MathOverflow is a question and answer site for professional mathematicians. 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 know (and am able to prove via Stone-Čech compactification) that the following is correct:

Theorem: A metric space is compact if and only if its space of bounded, continuous, real-valued functions is separable in the uniform topology.

I use it in a paper for readers who are presumably not familiar with this kind of topology, so I cannot call it "obvious" or "well-known". I would be thankful for a name and/or good reference to cite this theorem!

share|cite|improve this question
I don't have a reference, but I'd suggest this argument: if $X$ is not compact, there is an infinite closed discrete subset $D$ of $X$. For every $A \subset D$ choose a continuous function $f_A \colon X \to [0,1]$ such that $f_A|_A = 1$ and $f_A|_{D \setminus A} = 0$ (by Urysohn). This gives an uncountable family of continuous bounded functions such that $\|f_A - f_B\| = 1$ whenever $A \neq B$. Alternatively, embed $\ell^\infty$ using a similar trick. If $X$ is compact then $C(X)$ is separable by Stone-Weierstrass. – Theo Buehler Jan 9 '13 at 12:29
Agreed, it is easy to prove. If you don't get a good reference, perhaps searching the index in Dunford & Schwartz will work. They generally have collected all results of this type. – Gerald Edgar Jan 9 '13 at 14:07
@Gerald: Thanks, I'll check Dunford & Schwartz @Theo: This proof is more elementary than mine, thank you. – Wolfgang Loehr Jan 9 '13 at 15:11
I think this result is also in the book A Course in Functional Analysis by John Conway. – Joseph Van Name Jan 9 '13 at 18:03
up vote 8 down vote accepted

The result does appear in Dunford/Schwartz, Linear Operators Part I (page 437), but is only stated as an exercise.

Edit after @JosephVanName' comment: Conway's Functional Analysis has the result for completely regular spaces as Theorem 6.6 (page 140).

share|cite|improve this answer
Thank you very much! I think one can cite exercises in Dunford & Schwartz ;-) – Wolfgang Loehr Jan 9 '13 at 16:57
I don't see why you wanted this to be a comment. It seems to be a fine answer. – S. Carnahan Jan 9 '13 at 22:35
Well, a proof should not only convince the reader that a statement is true, but also explain why it is true. For the former, an appeal to authority (like a reference to an exercise in D&S) is sufficient, but the latter requires pointing the reader to a full proof. That's why I did not think my first reply was a full answer. (Also, the references were suggested by others and I just checked them, so it seemed wrong to harvest the rep for them.) – Christian Clason Jan 10 '13 at 8:31

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.