# Compact Hausdorff space is metrizable if there countable separating continuous functions

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, $$B(x;r):=\{y\in X| d(x,y). I know $$B(x;r)\supset \bigcap_{n \in \mathbb{Z}_+} \{y \in X |f_n(x)-f_n(y)|, but right term is not open.

How to prove this proposition?

• The last right term is a countable union of opens, so might not be open. But it indeed contains an open: any open in the union would do. – awllower Apr 13 at 7:42
• @awllower Why do you know last right term is countable union? I think it is countable intersection. – B.T.O Apr 13 at 7:56
• @B.T.O I wad referring to the question without edits. Sorry for the horrible mistake. – awllower Apr 13 at 10:22

## 1 Answer

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.