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Math Help - proof Haussdorff space, when i have open surjection and grapf closed subset...

  1. #1
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    proof Haussdorff space, when i have open surjection and graph closed subset...

    Hello. I need help with this:
    Let f:X\rightarrow Y be open surjection such that graph \{(x,f(x)):\mbox{x \in X\}} is closed subset of X\times Y. Prove that Y is Hausdorff.
    Thanks.
    Last edited by tom007; April 19th 2011 at 09:05 AM.
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  2. #2
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    Quote Originally Posted by tom007 View Post
    Hello. I need help with this:
    Let f:X\rightarrow Y be open surjection such that graph {(x,f(x)):x in X} is closed subset of X\times Y. Prove that Y is Hausdorff.
    Thanks.
    Let y_1,y_2\in Y be distinct, and put S=\{(x,f(x)):x\in X\}. Then there are x_1,x_2\in X distinct with f(x_1)=y_1,f(x_2)=y_2, and therefore (x_1,f(x_2))\in S^c=\{(x,y):f(x)\neq y\}. Since S^c is an open subset of X\times Y, then there must be open sets U\subseteq X,V\subseteq Y such that U\times V\subseteq S^c, where x_1\in U and f(x_2)\in V; this is because \{U\times V:U\in\tau_X,V\in\tau_Y\} is a basis for X\times Y. Notice that f(U) and V must be disjoint, because otherwise we would have (x,f(x))\in U\times V\subseteq S^c, a contradiction. So f(U),V are disjoint neighborhoods of f(x_1)=y_1,f(x_2)=y_2, respectively. It follows that Y is Hausdorff.

    EDIT: The latex software is acting up, but you can quote my post to see the code.
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  3. #3
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    Quote Originally Posted by tom007 View Post
    Hello. I need help with this:
    Let f:X\rightarrow Y be open surjection such that graph {(x,f(x)):x in X} is closed subset of X\times Y. Prove that Y is Hausdorff.
    Thanks.
    Note that it suffices to prove that the diagonal \Delta_Y\subseteq Y\times Y is closed, or that Y\times Y-\Delta_Y is open. Prove though that the map f\oplus\text{id}_Y:X\times Y\to Y\times Yx,y)\mapsto (f(x),y) is open since f is. Then, since f is a surjection (and thus so is f\oplus \text{id}_Y) we have that \left(f\oplus\text{id}_Y\right)\left(X\times Y-\Gamma_f\right)=Y\times Y-\left(f\oplus\text{id}_Y\right)\left(\Gamma_f\righ t). But, this latter set must be open (since X\times Y-\Gamma_f is open by assumption and f\oplus\text{id}_Y is open) but it's easy to show that \left(f\oplus\text{id}_Y\right)\left(\Gamma_f\righ t)=\Delta_Y. The conclusion then follows.
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    wow.. guys.. thanks a lot!
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