Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am very familiar with the proof of the following statement: If $X$ is a compact Hausdorff space such that the Banach algebra $C(X)$ is separable, then $X$ is metrizable.

Can this be used to prove a more generalized version of this statement with the set $C_b(X)$ of all continuous bounded functions on $X$? Namely, if $X$ is a completely regular Hausdorff space such that $C_{b}(X)$ is separable, then $X$ is a compact metrizable space.

share|cite|improve this question
up vote 8 down vote accepted

$C_b(X)$ is isomorphic and homeomorphic to $C(\beta X)$, where $\beta X$ is the Čech-Stone compactification of $X$, which exists, as $X$ is completely regular and Hausdorff. The isomorphism is of course given by mapping a bounded continuous real-valued function $f$, which has a compact codomain, essentially, to its (unique) Čech-Stone extension to $\beta X$. This also preserves the sup norm, as we can use $[\inf{f},\sup{f}]$ as the codomain of both $f$ and its extension. So it's an isometry between the function spaces.

Now, if the former is separable, so is the latter, and then we can apply your theorem to conclude that $\beta X$ is metrizable, but this only happens if $X$ was already compact metrizable (and thus $X = \beta X$). See this question for that.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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