Compact Hausdorff space is metrizable if there countable separating continuous functions

Question

Proposition: Let $X$ be a compact Hausdorff space. Suppose there are countable real valued continuous functions $\{f_n\}_{n \in \mathbb{Z}_+}$ separating $X$ i.e. for all $x, y \in X$ with $x \neq y$, $\exists k:=k(x,y) \in \mathbb{Z}_+$, $f_k(x)\neq f_k(y)$. Let $$ d(x,y):=\sum_{n=1}^\infty \frac{\min\{|f_n(x)-f_n(y)|, 1\}}{2^n} $$ Then $X$ is metrizable by $d$.

I want to prove that, for all open set $U$ and $x \in U$, there exists $B(x;r)$ s.t. $B(x;r)\subset U$ and for all $B(x;r)$, there exists an open set $U$ s.t. $U\subset B(x;r)$. Here, $B(x;r):=\{y\in X| d(x,y)<r\}$. I know $B(x;r)\supset \bigcap_{n \in \mathbb{Z}_+} \{y \in X |f_n(x)-f_n(y)|<r\}$, but right term is not open.

How to prove this proposition?

Answer 1

I'll denote $\Bbb Z^+$ by $\omega$.

$\mathbb{R}^{\omega}$ (in the product topology) is metrisable by the metric $$D((x_n), (y_n))=\sum_{n \in \omega} \frac{\min(|x_n-y_n|, 1)}{2^n}$$ as is well-known, e.g. see my answer here.

Then from the $f_n$ we define $F: X \to \mathbb{R}^\omega$ by $F(x)=(f_n(x))_{n \in \omega}$ and note that $F$ is continuous as $\pi_n \circ F = f_n$ is continuous for all $n$ and where $\pi_n$ is the projection onto the $n$-th coordinate. This follows from the characterisation of the product topology as the smallest topology that makes all pprojections continuous, and is a standard fact proved in many text books.

The fact that the $f_n$ separate points means exactly that $F$ is injective (1-1).

So $F: X \to F[X]$ is a continuous bijection between a compact space and a Hausdorff space (metric implies Hausdorff) and so $X$ is homeomorphic to $F[X]$ and the pulled-back metric of $F[X]\subseteq (\mathbb{R}^\omega, D)$ to $X$ is exactly $d(x,y)=D(F(x), F(y))$ and as $D$ is a metric for $F[X]$ and $F$ is a homeomorphism, $d$ (i.e. your metric on $X$) is a metric for $X$, as required.

E.g. $B_d(x,r) = F^{-1}[B_D(F(x),r)]$ so $d$-open balls are open and inverse images of a base under a homeomorphism form a base etc.

Answer 2

Here's another proof, for fun.

Fix $x\in X$. The function $d_x(y)=d(x,y)$ is continuous as the partial sums $$\sum_{n=1}^N\frac{\min\{|f_n(x)-f_n(y)|, 1\}}{2^n}$$ converge uniformly to $d_x(y)$ and are each continuous. It follows that $\epsilon$-balls $B_\epsilon(x):=\{y\in X: d(x,y)<\epsilon \} $ are open, so all that remains to be shown is that each neighborhood $U$ of $x$ contains some $\epsilon$-ball about $x$. Suppose not, in which case $d_x(U^c)$ would contain some sequence $\alpha_n\to 0$. Consider then some sequence $\{x_n\}\subset U^c$ with $d_x(x_n)=\alpha_n$, and, noting $X$ is compact, reduce to a convergent subsequence with $x_n\to y$. Then, $d_x(y)=\lim_n d_x(x_n)=0$ so $x=y$ as $d$ is a metric. This implies $x_n$ is eventually in every neighborhood of $x$; in particular, eventually in $U$, a contradiction.