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Math Help - continuous function and graph

  1. #1
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    continuous function and graph

    Hello,

    i have to show following statement:

    If X,Y are Hausdorff spaces with Y compact. f:X->Y some map.
    When the Graph Gr f:={(x,f(x)) \in X x Y: x  \in X}
    is closed in X x Y, then f must be continuous.


    I'm not sure, why i need compactness of Y.

    Let U be an open subset of Y. If A:=f^(-1) (U) is empty, there is nothing to show.
    So we can assume that A is not empty.
    Now i don't know how i can continue the proof...

    have someone a idea for me?

    Thank you
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Sogan View Post
    Hello,
    I'm not sure, why i need compactness of Y.
    You need Y to be compact, not Hausdorff.

    If Y is compact, the projection p:X\times Y\rightarrow X is closed.

    Then, for every F\subset Y closed,

    f^{-1}(F)=p[(X\times F)\cap \Gamma_f]

    is closed.

    Regards.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    You need Y to be compact, not Hausdorff.

    If Y is compact, the projection p:X\times Y\rightarrow X is closed.

    Then, for every F\subset Y closed,

    f^{-1}(F)=p[(X\times F)\cap \Gamma_f]

    is closed.

    Regards.
    You're obviously right, but the reasoning is very (intentionally?) vague.

    If C\subseteq Y is closed then X\times C is closed in X\times Y with the product topology. Now, note then \Gamma_f\cap\left(X\times C\right) is a closed subset of the subspace \Gamma_f, but by continuity this implies that \Gamma_f\cap\left(X\times C\right) is compact. Thus, since the canonical injection onto X, \pi_X is continuous we may conclude that \pi_X\left(\Gamma_f\cap\left(X\times \right)\right)\subseteq X is compact, and since X is Hausdorff it follows that \pi_X\left(\Gamma_f\left(X\times C\right)\right)\subseteq X is closed. But, using the above poster's correct formula we note that f^{-1}\left(C\right)=\pi_X\left(\Gamma_f\cap\left(X\ti  mes C\right)\right) it follows that the preimage under f of every closed C\subseteq Y is closed and thus f continuous.
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Drexel28 View Post
    You're obviously right, but the reasoning is very (intentionally?) vague.

    Yes, intentionally vague. The idea was to provide a clue.

    Regards.
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