[Munkres Topology] 21. The Metric Topology (continued)

 This article is one of the posts in the Textbook Commentary Project.

Subspace of metric spaces behave the way one would wish them to; if $A$ is a subspace of the topological space $X$ and $d$ is a metric for $X$, then the restriction of $d$ to $A\times A$ is a metric for the topology of $A$.

About order topologues, some are metrizable, and others are not.

The Hausdorff axiom is satisfied by every metric topology. If $x$ and $y$ are distince points of the metric space $(X, d)$, we let $\epsilon=\frac{1}{2}d(x,y)$; then the triangle inequality implies that $B_d(x,\epsilon)$ and $B_d(y,\epsilon)$ are disjoint.

The countable products of metrizable spaces are metrizable.

Theorem 21.1 ($\epsilon-\delta$ condition). Let $f:X\rightarrow Y$; let $X$ and $Y$ be metrizable with metrices $d_X$ and $d_Y$, respectively. Then continuity of $f$ is equivalent to the requirement that given $x\in X$ and given $\epsilon >0$, there exists $\delta>0$ such that $$d_X(x,y)<\delta  \Rightarrow d_Y(f(x),f(y))<\epsilon.$$

Proof. 

[$\rightarrow$] Suppose that $f$ is continuous. Given $x$ and $\epsilon$, consider the set $$f^{-1}(B(f(x).\epsilon)),$$ which is open in $X$ and contains the point $x$. It contains some $\delta$-ball $B(x,\delta)$ centered at $x$. If $y$ is in this $\delta$-ball, then $f(y)$ is in the $\epsilon$-ball centered at $f(x)$, as desired.

[$\leftarrow$] Suppoe that the $\epsilon-\delta$ condition is satisfied. Let $V$ be open in $Y$; we show that $f^{-1}(V)$ is open in $X$. Let $x$ be a point of the set $f^{-1}(V)$. Since $f(x)\in V$, there is an $\epsilon$-ball $B(f(x),\epsilon)$ centered at $f(x)$ and contained in $V$. By the $\epsilon-\delta$ conditino, there is a $\delta$-ball $B(x,\delta)$ centered at $x$ such that $f(B(x,\delta))\subset B(f(x),\epsilon)$. Then $B(x,\delta)$ is a neighborhood of $x$ contained in $f^{-1}(V)$, so that $f^{-1}(V)$ is open, as desired.

Lemma 21.2 (The sequence lemma). Let $X$ be a topological space; let $A\subset X$. If there is a sequence of points of $A$ converging to $x$, then $x\in \bar{A}$; the converse holds if $X$ is metrizable.

Proof. [$\Rightarrow$] Suppose that $x_n\rightarrow x$, where $x_n\in A$. Then every neighorhood $U$ of $x$ contains a point of $A$, so $x\in \bar{A}$ by theorem 17.5. [$\Leftarrow$] Suppose that $X$ is metrizable and $x\in \bar{A}$. Let $d$ be a metric for the topology of $X$. For each positive integer $n$, take the neighborhood $B_d(x,1/n)$ of radius $1/n$ of $x$, and choose $x_n$ to be a point of its intersection with $A$. We assert that the sequence $x_n$ converges to $x$: Any open set $U$ containing $x$ contains an $\epsilon$-ball $B_d(x,\epsilon)$ centered at $x$; if we choose $N$ so that $1/N<\epsilon$, then $U$ contians $x_i$ for all $i\ge N$.

Theorem 21.3. Let $f:X\rightarrow Y$. If the function $f$ is continuous, then for every convergent sequence $x_n\rightarrow x$ in $X$, the sequence $f(x_n)$ converges to $f(x)$. The converse holds if $X$ is metrizable.

Proof. 

[$\Rightarrow$] Assume that $f$ is continuous. Given $x_n\rightarrow x$, we wich to show that $f(x_n)\rightarrow f(x)$. Let $V$ be a neighborhood of $f(x)$. Then $f^{-1}(V)$ is a neighborhood of $x$, and so there is an $N$ such that $x_n\in f^{-1}(V)$ for $n\ge N$. Then $f(x_n)\in V$ for $n\ge N$.

[$\Leftarrow$] Assume that the convergent sequence condition is satisfied. Let $A$ be a subset of $X$; we show that $f(\bar{A})\subset \bar{f(A)}$ (Theorem 18.1.(2)). If $x\in\bar{A}$, then there is a sequence $x_n$ of points of $A$ converging to $x$ (by the preceding lemma). By assumption, the sequcence $f(x_n)$ converges to $f(x)$. Since $f(x_n)\in f(A)$, the preceding lemma implies that $f(x)\in \bar{f(A)}$. (Note that metrizability of $Y$ is not needed.) Hence $f(\bar{A})\subset \bar{f(A)}$, as desired.

Metric condition is more than sequence convergence. 'Countable basis' is enough.

Lemma 21.4. The addition, subtraction, and multiplication operations are continuous functions from $\mathbb{R}\times \mathbb{R}$ into $\mathbb{R}$; and the quotient operation is a continuous function from $\mathbb{R}\times (\mathbb{R}-\{0\})$ into $\mathbb{R}$.

Theorem 21.5. If $X$ is a topological space, and if $f,g:X\rightarrow \mathbb{R}$ are continuous functions, then $f+g$, $f-g$, and $f\cdot g$ are continuous. If $g(x)\ne 0$ for all $x$, then $f/g$ is continuous.

Proof. The map $h:X\rightarrow \mathbb{R}\times \mathbb{R}$ defined by $$h(x)=f(x)\times g(x)$$ is continuous, by Theorem 18.4.. The function $f+g$ equals the composite of $h$ and the addition operation $$+:\mathbb{R}\times \mathbb{R}\rightarrow \mathbb{R};$$ therefore $f+g$ is continuous. Similar arguments apply to $f-g$, $f\cdot g$, and $f/g$.

Finally, we come to the notion of uniform convergence.

Definition. Let $f_n:X\rightarrow Y$ be a sequence of functions from the set $X$ to the metric space $Y$, Let $d$ be the metric for $Y$. We say that the sequence $(f_n)$ converges uniformly to the function $f:X\rightarrow Y$ if given $\epsilon>0$, there exsits an integer $N$ such that $$d(f_n(x),f(x))<\epsilon$$ for all $n>N$ and all $x$ in $X$.

Uniformity of convergence depends not only on the topology of $Y$ but also on its metric. We have the following theorem about uniformly convergent sequences: 

Theorem 21.6 (Uniform limit theorem). 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.

Proof. Let $V$ be open in $Y$; let $x_0$ be a point of $f^{-1}(V)$. We with to find a neighborhood $U$ of $x_0$ suvh that $f(U)\subset V$.

Let $y_0=f(x_0)$. First choose $\epsilon$ so that the $\epsilon$-ball $B(y_0,\epsilon)$ is contained in $V$. Then, using uniform convergence, choose $N$ so that for all $n\ge N$ and all $x\in X$, $$d(f_n(x),f(x))<\epsilon/3.$$ Finally, using continuity of $f_N$, choose a neighborhood $U$ of $x_0$ such that $f_N$ carries $U$ into the $\epsilon/3$ ball in $Y$ centered at $f_N(x_0)$.

We claim that $f$ carries $U$ into $B(y_0,\epsilon)$ and hence into $V$, as desired. For this purpose, note that if $x\in U$, then $$d(f(x),f_N(x))<\epsilon/3\quad (\mbox{by choice of }N),$$ $$d(f_N(x),f_N(x_0))<\epsilon/3\quad (\mbox{by choice of }U),$$ $$d(f_N(x_0),f_N(x_0))<\epsilon/3\quad (\mbox{by choice of }N).$$

Adding and using the triangle unequality, we see that $d(f(x),f(x_0))<\epsilon$, as desired.

Let us remark that the notion of uniform convergence is related to the definition of the uniform metric, which we gave in the preceding section. Consider, for example, the space $\mathbb{R}^X$ of all functions $f:X\rightarrow \mathbb{R}$, in the uniform metric $\bar{\rho}$. It is not difficult to see that a sequence of functions $f_n:X\rightarrow \mathbb{R}$ converges uniformly to $f$ if and only if the sequence $(f_n)$ converges to $f$ when they are considered as elements of the metric space $(\mathbb{R}^X,\bar{\rho})$.