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Math Help - Compact spaces

  1. #1
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    Compact spaces

    Hi, I would like some help with this problem if possible. Been trying to solve it for a while but haven't really got anything meaningful...

    Let X be a topological space, Y a compact topological space and p:X\to Y such that p is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that \forall y \in Y, ~ p^{-1}(\{y\}) is compact.

    Prove that X is compact.

    Thanks for any help.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Defunkt View Post
    Hi, I would like some help with this problem if possible. Been trying to solve it for a while but haven't really got anything meaningful...

    Let X be a topological space, Y a compact topological space and p:X\to Y such that p is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that \forall y \in Y, ~ p^{-1}(\{y\}) is compact.

    Prove that X is compact.

    Thanks for any help.
    Have you considered that X=\bigcup_{y\in Y}p^{-1}(\{y\})?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Have you considered that X=\bigcup_{y\in Y}p^{-1}(\{y\})?
    Yes I have, but is it true that *any* union of compact sets is compact?

    It is trivially correct for a finite union, but I don't know if it is for inifinte.
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  4. #4
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    Have you considered the Finite Intersection Property?
    A topological space is compact if and only if every collection of subsets having the FIP has itself a nonempty intersection.
    You are given that Y is compact.
    You are given that p is a closed mapping.

    Can you use the FIP to prove that X is compact?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Have you considered the Finite Intersection Property?
    A topological space is compact if and only if every collection of subsets having the FIP has itself a nonempty intersection.
    You are given that Y is compact.
    You are given that p is a closed mapping.

    Can you use the FIP to prove that X is compact?
    I did try approaching this way, seemed rational as the mapping is closed, but could not really finish it:

    take some collection \mathcal{S} of closed sets in X that has the FIP. Then p maps it to a collection of closed sets \mathcal{T} = \{p(S) : S \in \mathcal{S}\} in Y, and we know that a closed subset of a compact space is compact itself, so for every T \in \mathcal{T}, T is compact...

    Does Y being compact also imply that \mathcal{T} also has FIP (this seems to be correct intuitively...)? And also, if that is true, then isn't p^{-1}(\bigcap_{T \in \mathcal{T}}T) \subset \bigcap_{S \in \mathcal{S}}S which would mean we are done? :\

    I know this is not much, but I'm just having a brainfart, so excuse me.
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  6. #6
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    It means that \mathcal{T} has the FIP. You may need to prove that.
    Because Y is compact then {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }.
    You know that p is onto.
    Does that mean that the intersection of the sets in \mathcal{S} is not empty?
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  7. #7
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    Quote Originally Posted by Plato View Post
    It means that \mathcal{T} has the FIP. You may need to prove that.
    Because Y is compact then {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }.
    You know that p is onto.
    Does that mean that the intersection of the sets in \mathcal{S} is not empty?
    Well, since we know \exists t \in \bigcap\limits_{v \in \mathcal{T}} v we get that \forall S \in \mathcal{S}, ~t \in p(S) and p being onto means \forall S \in \mathcal{S} ~ \exists s \in S : p(s) = t

    Now, if |p^{-1}(t)| = 1 we are done, otherwise, we (I) seem to be in a problem. I guess we need to use that p^{-1}(\{t\}) is compact at this point, but I don't see how...
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  8. #8
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    Frankly I have no idea what all of that means.
    Here is what we must do.
    Given any collection of closed subset of X having the FIP then that collection must have a non-empty interection.
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  9. #9
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    Quote Originally Posted by Plato View Post
    Frankly I have no idea what all of that means.
    Here is what we must do.
    Given any collection of closed subset of X having the FIP then that collection must have a non-empty interection.
    I realize that, but I still can't seem to solve it.

    Given a collection of closed subsets in X that has the FIP, its image (of p) also has the FIP since p is closed. Now, since Y is compact, there exists some t \in \bigcap_{T \in \mathcal{T}}T where \mathcal{T} is the collection of images of p. But we may have that p^{-1}(\{t\}) contains more than one element "spread" over the collection of closed subsets in X, and then I don't see how we are sure that there is an element in the intersection of that collection.

    Sorry for being a bother, I just can't seem to see how to get the desired result. I, of course, appreciate your help.
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  10. #10
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    Quote Originally Posted by Defunkt View Post
    I realize that, but I still can't seem to solve it.

    Given a collection of closed subsets in X that has the FIP, its image (of p) also has the FIP since p is closed. Now, since Y is compact, there exists some t \in \bigcap_{T \in \mathcal{T}}T where \mathcal{T} is the collection of images of p. But we may have that p^{-1}(\{t\}) contains more than one element "spread" over the collection of closed subsets in X, and then I don't see how we are sure that there is an element in the intersection of that collection.

    Sorry for being a bother, I just can't seem to see how to get the desired result. I, of course, appreciate your help.
    This is the last response I will give you.
    The remainder of the proof is extremely messy.
    Suppose that  \bigcap_{T \in \mathcal{T}}T is empty.
    For each point in p^{-1}(\{t\}) is not in one of those sets.
    So we cover p^{-1}(\{t\}) with finite collect of open sets that contains no point is a finite sub-collection of \mathcal{T}.
    That is not possible.
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  11. #11
    MHF Contributor Drexel28's Avatar
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    I'm sorry. I misread the question.

    Quote Originally Posted by Defunkt View Post
    Yes I have, but is it true that *any* union of compact sets is compact?

    It is trivially correct for a finite union, but I don't know if it is for inifinte.
    No. \mathbb{R}=\bigcup_{n\in\mathbb{N}}\left[-n,n\right]
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  12. #12
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    Quote Originally Posted by Plato View Post
    This is the last response I will give you.
    The remainder of the proof is extremely messy.
    Suppose that  \bigcap_{T \in \mathcal{T}}T is empty.
    For each point in p^{-1}(\{t\}) is not in one of those sets.
    So we cover p^{-1}(\{t\}) with finite collect of open sets that contains no point is a finite sub-collection of \mathcal{T}.
    That is not possible.
    Thanks, I understand now.
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