Results 1 to 7 of 7

Math Help - Topology question

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

    Topology question

    Hi, I would appreciate some guiding with this question. Thanks.


    Let X be some topological space and Y a compact, Hausdorff topological space. Let f:X \to Y and define \Gamma_f = \{(x, f(x)) : x \in X\}.

    Show that if \Gamma_f is a closed subset of X \times Y with the product topology, then f is continuous.
    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
    21
    Quote Originally Posted by Defunkt View Post
    Hi, I would appreciate some guiding with this question. Thanks.


    Let X be some topological space and Y a compact, Hausdorff topological space. Let f:X \to Y and define \Gamma_f = \{(x, f(x)) : x \in X\}.

    Show that if \Gamma_f is a closed subset of X \times Y with the product topology, then f is continuous.
    What have you tried?

    P.S. Y need not be Hausdorff.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Theorem: Let X be any space and Y be compact. Then, if \Gamma_f\subseteq X\times Y is closed we have that f:X\mapsto Y is continuous.

    Proof:

    Lemma: Let Y be compact, then \pi:X\times Y\mapsto X is a closed.
    Proof: Let E\subseteq X\times Y be closed. We prove that \pi(E) is closed by proving that it's compliment is open. So, let x\notin\pi(E). Then, it follows that \left(x\times Y\right)\cap E=\varnothing, and for each y\in Y (x,y) has a basic open neighborhood G_x(y)\times F_y\subseteq X\times Y-E. Since the tube x\times Y is compact (since it's homeomorphic to Y) there is a finite subcovering \left\{U_{y_1},\cdots,U_{y_m}\right\}. The neighborhood U=\bigcap_{k=1}^{m}U_x(y_k) is clearly disjoint from \pi(E). The conclusion follows. \blacksquare

    Now, it is easily proved that since \pi:X\times Y\mapsto X is closed so is \pi\mid \Gamma_f:\Gamma_f\mapsto X. But, it is clearly a bijective continuous mapping and since it's closed it's also a homeomorphism. The problem follows after considering the following lemma

    Lemma: A mapping f:X\mapsto Y is continuous if \pi\mid \Gamma_f:\Gamma_f\mapsto X is a homeomorphism.
    Proof: We must merely note that f=\pi_Y\circ\left(\pi_X\mid \Gamma_f\right)^{-1}.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Drexel28 View Post
    What have you tried?

    P.S. Y need not be Hausdorff.
    I think I came up with a solution. Tell me if you see anything wrong please.

    I want to show that f is continuous, ie. that if V \subset Y is open then f^{-1}(V) is open as well.

    Let V be an open subset of Y. Let U = V^c, then U is closed, and so is X \times U. Since \Gamma_f is closed as well, \Gamma_f \cap (X \times U) is also closed.

    From a previous result, we know that the projection \pi_x:X \times Y \to X is a closed mapping when Y is compact. Therefore, \pi_x(\Gamma_f \cap (X \times U) ) is closed.

    However, note that \pi_x(\Gamma_f \cap (X \times U) ) is exactly the set of points in X which f does not map into V, and it is closed, therefore f^{-1}(V) is open and we are done.

    I know it is a bit informal, but that is simply the outline.

    Thanks again for helping.
    Last edited by Defunkt; March 1st 2010 at 03:16 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Defunkt View Post
    I think I came up with a solution. Tell me if you see anything wrong please.

    I want to show that f is continuous, ie. that if V \subset Y is open then f^{-1}(V) is open as well.

    Let V be an open subset of Y. Let U = V^c, then U is closed, and so is X \times U. Since G is closed as well, G \cap (X \times U) is also closed.

    From a previous result, we know that the projection \pi_x:X \times Y \to X is a closed mapping when Y is compact. Therefore, \pi_x(G \cap (X \times U) ) is closed.

    However, note that \pi_x(G \cap (X \times U) ) is exactly the set of points in X which f does not map into V, and it is closed, therefore f^{-1}(V) is open and we are done.

    I know it is a bit informal, but that is simply the outline.

    Thanks again for helping.
    I'm having a little trouble following it, but I think you are saying that \left[f^{-1}(V)\right]'=\pi_X\left(\Gamma_f\cap\left(X\times V'\right)\right). If you can prove that I'll believe you.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2009
    From
    Israel
    Posts
    976
    Quote Originally Posted by Drexel28 View Post
    I'm having a little trouble following it, but I think you are saying that \left[f^{-1}(V)\right]'=\pi_X\left(\Gamma_f\cap\left(X\times V'\right)\right). If you can prove that I'll believe you.
    \Gamma_f \cap (X \times V') = \{(x, f(x)) : x \in X\} \cap \{(x,y) : y \notin V\} = \{(x, f(x)) : f(x) \notin V\} =  \{(x, f(x)) : x \notin f^{-1}(V)\}

    Sorry for being messy, by the way I mistakenly wrote G in my first answer instead of \Gamma_f, but I fixed it now.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Defunkt View Post
    \Gamma_f \cap (X \times V') = \{(x, f(x)) : x \in X\} \cap \{(x,y) : y \notin V\} = \{(x, f(x)) : f(x) \notin V\} =  \{(x, f(x)) : x \notin f^{-1}(V)\}

    Sorry for being messy, by the way I mistakenly wrote G in my first answer instead of \Gamma_f, but I fixed it now.
    Then I would agree!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Topology question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: August 3rd 2011, 05:13 AM
  2. Topology question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: February 6th 2009, 10:00 PM
  3. Topology question
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 20th 2009, 10:53 AM
  4. Topology Question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 19th 2009, 12:01 AM
  5. Topology Question
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 16th 2008, 04:33 AM

Search Tags


/mathhelpforum @mathhelpforum