ must have distance zero which makes $h(X)$ close
to $h(Y)$.
\endproof
 
This result is particularly useful for working with binary operations on
strongly continuous functions.  Some calculations of this type are used
to prove the following list of useful facts about strong continuity.


\begin{theorem}
\label{facts}
If $A$, $B$ and $C$ are arbitrary metric spaces, and $f$ and $g$ are strongly
continuous, the following are true.

\gap
(a) If $f:A\to B$ is constant, it is strongly continuous.

\gap
(b) If $f:A\to B$ and $g:B\to C$ then $h:A\to C=g\circ f$ is strongly
continuous.

\gap
(c) If $f:A\to A$ is the identity map, it is strongly continuous.

\gap
If $A$ is an arbitrary metric space, $B$ is a vector space, and $f:A\to B$ 
and $g:A\to B$ are strongly continuous, the following are true.

\gap
(d) For any scalar $c$, $c\mdot f$ is strongly continuous.

\gap
(e) $f+g$ is strongly continuous.

\gap
(f) $f-g$ is strongly continuous.

\gap
If $A$ is an arbitrary metric space, $B$ is an arbitrary subfield of 
the complex numbers,
and $f:A\to B$ and $g:A\to B$ are strongly continuous, the following
are true.

\gap
(g) $f\mdot g$ is strongly continuous.

\gap
(h) $1/f$ is strongly continuous if for some $\epsilon>0$ 
and every $a\in A$, $d(f(a),0) >\epsilon$.  

\gap
(i) $f/g$ is strongly continuous if for some $\epsilon>0$ and
every $a\in A$, $d(g(a),0) >\epsilon$.
\end{theorem}

\proof

(a) If $f$ is constant, its image in $B$ is the single point, $b$.  
Any two nonempty subsets of $b$ are close, so the image of close sets
under $f$ must be close.

(b) For two arbitrary close sets $\alpha\subset A$ and $\beta\subset A$, 
$f(\alpha)$ and
$f(\beta)$ must be close, since $f$ is strongly continuous.  Since $g$ is 
also strongly continuous, $g(f(\alpha))$ must be close to $g(f(\beta))$.
This means that $\alpha$ close to $\beta$ implies $h(\alpha)$ close to
$h(\beta)$.

(c) For two arbitrary close sets, $\alpha\subset A$ and $\beta\subset A$,
$f(\alpha)=\alpha$, so it is close to $f(\beta)=\beta$.

\gap

(d) $c\mdot f$ is $h\circ f$ where $h(x)=c\mdot x$.  Because of part(b),
showing that $h$ is strongly continuous is sufficient.  

Let $\alpha\subset B$ and $\beta\subset B$ be arbitrary close sets.
There must be
totally bounded sequences $a_n\in \alpha$ and $b_n\in\beta$ such that
$\lim_{n\to\infty}d(a_n,b_n)=0$.
  $d(h(a_n),h(b_n))=d(c\mdot a_n,c\mdot b_n)=c\mdot d(a_n,b_n)$, so
$\lim_{n\to\infty}d(h(a_n),h(b_n))=0$.  $h(\alpha)$ must be close to
$h(\beta)$, making $h$ is strongly continuous.

\gap

To prove strong continuity for the binary operations begin by defining
$h:A\times A\to B\times B$ (where $B$ is a vector space in (e) and (f)
but a field in (g)) so that $h(a_1,a_2)=(f(a_1),g(a_2))$. The functions from
$B\times B$ to $B$ $p(x,y)=x+y$ and $t(x,y)=x\mdot y$ can be constructed, and
showing that $p$ and $t$ are strongly continuous is sufficient to show that
$p\circ h=f+g$ and $t\circ h=f\mdot g$ are strongly continuous.

(e) Given two arbitrary close sets $\alpha\subset B\times B$ and 
$\beta\subset B\times B$, there must be totally bounded sequences
$a_n=(x_{a_n},y_{a_n})\in \alpha$ and
 $b_n=(x_{b_n},y_{b_n})\in\beta$ so that
$\lim_{n\to\infty} d(x_{a_n},x_{b_n})+d(y_{a_n},y_{b_n})=0$.

\gap

$d(x_{a_n},x_{b_n})+d(y_{a_n},y_{b_n})  = 
\left\| x_{a_n}-x_{b_n}\right\| + \left\| y_{a_n}-y_{b_n}\right\|
\ge \left\| (x_{a_n}-x_{b_n}) + (y_{a_n}-y_{b_n})\right\|$

$ =  \left\| (x_{a_n}+y_{a_n})-(x_{b_n}+y_{b_n})\right\| =
 d(p(x_{a_n},y_{a_n}),p(x_{b_n},y_{b_n})).$

\gap

Thus, $\lim_{n\to\infty}d(p(a_n),p(b_n))=0$ so $p(\alpha)$ is close to 
$p(\beta)$ and $p$ is strongly continuous.

(f) $f-g=f+(-1)\mdot g$.  If $g$ is strongly continuous, then $(-1)\mdot g$
is as well by part (d), and if $f$ is also strongly continuous 
$f+(-1)\mdot g$ must be because of part (e).

\gap

(g) Given two arbitrary close sets $\alpha\subset B\times B$ and 
$\beta\subset B\times B$, there must be totally bounded sequences
$a_n=(x_{a_n},y_{a_n})\in \alpha$ and
 $b_n=(x_{b_n},y_{b_n})\in\beta$ so that
 $\lim_{n\to\infty}\max(d(x_{a_n},x_{b_n}),d(y_{a_n},y_{b_n}))=0$.
This implies that $\lim_{n\to\infty}d(x_{a_n},x_{b_n})=0$ and
$\lim_{n\to\infty}d(y_{a_n},y_{b_n})=0$.

\begin{eqnarray*}
d(t(a_n),t(b_n)) & = &
\left\| x_{a_n}\mdot\, y_{a_n} - x_{b_n}\mdot\, y_{b_n}\right\| \\
& = & \left\| (x_{a_n}-x_{b_n})\mdot\, (y_{a_n}-y_{b_n})+
x_{b_n}\mdot\, (y_{a_n}-y_{b_n})+y_{b_n}\mdot\, (x_{a_n}-y_{b_n})\right\| \\
& \le & \left\| x_{a_n}-x_{b_n}\right\| \mdot \left\| y_{a_n}-y_{b_n}\right\|
+\left\| x_{b_n}\right\| \mdot \left\|y_{a_n}-y_{b_n}\right\|
+\left\| y_{b_n}\right\| \mdot \left\|x_{a_n}-x_{b_n}\right\| \\
& = & d(x_{a_n},x_{b_n})\mdot d(y_{a_n},y_{b_n})
+\left\|x_{b_n}\right\| \mdot d(y_{a_n},y_{b_n})
+\left\|y_{b_n}\right\| \mdot d(x_{a_n},x_{b_n}).
\end{eqnarray*}

Since $\lim_{n\to\infty}d(x_{a_n},x_{b_n})=0$,
$\lim_{n\to\infty}d(y_{a_n},y_{b_n})=0$ and both $x_{b_n}$ and $y_{b_n}$ are
bounded, $\lim_{n\to\infty}d(t(a_n),t(b_n))=0$.  This makes $t(\alpha)$
close to $t(\beta)$, so $t$ must be strongly continuous.

\gap

(h) Given two arbitrary close sets $\alpha$ and $\beta$, there must be
two sequences $a_n\in\alpha$ and $b_n\in\beta$ which are both totally
bounded and so that $\lim_{n\to\infty}d(a_n,b_n)=0$.

$$d\left({1\over f(a_n)},{1\over f(b_n)}\right)=
\left\|{1\over f(a_n)}-{1\over f(b_n)}\right\| =
\left\| {f(b_n)-f(a_n)}\over{f(b_n)\mdot f(a_n)}\right\|.$$

If $\left\| f(x)\right\|>\epsilon$ for all $x$,

$$d\left({1\over f(a_n)},{1\over f(b_n)}\right)
\le {d(f(a_n),f(b_n))\over \epsilon^2}$$

so $1\over f(\alpha)$ must be close to
 $1\over f(\beta)$ and $1/f$ is strongly continuous.

\gap

(i) $f/g=f\mdot\,(1/g)$.  If $g$ is strongly continuous and $d(g,0)>\epsilon$
for some $\epsilon>0$,
 $1/g$ must be strongly continuous by (h), and so $f\mdot\,(1/g)$ is
strongly continuous by (g).
\endproof
   
It is now easy to construct some examples of strongly continuous
functions.  For instance, all polynomials in ${\bf R}$, ${\bf Q}$, or
${\bf C}$ are strongly continuous, since polynomials are sums of terms
constructed from products of the identity with itself and constants, and
both sums and products preserve strong continuity.

The function $1/x$ in either ${\bf R}-\{0\}$ or 
${\bf Q}-\{0\}$ is continuous but not
strongly continuous.  This can be easily seen by observing $1/x$ on any bounded
region around the origin becomes arbitrarily large, while strongly continuous
functions of bounded sets are bounded.  $1/x$ is strongly continuous on any
closed subset of ${\bf R}-\{0\}$, including all sets of the form 
$\left(-\infty,-\epsilon\right]\cup\left[\epsilon,\infty\right)$, however.

By similar reasoning, rational functions can be seen to be strongly continuous
where they have no vertical asymptotes.  This is true of functions like
$x\mdot\,(x-1)/ x$ even though its denominator is not bounded away
from zero, because ${x\mdot\,(x-1)/ x}=(x-1)$ on its entire domain and
$(x-1)$ is strongly continuous on any subset of ${\bf R}$.  In general, 
a strongly continuous function is strongly continuous on any restriction of
its domain, so functions like $f(x)=\sin(x)$, $x\in{\bf Q}$ are strongly
continuous because the extension of $f$ to ${\bf R}$ is strongly continuous,
even though determining the strong continuity of $f$ on ${\bf Q}$ directly 
is not easy.

The notion of a {\bf strong homeomorphism}, meaning a bijective strongly
continuous function with a strongly continuous inverse, is also possible.
The unit circle, $\{(x,y):x^2+y^2=1\}$ and the square 
$\{(x,y):\left|x\right|+\left|y\right|=1\}$
are strongly homeomorphic by the map 
$f(x,y)=(x,y)\mdot\, {\left|x\right| + \left|y\right|\over\sqrt{x^2+y^2}}$,
for example, and in general every homeomorphism between two complete
spaces is a strong homeomorphism.  There are homeomorphisms which are not
strong, however.  In particular, total boundedness is invariant under strong
homeomorphisms so any homeomorphism $h:{\bf R}\leftrightarrow (0,1)$ cannot
be strong.

\subsection{Continuity at a Point and Uniform Continuity}

The usual definitions of continuity are local, in the sense that they can be
written so that each point in the domain of a function is tested independently
for continuity.  It is then possible to speak of functions which are continuous
on parts of their domain, continuous everywhere but at a specific point, and
so on.  Ideally strong continuity would also have this ability, so it may
be worthwhile to look for definitions of strong continuity at a specific point.

\begin{definition}[proposed]
A function $f:A\to B$ is {\bf strongly continuous at a point} $x\in A$ if and
only if for any set $\alpha\subset A$ which is close to $x$, $f(\alpha)$
is close to $f(x)$.
\end{definition}

This resembles the definition of strong continuity, with the additional
condition that one of the sets must be the point we are testing.  Requiring
that the set $\alpha$ be close to $x$, however, is just asking that 
$d(x,\alpha)=0$, or equivalently that there is a sequence $a_n\in\alpha$ which
converges to $x$.  If $f$ is strongly continuous at $x$, any subsequence of
$f(a_n)$ must have distance zero from $f(x)$, which is true exactly when
$f(a_n)$ converges to $f(x)$.  This is just the definition for continuity at
a point, so {\it our proposed definition of strong continuity at $x$
 is equivalent to the usual continuity at $x$.}

In fact there can be no local definition of strong continuity such that
making $f$ strongly continuous at every point forces $f$ to be strongly
continuous.  In example \ref{step}, every point in the domain has an interval
about it where $f$ is constant, so any generalization of continuity
at a point must either claim that some functions which are constant close to
$x$ are still not continuous at $x$, or claim that this function is continuous
at every point.  In essence, one thing that the definition of strong
continuity gives up is the idea that any function which fails to be continuous
fails at some specific point in the domain.  I will still use the notion
of continuity at a point occasionally, but with the implicit warning that
continuity at every point implies continuity in the usual sense, but does
not imply strong continuity.  The term {\bf pointwise continuous} can be used
as a synonym for continuous to  emphasize this distinction.

Independent of any mathematical difficulties related to the definitions of
closeness and
strong continuity, there is the more aesthetic question of why the requirement
of total boundedness is included for close sets.  This appears in some ways
to be an artificial addition, and perhaps invites the question of whether
results such as theorem \ref{bounded} are at all significant mathematically
or simply built intentionally into the definition.  It is no difficult matter
to make boundedness invariant by including requirements about
boundedness explicitly in the definitions. In fact, theorem \ref{bounded}
does not use the boundedness of close sets in any important way, and is equally
true when rewritten as

\begin{theorem}
\label{boundagain}
If $A$ is totally bounded and $f:A\to B$ takes any two sets with distance zero
from each other to images which also have distance zero from each other,
then $f(A)$ is totally bounded.
\end{theorem}

The proof is exactly analogous to that of theorem \ref{bounded}.

If total boundedness is not included in the definition specifically to make
the images of bounded sets under strongly continuous functions bounded, why
is it included?  Most of the properties of strong continuity which were 
proven in the theorems so far are equally valid without the restriction of
total boundedness (the exceptions are theorems \ref{closures} and
\ref{completeness}, and theorem \ref{facts} part (g).) In fact, by excluding
total boundedness from the definition of ``close'', yet another strong form
of continuity can be defined, and it turns out to be exactly equivalent to
uniform continuity.  

\begin{theorem}
\label{uniform}
A function $f:A\to B$ takes pairs of sets with distance zero to images with
distance zero if and only if it is uniformly continuous.
\end{theorem}

\proof
The function $f$ takes pairs of sets with distance zero to images with distance
zero if and only if for any two sequences $a_n\in A$ and $b_n\in B$ such that
$\lim_{n\to\infty}d(a_n,b_n)=0$ in $A$,
 $\lim_{n\to\infty}d(f(a_n),f(b_n))=0$ in $B$, because
any pair of sets $\alpha$ and $\beta$ have distance zero exactly when they
contain sequences $a_n\in \alpha$ and $b_n\in\beta$ so that
$\lim_{n\to\infty}d(a_n,b_n)=0$.

Suppose $f$ is not uniformly continuous.  There is some $\epsilon>0$ such that
no $\delta>0$ exists which forces $d(f(x),f(y))<\epsilon$ for every $x$ and
$y$ where $d(x,y)<\delta$.  This means that for any $\delta>0$ there are some
$a_1$ and $b_1$ such that $d(a_1,b_1)<\delta$, but
$d(f(a_1),f(b_1))\ge \epsilon$. Since $\delta$ is arbitrary,
there must also be $a_n$ and $b_n$ so $d(a_n,b_n)<\delta/n$ but 
$d(f(a_n),f(b_n))\ge \epsilon$ for any $n$.  Thus $a_n$ and $b_n$ form a 
pair of sequences where $\lim_{n\to\infty}d(a_n,b_n)=0$ but
$\lim_{n\to\infty}d(f(a_n),f(b_n))\ne 0$.

Conversely, suppose there is some pair of sequences $a_n$ and $b_n$ so that
$\lim_{n\to\infty}d(a_n,b_n)=0$ but $\lim_{n\to\infty}d(f(a_n),f(b_n))\ne 0$.
There must be some $\epsilon>0$ such that for any $N$ there is an $n>N$ where
$d(f(a_n),f(b_n))\ge \epsilon$. For any $\delta>0$ there must also be some $N$
 such that for any $n>N$, $d(a_n,b_n)<\delta$.  This implies that there is
no $\delta>0$ such that $d(a_n,b_n)<\delta$ implies
 $d(f(a_n),f(b_n))<\epsilon$, so $f$ cannot be uniformly continuous.
\endproof

Strong continuity is now revealed to be weaker than uniform continuity and
stronger than pointwise continuity, but has features of both.  On complete
spaces, strong continuity and pointwise continuity are equivalent while on
totally bounded spaces strong continuity and uniform continuity are 
equivalent.  Of course, spaces which are both complete and totally bounded
make all three equivalent, which agrees with the result from basic analysis
that continuous functions on compact spaces are uniformly continuous.  This
observation also strengthens the analogy between compactness for continuous
functions and boundedness for strongly continuous functions, since in either
case both conditions taken together imply uniform continuity.

\endsection
%-------------------------------------------------------------------------
\interlude{Number Systems}

In the last chapter the idea of continuity was extended in such a way that
functions on the rational numbers (as well as some other sets such as the
algebraic numbers) would behave much like functions on the real numbers.
Ideally, one might expect, this sort of thing could be done with other
concepts from analysis and topology until a structure was in place which
made the rational numbers just like the real numbers in all important respects.
Any philosophical queasiness about uncountable cardinalities could then be
sidestepped in good conscience by concentrating on the rational numbers, or
perhaps the algebraic numbers, or even some more general but still countable
number system.

A limited form of this plan is the essential purpose of this thesis, but
it is undertaken with the realization that such attempts cannot be completely
successful.  The set of real numbers is different from its countable subsets,
and those differences are not trivial or accidental.  Complete spaces have
properties which cannot be effectively mimicked in their incomplete subsets,
and every countable metric space containing the rational numbers must be
incomplete.

In usual treatments of continuity, for instance, there are generally theorems
presented which claim that continuous functions on closed intervals attain
minimum and maximum values, and if they are positive somewhere and negative
elsewhere, they must assume the value zero in between.  Such statements are
simply {\it not true} on incomplete spaces, even when stronger sorts of
continuity are required. If a function on the rational numbers, for instance,
assumes a maximum at point $x$, a new space ${\bf Q}-\{x\}$ can be constructed
which is dense in ${\bf Q}$, and has a function identical to the first one
except that it doesn't attain its maximum at $x$.  It is also easy to find
functions which look obviously continuous on ${\bf Q}$ but don't have the
expected zeros, such as $f(x)=x^2-3$.

A useful way to think of this problem is that number systems are created
sometimes in order to allow certain operations to be well defined, and 
other times to allow certain sorts of equations to have solutions.
The natural numbers can be defined as the smallest set which contains $1$ and
allows addition between any two of its elements.  If the condition is then
added that all equations of the form $x+a=b$ can be solved for $x$, the
negative integers and zero must be included.

Multiplication is already well defined on the integers, but the equations
$a\mdot x +b=0$ do not always have solutions.  Solving them requires the 
rational numbers.  The search for solutions to equations like
$\sum a_nx^n=0$ leads to the algebraic and complex algebraic numbers.

Each of the number systems just listed is countable.  This is a general
fact about numbers created to solve specific sets of equations.  If 
a notation is given which can represent all equations in the set, then
the size of the set of acceptable equations must be countable.  If the
domain of the free variable is assumed to be countable, and it is found
not to provide a solution for some of the equations, at most a countable 
number of new elements must be added to it to make them solvable.  Now
there may be new equations that can be formed from the new elements in the
domain, and these may require additional elements to be added, but this
process only has to be repeated a countable number of times, so the final
set produced will still be countable.

As a result, any countable list of functions can be extended to
a countable domain over which
theorems about intermediate and extreme values will be true.  For example,
every polynomial in the algebraic numbers has a maximum and minimum value
on any closed, bounded interval.  If the polynomial is positive at $x=a$ and
negative at $x=b$ then it will be zero on $[a,b]$.

The set of continuous functions on ${\bf R}$, or even the set of strongly
continuous functions on ${\bf Q}\cap [0,1]$ are not countable. There is not
any countable set over which they will assume their mean or extreme
values.  In fact, this provides a new definition of the real numbers
analogous to the previous definitions of other number systems.

\beginth
The real numbers are the smallest ordered set which contains the rational
numbers and in which every strongly continuous function that takes on
both positive and negative values must somewhere be zero.
\endth

The real numbers are obviously a set which meets these demands, but is it the
smallest?  This can be shown to be true by constructing a pair of sequences
of rational numbers which converge to some real number $x_0$,
$l_n$ converging from below and $r_n$ converging from above.
This can be done for any $x_0$.  A function $f$ can now be assembled by taking
$f(l_n)=-1/n$ and connecting these points by linear segments, then taking
$f(r_n)=1/n$, connecting these by linear segments and making $f(x)=1$ for
large $x$ and $f(x)=-1$ for small $x$.  $f$ is strongly continuous, and
if $x_0$ is not in the domain of $f$ then $f$ assumes both positive and
negative values but is never zero.  If $f$ is required to have a zero
then $x_0$ must be in the domain of $f$.  Since $x_0$ can be any real number,
any set which meets our definition must contain every real number.
 
\endsection
%--------------------------------------------------------------------------
\def\Dx {{d\over dx}}

\section{Differentiation}

\subsection{Derivatives}

Derivatives are usually only calculated for functions on complete spaces such
as ${\bf R}^n$ or $[0,1]\subset{\bf R}$.  In fact, the usual definitions and
techniques for differentiation are often applicable to other spaces, such as
${\bf Q}$.  The one dimensional derivative
$$ f^\prime(x)=\lim_{t\to 0}{f(x+t)-f(x)\over x-t} $$
is well defined for many functions on the rational numbers, and often gives
reasonable values for derivatives.  Rational polynomials, for example, are all
differentiable on all of ${\bf Q}$ by this definition, and
$\Dx x^n=nx^{n-1}$.  Also,

\begin{eqnarray*}
(f+g)^\prime      & = & f^\prime + g^\prime, \\
(f\mdot g)^\prime & = & f\mdot g^\prime + g\mdot f^\prime, \\
(f\circ g)^\prime & = & (f^\prime\circ g)\mdot g^\prime,\qquad\mbox{and} \\
\left({1\over f}\right)^\prime & = & {-f^\prime\over f^2},
\end{eqnarray*}
whenever $f$ and $g$ are differentiable over appropriate domains. None of
these facts depends on details of the number system such as completeness, so
they will be true on any dense subset of ${\bf R}$ where the arithmetic
operations are defined.

Difficulties with differentiation on incomplete spaces do occur for other
reasons, however.  This definition of differentiation calculates derivatives
for functions which look like they shouldn't be differentiable.

\begin{example}
\label{step2}
The function defined in example \ref{step},
$$
f(x)=
\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
                          1&x^2<2;\cr 0&x^2>2.\cr}}\right.
$$
is not only continuous on the rational numbers, it is differentiable,
and $f^\prime(x)=0$ for all $x$.
\end{example}

There are also functions from an incomplete set to itself which fail to be
differentiable because although all of their values are in the set, the
derivative at some point is not.

\begin{example}
\label{incomplete}
Construct a function $f: [-1,1]\cap {\bf Q}\to {\bf Q}$ by setting
$f({1\over n})={s_n\over n}$ and $f({-1\over n})={-s_n\over n}$ where
 $s_1=1$, $s_2=1.4$, $s_3=1.41$, and
$\lim_{n\to\infty}s_n{}^2=2$. The values of $f(x)$ for $x\ne {1\over n}$ can
be found by connecting these points with straight line segments and setting
$f(0)=0$.

The derivative at $x=0$ is $\lim_{n\to\infty}s_n$, which does not exist in
${\bf Q}$.
\end{example}

Failures of this second sort are a fundamental property of incomplete sets.
Essentially the procedure used in example \ref{incomplete} can be used in any
incomplete field to produce a function whose derivative at zero would be the
(nonexistent) limit of some Cauchy sequence $s_n$. These examples must 
either be accepted as not differentiable, or their derivatives must be
considered to be functions into a superset of their codomains.

Example \ref{step2} demonstrates a problem with differentiation which is much
harder to ignore. Much of the usefulness of the derivative comes from the fact
that functions can be categorized as constant, increasing
 or decreasing by finding
whether the derivatives are zero, positive, or negative.  The calculation of
integrals as antiderivatives assumes that two functions with the same 
derivative must differ by a constant.
 
\subsection{Strong Differentiability}

In order to eliminate this problem, a new condition must be imposed on 
differentiable functions. One such restriction is as follows.

\begin{definition}
A function $f$ with domain $A$ is {\bf strongly differentiable} if and only if 
$f^\prime=lim_{t\to 0}{f(x+t)-f(x)\over t}$ exists for all $x\in A$ and
the function
$$
g(x,t)=\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
				{f(x+t)-f(x)\over t} & t\ne 0\cr
				f^\prime(x) & t=0\cr}}\right.
$$
is strongly continuous.
\end{definition}

This notion is stronger than differentiability, even on complete sets.
In particular, it forces $f^\prime$ to be continuous.  In fact, on closed
subsets of ${\bf R}$, that additional condition is sufficient.

\begin{theorem}
\label{c-one}
A function $f:[a,b]\subset{\bf R}\to {\bf R}$ is strongly differentiable 
if and only if it has a continuous derivative everywhere on $[a,b]$.
\end{theorem}

\proof
If $f$ is strongly differentiable, its derivative must exist everywhere, and
since $f^\prime(x)=g(x,0)$ and $g$ is continuous, $f^\prime$ must be 
continuous as well.

Since $g$ is defined on a closed subset of ${\bf R}^2$, it will be strongly 
continuous whenever it is continuous. But $g$ will be continuous at $(x_0,y_0)$
if for any $\epsilon>0$ there is some $\delta>0$ so that for any $(x,t)$
where $d(x_0,x)+d(t_0,t)<\delta$, $d(g(x_0,t_0),g(x,t))<\epsilon$.  
If $t_0\ne 0$, then any $\delta<d(t_0,0)$ is sufficient, since the
function $g$ is continuous whenever $f$ is continuous and $t\ne 0$ by
the arithmetic properties of continuity.

If $t_0=0$ and $t\ne0$, $d(g(x_0,t_0),g(x,t))\le 
d(g(x_0,t_0),g(x_0,t))+d(g(x_0,t),g(x,t))$. There is some $\delta_1$ such that
$d(t_0,t)<\delta_1$ implies $d(g(x_0,t_0),g(x_0,t))<\epsilon/2$
 because $g$ must be continuous with respect to $t$ near 0 when 
$f$ is differentiable.  There must also be a
$\delta_2$ such that $d(x_0,x)<\delta_2$ implies
$d(g(x_0,t),g(x,t))<\epsilon/2$ because $g$ is continuous when $t\ne 0$.
Choosing $\delta=\min(\delta_1,\delta_2)$ forces
$d(g(x_0,t_0),g(x,t))<\epsilon$.

If $t_0=0$ and $t=0$, it is sufficient that $f^\prime$ is continuous.
\endproof
 
This definition is restrictive enough to force strongly differentiable
functions to have all the properties I am interested in, but it may be
somewhat too restrictive.  The function
$$
f(x)=\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
				x & x^2<2,\cr
				1\over x & x^2>2.\cr}}\right.
$$
on the rational numbers
is not strongly differentiable, but it is differentiable everywhere,
increases where its derivative is positive, decreases where its derivative
is negative, and behaves as expected under the fundamental theorem of calculus
(which is proven later and uses strong differentiability.)  In fact, for
all of these theorems, strong continuity and piecewise strong differentiability
is all that is required.  It should be noted, however, that strong continuity 
of $g(x_0,t)$ and $g(x,t_0)$ for fixed $x_0$ and $t_0$ is a weaker condition
than strong differentiability, and there are functions with that property
for which Theorem \ref{constant} does not hold.

\begin{theorem}
\label{constant}
If $f:A\to B$ is a strongly differentiable function, 
$A\subset {\bf R}$ is not the union of two subsets which are not close,
and $B\subset {\bf R}$, the following are true.

\gap

(a) If $f^\prime\ge 0$, $f$ is nondecreasing.

(b) If $f^\prime\le 0$, $f$ in nonincreasing.

(c) If $f^\prime=0$, $f$ is constant.

(d) If $f^\prime > 0$, $f$ is strictly increasing.

(e) If $f^\prime < 0$, $f$ is strictly decreasing.
\end{theorem}

\proof
A function is nondecreasing exactly when it is nondecreasing over every bounded
subset of its domain.  The same is true for nonincreasing, decreasing,
increasing, and constant functions.  It suffices to show results for arbitrary
bounded subsets of $A$.

(a) The function $g(x,t)=(f(x+t)-f(x))/t$ for $t\ne 0$ and $g(x,0)=f^\prime(x)$
is strongly continuous by hypothesis, and since its domain is bounded it must
also be uniformly continuous.  This means that for any $\epsilon>0$, there
is some $\delta>0$ so that $f(x+t)-f(x)/t>f^\prime(x)-\epsilon\ge -\epsilon$
whenever $|t|<\delta$, and this $\delta$ is independent of $x$.

If $f(a)=c$,  $f(a+\delta/2)>c-\epsilon\mdot \delta/2$.  This in turn
implies that $f(a+\delta/2+\delta/2)>f(a+\delta/2)-\epsilon\mdot\delta/2$, so
$(f(a+\delta)>c-\epsilon\mdot\delta$.  In general,
 if $f(a+x)>c-\epsilon\mdot x$,
$f(a+x+t)>c-\epsilon\mdot(x+t)$ whenever $0<t<\delta$. Applying this procedure
inductively shows that $f(a+x)>c-\epsilon\mdot x$ whenever $x>a$ as long as
for every point $a+x$ there is some point $a+x+t$ where $0<t<\delta$.  If there
were no such point, $A$ could be divided into subsets which weren't close by
choosing $A_1=\{y\in A:y\le a+x\}$ and $A_2=\{y\in A:y>a+x\}$.  Thus,
for any $a\in A$ and $a+x\in A$, $f(a+x)>f(a)-\epsilon\mdot x$.  Since
$\epsilon$ is arbitrary, this forces $f(a+x)>f(a)$.

The proof of (b) is analogous to that of (a).

(c) If $f^\prime=0$, then the function is nondecreasing by (a), and
nonincreasing by (b), so $f$ must be constant.

(d) If $f$ is nondecreasing, but not strictly increasing, there must be some
$x$ and $\delta$ so that for any $0<t<\delta$ $f(x+t)=f(x)$.  But then
$f^\prime(x+t)=0$, contrary to the assumption $f^\prime>0$.  (e) follows
by the same argument.
\endproof

The usual properties for arithmetic combination of differentiable functions
generalize in the obvious way.

\begin{theorem}
If $f_1$ and $f_2$ are strongly differentiable then so are $f_1+f_2$,
$f_1\mdot f_2$ and  $f_2\circ f_1$, and $1/f_1$ is strongly differentiable
wherever $f_1>\epsilon$ for some fixed positive $\epsilon$.
\end{theorem}

\proof
Assuming
$$
g_n(x,t)=\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
				{f_n(x+t)-f_n(x)\over t} & t\ne 0\cr
				f_n^\prime(x) & t=0\cr}}\right. ,
$$
if $f=f_1+f_2$ then $g=g_1+g_2$, so $g$ is strongly continuous, and 
$f_1+f_2$ is strongly differentiable.

If $f=f_1\mdot f_2$, then 
\begin{eqnarray*}
g(x,t) & = & {f_1(x+t)\mdot f_2(x+t)-f_1(x)\mdot f_2(x)\over t}, \\
       & = & f_1(x+t)\mdot\left( {f_2(x+t)-f_2(x)\over t} \right) +
	     f_2(x) \mdot \left( {f_1(x+t)-f_1(x)\over t} \right) \\
\end{eqnarray*}
whenever $t\ne 0$, so 
$g(x,t)=f_1(x+t)\mdot g_2(x,t) + f_2(x)\mdot g_1(x,t)$, which is
strongly continuous whenever $f_1$, $f_2$, $g_1$ and $g_2$ are
strongly continuous.  $g_1$ and $g_2$ are strongly continuous by hypotheses,
and $f_1$ must be strongly continuous since
$f_1(t)=t\mdot g_1(x_0,t)+c$ for any fixed $x_0$ if $c=f(x_0)$.  Since 
$g_1$, $t$ and $c$ are strongly continuous $f_1$ must be,
as must $f_2=t\mdot g_2(x_0,t)+f_2(x_0)$.

Since uniform continuity implies strong continuity and strong continuity
only requires that bounded sets have close images, showing that
$g$ is uniformly continuous on arbitrary bounded sets when $f=f_2\circ f_1$
is sufficient to show that $g$ is strongly continuous.

For any fixed bounded domain, $g_1$ and $g_2$ are uniformly continuous.
This means that for any $\epsilon_1>0$ and $\epsilon_2>0$ there are 
$\delta_1>0$ and $\delta_2>0$ such that $|p_1|<\epsilon$ when
$p_1=g_1(x,t)-f^\prime_1$ and  $t<\delta_1$, and similarly for $p_2$.

Since
\begin{eqnarray*}
g(x,t) & = & {f_1(f_2(x+t))-f_1(f_2(x))\over t} \\
       & = & {f_1(t\mdot\,(f_2^\prime(x)+p_2)+f_2(x))-f_1(f_2(x))\over t}\\
       & = & {t\mdot\,(f_2^\prime(x)+p_2)\mdot\,
		(f_1^\prime(b(x))+p_1)\over t} \\
       & = & f_1^\prime(f_2(x))\mdot f_2^\prime(x) +
		p_1\mdot f_2^\prime(x)+p_2\mdot f_1^\prime(f_2(x)) + 
		p_1\mdot p_2 \\
\end{eqnarray*}
(this las equality holds even for $t=0$),
 
$|g(x,t)-g(x_0,t_0)|<|f_1^\prime(f_2(x))\mdot f_2^\prime(x)
-f_1^\prime(f_2(x_0))\mdot f_2^\prime(x_0)|+
\epsilon\mdot|f_2^\prime(x)+f_1^\prime(f_2(x))|$ as long as 
$t$ and $t_0<\min(\delta_1,\delta_2)$. Since $f_1^\prime$ and $f_2^\prime$
are bounded and $f_1^\prime(f_2(x))\mdot f_2^\prime(x)$ is uniformly 
continuous, $g(x,t)$ must also be uniformly continuous wherever 
$t\le \min(\delta_1,\delta_2)/2$. It must also be uniformly continuous
wherever $t\ge \min(\delta_1,\delta_2)/2$ since $f$ is uniformly
continuous and $t$ is bounded away from 0.  Since a function which can be
divided into a finite number of uniformly continuous pieces must be 
uniformly continuous, $g$ uniformly continuous on bounded domains, and so
strongly continuous everywhere.
 
$1/f_1$ must be strongly differentiable wherever $f_1$ and $1/x$ are.
$f_1$ is strongly differentiable everywhere, and $1/x$ is whenever
$|x|>\epsilon$ for any $\epsilon>0$.
\endproof

Of course, since strong differentiability is an additional condition imposed
on the functions, rather than a change in the meaning of the derivative, the
derivatives of these functions can be calculated by the usual methods.
Also, any functions which are continuously differentiable on ${\bf R}$
are strongly differentiable when restricted to dense subsets of ${\bf R}$.
The usual warnings apply about functions such as $1/x$ which are only defined
on some open subset of ${\bf R}$, which sometimes must be restricted to
a closed set before they become strongly differentiable.
\endsection

%-----------------------------------------------------------------------
\interlude{Spaces of n Dimensions}
The work on differentiation in the previous chapter assumes that the sets
on which it operates are one dimensional.  Exactly what I even mean by that
may not be entirely clear.  Ideally, sets which are dense in ${\bf R}^n$
could be thought of as $n$-dimensional, and more generally, if some set were
dense in a more complicated space it would inherit the dimension of that space.

This approach has two problems.  First, it assumes a well developed theory
of the dimension of complete spaces, which is exactly the sort of knowledge
I am trying to avoid using for such definitions.  Even worse, the usual
methods for finding the dimensions of complete spaces already assign
dimensions to some of their dense subsets, and these are often different
than the ones I would like to assign.  In particular, sets which are 
totally disconnected often have dimension zero, and some methods for
calculating topological dimension make every countable set zero dimensional.

A new definition of dimension is necessary.  It cannot rely on standard
topological techniques, because this would make dimension a homeomorphism
invariant, but I would like the spaces ${\bf Q}$ and ${\bf Q}\times{\bf Q}$
to have different dimensions even though they are homeomorphic\footnote{The
fact that ${\bf Q}$ and ${\bf Q}\times{\bf Q}$ are homeomorphic
is not easy to demonstrate, and I am indebted to Dr. Kenneth Ross of the
University of Oregon for proving it to me.}.
Making dimension invariant under strong homeomorphism (strongly continuous
bijection) would seem desirable, however.

One way to go about this is to assign finite sets dimension zero, then
examine larger sets which are not the union of two sets unless those sets are
close (this is essentially a generalization of connectedness) using the
following method.
\begindef
A set $X$ has dimension $n+1$ if there is a set $Y\subset X$ of dimension $n$
such that any set $C$ which makes $Y$ not close to $X-C$ also separates
$X-C$ into subsets which are not close, but there are no sets of smaller 
dimension than $n$ which do this.
\enddef

This makes ${\bf Q}$ 1 dimensional, since any set which can be removed
to make a single point not close to the rest of ${\bf Q}$ must contain some
interval around the point, and this will then separate ${\bf Q}$ into two
sections which are not close.  It does have the unfortunate property of
assigning a space which appears to have dimension $n$ near one point and
dimension $m$ near another a dimension of $\min(n,m)$, so one stray point
makes the entire set zero dimensional.

\subsection{Differentiation in n-space}
Since there is now some evidence that dimension can still mean something in
this context, the question of how to take derivatives in spaces of higher than
one dimension can be pursued.  Ideally a generalization of the strategy
used in the last chapter of creating a function from the difference 
quotients and forcing it to be continuous as the difference becomes small
should work.  So perhaps the derivative of a function on an $n$ dimensional
vector space can be found by finding a linear transformation $D_{x,t}$
where $x$ and $t$ are vectors such that 
$${|f(x+t)-f(x)-D_{x,t}(t)|\over |t|}=0$$
for every $x$ and $t$ and insisting that the resulting set of linear 
transformations form a continuous function on $x\times t$.  Unfortunately,
in general this method does not uniquely determine $D$ at any point, so
constructing a unique continuous extension of it is not really a 
well defined operation.  In order to completely determine $D$ at each point
the difference quotient must involve $n$ points.  that is, the equations
$${|f(x+t_1)-f(x)-D_{x,t_1,\dots,t_n}(t_1)|\over |t_1|}=0$$
$${|f(x+t_2)-f(x)-D_{x,t_1,\dots,t_n}(t_2)|\over |t_2|}=0$$
$$\vdots$$
$${|f(x+t_n)-f(x)-D_{x,t_1,\dots,t_n}(t_n)|\over |t_n|}=0$$
must all be satisfied simultaneously.  This makes $D$ a function into the
set of linear transforms from a $n\mdot n+n$ dimensional space produced by 
taking the product of $n+1$ copies of the domain of $f$.  As long as 
$t_i\ne t_j$ for $i\ne j$ and $t_i\ne 0$ for all $i$, this function is well
defined.  Insisting that it be continuously extendable to those places where it
isn't defined and evaluating it where $t_i=0$ for all $i$ may give a
reasonable definition for the derivative.  If this is the case, it will
certainly have to be suplimented with notions like the matrix of
partial derivatives in order to be useful for calculations, since continuity
is difficult to demonstrate on such abstract spaces and the dimension of
the space where the difference quotient exists becomes large rapidly with
increasing $n$.
\endsection

%---------------------------------------------------------------------------
\def\cint {\,\,{\cal C}\hspace{-11pt}\int}
\def\lcint {{\cal\scriptscriptstyle C}\hspace{-7pt}\int}

\section{Integration}
When working with derivatives, it was clear that the fundamental definition
of the derivative of a function at a point extended without change to
sets which were not complete. Some new sorts of pathological behavior
had to be accounted for, but the basic calculation was mostly unexamined and
unmodified.

Integration does not generalize so easily.  The usual elementary definition
of the integral involves taking supremums and infimums over intervals, as well
as over sets of partitions. The notion of a supremum only makes sense in
a complete space, so integration by this process on sets like ${\bf Q}$ will
in general result in undefined values, even for functions where a reasonable
integral should exist.  In practice, integrals do not have to be computed
using supremums or arbitrary partitions.  If a function is integrable, any
sequence of partitions with characteristic length tending towards zero can
be evaluated on arbitrary points within their segments, and they will still
tend toward the correct value for the integral. Unfortunately, this process
will proceed and give values for the integrals of functions which are not
integrable in the usual sense.  Under these conditions, the exact method used
to select the partitions and sample points can change the calculated value
of the integral. Obviously, some method must exist for finding which functions
can be assigned a meaningful and unique integral.

One strategy for deciding which functions should be integrable is to select
a collection of partitions which can be assigned topological properties, and
insist that the approximations to the integral given for each partition form
a continuous function.  This eliminates from consideration those functions
for which small changes in the structure of the partition produce 
radically different values for the integal.

If $X$ is a subset of ${\bf R}$ which contains points $A$ and $B$ and where
$A\le x\le B$ for any $x\in X$, A partition of $X$ is a finite list of points
in $X$ which contains both $A$ and $B$.  Partitions can also be thought of as
being numbered so that if $P_0\dots P_N$ is a partition, 
$P_0\le P_1\le\cdots\le P_N$.  Partitions can be assigned a measure by defining
the length of the partition $P_N$
$$
|P_N|=\sqrt{\sum_{i=1}^N (P_i-P_{i-1})^2}.
$$
The distance between two partitions can be calculated by extending the one with
the smaller number of elements by adding points $B$ (which doesn't change its
length) until the two have the same number of elements.  The distance between 
them is 
$$
d(P_N,Q_N)=\sqrt{\sum_{i=1}^N \left((P_i-P_{i-1})-(Q_i-Q_{i-1})\right)^2}.
$$
This has the usual properties of a metric, and $d(P,Q)\le |P|+|Q|$ so the
length function can be thought of as $d(P,P_0)$ even though there is no actual
partition which acts as $P_0$.  I will artificially add the object $P_0$ to
the space of partitions, and give it the properties $d(P,P_0)=|P|$ and
$|P_0|=0$, even though it is not a partition.

A refinement of a partition is another partition which 
contains all the points of the original at least as many times, and
possibly others.
\begin{theorem}
If $Q_N$ is a refinement of $P_N$, then $|Q_N|\le |P_N|$.
\end{theorem}
\proof
If $Q_N$ contains only points which are in $P_N$ (even if it contains more
instances of some of them) then $|Q_N|=|P_N|$.
If not, $Q_N$ must be a refinement of $P^\prime_N$ where
$P^\prime_N$ is a refinement of $P_N$ which contains only points from $P_N$,
and therefore has the same length.  $Q_N$ must differ from $P^\prime_N$
only be having some points which are between adjacent points of
$P^\prime_N$.  If $Q_N$ has points $q_1\dots q_n$ between $P^\prime_i$ and
$P^\prime_{i-1}$ then 
$$(P^\prime_i-P^\prime_{i-1})=(P^\prime_i-q_n)+(q_n-q_{n-1})+\cdots+
(q_2-q_1)+(q_1-P^\prime_{i-1})$$
and so
$$(P^\prime_i-P^\prime_{i-1})^2\ge (P^\prime_i-q_n)^2+(q_n-q_{n-1})^2+\cdots+
(q_2-q_1)^2+(q_1-P^\prime_{i-1})^2$$
Substitution of this inequality into the expression for $|P^\prime_N|$ for
each pair of points which are separated by elements of $Q_N$ gives 
$|P^\prime_N|^2\ge |Q_N|^2$, so $|Q_N|\le |P_N|$.
\endproof

Although there is no partition of length zero, partitions can have arbitrarily
small lengths.  In fact, sequences of partitions have lengths which converge
to zero if and only if the values $\max_{all\ i}{|P_i-P_{i-1}|}$ converge
to zero.
\begin{theorem}
For any $\epsilon>0$ there is some $\delta>0$ such that any partition $P_N$
where ${|P_i-P_{i-1}|}<\delta$ whenever $1\le i\le N$ has $|P_N|<\epsilon$.
\end{theorem}
\proof
Suppose $P_N$ has $|P_i-P_{i-1}|<\delta$ whenever $1\le i\le N$. A new
partition $Q_M$ can be constructed from $P_N$ by removing points so that
$\delta\le|Q_i-Q_{i-1}|<2\mdot\delta$ whenever $1\le i\le M$ and $P_N$ is
a refinement of $Q_M$.  Each term in the expression for $|Q_N|$ is less than
$(2\mdot\delta)^2$, and there are at most 
$\lfloor {P_N-P_0\over \delta}\rfloor$ of them.
this makes 
$$|Q_M| < \sqrt{(2\mdot\delta)^2\mdot{P_N-P_0\over\delta}}$$
so choosing $\delta=\epsilon^2/4\mdot\,(P_N-P_0)$ forces
$|P_N|\le |Q_M|<\epsilon$.
\endproof

Of course, if there is a $|P_i-P_{i-1}|>\delta$ for some $\delta$, 
$|P_N|\ge\sqrt{\delta^2}$, so the condition is necessary as well as
sufficient.

The integral can now be defined in terms of continuity on the set of all 
partitions.
\begin{definition}
The {\bf integral by continuity} of a function $f:X\to {\bf R}$ is the value
which must be assigned to the point $P_0$ attached to the space of all partitions
over $X$, so that the function
$$g(P_N,f)=\sum_{i=0}^{N-1}f(P_i)\mdot\,(P_{i+1}-P_i)$$
is continuous at $P_0$, if a unique such value exists.  Otherwise the integral
 is undefined.
\end{definition}

Of course, $X$ must be a bounded subset of ${\bf R}$ in order for the space
of partitions to be defined, and $X$ must be dense in order for there to be
partitions arbitrarily close to $P_0$.  The
 integral by continuity can be written
$$\cint_A^Bf\qquad\mbox{or}\qquad\cint_A^Bf(x)\,dx$$
when the domain of integration is the region of the domain of $f$ bounded by
$A$ and $B$.

  In the definition, the function $f$ was given the real numbers
as codomain.  This is done as a matter of convenience of notation, as
$f$ can be assumed to go into whatever subset of ${\bf R}$ seems appropriate
without changing the content of the definition.  The value of the integral 
itself may also be required to lie in some proper subset of ${\bf R}$, in
particular, the function
$$F(x)=\cint_0^xf(t)\,dt$$
might reasonably be expected to have the same codomain as $f$.  As in the
case of differentiation, however, it is possible to construct functions
$f$ which have integrals $F$ that are only well defined into supersets of
the codomain of $f$ whenever that set is not complete. In this case,
as long as the codomain has the arithmetic properties of closure under
addition, subtraction and multiplication $g(P_N,f)$ is defined for
every partition except possibly $P_0$.  This means that the existence of 
the continuous integral depends on only one possibly undefined limit
process in the codomain, instead of the usual arbitrarily large number.

In the case where both the domain and codomain are complete (that is, closed
subsets of ${\bf R}$), the integral by continuity is applied to the same cases
where the Riemann integral works.  The two should agree if both are defined,
and also be defined for similar types of functions.  In fact, on intervals
in ${\bf R}$ the two are equivalent.
\begin{theorem}
\label{riemann}
A function $f:{\bf R}\to {\bf R}$ is Riemann integrable on $[A,B]$ if and
only if it is integrable by continuity.
\end{theorem}
\proof
If a function is Riemann integrable, for and $\epsilon>0$ there is some
$\delta>0$ such that for any partition with $\max|P_i-P_{i-1}|<\delta$
$$\left| \int f-\sum_{i=1}^Nf(x_i)\mdot\,(P_i-P_{i-1})\right|<\epsilon$$
where $x_i\in [P_{i-1},P_i]$.  This means that $g(P,f)$ will be continuous
at $P_0$ if and only if $g(P_0,f)=\int f$.

Conversely, if $f$ is integrable by continuity, there is some $\delta>0$
for each $\epsilon>0$ so that any partition with $\max|P_i-P_{i-1}|<\delta$
has sum $|g(P,f)-\lcint f|<\epsilon/4$.  Choose any partition $Q$ with
$\max|Q_i-Q_{i-1}|<\delta/2$.  A new partition $U$ can be constructed with one
more point than $Q$ by assigning $U_0=Q_0$, $U_{N+1}=Q_N$ and choosing
$U_i\in [Q_{i-1},Q_i]$ for $1\le i\le N$ such that 
$f(U_i)>\sup_{x\in[Q_{i-1},Q_i]}f(x)-\epsilon/4\mdot\,(B-A)$.
Now $g(U,f)>{\cal U}(U)-\epsilon/4$, where ${\cal U}(U)$ is the Riemann
 upper sum
of $U$, and a partition $L$ can be constructed by analogous methods so that
$g(L,f)<{\cal L}(L)+\epsilon/4$.  Since $\max|U_i-U_{i-1}|<\delta$ and
$\max|L_i-L_{i-1}|<\delta$, $|g(U,f)-g(L,f)|<\epsilon/2$, so
${\cal U}(U)-{\cal L}(L)<\epsilon$ for any $\epsilon$, and $f$ must be
Riemann integrable.
\endproof
  
Even in cases where the domain and codomain of $f$ are not complete,
 it is true that 
$$\cint_A^B f+\cint_B^Cf=\cint_A^C f$$
if all of these integrals exist, although cases can be constructed where
$f$ is not integrable on $[A,B]$ or $[B,C]$ but is integrable on $[A,C]$.

It is always true when addition and scalar multiplication are defined that if 
$f$ and $g$ are integrable,
$$\cint (f+g)=\cint f + \cint g\qquad\mbox{and}\qquad 
\cint(c\mdot f)=c\mdot\cint(f).$$

\subsection{The Fundamental Theorem of Calculus}
In complete spaces, the operations of differentiation and integration 
behave much like inverses of each other.  For cases where all of the
operations in question are defined,
$${d\over dx}\int_A^x f(t)\, dt=f(x)
\qquad\mbox{and}\qquad
\int_A^Bf^\prime=f(B)-f(A).$$
Ideally this should also be true for more general domains.  Some care must be
taken, however, since the mere existence of derivatives and integrals is not
always sufficient.
\begin{example}
\label{badftc}
Suppose $f:{\bf Q}\to{\bf Q}$ is defined by
$$
f(x)=
\left\{ \vcenter{\halign{{$# $}\qquad &if {$# $}\cr
                          1&x^2<2;\cr 0&x^2>2.\cr}}\right.
$$
This function is differentiable everywhere, and $f^\prime=0$. This makes
$$\cint_0^2f^\prime=\cint_0^20=0\ne f(2)-f(0).$$
\end{example}

As might be expected, something stronger than the existance of derivatives
at every point is necessary here.  For arbitrary differentiable functions
taking a derivative loses not only information about the value of $f(A)$,
but also the magnitude of any steplike features such as those in example
\ref{badftc}. Basically, since not all functions with derivative 0 everywhere
differ by only a constant, integration cannot uniquely reconstruct them.
Functions which are strongly differentiable are determined up to a constant by
their derivatives, by Theorem \ref{constant}, so the fundamental theorems
should apply to them.
\begin{theorem}
\label{ftc1}
If $f$ is integrable by continuity on $X$, $A$ is the least element of $X$
 and $f$ is continuous on $X$, then
$${d\over dx}\cint_A^xf=f(x)$$
whenever $x\in X$.
\end{theorem}
\proof
For any $\epsilon>0$ there must be some $\delta_0>0$such that for any
partition of $X$, $P_N$ such that $\max|P_i-P_{i-1}|<\delta_0$,
$$-\epsilon+\sum_{i=0}^{N-1} f(P_i)\mdot\,(P_{i+1}-P_i)
<\cint\limits_Xf<
\epsilon+ \sum_{i=0}^{N-1} f(P_i)\mdot\,(P_{i+1}-P_i).
$$
In particular, $P_i=A+i\mdot\delta$ for $i<N$ and $P_N=\max(X)$ is such a 
partition if $\delta<\delta_0$.

In order to be integrable by continuity, there must be some $M$ so that 
$|f(x)|<M$ for any $x\in X$ (otherwise, by selecting points in the partition 
where $f$ is arbitrarily large, partitions of arbitrarily small mesh could
have arbitrarily large sums), so $\epsilon_2\ge\epsilon+M\mdot\delta$ can
be made arbitrarily small by selecting $\epsilon$ and $\delta$ to be small.
If $n=\lfloor (x-A)/\delta\rfloor$ then 
$$-M\mdot\delta+\cint_A^xf<\cint_A^{A+n\mdot\delta}f\le\cint_A^xf$$ 

so
$$-\epsilon_2+\sum_{i=0}^{n-1}\delta\mdot f(A+i\mdot\delta)
<  \cint_A^xf
< \epsilon_2+\sum_{i=0}^{n-1}\delta\mdot f(A+i\mdot\delta)
$$

and
$$-\epsilon_2+\sum_{i=0}^{n}\delta\mdot f(A+i\mdot\delta)
<  \cint_A^{x+\delta}f
< \mdot\epsilon_2+\sum_{i=0}^{n}\delta\mdot f(A+i\mdot\delta)
$$

Using this to approximate the derivative,
 
$${1\over\delta}\mdot\left(
{\sum_{i=0}^{n}\left({-\epsilon_2\over n}+\delta\mdot f(A+i\mdot\delta)\right)-
\sum_{i=0}^{n-1}\left({\epsilon_2\over n-1}+\delta\mdot f(A+i\mdot\delta)
\right)}\right)$$
$$<{1\over\delta}\mdot\Big({\cint_A^{x+\delta}f-\cint_A^xf}\Big)<$$
$${1\over\delta}\mdot\left(
{\sum_{i=0}^{n}\left({\epsilon_2\over n}+\delta\mdot f(A+i\mdot\delta)\right)-
\sum_{i=0}^{n-1}\left({-\epsilon_2\over n-1}+\delta\mdot f(A+i\mdot\delta)
\right)}\right)
$$

which simplifies to
$$
{-{\epsilon_2\over n-1}+\delta\mdot f(A+n\mdot\delta)\over\delta}
<{1\over\delta}\mdot\Big({\cint_A^{x+\delta}f-\cint_A^xf}\Big)<
{{\epsilon_2\over n-1}+\delta\mdot f(A+n\mdot\delta)\over\delta}
$$

and
$$
-{\epsilon_2\over\delta\mdot\,(n-1)}+f(A+n\mdot\delta)
<{1\over\delta}\mdot\Big({\cint_A^{x+\delta}f-\cint_A^xf}\Big)<
{\epsilon_2\over\delta\mdot\,(n-1)}+f(A+n\mdot\delta).
$$

If $\delta$ is made small, then $\delta\mdot\,(n-1)\approx(x-A)$ and since
$f$ is continuous $f(A+n\mdot\delta)\approx f(x)$. Thus, for fixed
$x$ and $\epsilon_2$, as $\delta\to 0$

$$
-{\epsilon_2\over(x-A)}+f(x)
<{d\over dx}\cint_A^xf<
{\epsilon_2\over(x-A)}+f(x).
$$

and since $\epsilon_2$ can be arbitrarily small,
$${d\over dx}\cint_A^xf=f(x)$$

for any $x>A$. 

For the case $x=A$, notice that $f$ is continuous, so for any 
$\epsilon>0$ there is some $\delta>0$ such that 
$|f(x+\delta)-f(x)|<\epsilon$.  This means
$$\left|\cint_A^{A+\delta}f-f(x)\mdot\delta\right|<\epsilon\mdot\delta.$$

Varying $\delta$ with $\epsilon$ fixed gives
$$\left|\lim_{\delta\to0}
\left({1\over\delta}\mdot\cint_A^{A+\delta}f\right)-f(A)\right|
<\epsilon$$ 

and since $\epsilon$ is arbitrary, this makes
$${d\over dx}\cint_A^xf=f(x)$$
true for $x=A$.
\endproof

This theorem is not the one which requires the extra conditions; the only
restriction on $f$ is that it be continuous (not even strong continuity
is necessary.)  Strong differentiability, and with it strong continuity,
are required for the other theorem of the pair.

\begin{theorem}
If $f$ is strongly differentiable on $X\cap[A,B]$ and $f^\prime$ is
integrable by continuity on $X\cap[A,B]$ then
$$\cint_A^Bf^\prime=f(B)-f(A).$$
\end{theorem}
\proof
Since $f^\prime$ is integrable by continuity, for any $\epsilon>0$ there
is some $\delta_1>0$ such that for any $0<\delta<\delta_1$ 
$$-\epsilon+\sum_{i=0}^{N}\delta\mdot f^\prime(A+i\mdot\delta)
<\cint_A^Bf^\prime<
\epsilon+\sum_{i=0}^{N}\delta\mdot f^\prime(A+i\mdot\delta)$$
where $N=\lfloor (B-A)/\delta\rfloor$.  Also, since $f$ is strongly
differentiable and its domain is bounded, the approximation function $g$ for
$f^\prime$ must be uniformly continuous.  This means that there must be some 
$\delta_2>$ so that for any $\delta<\delta_2$ 
$$-\epsilon+{f(A+i\mdot\delta+\delta)-f(A+i\mdot\delta)\over \delta}
<f^\prime(A+i\mdot\delta)<
\epsilon+{f(A+i\mdot\delta+\delta)-f(A+i\mdot\delta)\over \delta}$$
where $\delta_2$ does not depend on $A+i\mdot\delta$.  Combining these 
results gives
$$\left|\cint_A^Bf^\prime-
\sum_{i=0}^{N}{\delta\mdot(f(A+i\mdot\delta+\delta)-f(A+i\mdot\delta))
\over \delta}\right|
<\sum_{i=0}^N\delta\mdot\epsilon+\epsilon$$

or, since $N\mdot\delta\le(B-A)$,
$$
\left|\cint_A^Bf^\prime - (-f(A)+f(A+N\mdot\delta+\delta))\right|
<\epsilon\mdot(B-A)+\epsilon$$
Since $\epsilon$ is arbitrary, the term on the left must be 0, and since $f$ is
continuous and $B\le A+N\mdot\delta+\delta\le B+\delta$ this must mean
$$\cint_A^Bf^\prime=f(B)-f(A).$$
 \endproof

These results allow the usual techniques of integration to be used.  If a 
strongly differentiable function $F$ can be found which has derivative
$F^\prime=f$, then the integral of $f$ can be found directly by evaluating $F$.
The formula for integration by parts follows directly from the product
rule for differentiation.

\endsection
%-------------------------------------------------------------------------
\interlude{More Sets of Numbers}
The past few chapters have gone through arcane contortions to produce
definitions of the operations of calculus which make sense on sets of
numbers which aren't complete.  This seems like a reasonable project, but
in execution something seems to be missing; for both integration and 
differentiation there is a remark buried away in the text to the effect
that these operations still cannot be guaranteed to be defined unless they
are taken to give arbitrary real numbers as values.  Certainly the
processes used to approximate these operators as presented now give meaningful
values for any closeness of approximation desired, but at the final step there
is still a limit process, which means that the correct value can still fall
anywhere in the real numbers, not just in some incomplete subset.
It is perhaps a nice thing to have definitions which fail only in predictable
ways, and to know that when they do converge to a solution that the solution 
will have reasonable properties, but we are still left with the fundamental
problem that sometimes functions which look like they should have derivatives
or integrals somehow simply do not.

My response to this is to draw an analogy to the algebraic numbers.  When
working in the rational numbers often there are polynomials which give
every impression of having roots, but where the roots are not any rational
number.  This situation can be dealt with either by appealing to the 
set of real numbers, with its attendant abilities and difficulties, or by
constructing a new set which contains the rationals as well as new elements
for all the missing solutions.  This set is more difficult to work with
than the rationals, and its elements are harder to characterize, but it
is still countable, and its elements can be defined strictly in terms of their
algebraic properties.  The tendency for reasonable looking operations to
be undefined is a general property of incomplete spaces, but for any specific
set of operations an incomplete set can be found which suffices for them.

The temptation here is to simply assert that there is some countable set
of numbers which is closed under differentiation and integration, and perhaps
to make a few conjectures about its structure.  Unfortunately, example
\ref{incomplete} had already shown that this is not true.  There are
functions defined over the rational numbers which take any one element of
the reals as a derivative somewhere.  What has gone wrong?

Integration and differentiation are not operations on numbers.  They are
operators on functions.  The correct space to try to enlarge in order
to make these processes closed is a space of functions.  If we are to
consider all possible functions from the rational numbers to themselves,
there are already an uncountable number of them and it is no surprise that
in order to make the set closed under some arbitrary operation another
uncountable number must be added.  Since some of these functions bring
with them new points in their ranges, the underlying space of numbers
can be expected to become large.

If the idea of closure of a countable set under derivatives and integrals is
to be saved, the set to first consider is the set of functions on which they
act.  We can begin by choosing an essentially arbitrary set of numbers on
which these functions will be provisionally defined.  It will probably need
to be expanded as functions are added, but as long as it is clear how the
existing functions should be extended to their new domains this does not
necessarily present a problem.  If we want this set to be countable when we
are finished it should be countable when we start, of course, and it should
have the sort of ordering properties and arithmetic operations that allow
useful calculations.  If these conditions are satisfied for the initial set
of numbers, they will still be true when all extensions are made, since
closing the set under algebraic operations should be straightforward and
as long as only a countable number of functions are added to the space of 
functions, even if each one required the addition of a countable number of
points to the set of numbers, that set will still be countable.

My inclination is to use the rational numbers as an initial set, occasionally
restricting the domains of functions to intervals in the rationals to avoid
division by zero and keep algebraic operations from making functions unbounded.
This done, an initial set of functions should be chosen.  A very large set
could plausibly be used on the assumption that it would be more likely to
contain every useful function when the process was finished, but it is also
possible to begin with a very small set based on the philosophy that this
will allow easier analysis of the final result by algebraic methods.  If both
the set of operations and the set of initial objects which they act on to
generate the final set are small and well understood, this leaves more
opportunities to develop powerful tools to examine the structure of the final 
result.  With this in mind, I propose using the set $\{1,x\}$.
(In fact, $x$ is a redundant member, since $\int1=x$, but I am including it
because it allows me to do things in an order I fine easier.)

This set of functions can now be closed under the operations of pointwise
addition, additive inverse, multiplication and multiplicative inverse which
produces the set of all rational functions.  It can be closed as well under
differentiation (which adds no new members, since the derivative of a rational
function is another rational function) and integration, expressed as the
unary operator
$$I(f(x))=\int_0^x f(t)\, dt$$
with the usual convention that the negative of the integral is taken and the
bounds reversed when $x<0$.  This last step has added new functions to
the set, and with them come new numbers.  The function
$$\int {1\over x^2+1}dx$$
for instance, is not a rational function and most of its values are not
rational numbers.

Once some initial analysis has been done to demonstrate that these operations
can actually be expected to be well defined, these sets, both the numbers and
the functions can be manipulated by the purely formal methods characteristic
of algebra.  No questions need necessarily be asked about questions such as
convergence of limits, and the sets do not need to be treated as topological
spaces.  They can be understood strictly in terms of the interactions between
the operations given; addition commutes with integration, while multiplication
does not.  Differentiation and multiplication are not commutative, but there
is a more complex symbolic relationship, and so forth.

In fact, this is not the only interesting countable set built by these
 operations.  Just as the rational numbers are closed under
multiplication and addition, but not all polynomials constructed from
them by these operations have solutions, this set does not necessarily
contain the solutions to all equations created from the given operations.
It is reasonable to consider the set of all functions which can solve
an equation of the form
$${\cal O}(f)=0$$
where ${\cal O}$ is any operator constructed by finite combination of 
addition, subtraction, multiplication, division and members of the set of
functions itself which can act meaningfully on a single other function.
These operators are the logical generalization of the polynomials on the
rational numbers to this space of functions, and there can only be
countably many of them.  If each one is only required to have a single,
or even a countable number of solutions then the set of functions produced
and the set of numbers they are defined over will both remain countable.

There are some difficulties with this notion.  If any possible functions which
satisfy these operator equations pointwise are allowed, there are not only a
countable number of solutions to each one.  ${\cal O}(f)=f^2-1$, for instance,
has solutions $f=1$ and $f=-1$, but also an uncountable number of hybrid
choices which take on one value some places and the other elsewhere.  Also,
$f^\prime=0$  has as its solutions every possible constant function.  Some
method for selecting from these possibilities must be found before these
sets can be constructed.

If these difficulties are avoided, this procedure will produce a pair of
sets, one of numbers and one of functions, whose properties can be described
algebraically, and which are countable.  The set of functions will be a
neglidgable subset of the analytic functions with coefficients from the
set of numbers.  By the standards of analysis, such a set would be startlingly
small.  None the less, it contains a huge number of the functions which are
actually used, including the solutions to all one dimensional differential
equations, and essentially every function used outside of pure mathematics is
at least constructed from members of it piecewise.

\endsection
%----------------------------------------------------------------
\end{document}
\end
