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I have a little problem with this Theorem 3.8 c) If $X$ is a compact topological space and if some sequence $\{f_n\}$ of continuous functions separates points on $X$, then $X$ is metrizable.

The big deal is that in the proof Rudin constructs a metric $d(x, y)=\sum_{n=1}^{\infty}2^{-n}|f_n(x)-f_n(y)|$, and this metric induces a topology denoted by $\tau_{d}$ and I have to prove that $\tau_{d}\subseteq \tau$ so in order to do that I must take a basic $B_r(x) \in \tau_{d}$ and prove that there exists an open set $U \in \tau$ such that $U\subseteq B_r(x)$ but I don't know exactly how to do it, may you give any hint please? I'm really stuck.

I just have some ideas with weak topology but I'm not sure if it works.

Gary
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sigma7541
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  • You need uniform boundedness of $f_n$'s.Otherwise $d$ may not be a metric. – Kavi Rama Murthy Dec 28 '21 at 23:13
  • @KaviRamaMurthy easily fixed y taking $d(x,y)= \sum_n \frac{1}{2^n}\min(|f_n(x)-f_n(y)|,1)$ instead. Not an essential flaw. The sum then converges absolutely and uniformly (and so is continuous by standard facts). – Henno Brandsma Dec 28 '21 at 23:15
  • @HennoBrandsma Of course, I know how to fix that. But if you find a mistake you should point it out before posting an answer. – Kavi Rama Murthy Dec 28 '21 at 23:25

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Because $d$ is continuous as a function on $X \times X$, $\tau_d \subseteq \tau$ is immediate:

$B_r(x)=(d_x)^{-1}[(-\infty,r)]$ where (for any fixed $x$) the map $d_x: y \to d(x,y)$ is continuous from $X$ to $\Bbb R$.

Henno Brandsma
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