Let $X$ be a compact metric space. Prove that $C[X]$ contains a countable subset that separates points.

Question

My proof seems way too trivial, so I am really doubting it.

Equip $X$ with any norm $\|\cdot\|_X$ (note that $\|\cdot\|_X : X \rightarrow \Bbb R$ is continuous). Since $\Bbb Q \setminus \{0\}$ is countable, we may enumerate it; i.e. $\Bbb Q \setminus \{0\} = \{q_1, q_2, \ldots \}$. Then define $f_n: X\to \mathbb{R}$ by $f_n(x) = q_n\|x\|_X, \forall n \in \Bbb N$.

Each $f_n$ is continuous on X as $\|\cdot\|_X$ is continuous. Take $x,y \in X$ such that $x \neq y$. Then: $$|f_n(x) - f_n(y)| = |q_n| \cdot|\|x\|_X - \|y\|_X| > 0$$ since $\|x\|_X \neq \|y\|_X$ since $x \neq y$ (and since $q_n \neq 0, \forall n \in \Bbb N$).

Thus, $\forall x,y \in X$ for which $x \neq y, \exists n \in \Bbb N$ such that $f_n(x) \neq f_n(y)$. So $\{f_n\}_{n \in \Bbb N}$ separates points.

If I am way off the mark, hints only please!

Answer 1

A compact metric space is separable so let $D$ be a countable dense subset of $(X,d)$.

Define $f_q \in C(X)$ as $f_q(x) = d(x,q)$ for $q \in D$.

If $x, y \in X$ with $x \ne y$ then pick a sequence $(q_n)_n$ in $D$ which converges to $x$. Then $(q_n)_n$ doesn't converge to $y$ so there exists $\varepsilon > 0$ and a subsequence $(q_{p(n)})_n$ such that $d(y, q_{p(n)}) \ge \varepsilon$. There exists $n \in \mathbb{N}$ such that $d(x, q_{p(n)}) < \varepsilon$ so $f_{q_{p(n)}}(x) < \varepsilon$ and $f_{q_{p(n)}}(y) \ge \varepsilon$. Hence $f_{q_{p(n)}}(x) \ne f_{q_{p(n)}}(y)$.

Therefore $\{f_q\}_{q \in D}$ is a countable subset of $C(X)$ which separates points.

Answer 2

It is very wrong to say that $$x\neq y\implies\|x\|\neq \|y\|$$

Note that the way $f_n$ is defined, whether or not $f_n(x)=f_n(y)$ does not actually depend on $n$ so it is unlikely that this family will be useful.

Finally, and that's actually a hint, you haven't used the fact that $X$ is compact.