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Let $(X,d)$ be a compact metric space. Show that $C(X,\mathbb{R})$ is a separable metric space (space of continuous functions from $X$ to $\mathbb{R}$).

I first showed that if $(X,d)$ is compact, then it must be separable, so we have a dense subset $\{x_{1},x_{2},...\}$ which is countable of $X$. Then, I'm not so sure on how to move forward. I was thinking of using the Stone Weierstrass Theorem for the set of functions:


Where $f_{n}(x)=d(x,x_{n})$ for $x \in X$. Then, this implies that the above set is dense in $C(X,\mathbb{R})$ and countable, so $C(X,\mathbb{R})$ is separable if $F$ is a unital separating subalgebra.

Clearly $F$ is unital, but I'm not sure on how to show it is separating and a subalgebra of $C(X,\mathbb{R})$ (it is a subset of the former set since the distance function is continuous). How would one proceed with this step?

Thank you for your help.


marked as duplicate by Alex M., YuiTo Cheng, cmk, José Carlos Santos, воитель Jun 18 at 21:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Let $F$ as you said. Let $\mathbb R[F]$ the $\mathbb R$-subalgebra generated by $F$.

We want to use Stone-Weierstrass theorem on the latter (rather than $F$) and show that it is dense. This will suffice for a proof that $C(X,\mathbb R)$ is separable, since $\mathbb Q[F]$ is countable and dense in $\mathbb R[F]$.

$\mathbb R[F]$ contains $1$ and it is obviously an algebra. Let's show that it separates points.

Let $x\ne y\in X$. Since $\{x_n\}_{n\in\mathbb N}$ is dense, there must exist $x_m$ such that $d(x,x_m)\le \frac13 d(x,y)$. Forcibly, it cannot hold $d(y,x_m)=d(x,x_m)$. If it held, then $$d(x,y)\le d(x,x_m)+d(y,x_m)\le \frac23 d(x,y)$$ absurd. So the function $f_m$ separates $x$ and $y$.

Stone-Weierstrass can therefore be used on $\mathbb R[F]$, completing the proof.


  1. How is $\mathbb R[F]$ defined? Either the intersection of all the $\mathbb R$-subalgebras of $C(X,\mathbb R)$ which contain $F$ or, equivalently, as the $\mathbb R$-vector subspace of $C(X,\mathbb R)$ generated by the products of finitley many elements of $F$.

  2. How is $\mathbb Q[F]$ defined? Either the intersection of all the $\mathbb Q$-subalgebras of $C(X,\mathbb R)$ that contain $F$ or, as above, the $\mathbb Q$-vector subspace of $C(X,\mathbb R)$ generated by the products of finitely many elements of $F$.

  3. Why is $\mathbb Q[F]$ dense in $\mathbb R[F]$? It is rather easy, actually, but the notation is a bit tedious.

    If $g\in\mathbb R[F]$, then there exist $k\in\mathbb N,\ g_1, \cdots, g_k\in F$ and a finite set $S\subseteq \mathbb N^k$ such that $$g=\sum_{(n_1,\cdots,n_k)\in S} \lambda_{n_1,\cdots,n_k}g_1^{n_1}\cdots g_k^{n_k}$$ for some $\lambda_{n_1,\cdots,n_k}\in\mathbb R$.

    Now, if you approximate each $\lambda_{n_1,\cdots,n_k}$ with rationals $$\alpha_{n_1,\cdots,n_k}^{(t)}\stackrel{t\to\infty}{\longrightarrow}\lambda_{n_1,\cdots,n_k}$$ and call $$g^{(t)}=\sum_{(n_1,\cdots,n_k)\in S} \alpha^{(t)}_{n_1,\cdots,n_k}g_1^{n_1}\cdots g_k^{n_k}\in\mathbb Q[F]$$ you get $$\Vert g-g^{(t)}\Vert_\infty=\left\Vert \sum_{(n_1,\cdots,n_k)\in S} (\lambda_{n_1,\cdots,n_k}-\alpha_{n_1,\cdots,n_k}^{(t)})g_1^{n_1}\cdots g_k^{n_k}\right\Vert_\infty\le\\\le \left(\sum_{(n_1,\cdots,n_k)\in S}\Vert g_1^{n_1}\cdots g_k^{n_k}\Vert_\infty\right)\cdot\max_{(n_1,\cdots,n_k)\in S}\left\vert\lambda_{n_1,\cdots,n_k}-\alpha^{(t)}_{n_1,\cdots,n_k}\right\vert\stackrel{t\to\infty}{\longrightarrow}0$$

  • $\begingroup$ Perfect. How do we define the $\mathbb{R}$-subalgebra generated by $F$? The intersection of all subalgebras that contain $F$? As well, how do we know that the $\mathbb{Q}$-subalgebra is dense in the $\mathbb{R}$-subalgebra? Intuitively I think it would be because of $\mathbb{Q}$ is dense in $\mathbb{R}$, but how would the argument be made rigorous? Thanks! $\endgroup$ – arcbloom Dec 15 '15 at 5:00
  • $\begingroup$ @fogvajarash, I added something. If you have further perplexities, don't hesitate to ask! $\endgroup$ – user228113 Dec 15 '15 at 5:40

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