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\centerline {\bf METRIC SPACES}
\medskip
In this note we discuss various properties of metric spaces. 

\bigskip
\centerline {\bf \S1. Metric Spaces}
\smallskip
A metric space is a topological space where the topology is 
determined by a distance function. 

{\bf Definition.} A metric space $(\x, d)$ is a nonempty 
set $\x$ with a real-valued function $d$ defined on $\x\times\x$  
such that for any $x, y, z\in\x$ 

(a) $d(x, y)\ge 0$ and $d(x, y)=0$ if and only if $x=y$; 

(b) $d(x, y)=d(y, x)$; 

(c) $d(x, y)\le d(x, z)+d(z, y)$. 

\nin 
The function $d$ is called a metric. 
\smallskip

The purpose of introducing metrics is to describe convergence. 

{\bf Definition.} A sequence $\{x_n\}$ in metric space $(\x, d)$ is 
convergent to $x\in\x$ if $\lim_{n\to\infty}$ 
$ d(x_n, x)=0$. 
 A sequence $\{x_n\}$ is Cauchy if 
$ \lim_{m, n\to\infty} d(x_n, x_m)=0$. 

{\bf Definition.} A metric space $(\x, d)$ is complete if 
any Cauchy sequence is convergent. 

\smallskip 
\demo{Remark} 
A metric space is the generalization of the Euclidean space. 
Let $\Bbb R$ denote the collection of all real numbers and set 
$d(x, y)=|x-y|$ for any $x, y\in \Bbb R$. Then $(\Bbb R, d)$ 
is a complete metric space. 
\enddemo 

\demo{Example} Let $C[a, b]$ denote the collection of all 
continuous functions in $[a, b]\subset \Bbb R$. Set 
$$d(x, y)=\max_{a\le t\le b}|x(t)-y(t)|\quad\hbox{for any}\ 
x, y\in C[a, b].$$ 
Then $(C[a, b], d)$ is a metric space. In $C[a, b]$, a sequence $\{x_n\}$ 
convergent to $x$ is equivalent to the fact that $\{x_n(t)\}$ 
is uniformly convergent to $x(t)$ in $[a, b]$.  
In fact $(C[a, b], d)$ is complete. 
\enddemo 

\smallskip
Now we use metric to define a topology. In the following we 
always assume that $(\x, d)$ is a metric space. 

{\bf Definition.} A set $O\subset \x$ is open if for any $x\in O$ there 
exists an $r>0$ such that $y\in O$ for any $y\in \x$ with $d(x, y)<r$. 

\proclaim{Proposition} The set $\x$ and $\phi$ are open; the intersection of 
any two open sets is open; and the union of any 
collection of open sets is open. 
\endproclaim 

{\bf Definition.} A set $F\subset \x$ is closed if $F^c$ is open. 

\proclaim{Proposition} A set $F\subset \x$ is closed if and only if for 
any sequence $\{x_n\}\subset F$ which is convergent in $\x$ the limit 
$x$ belongs to $F$. 
\endproclaim 

Suppose $(\x, d)$ and $(\y, \rho)$ are two metric spaces.

{\bf Definition.} A mapping $A: \x\to \y$ is continuous at $x_0\in\x$ if
for any $\varepsilon>0$ there exists a $\delta>0$ such that
$d(x, x_0)<\delta$ implies $\rho(A(x), A(x_0))<\varepsilon$.

\proclaim{Proposition} 
A mapping $A: \x\to \y$ is continuous at $x_0\in\x$ if 
and only if for any open set $O\subset \y$ containing $A(x_0)$ the set
$A^{-1}(O)$ is open $\x$. 
\endproclaim 

\bigskip 
\centerline{\bf \S2. Compactness} 
\smallskip 

We always assume that $(\x, d)$ is a metric space. For any $x_0$ and $r>0$ 
we denote 
$$B(x_0; r)=\{x\in \x; d(x, x_0)<r\}.$$ 
A subset $A\subset \x$ is bounded if $A\subset B(x_0; r)$ 
for some $x_0$ and $r>0$. 

In $\Bbb R^n$ any bounded infinite set contains a convergent subsequence. 
This is not true in the general metric space. 

\demo{Example} In $C[0, 1]$ consider the sequence  
$$x_n(t) = \cases
0 &\text{ for } t\ge \frac 1n\\
1-nt &\text{ for  } t\le \frac 1n \endcases$$
for $n=1, 2, \cdots$. Obviously $\{x_n\}\subset B(0; 1)$ where $0$ denotes 
the identically zero function. However $\{x_n\}$ has no convergent subsequence 
in $C[0, 1]$. 
\enddemo 

\smallskip 
{\bf Definition.} In a metric space $(\x, d)$, a subset $M$ is compact if 
any open covering of $M$ has a finite subcovering. 

\proclaim{Theorem} In a metric space $(\x, d)$, a subset $M$ is compact if 
and only if $M$ is closed and any sequence in $M$ has a convergent 
subsequence. 
\endproclaim 
\demo{Proof} $\Rightarrow$ Suppose $M$ is compact. First we prove 
that 
$M^c$ is open. For any $x_0\in M^c$ we have 
$M\subset \cup_{x\in M}B(x; d(x, x_0)/2)$. By compactness there 
exist $x_1, \cdots, x_n\in M$ such that 
$M\subset \cup_{i=1}^nB(x_i; d(x_i, x_0)/2)$. 
Take $\delta=\min_{1\le i\le n}d(x_i, x_0)/2>0$. Then $M\cap B(x_0, \delta)= 
\phi$. Hence $M^c$ is open. 

Next we take any sequence $\{x_n\}\subset M$. We 
prove by contradiction that $\{x_n\}$ has a convergent subsequence. 
Suppose the sequence $\{x_n\}\subset M$ has no convergent subsequences. 
We may assume $x_n\neq x_m$ for $n\neq m$. For any $n\in \Bbb N$ the set 
$S_n=\{x_1, \cdots, x_{n-1}, x_{n+1}, \cdots\}$ is closed since it does 
not have convergent subsequence. Hence $\x\backslash S_n$ is open. 
Note $\cup_{n=1}^\infty (\x\backslash S_n)=\x\backslash \cap _{n=1}^\infty 
S_n=\x\supset M$. By the compactness there exists an $N\in \Bbb N$ such that 
$\cup_{n=1}^N\x\backslash S_n\supset M$, i.e., 
$\x\backslash \{x_{N+1}, x_{N+2}, \cdots\}\supset M$. This is 
impossible since $x_{N+1}\subset M$ while 
$x_{N+1}\notin \x\backslash \{x_{N+1}, x_{N+2}, \cdots\}$. Contradiction. 

$\Leftarrow$ Suppose $M$ is closed and that any sequence in $M$ has 
a convergent subsequence. We first prove that 
for any $\varepsilon>0$ there exist finitely
many points $x_1, \cdots, x_k$ in $M$ for some $k=k(\varepsilon)$
such that
$M\subset \cup_{i=1}^kB(x_i; \varepsilon)$.
 
If it is not true, then there
exists an $\varepsilon_0>0$ satifying the following property:
 
for some $x_1\in M$ there exists an $x_2\in M\backslash B(x_1; \varepsilon_0)$;
 
for $\{x_1, x_2\}\subset M$ there exists an
$x_3\in M\backslash B(x_1; \varepsilon_0)\cup B(x_2; \varepsilon_0)$;
 
$\cdots\cdots$
 
for $\{x_1, \cdots, x_n\}\subset M$ there exists an
$x_{n+1}\in M\backslash \cup_{i=1}^n B(x_i; \varepsilon_0)$;
 
$\cdots\cdots.$
 
\nin The sequence $\{x_n\}\subset M$ satisfies $d(x_n, x_m)\ge
\varepsilon_0$ for $n\neq m$. It could not have convergent subsequence.
This is a contradiction.
 
Take any open covering $\{G_\lambda\}$ of $M$. Suppose it has no 
finite subcovering.   
For any $n\in \Bbb N$ there exist 
$x_1^{(n)}, \cdots, x_{k(n)}^{(n)}\in M$ such that $M\subset 
\cup_{i=1}^{k(n)}B(x_i{(n)};  1/n)$.  
Hence for any $n\in \Bbb N$ there exists a $y_n\in 
\{ x_1^{(n)}, \cdots, x_{k(n)}^{(n)}\}\subset M$ such that 
$B(y_n; 1/n)$ cannot be covered by finitely many $G_\lambda$. 
By assumptions there exists a convergent 
subsequence 
$\{y_{n_k}\}$ with limit $y\in M$. We may assume $y\in G_{\lambda_0}$. 
Since $G_{\lambda_0}$ is open there exists a $\delta>0$ such that 
$B(y; \delta)\subset G_{\lambda_0}$. For such $\delta$ we may take 
$k$ large such that 
$B(y_{n_k}; 1/n_k)\subset B(y, \delta)\subset G_{\lambda_0}$. This 
contradicts the assumption that 
every $B(y_n; 1/n)$ cannot be covered by finitely many 
$G_\lambda$. $\bl$ 
\enddemo 


\bigskip
\centerline{\bf \S3. Continuous Mappings on Compact Spaces}
\smallskip

Next we discuss properties of continuous mappings on compact metric space.
We state some results without proof. 
 
\proclaim{Theorem} Suppose $A$ is a continuous mapping of a 
compact metric space $\x$ into a metric space $\y$. Then $f(\x)$ 
is compact.  
\endproclaim 
 
\proclaim{Corollary} Suppose $f$ is a continuous function in a 
compact metric space $\x$. Set
$$M=\sup_{x\in \x}f(x), \quad m=\inf_{x\in \x}f(x). $$
Then there exist points $p, q\in \x$ such that
$f(p)=M$ and $f(q)=m$.
\endproclaim 
 
\smallskip 
{\bf Definition.}
Suppose $A$ is a mapping: $\x\to \y$.
$A$ is uniformly continuous in $\x$ if
for any $\varepsilon>0$ there exists a $\delta>0$ such that
$$\rho(A(x_1), A(x_2))<\varepsilon\quad\text{if}\
d(x_1, x_2)<\delta, \ \text{and}\ x_1, x_2\in \x.$$
 
\proclaim{Theorem} Suppose $A$ is a continuous mapping from a 
compact metric space $\x$ into a metric space $\y$. Then  $A$
is uniformly continuous.
\endproclaim 

\medskip
 
Next we generalize the notion of continuous functions in bounded closed 
intervals. Suppose $(M, \rho)$ is a compact metric space. 
$C(M)$ denotes the collection of all continuous functions: 
$M\to \Bbb R$. Define 
$$d(u, v)=\max_{x\in M}|u(x)-v(x)|\quad\hbox{for any}\ u, v\in C(M).$$ 

\proclaim{Lemma} $(C(M), d)$ is a complete metric space. 
Moreover the convergence in $(C(M), d)$ is equivalent to the
uniform convergence.
\endproclaim 

{\bf Definition.} Suppose $\{f_n\}$ is a sequence in $C(M)$.
 
(a) $\{f_n\}$ is uniformly bounded if for some $M>0$
there holds $|f_n(x)|\le M$ for any $x\in M$ and any integer $n$;
 
(b) $\{f_n\}$ is equicontinuous if for any $\varepsilon>0$ there exists a
$\delta(\varepsilon)>0$ such that
$|f_n(x_1)-f_n(x_2)|<\varepsilon$ for any integer $n$
and $x_1, x_2\in M$ with $\rho(x_1, x_2)<\delta$.
 
\proclaim{Arzela-Ascoli Theorem} Suppose $\{f_n\}$ is a sequence in $C(M)$.
Then $\{f_n\}$ has a convergent subsequence (with respect to the metric
$d$ defined above) if and only if $\{f_n\}$ is
uniformly bounded and equicontinuous.
\endproclaim


\bigskip
\centerline {\bf \S4. Contraction Mapping Theorem}
\smallskip
Suppose $(\x, d)$ is a metric space and $A: \x\to\x$ is a mapping.
In lots of cases we need to investigate the
solvability of $Ax=x$ in $\x$, i.e., whether
$A$ has a fixed point in $\x$.
 
{\bf Definition.} The mapping $A: (\x, d)\to (\x, d)$ is a contraction
mapping if for some $\alpha\in (0, 1)$ there holds
$d(Ax, Ay)\le \alpha d(x, y)$ for any $x, y\in \x$.
 
\demo{Remark} Contraction mappings are always continuous.
\enddemo
 
The following theorem is the simplest and most widely
used existence result in functional analysis.
 
\proclaim{Contraction Mapping Theorem}
Suppose $(\x, d)$ is a complete metric space
and  $A: (\x, d)\to (\x, d)$ is a contraction
mapping. Then $A$ has a unique fixed point on $\x$.
\endproclaim
 
\demo{Proof} For any $x_0\in \x$ set $x_{n+1}=Ax_n$ for
$n=0, 1, 2, \cdots$. We claim that $\{x_n\}$ is a
Cauchy sequence. To see this we calculate
$$d(x_{n+1}, x_n)=d(Ax_n, Ax_{n-1})\le \alpha d(x_n, x_{n-1})\le
\cdots\le \alpha^n d(x_1, x_0).$$
Hence for any $n, p\in\Bbb N$, there holds
$$\align
d(x_{n+p}, x_n)\le& \sum_{i=1}^pd(x_{n+i}, x_{n+i-1})\le \sum_{i=1}^p
\alpha^{n+i-1}d(x_1, x_0)\\
\le & \sum_{i=1}^\infty \alpha^{n+i-1}d(x_1, x_0)=\frac{\alpha^n}{1-\alpha}
d(x_1, x_0). \endalign$$
Hence $\{x_n\}$ is a
Cauchy sequence, and convergent to $x^*$ in $\x$. Taking the limit
in $Ax_n=x_{n+1}$ we conclude $Ax^*=x^*$ by continuity of $A$.
 
If $x^{**}$ also satisfies $Ax^{**}=x^{**}$, then
$d(x^*, x^{**})=d(Ax^*, Ax^{**})\le \alpha d(x^*, x^{**}).$
This implies $x^*=x^{**}$. $\bl$
\enddemo
 
As an application we discuss the initial value
problem for the following ordinary differential equation
$$
\frac{dx}{dt}=f(t, x), \  \
x(0)=\xi\tag1$$
where $f$ is a continuous function in
$[-a, a]\times[\xi-b, \xi+b]$ for some $a, b\in (0, \infty)$.
We consider its equivalent integral equation
$$x(t)=\xi+\int_0^tf(s, x(s))ds$$
and view it as a fixed point problem.

\smallskip
For some $h\in (0, a]$, consider
$$\x=\{x\in C[-h, h]; \max_{-h\le t\le h}|x(t)-\xi|\le b\}.$$
$\x$ is a closed subset in $ C[-h, h]$ and hence a complete metric
space. Set
$$Ax(t)=\xi+\int_0^tf(s, x(s))ds\quad\hbox{for any}\  x\in\x.$$
Obviously $Ax\in C[-h, h]$. First we need to check that
$Ax\in \x$ for any $x\in \x$. To do this we have for $|t|\le h$
$$|Ax(t)-\xi|=|\int_0^tf(s, x(s))ds|\le \int_0^h|f(s, x(s))|ds.$$
We assume that
$$M=\sup_{|t|\le a, |x-\xi|\le b}|f(t, x)|.\tag2$$
Then
$$\max_{|t|\le h}|Ax(t)-\xi|\le hM.$$
Hence $Ax\in\x$ for any $x\in\x$ if we choose $h\le a$ such that
$hM\le b$. Next for any $x, y\in\x$ we consider
$$Ax(t)-Ay(t)=\int_0^tf(s, x(s))ds-\int_0^tf(s, y(s))ds.$$
We assume for some positive constant $L$
$$|f(t, x_1)-f(t, x_2)|\le L|x_1-x_2|\tag3$$
for any $|t|\le a$ and $|x_1-\xi|\le b$, $|x_2-\xi|\le b$.
Then
$$\max_{|t|\le h}|Ax(t)-Ay(t)|\le h\max_{|t|\le h}
|f(t, x(t))-f(t, y(t))|\le hL\max_{|t|\le h}|x(t)-y(t)|.$$
Hence $A$ is a contraction mapping if $hL<1$. We may apply
the Contraction Mapping Theorem if we choose
$$h\le a, \ \ h\le \frac bM,\ \ h<\frac 1L.\tag4$$

\proclaim{Theorem} The initial value problem (1) has a unique 
solution $x(t)$ in $[-h, h]$ for $h$ in (4) if 
$f\in C([-a, a]\times[\xi-b, \xi+b])$
satisfies (2) and (3).
\endproclaim


 




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