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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.

Martin Sleziak
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Sam
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    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. – Henno Brandsma Feb 06 '11 at 09:34
  • @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. – Sam Feb 06 '11 at 10:12
  • 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. – t.b. Feb 06 '11 at 10:26
  • @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. – Sam Feb 06 '11 at 10:27
  • 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. – t.b. Feb 06 '11 at 10:42
  • While looking up something else in Kelley's General Topology I stumbled over his section on Arzelà-Ascoli. He proves converses in various degrees of generality (e.g. replacing Hausdorff by regularity & equicontinuity by something he calls even continuity). E.g. Theorem 21 on p. 236 states that a family of continuous functions from a regular locally compact space to a regular Hausdorff space is compact in the compact-open topology if and only if 1. it is closed, 2. the pointwise evalutations have compact closure and 3. it is evenly continuous. Maybe that helps answering your question. – t.b. Jul 14 '11 at 18:36
  • @Theo: Thank you for this reference. I will make sure to have a look at it! :) – Sam Jul 26 '11 at 21:12
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    @t.b. When you look carefully at Dugundji proof, you notice that it used the compact-open topology where compact subspaces are all suppose to be Hausdorff by definition ... – brunoh Jul 20 '14 at 11:23

1 Answers1

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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.

Sam
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