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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.

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5  
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 6 '11 at 9:34
    
@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 6 '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 6 '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 6 '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 6 '11 at 10:42

1 Answer 1

up vote 1 down vote accepted

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.

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