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While reading this post, I was a bit confused by the following :

If $X$ is a compact metric space, then by Urysohn's Lemma and Stone-Weierstrass, the continuous functions $C(X)$ on $X$ are separable and hence the result follows as $C_{c}(X) = C(X)$.

I understand the use of Stone-Weierstrass because polynomial with rational coefficient are countable and are dense in polynomials with real coefficients which are dense (by stone-Weierstrass) in $C(X)$ when $X$ is compact. But I don't understand where Urysohn's Lemma comes into play.

edamondo
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    "polynomials with real coefficients which are dense (by stone-Weierstrass) in C(X) when X is compact" this is wrong in an important way: You're given just that $X$ is a compact metric space, hence there's no such thing as a polynomial on $X$. – David C. Ullrich Mar 28 '21 at 15:33

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Since $X$ is compact metric, it has a countable basis $\{U_n\}_{n\in {\mathbb N}}$ of open sets.

For each pair $(n, m)$ such that $\overline U_n\subseteq U_m$, you can use Urysohn to pick some $f_{n, m}$ in $C(X)$ such that $f_{n, m}=1$ on $U_n$, and $f_{n, m}=0$ on $X\setminus U_n$.

The subalgebra $A\subseteq C(X)$ generated by all of the $f_{n, m}$ separates points of $X$, so it is dense by Stone-Weierstrass. Now you can finish your argument by pretending that $A$ is the algebra of polynomials.

Ruy
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  • Thank you. There is a technical point I don't get. The algebra generated by a countable number of elements is always countable? – edamondo Mar 28 '21 at 18:24
  • No. You must first consider the generated $\mathbb Q$-algebra, which is countable. It's just like in your argument, where rational polynomials play an intermediate role. – Ruy Mar 28 '21 at 19:22
  • When we say the subalgebra $A$ generated by all of the $f_{n,m}$ it means $\mathbb{R}$-linear combinations? So if I understand correctly the $\mathbb{Q}$-algebra is dense in the $\mathbb{R}$-algebra which is dense in $C(X)$? – edamondo Apr 01 '21 at 10:28
  • Yes, that is correct. – Ruy Apr 01 '21 at 12:18