\magnification 2300 \parindent 0pt \parskip 10pt \pageno=23 \hoffset -0.1truein \hsize 6.9truein \vsize 9.3truein \def\u{\vskip -10pt} \def\v{\vskip -5pt} 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