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?
Best Answer
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)?