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For a compact Hausdorff space $S$, show that the following are equivalent:

(a) $S$ is metrizable.

(b) $S$ has a countable base.

(c) $C(S)$ is separable, where $C(S)$ denotes the space of all continuous functions on $S$.

(a)$\implies$(b) This is trivial. For every $N\in\mathbb Z_{>0}$, take the finite $1/N$-net of $S$ since $S$ is metrizable and compact. Then for each $1/N$-net, we take the countable balls with diameter $1/M$ where $M$ ranges over $\mathbb Z_{>0}$. And we take $N$ to infinite. Thus (a) implies (b).

(b)$\implies$(c) This is again trivial. Since compactness together with Hausdorff implies normal, then we can easily construct a countable dense set of $C(S)$ by invoking Tietze extension theorem.

(c)$\implies (a)$ Suppose $\{f_n\}_{n=1}^\infty$ is a dense subset of $C(S)$ and we construct a metric: $$ d(x,y):=\sum_{n=1}^\infty 2^{-n}\min\{|f_n(x)-f_n(y)|,1\}. $$ How to show that the topology of $S$ is induced by this metric?

Bach
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    https://math.aalto.fi/opetus/old_specfuna/pruju/sect11.pdf seems relevant? – kahen Jun 01 '20 at 00:26
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    Are you sure $(b)\implies (c)$ is trivial? I have only seen it proven using Stone-Weierstrauss in combination with an extension theorem – K.Power Jun 01 '20 at 00:43
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    @K.Power It's the old under-confident maths student's mantra: "If I can do it, it's trivial." – user793679 Jun 01 '20 at 02:23
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    Classic fact: if $S$ is compact Hausdorff and $d$ is a metric function on $S$ such that $d$ is continuous on $S \times S$, then $d$ induces the topology on $S$. So all you have to do is show $d$ is continuous. – Henno Brandsma Jun 01 '20 at 06:24
  • Maybe easier to show $(c) \implies (b),$ perhaps by showing $[\neg (b)]\implies \neg (c).$ And $(b)\implies (a)$ because a 2nd-countable Tychonoff space is homeomorphic to a subspace of the metrizable space $[0,1]^{\Bbb N}$. – DanielWainfleet Jun 01 '20 at 14:11

1 Answers1

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Show that $d$ is continuous on $S \times S$. Also show that $d$ is actually a metric on $S$ (there is something to be shows for $d(x,y)=0 \to x=y$, e.g.)

Let $\mathcal{T}_d$ be the topology on $S$ that is generated by $d$ (i.e. in which the balls $B_d(x,r)$ are open for all $x \in S$ and all $r>0$), and $\mathcal{T}$ the given topology on $S$. As the continuity of $d$ implies that

$$\forall x \in X: \forall r>0: B_d(x,r)= d_x^{-1}[(-\infty, r)] \text{ where } d_x: S \to \Bbb R \text{ is defined by } d_x(y)=d(x,y)$$

we have that by continuity of $d$ and hence all maps $d_x$, that $\mathcal{T}_d \subseteq \mathcal{T}$ and so the identity map $(S,\mathcal{T}) \to (S, \mathcal{T}_d)$ is a continuous bijection from a compact space to a Hausdorff space and hence closed and a homeomorphism, so $\mathcal{T}=\mathcal{T}_d$ and $d$ induces the topology on $S$. Now just fulfill the promise in the first sentence and you're done.

I also don't agree that (b) to (c) is "trivial". Give an actual proof! Or maybe you already covered a version of Stone-Weierstraß? (a) to (b) could use a more fleshed out version of the argument as well (as the theorem I proved in this answer, which is more general).

Henno Brandsma
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