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I am studying measure theory, topology, and functional analysis in mathematics. Let $X$ be a compact topological space. We assume that there exists a countable family $\{f_n\: X \to \mathbb C: n\in\mathbb{N}\}$ of continuous complex-valued functions on $X$ that {\em separates points} in the following sense: for all $x,y\in X$ with $x\neq y$ there exists $n\in \mathbb{N}$ such that $f_n(x)\neq f_n(y)$. I have figured that wlog we may assume that $|f_n| \leq 1$ for all $n \in \mathbb N$ (take $\frac{f_n}{1 + |f_n|}$). I want to show that then $X$ is metrizable, i.e., there exists a metric $d$ on $X$ that induces the given topology on $X$.

I have shown that the function $$ d(x,y) := \sum_{n=1}^\infty \frac{1}{2^n} |f_n(x)-f_n(y)|$$ for $x,y\in X$ is a metric on $X$. I also know that if $g\: Y\to Z$ is a continuous bijection between a compact space $Y$ and a Hausdorff space $Z$, then $g$ is a homeomorphism.

Let $\tau$ denote the original topology and let $\tau_d$ be the topology induced by $d$. I have read this post on proving that $\tau_d \subset \tau$, but I am not sure how $d$ is a continuous function on $X \times X$. Could you clarify this? Also, I am assuming that the answer there uses the fact that, for any fixed $x \in X$, the map $y \mapsto (x,y)$ is also continuous; why is this true? I am also having trouble proving the other direction that $\tau \subseteq \tau_d$; I also read this post and I am lost on how $B(x;r)\supset \bigcap_{n \in \mathbb{Z}_+} \{y \in X |f_n(x)-f_n(y)|<r\}$ and how there is an open set in $\tau$ contained in $B(x,r)$.

Thanks so much for all your help!

Steven Clontz
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love and light
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2 Answers2

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Consider $Y := \mathbb C^{\mathbb N}$ with the product topology. Of course, $Y$ is metrizable.

Define $f: X \rightarrow Y, f(x) := (f_n(x))_{n \in \mathbb N}$.

Since each $f_n$ is continuous, $f$ is continuous. Since $(f_n)_n$ separates points, $f$ is injective. Hence $f^\prime = f : X \rightarrow f(X)$ is continuous and bijective. Since $X$ is compact and $Y$ is $T_2$, $f^\prime$ is closed, hence a homeomorphism. Thus, $X$ is homeomorphic to $f(X) \subset Y$ and therefore metrizable.

Note that we didn't use separability of $X$, although, of course, $X$ turns out to be separable.

Ulli
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Let $x_0\in X$ and $\epsilon>0$. Let $N\in\Bbb{N}^+$ such that $\sum_{n=N+1}^\infty 2^{-n}<\frac{\epsilon}{4}$. Since $f_1,\ldots,f_N$ are $\tau$-continuous, the sets $U_n=\{x\in X\colon |f_n(x)-f_n(x_0)|<\frac{\epsilon}{2}\}$ are $\tau$-open. Let $U=\bigcap_{n=1}^N U_n$. Then $U$ is $\tau$-open, $x_0\in U$ and for all $x\in U$ $$ d(x,x_0)=\sum_{n=1}^\infty 2^{-n}|f_n(x)-f_n(x_0)|\le \sum_{n=1}^N 2^{-n}\frac{\epsilon}{2}+\sum_{n=N+1}^\infty 2^{-n}\cdot 2< \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon $$ Hence $x_0\in U\subseteq B_d(x_0,\epsilon)$. This proves that $\tau_d\subseteq\tau$.

Now consider the identity map $\mathbf{1}\colon (X,\tau)\to(X,\tau_d)$. Since $\tau_d\subseteq\tau$, the map $\mathbf{1}$ is continuous. Let $C$ be a closed subset of $(X,\tau)$. Since $(X,\tau)$ is compact, $C$ is compact in the subset topology from $(X,\tau)$. Since $\mathbf{1}$ is continuous, $\mathbf{1}(C)=C$ is compact in the subset topology from $(X,\tau_d)$. since $(X,\tau_d)$ is a Hausdorff space, $C$ is a closed subset of $(X,\tau_d)$. This proves that $\tau\subseteq\tau_d$.

Chad K
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