0

Let $(X, d)$ be a compact metric space. Is $C(X, \mathbb{C})$ separable? I've already checked an answer, but I think that my problem is that my continuous functions take value in $\mathbb{C}$. Does this change anything in separability? If not, can I follow the same proof or I have to change something?

Nick
  • 97
  • 1
    It's actually a criterion: for a compact Hausdorff topological space $X$, $C(X)$ is separable iff $X$ is a metric space, see here: https://math.stackexchange.com/a/1331353/1104384 – Bruno B Oct 24 '23 at 15:50
  • By the way, one-line questions like this usually won't cut it on this forum, you're kind of expected to give at least some context as for what you've tried even if it failed, and/or research you've done on the subject, or some other info that you think would help us help you, like for example if there's an argument that you're forbidden/you're limiting yourself on its usage, like a limit without l'Hôpital's rule. – Bruno B Oct 24 '23 at 15:56
  • I'm not sure, but I think that answer refers to continuous functions to $\mathbb{R}$. What can we say about continuous function to $\mathbb{C}$? – Nick Oct 24 '23 at 16:43
  • N.B. the proof that I know in the case of $\mathbb{R}$ uses the Stone - Weiestrass theorem. – Nick Oct 24 '23 at 17:10
  • Well, there's a Stone-Weierstrass theorem for $\mathbb{C}$, all you add is requiring stability by complex conjugation. The real version then gives you the complex version simply by noting that real and imaginary parts of your functions are then part of your complex algebra, and you apply the real SW theorem on the real algebra generated by these functions. – Bruno B Oct 24 '23 at 17:17
  • Yes, and I think that algebra is stable by conjugation. But then can I follow the next step (considering linear combination of rational number)? Are my combination complex? – Nick Oct 24 '23 at 17:22
  • You can consider coefficients in $\mathbb{Q}(i)$, aka rational real part and rational imaginary part. – Bruno B Oct 24 '23 at 17:24
  • Okay, so my ideas were right. Thank you a lot! – Nick Oct 24 '23 at 17:25

0 Answers0