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?

  • $\begingroup$ 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
  • 2
    $\begingroup$ @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
  • 1
    $\begingroup$ @B.T.O I wad referring to the question without edits. Sorry for the horrible mistake. $\endgroup$ – awllower Apr 13 at 10:22

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.