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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?

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  • $\begingroup$ 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. $\endgroup$ – awllower Apr 13 at 7:42
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    $\begingroup$ @awllower Why do you know last right term is countable union? I think it is countable intersection. $\endgroup$ – B.T.O Apr 13 at 7:56
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    $\begingroup$ @B.T.O I wad referring to the question without edits. Sorry for the horrible mistake. $\endgroup$ – awllower Apr 13 at 10:22
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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.

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