Let $X$ be a metrizable compact space. I want to show that $C(X, \mathbb{C})$ is separable in the uniform topology.

Attempt: By our assumption, $X$ is separable, so we can pick a countable dense subset $\{x_n:n \geq 1\}$. Let $d$ be a metric inducing the topology on $X$. Define for $n \geq 1$

$$d_n: X \to \mathbb{R}: x \mapsto d(x,x_n)$$

Let $$\mathcal{A}= \bigcup_{n=1}^\infty \mathbb{C}[d_1, \dots, d_n]$$ where $\mathbb{C}[d_1, \dots, d_n]$ is the set of complex polynomials in the functions $d_1, \dots, d_n$.

It is easily checked that $\mathcal{A}$ is a subalgebra of $C(X, \mathbb{C})$, since $\mathcal{A}$ is closed under addition, multiplication and scalar multiplication. Moreover, $1 \in \mathbb{C}[d_1]\subseteq \mathcal{A}$ so the algebra is unital. Clearly $\mathcal{A}$ is closed under complex conjugation, since the functions $d_1, d_2, \dots$ are all real-valued. We know that $\mathcal{A}$ separates the functions of $C(X, \mathbb{C})$:

If $x \neq y$ with $x, y \in X$. Use density to choose $n \geq 1$ with $d(x_n,x) < d(x,y)/2$. Then we have $d(x_n,x) \neq d(x_n,y)$. Otherwise $d(x_n,x) = d(x_n,y)$ and we get $d(x,y) \leq d(x,x_n) + d(x_n,y) = 2d(x,x_n)<d(x,y)$ which is impossible. Thus $d_n(x) \neq d_n(y)$ so our algebra separates the points.

By Stone-Weierstrass, $\mathcal{A}$ is dense in $C(X, \mathbb{C})$. Let $D$ be a countable dense subset of $\mathbb{C}$, for example $D= \mathbb{Q}+ i \mathbb{Q}$. Any element of $\mathcal{A}$ can be approximated by an element in $$\mathcal{B}:= \bigcup_{n=1}^\infty D[d_1, \dots, d_n]$$ and we conclude that $\mathcal{B}$ is dense in $C(X, \mathbb{C})$. Clearly $\mathcal{B}$ is countable, and thus we conclude the proof. $\quad \square$

Is this proof correct?

  • $\begingroup$ I think that I did not carefully read your proof. Sorry. I will delete my comment. $\endgroup$ – Stinking Bishop May 22 at 21:52
  • $\begingroup$ No worries. Your help is appreciated anyway! $\endgroup$ – user745578 May 22 at 21:54
  • $\begingroup$ Stone-Weierstraß is not the only way.. but if it's an allowed tool, go for it. $\endgroup$ – Henno Brandsma May 22 at 23:59
  • $\begingroup$ @HennoBrandsma Are there maybe other approaches that are easier/more elementary? $\endgroup$ – user745578 May 23 at 8:20

The idea of the proof is fine, and standard, I think. Maybe you might want to add details about why your $\mathcal{B}$ is dense, so why replacing the coefficients of the members of the algebra by a dense subset is OK. Maybe you're relying on an earlier lemma? The countability might warrant a few words too, depending on the target audience (how much set theory do they know)?

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  • $\begingroup$ Well, take a polynomial $p = \sum_{i_1, \dots, i_n} a_{i_1, \dots, i_n} d_1^{i_1}\dots d_n^{i_n}$ in $d_1, \dots, d_n$. We approximate this with an element in $\mathcal{B}$. Let $m$ be the amount of terms in this polynomial. Given $\epsilon > 0$, choose elements $b_{i_1, \dots, i_n} \in D$ with $|b_{i_1, \dots, i_n}-a_{i_1, \dots, i_n}| < \epsilon/m$. Then $|\sum a_{i_1, \dots, i_n} d_1^{i_1}\dots d_n^{i_n}- \sum b_{i_1, \dots, i_n} d_1^{i_1}\dots d_n^{i_n}| < \epsilon$ and this shows that an element of $\mathcal{A}$ can be approximated by an element in $\mathcal{B}$. $\endgroup$ – user745578 May 23 at 8:08
  • $\begingroup$ Since an element in $C(X)$ can be approximated by an element in $\mathcal{A}$ and an $\epsilon/2$-argument, we see that $\mathcal{B}$ is dense in $C(X)$. Countability of $\mathcal{B}$ is obvious (countable union of countable sets is countable) $\endgroup$ – user745578 May 23 at 8:09
  • $\begingroup$ What do you think? Would that be enough details? $\endgroup$ – user745578 May 23 at 8:10
  • $\begingroup$ @user745578 In the sup norm (the $d_i$ are functions so we have to estimate for all $x \in X$)) we need uniform continuity (which is true, but shows the argument is not yet complete).. $\endgroup$ – Henno Brandsma May 23 at 8:11
  • $\begingroup$ $X$ is compact so uniform continuity is garantueed? $\endgroup$ – user745578 May 23 at 8:12

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