The more general version of this theorem in Munkres' 'Topology' (p. 290 - 2nd edition) states that

Given a locally compact Hausdorff space $X$ and a metric space $(Y,d)$; a family $\mathcal F$ of continuous functions has compact closure in $\mathcal C (X,Y)$ (topology of compact convergence) if and only if it is equicontinuous under $d$ and the sets

$$ \mathcal F _a = \{f(a) | f \in \mathcal F\} \qquad a \in X$$

have compact closure in $Y$.

Now I do not see why the Hausdorff condition on $X$ should be necessary? Why include it then? Am I maybe even missing something here (and there are counterexamples)?

btw if you are looking up the proof: Hausdorffness is needed for the evaluation map $e: X \times \mathcal C(X,Y) \to Y, \, e(x,f) = f(x)$ to be continuous. But the only thing really used in the proof is the continuity of $e_a: \mathcal C(X,Y) \to Y, \, e_a(f) = f(a)$ for fixed $a \in X$.

Cheers, S.L.

  • 6
    $\begingroup$ If we omit Hausdorff, what is the definition of local compactness? Every point has a compact neighbourhood, every point has an open neighbourhood with compact closure, every point has a neighbourhood base of open sets with compact closure? , etc. These are the same for Hausdorff spaces, not in general. This makes the combination locally compact + Hausdorff very common, and it also implies Tychonoff (completely regular), ensuring that there are continuous functions to $\mathbb{R}$ (e.g.) at all. If looking for counterexamples, I think indiscrete spaces will work, if you allow those as loc.cpt. $\endgroup$ – Henno Brandsma Feb 6 '11 at 9:34
  • $\begingroup$ @Henno: By Munkres' definition a space is locally compact if for every point $x$ there is a compact set containing a neighborhood of $x$. (which is a bit strange for a local property) Assuming the space to be Hausdorff, this condition indeed gets much nicer. @Theo: Thanks for the reference. I'll have a look at it. $\endgroup$ – Sam Feb 6 '11 at 10:12
  • $\begingroup$ The Hausdorff condition is not necessary, you can also drop local compactness altogether. A proof can be found e.g. in Dugundji, Topology, p.267. I removed my previous comment because I managed to confuse myself. $\endgroup$ – t.b. Feb 6 '11 at 10:26
  • $\begingroup$ @Theo: Dugundji only seems to proof one direction. I don't believe local compactness can be dropped for the other implication (would be very strange). Munkres original statement is something like: X a space, Y metric then the two condiditions on the family imply compactness. The converse is true if X is loc. cpt. H'dorff. $\endgroup$ – Sam Feb 6 '11 at 10:27
  • $\begingroup$ You're right. On the other hand, the direction proved by Dugundji is the more important one and it's good to know that no local compactness is needed for that. $\endgroup$ – t.b. Feb 6 '11 at 10:42

I think this question has been already been answered through the helpful comments. So thanks to Henno Brandsma and t.b.! This is just to finally tick it off.

My conclusion: It seems that $X$ being Hausdorff is rather a matter of convenience (maybe to avoid issues with the definition of local compactness for non-Hausdorff spaces, as pointed out in the comments), than a necessary condition.

Also this version of the theorem seems quite general enough for most uses.


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.