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This question arose when I was reading a book on topology. In fact, it is a book on locally compact groups, hence only briefly discusses the topic. And the question is that, while in the proposition of the theorem, it is stated that when a map $\phi$ maps a topological space $X$ to another $Y$, and when $X$ is Hausdorff, $Y$ is compact and the graph of $\phi$ is closed, then $\phi$ is continuous, is it reallly necessary to include the condition that $X$ is Hausdorff? Since I see no reason, and I appear to have a proof without using the condition, I would like to know the answer.

A proof:
For any closed subspace C of $Y$, the pre-image D ought to be closed. For any element $a$ in the complement of D, we can use the compactness to show that there is a finite number of open sets in $Y$ such that the union of them covers the C, and then the corresponding open sets in $X$ is an open neighborhood of $a$ which has an empty intersection with D; so D is closed.
P.S. in the proof, those open sets are obtained by the condition that the graph is closed and that (a,c) is not in the graph for any c in C.

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Why would the pre-images of the open cover of $C$ be disjoint from $D$? –  Alon Amit Jun 14 '11 at 1:27
    
(1) Why is $C$ compact; (2) Why are the "corresponding sets" open? (you are trying to show $\phi$ is continuous) –  jspecter Jun 14 '11 at 1:28
    
@jspector: Since C is a closed subspace of a compact space, C is compact. As for the reason the corresponding stes are open, this results from the definition of them. In fact, as I said, ($a$,$c$) is not in the graph of $\phi$, hence there are open sets U and V such that UxV(the cartesian product of U and V) is an open neighborhood of ($a$,$c$) disjoint from the graph of $\phi$; then there is a finite number of U and V such that the union of V covers C and the intersection of U is an open neighborhood of $a$. Hope this helps explain the rushed question. –  awllower Jun 14 '11 at 1:46
    
yes. I messed up. –  jspecter Jun 14 '11 at 1:49
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While I have not really followed your proof in detail, it seems to me that your suspicion is correct. Indeed, my edition of Bourbaki, Topologie Générale I, p. TG I.113, exercise 14 to paragraph 10 asserts the following: Let $Y$ be locally compact (Hausdorff) and let $X$ be an arbitrary topological space. In order for $f$ to be continuous it is necessary and sufficient that the graph of $f$ be closed in $X \times Y'$ for all compact spaces $Y'$ containing $Y$ as a subspace. Now if $Y$ is compact, it is closed in all compact spaces containing it. (For Bourbaki compact includes Hausdorff). –  t.b. Jun 14 '11 at 3:39

1 Answer 1

up vote 8 down vote accepted

You do not need Hausdorffness on $X$. Let $\pi$ be the projection of $X \times Y$ onto $X$. As $Y$ is compact (we just need compact, not Hausdorff) the map $\pi$ is closed. See here or here e.g. It's due to Kuratowski, I believe.

Now, if $f: X \mapsto Y$ has the property that its graph $G_f = \{ (x,f(x)) \mid x \in X \}$ is closed in $X \times Y$, and $Y$ is compact, then $f$ is continuous: let $C$ be a closed subset of $Y$. Then $$f^{-1}[C] = \pi[(X \times C) \cap G_f]$$ and as $X \times C$ is closed in $X \times Y$, and so is $G_f$, we have written $f^{-1}[C]$ as the image of a closed set under a closed map, hence is closed. As $C$ was arbitary, $f$ is continuous (inverse image of closed is closed).

So we need no separation axioms at all.

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this is found as an exercise 7&8, page 171 in Munkres, Topology. –  Leon Lampret Jun 26 '11 at 20:46

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