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%\huge
\centerline{\large \bf Logic Notation: Convergence and Continuity}
\vskip20pt {\sc Reference: } This is copied from the book
\emph{Fundamentals of Abstract Analysis} by Andrew Gleason.
\vskip 20pt
{\bf Convergence of a Sequence}
\medskip
1. \ A sequence $x_n$ of real numbers is said to be \emph{increasing} (or
\emph{monotone increasing}) if
$$
(\forall m,\, n) \ (m>n) \Rightarrow x_m \ge x_n.
$$
\medskip
2. \ Let $z_n$ be a sequence of complex numbers. The sequence is said to
{\it converge to} $z$, in symbols, $z_n \to z$, if
$$
(\forall \epsilon>0)\ (\exists N\in\N)\ (\forall n>N)
\ \abs{z_n-z}<\epsilon.
$$
\medskip
3. \ Let $z_n$ be a sequence of complex numbers. The sequence is said to
{\it converge} if
$$
(\exists z\in\C)\ (\forall \epsilon>0)\ (\exists N\in\N)\ (\forall n>N) \
\ \abs{z_n-z}<\epsilon.
$$
In greater detail:
Since this begins with an existential quantifier, we must move first
by choosing a number $z$. Since the next quantifier is universal, our
opponent moves next by choosing a positive number $\epsilon$. The
opponent will presumably make the best possible move and choose
$\epsilon$ so that
$$
(\exists N\in\N)\ (\forall n>N) \ \ \abs{z_n-z}<\epsilon
$$
is false (if possible). Now it is our move to choose a number $N$
with a knowledge of the previous moves, that is, $N$ may (and surely
will) depend on both $z$ and $\epsilon$. Finally our opponent chooses
a number $n>N$ and the burden is on us to prove the inequality
$\abs{z_n-z}<\epsilon$.
\medskip
\vskip 20pt
{\bf Chess: Checkmate}
\medskip
4. \ Using this language, for a chess game, the usual ``white mates in two
moves'' can be thought of as:
\begin{center}
($\exists$ white move)\ ($\forall$ black moves) \ ($\exists$ white
move) \ ($\forall$ black moves) \ black is checkmated
\end{center}
where of course ``black is checkmated'' means ``white can capture
black's king''.
\vskip 20pt
\newpage %\vskip 20pt
{\bf Continuous Functions}
\medskip
5. \ Let $S$ and $T$ be metric spaces with metrics
$d_S(\cdot,\,\cdot)$ and $d_T(\cdot,\,\cdot)$ and $f:S \to T$. Then
$f$ is {\it continuous at a point $p\in S$} if
$$
(\forall \epsilon>0)\ (\exists \delta>0)\ (\forall q\in S)
\ d_S(p,\,q)<\delta \Rightarrow d_T(f(p),\,f(q))<\epsilon.
$$
$f$ is continuous on the whole set $S$ if it is continuous at each
point of $S$. More formally,
$$
(\forall p\in S) \ (\forall \epsilon > 0) \ (\exists \delta>0)
(\forall q\in S) \ \ d_S(p,\,q)<\delta \Rightarrow
d_T(f(p),\, f(q)) <\epsilon.
$$
%\smallskip
We restate this in terms of sequences. Say $p\in S$. Then $f$ is
continuous at $p$ if and only for every sequence $x_n\to p$ then
$f(x_n) \to f(p)$. That is
$$
\lim f(x_n) = f(\lim x_n)
$$
\smallskip
We can also restate the definition of continuity in terms of balls
$B_S(p,\,\delta)$ and $B_T(s,\,\epsilon)$ in $S$ and $T$,
respectively:
$$
(\forall p\in S)\ (\forall \epsilon>0)\ (\exists \delta>0)
\ \ f(B_S(p,\,\delta)) \subseteq B_T(f(p),\,\epsilon).
$$
%\smallskip
5'. {\sc Equivalent Definition of Continuity} \\ It is interesting
that one can describe ``continuity'' in a dramatically different way
without using ``limit'' or explicitly referring to the metric. As a
consequence, this is used as the \emph{definition} of continuity in
more general topological spaces that are not metric spaces. It also
simplifies many proofs.
\smallskip
Some notation: Let $f:S\to T$. If $S_0$ is a subset of $S$, denote by
$f(S_0)$ the set of all image points of $S_0$ under the function $f$,
so
$$
f(S_0)= \{t\in T : t=f(s)\text{\ for some\ } s \in S_0\}.
$$
Similarly, if $T_0$ is a subset of $T$, denote by $f^{-1}(T_0)$ the
set of all points in $S$ whose image is in $T_0$:
$$
f^{-1}(T_0) = \{s\in S : f(s)\in T_0\}.
$$
$f^{-1}(T_0)$ is called the {\it preimage} of $T_0$. It can happen
that no points in $S$ have their image in $T_0$. A simple example is
the map $f:\R\to \R$ defined by $f(s)=s^2$ and $T_0 = \{ t\in \R : t <
0\}$. Then the preimage of this $T_0$ is empty since the square of any
real number is not negative.
\medskip
{\sc Caution } The operation $f^{-1}$ applied to subsets of $T$
behaves nicely. It preserves inclusions, unions, intersections and
differences of sets. However the operation $f$ applied to subsets of
$S$ is more complicated.
Note also that if $f:S\to T$, while
$$
f^{-1}(f(S_0)) \supset S_0 \text{\ \ for \ } S_0\subset S \qquad
\text{and}\qquad
f(f^{-1}(T_0)) \subset T_0 \text{\ \ for \ } T_0\subset T,
$$
equality often does not hold. Here is a (non-pathological) example:
\medskip
\parbox{10cm}{
Let $f: \R\to \R$ be $f(x)=2x^2 + 1$ and use the standard notation
$[a,b] = \{a\le x \le b\}$. Since $f$ is not one-to-one, then two
different sets can have the same images
$$
f([0,\,1])= f([-1,\,1])= [1,\,3].
$$
while because $f$ is not onto, two different sets can have the same
preimages
$$
f^{-1}([0,\,3]) = f^{-1}([1,\,3]) = [-1,\,1],
$$}
\hskip25pt
\parbox{3.5cm}{\includegraphics[scale=.70]{preimage.eps}
\hspace*{0.9cm}
$f(x)=2x^2 + 1$}
\smallskip
Using these we obtain the examples:
$$
f^{-1}(f([0,\,1])) = f^{-1}([1,\,3]) = [-1,\,1] \qquad\text{and} \qquad
f(f^{-1}([0,3])) = f([-1,\,1]) = [1,\,3].
$$
\smallskip
{\sc Theorem } \emph{Let $S$, $T$ be metric spaces and $f:S \to T$. Then
$f$ is continuous on $S$ $\Longleftrightarrow$ for any open set $G \subset
T$, the set $f^{-1}(G)$ is open. That is, the preimage of an open set
is open.}
\medskip
{\sc Proof: } $\Longrightarrow$ Say $f$ is continuous and we are given an
open set $G\in T$. If $f^{-1}(G)$ is the empty set, there is nothing
to prove. Thus, say $p \in f^{-1}(G)$. This means there is a point
$q\in G$ with $f(p)=q$. We need to find a ball $U:=B_S(p,\,\delta)$
around $p$ so that $f(U)\subset G$.
Since $G$ is open, it contains some ball $V:=B_T(q,\,\epsilon)$. By
the continuity of $f$ there is a $\delta>0$ so that $f(U)\subset V$.
Thus the open set $U$ is in the preimage of $G$. $\Box$
\smallskip
$\Longleftarrow$. Say the preimage of any open set $G\subset T$ is open. To
show that $f$ is continuous at every point $p\in S$.
Given any $\epsilon>0$, let $G:= B_T(f(p),\,\epsilon)$. We need to
find a $\delta>0$ so that image of $B_S(p,\,\delta) \subset
B_T(f(p),\,\epsilon)$. But since the preimage of $G$ is open, it
contains some small ball $B_S(p,\,\delta)$ around $p$. $\Box$
\medskip
Using that a set is closed if and only if its complement is open and
that $f^{-1}(T_0^c)=[f^{-1}(T_0)]^c$ for every subset $T_0\subset T$,
we can use closed sets instead of open sets to varify continuity.
That is,
\medskip
{\sc Corollary } \emph{Let $S$, $T$ be metric spaces and $f:S \to T$.
Then $f$ is continuous on $S$ $\Longleftrightarrow$ for any closed
set $C \subset T$, the set $f^{-1}(C)$ is closed. That is, the preimage
of a closed set is closed.}
\bigskip
6. \ In general if $f$ is continuous at every point $p$ of a metric
space, the choice of $\delta$ will depend on both $\epsilon$ and the
particular point $p$. If given $\epsilon$ we can find a $\delta$ that
works simultaneously for every point $p$ then the function is said to
be {\it uniformly continuous}. More formally
$$
(\forall \epsilon>0)\ (\exists \delta>0) \ (\forall p,\,q\in S)
\ \ d_S(p,\,q) <\delta\ \Rightarrow\ d_T(f(p),\,f(q)) < \epsilon).
$$
Equivalently:
$$
(\forall \epsilon>0)\ (\exists \delta>0) \ (\forall p\in S)
\ \ f(B_S(p,\,\delta)) \subseteq B_T(f(p),\,\epsilon).
$$
\vskip 20pt
{\bf Convergence of a Sequence of Functions}
\medskip
If $S$ is a set, $(T,\,d)$ is a metric space and $\{f_n\}: S\to T$ is
a sequence of functions, the next two definitions concern the
convergence of the $\{f_n\}$ to a function $g$.
\medskip
7. \ $\{f_n\}$ is said to converge \emph{pointwise} to $g$ if for all
$p\in S$ we have $f_n(p) \to g(p)$. In greater detail
$$
(\forall p\in S) \ (\forall \epsilon > 0) \ (\exists N\in \N)
(\forall n>N) \ \ d(f_n(p),\, g(p)) <\epsilon.
$$
\medskip
9.. \ \ $\{f_n\}$ is said to converge to $g$ \emph{uniformly on $S$} if
$$
(\forall \epsilon > 0) \ (\exists N\in \N) \
(\forall n>N) \ (\forall p\in S) \ \ d(f_n(p),\, g(p)) <\epsilon.
$$
For pointwise convergence the choice of $N$ can depend on $p$, while
for uniform convergence the same $N$ works simultaneously for all
$p\in S$.
For example, if $S=\{0 < x < 1\}$ then $f_n(x):=x^n$ converges
pointwise but not uniformly to $g(x):=0$. However if $S:=\{0