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$X$ is a compact metric space, then $C(X)$ is separable, where $C(X)$ denotes the space of continuous functions on $X$. How to prove it?

And if $X$ is just a compact Hausdorff space, then is $C(X)$ still separable? Or if $X$ is just a compact (not necessarily Hausdorff) space, then is $C(X)$ still separable?

Please help me. Thanks in advance.

Syzygy Yuan
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David Chan
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    this result is not trivial: If X is a compact $T_{2}$ space $X$, then $C(X)$ is separable iff there is a metric $X\times X\rightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm. – Matematleta Jun 19 '15 at 13:21

1 Answers1

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Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.

There is a natural embedding $x\in X\to \delta _x\in \mathcal{M}(X)$ (more precisely in the unit ball of $\mathcal{M}(X)$). This is an homeomorphism for the weak*-topology of $\mathcal{M}(X)$. If $C(X)$ is separable then $(\mathcal{M}(X), w*)$ have a compact metrizable unit ball. So $X$ is metrizable.

For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: x\in X \to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.

Martin Argerami
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Patissot
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    Do you know by any chance if this theorem has a name? – Anguepa Oct 27 '18 at 16:49
  • @Anguepa I know it is late but for anyone who is wondering it is called Riesz Theorem – Sorfosh Sep 19 '19 at 18:16
  • @Sorfosh I can`t find any reference to it here. Do you have a reference by any chance? – Anguepa Sep 21 '19 at 04:19
  • @Anguepa Royden, Real Analysis Section 12.3 page 251 – Sorfosh Sep 21 '19 at 04:55
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    In the fourth edition (Royden & Fitzpatrick, p. 251) the same theorem is attributed to Borsuk. – Alp Uzman Mar 12 '21 at 02:33
  • The fact that the embedding $x\to\delta_x$ is a homeomorphism is not obvious and follows from the existence of partitions of unity (which in turn follows from the Urysohn lemma). – No-one Nov 01 '22 at 21:04
  • The space $C(X,\mathbb{C})$ is separable too? – N230899 Oct 24 '23 at 14:54
  • Can I ask why it's necessary to explicitly put $1$ in as a generator? Isn't the conclusion of Stone-Weierstrass that if the subalgebra is not dense in C(X) then it's dense in a subset of C(X) which have a common root. But the functions $d_n$ can't share a root because $d_n(x) = 0$ iff $x = x_n$. – Jeremy Weissmann Aug 03 '24 at 17:13