3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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)?

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.