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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!

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    $\begingroup$ 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. $\endgroup$ Jan 9, 2013 at 12:29
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    $\begingroup$ 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. $\endgroup$ Jan 9, 2013 at 14:07
  • $\begingroup$ @Gerald: Thanks, I'll check Dunford & Schwartz @Theo: This proof is more elementary than mine, thank you. $\endgroup$ Jan 9, 2013 at 15:11
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    $\begingroup$ I think this result is also in the book A Course in Functional Analysis by John Conway. $\endgroup$ Jan 9, 2013 at 18:03

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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).

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    $\begingroup$ Thank you very much! I think one can cite exercises in Dunford & Schwartz ;-) $\endgroup$ Jan 9, 2013 at 16:57
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    $\begingroup$ I don't see why you wanted this to be a comment. It seems to be a fine answer. $\endgroup$
    – S. Carnahan
    Jan 9, 2013 at 22:35
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    $\begingroup$ 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.) $\endgroup$ Jan 10, 2013 at 8:31
  • $\begingroup$ Re, although you're stuck with the reputation you've already earned, you can always mark an answer as Community Wiki if you don't want to earn (further) reputation from it. Another approach is to find a problem you think deserves attention, and place an appropriate bounty on it. $\endgroup$
    – LSpice
    Nov 26, 2022 at 22:44

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