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21.  Metric spaces (continued).

Lemma:  If $d$ is a metric on $X$ and $A \subset X$, then $d|_{A \times A}$ is a metric for the subspace topology on $A$.

Some order topologies are metrizable, some are not.

Thm 20.3$'$:  Suppose $d_X$ and $d_Y$ are metrics on $X$ and $Y$,
respectively.  Then 

\vskip 5pt
\centerline{$d((x_1, y_1), (x_2, y_2)) = max\{d_X(x_1, x_2),
d_Y(y_1, y_2)\}$}
\vskip -5pt

 is a metric which induces the product topology on
$X
\times Y$.

Note the generalization of this metric to countable products does not induce the product topology on countable products.  See Thm 20.5 for a metric which does induce the product topology on $R^\omega$.

Lemma:  If $X$ is a metric space, then $X$ is Hausdorff.

\eject
Thm 21.2:  Let $f: X \rightarrow Y$;  let $X$ and $Y$ be metrizable
with metrics $d_X$ and $d_Y$, respectively.  Then $f$ is continuous
if and only if for every $x \in X$ and \break
for every $\epsilon > 0$,
there exists a $\delta > 0$ such that 

\centerline{$d_X(x, y) < \delta$ implies
$d_Y(f(x), f(y)) < \epsilon$.}

Note:  $d_X(x, y) < \delta$ implies $d_Y(f(x), f(y)) < \epsilon$.
is equivalent to $f(B_X(x, \delta)) \subset B_Y(f(x), \epsilon)$

Lemma 21.2 (the sequence lemma):  Let $X$ be a topological space, $A \subset X$.  If there exists a sequence of points in $A$ which converge to $x$, then $x \in \overline A$

If $X$ is metrizable,  $x \in \overline A$ implies there exists a sequence of points in $A$ which converge to $x$.

Defn:  $X$ is said to have a {\bf countable basis at the point $x$}
if there exists a countable collection ${\cal B} = \{B_n ~|~ n \in
Z_+\}$ of
neighborhoods of $x$ such that if $x \in U^{open}$ implies there
exists a $B_i \in {\cal B}$ such that $B_i \subset U$

$X$ is {\bf first countable} if $X$ has a countable basis at each of its points.

Lemma:  A metrizable space is first countable.

Lemma:  If $X$ is first countable,  $x \in \overline A$ implies there exists a sequence of points in $A$ which converge to $x$.

Thm 21.3:  Let $f: X \rightarrow Y$.  If $f$ is continuous, then for every convergent sequence $x_n \rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$.

Suppose $X$ is first countable. If for every convergent sequence $x_n
\rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$,
then $f$ is continuous.

Lemma 21.4:  

$+: R \times R \rightarrow R$, $+(x, y) = x+y$ is continuous.

$-: R \times R \rightarrow R$, $-(x, y) = x-y$ is continuous.


$\cdot: R \times R \rightarrow R$, $\cdot(x, y) = xy$ is continuous.


$\div: R \times (R- \{0\} \rightarrow R$, $\div(x, y) = x/y$ is
continuous.


\eject

Thm 21.5:  If $X$ is a topological space, and if \break $f, g: X
\rightarrow R$ are continuous, then $f + g$, $f - g$, $f \cdot g$ are
continuous.  If $g(x) \not= 0$ for all $x$, then $f/g$ is continuous.

Defn:  Let $f_n:  X \rightarrow Y$ be a sequence of functions from
the set $X$ to the topological space $Y$.  Then the sequence of
functions $(f_n)$ {\bf converges} to the function $f: X \rightarrow
Y$ if the sequence of points $(f_n(x))$ converges to the point $f(x)$
for all $x \in X$.

Defn:  Let $f_n:  X \rightarrow Y$ be a sequence of functions from
the set $X$ to the metric space $Y$.  The sequence of functions
$(f_n)$ {\bf converges uniformly} to the function $f: X \rightarrow
Y$ if for all $\epsilon > 0$, there exists an integer $N$ such that
$n > N, ~x \in X$ implies $d_Y(f_n(x), f(x)) < \epsilon$.

Thm 21.6 (uniform limit theorem): \hfil \break
Let $f_n:  X \rightarrow Y$ be a
sequence of continuous functions from the topological space $X$ to
the metric space $Y$.  If $(f_n)$ {converges uniformly} to  $f$,
then $f$ is continuous.

\end
p. 127:  3, 11

11.) $d'(x, y) + d'(y, z)  = {d(x, y) \over 1 + d(x, y)} + {d(y, z)
\over 1 + d(y, z)} = {d(x, y)[1 + d(y, z)] + d(y, z)[1 + d(x, y)]
\over [1 + d(x, y)][ 1 + d(y, z)]} = {d(x, y) + d(x, y)d(y, z) + d(y,
z) + d(y, z)d(x, y)] \over 1 + d(x, y) + d(y, z) + d(x, y)d(y, z)}$



12.) $d'(x, y) - d'(y, z) $

$= {d(x, y) \over 1 + d(x, y)} - {d(y, z) \over 1 + d(y, z)}$

$= {d(x, y)[1 + d(y, z)] - d(y, z)[1 + d(x, y)] \over [1 + d(x, y)][
1 + d(y, z)]}$

$= {d(x, y) + d(x, y)d(y, z) - d(y, z) - d(y, z)d(x, y) \over 
1 + d(x, y) + d(y, z) + d(x, y)d(y, z)}$

$= {d(x, y) - d(y, z) \over 
1 + d(x, y) + d(y, z) + d(x, y)d(y, z)}$

$\leq  {d(x, z) \over 1 +  d(x, y) + d(y, z)}  $

$\leq  {d(x, z) \over 1 + d(x, z)}$

$d(x, y) \leq d(x, z)  + d(y, z)$

$d(x, y) - d(y, z) \leq d(x, z)$
\eject