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A continuous map $f\colon X \to Y$ of locally compact spaces is called proper if for any compact $C\subset Y$ the preimage $f^{-1}(C)$ is compact. My question is: How I can prove that: If $X$ is compact, then any map $f\colon X\to Y$ is proper?

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It appears that locally compact includes the Hausdorff condition for you. Hint: Observe that compact subsets of a Hausdorff space are closed and continuity can be defined via closed sets. // What does this have to do with finite groups and abelian groups? –  Martin May 5 '13 at 13:21
    
@Martin: I do not speak about groups!. –  DER May 5 '13 at 13:23
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But you used the tags (finite-groups) and (abelian-groups). I asked why :-) –  Martin May 5 '13 at 13:24
    
@Martin: This is from the precedent question!. –  DER May 5 '13 at 13:25

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up vote 4 down vote accepted

Assuming $Y$ is Hausdorff, if $C$ is compact then it is closed. So $f^{-1}(C)$ is closed and any closed subspace of a compact space is compact.

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I think that a compact $C ⊂ Y$ need not be closed if $Y$ is not hausdorff. –  k.stm May 5 '13 at 13:24
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@K.Stm. You are correct, I'll edit. –  pritam May 5 '13 at 13:27

Hint: Assume $Y$ to be hausdorff. Then use that preimages under a continuous map of closed subsets are closed.

As pointed out by StefanH. and Martin in the comments, what you really need is compact subsets of $Y$ to be closed therein, which holds in hausdorff spaces. Locally compact spaces are usually assumed to be hausdorff.

Here’s an example where the criterion fails if $Y$ if compact subsets of $Y$ are not necessarily closed:

Take $Y = \{0,1\}$ with the trivial topology. And let $f : [0..1] → Y$ be the indicator function on $(0..1)$. Since the only open subsets of $Y$ are $∅$ and $Y$ itself, clearly $f$ is continuous, but $f^{-1}(\{1\}) = (0..1)$ is not compact, whereas singletons are definitely compact.

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Nice answer. Is "concrete topology" = "trivial topology" = "indiscrete topology", i.e., the only open sets are the empty set and the full space? –  Martin May 5 '13 at 13:39
    
@Martin – trivial topolgy, yes. I thought it was called “concrete”. –  k.stm May 5 '13 at 13:42
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It was rather clear from the context, but I've never seen it used that way. Thanks. –  Martin May 5 '13 at 13:44
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Actually, $Y$ Hausdorff is not necessary. What you need is the property that compact subsets are closed, and there are non-Hausdorff spaces where this holds. –  Stefan Hamcke May 5 '13 at 14:04
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An easy example of a non-Hausdorff space in which compact sets are closed is the cocountable topology on an uncountable set. The compact sets are precisely the finite sets. –  Martin May 5 '13 at 14:14

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