Thread: proof Haussdorff space, when i have open surjection and grapf closed subset...

Hello. I need help with this:
Let be open surjection such that graph is closed subset of . Prove that Y is Hausdorff.
Thanks.

Originally Posted by tom007

Hello. I need help with this:
Let be open surjection such that graph {(x,f(x)):x in X} is closed subset of . Prove that Y is Hausdorff.
Thanks.

Let be distinct, and put . Then there are distinct with , and therefore . Since is an open subset of , then there must be open sets such that , where and ; this is because is a basis for . Notice that and must be disjoint, because otherwise we would have , a contradiction. So are disjoint neighborhoods of , respectively. It follows that is Hausdorff.

EDIT: The latex software is acting up, but you can quote my post to see the code.

Originally Posted by tom007

Hello. I need help with this:
Let be open surjection such that graph {(x,f(x)):x in X} is closed subset of . 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.

wow.. guys.. thanks a lot!