Results 1 to 12 of 12

Thread: Compact spaces

  1. #1
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976

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

    Prove that $\displaystyle X$ is compact.

    Thanks for any help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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 $\displaystyle X$ be a topological space, $\displaystyle Y$ a compact topological space and $\displaystyle p:X\to Y$ such that $\displaystyle p$ is continuous, onto and closed (maps any closed set in X to a closed set in Y), and such that $\displaystyle \forall y \in Y, ~ p^{-1}(\{y\})$ is compact.

    Prove that $\displaystyle X$ is compact.

    Thanks for any help.
    Have you considered that $\displaystyle X=\bigcup_{y\in Y}p^{-1}(\{y\})$?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Drexel28 View Post
    Have you considered that $\displaystyle X=\bigcup_{y\in Y}p^{-1}(\{y\})$?
    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    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 $\displaystyle Y$ is compact.
    You are given that $\displaystyle p$ is a closed mapping.

    Can you use the FIP to prove that $\displaystyle X$ is compact?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    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 $\displaystyle Y$ is compact.
    You are given that $\displaystyle p$ is a closed mapping.

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

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

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

    I know this is not much, but I'm just having a brainfart, so excuse me.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    It means that $\displaystyle \mathcal{T}$ has the FIP. You may need to prove that.
    Because $\displaystyle Y$ is compact then $\displaystyle {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }$.
    You know that $\displaystyle p$ is onto.
    Does that mean that the intersection of the sets in $\displaystyle \mathcal{S}$ is not empty?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Plato View Post
    It means that $\displaystyle \mathcal{T}$ has the FIP. You may need to prove that.
    Because $\displaystyle Y$ is compact then $\displaystyle {\exists t \in \bigcap\limits_{v \in \mathcal{T}} v }$.
    You know that $\displaystyle p$ is onto.
    Does that mean that the intersection of the sets in $\displaystyle \mathcal{S}$ is not empty?
    Well, since we know $\displaystyle \exists t \in \bigcap\limits_{v \in \mathcal{T}} v$ we get that $\displaystyle \forall S \in \mathcal{S}, ~t \in p(S)$ and $\displaystyle p$ being onto means $\displaystyle \forall S \in \mathcal{S} ~ \exists s \in S : p(s) = t$

    Now, if $\displaystyle |p^{-1}(t)| = 1$ we are done, otherwise, we (I) seem to be in a problem. I guess we need to use that $\displaystyle p^{-1}(\{t\})$ is compact at this point, but I don't see how...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    Frankly I have no idea what all of that means.
    Here is what we must do.
    Given any collection of closed subset of $\displaystyle X$ having the FIP then that collection must have a non-empty interection.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    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 $\displaystyle X$ 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 $\displaystyle t \in \bigcap_{T \in \mathcal{T}}T$ where $\displaystyle \mathcal{T}$ is the collection of images of p. But we may have that $\displaystyle p^{-1}(\{t\})$ 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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    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 $\displaystyle t \in \bigcap_{T \in \mathcal{T}}T$ where $\displaystyle \mathcal{T}$ is the collection of images of p. But we may have that $\displaystyle p^{-1}(\{t\})$ 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 $\displaystyle \bigcap_{T \in \mathcal{T}}T$ is empty.
    For each point in $\displaystyle p^{-1}(\{t\})$ is not in one of those sets.
    So we cover $\displaystyle p^{-1}(\{t\})$ with finite collect of open sets that contains no point is a finite sub-collection of $\displaystyle \mathcal{T}$.
    That is not possible.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22
    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. $\displaystyle \mathbb{R}=\bigcup_{n\in\mathbb{N}}\left[-n,n\right]$
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    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 $\displaystyle \bigcap_{T \in \mathcal{T}}T$ is empty.
    For each point in $\displaystyle p^{-1}(\{t\})$ is not in one of those sets.
    So we cover $\displaystyle p^{-1}(\{t\})$ with finite collect of open sets that contains no point is a finite sub-collection of $\displaystyle \mathcal{T}$.
    That is not possible.
    Thanks, I understand now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Non-compact metric spaces
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: Jun 29th 2011, 08:50 PM
  2. [SOLVED] locally compact spaces
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 3rd 2011, 03:28 AM
  3. Compact spaces - Union and Intersection
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Mar 9th 2010, 01:19 PM
  4. Metric spaces - continuity / compact
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: Jun 29th 2009, 06:04 PM
  5. Compact Metric Spaces
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 15th 2008, 07:15 PM

Search Tags


/mathhelpforum @mathhelpforum