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

• Apr 18th 2011, 01:53 PM
tom007
proof Haussdorff space, when i have open surjection and graph closed subset...
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph $\displaystyle \{(x,f(x)):\mbox{x \in X\}}$ is closed subset of $\displaystyle X\times Y$. Prove that Y is Hausdorff.
Thanks.
• Apr 18th 2011, 06:12 PM
hatsoff
Quote:

Originally Posted by tom007
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph {(x,f(x)):x in X} is closed subset of $\displaystyle X\times Y$. Prove that Y is Hausdorff.
Thanks.

Let $\displaystyle y_1,y_2\in Y$ be distinct, and put $\displaystyle S=\{(x,f(x)):x\in X\}$. Then there are $\displaystyle x_1,x_2\in X$ distinct with $\displaystyle f(x_1)=y_1,f(x_2)=y_2$, and therefore $\displaystyle (x_1,f(x_2))\in S^c=\{(x,y):f(x)\neq y\}$. Since $\displaystyle S^c$ is an open subset of $\displaystyle X\times Y$, then there must be open sets $\displaystyle U\subseteq X,V\subseteq Y$ such that $\displaystyle U\times V\subseteq S^c$, where $\displaystyle x_1\in U$ and $\displaystyle f(x_2)\in V$; this is because $\displaystyle \{U\times V:U\in\tau_X,V\in\tau_Y\}$ is a basis for $\displaystyle X\times Y$. Notice that $\displaystyle f(U)$ and $\displaystyle V$ must be disjoint, because otherwise we would have $\displaystyle (x,f(x))\in U\times V\subseteq S^c$, a contradiction. So $\displaystyle f(U),V$ are disjoint neighborhoods of $\displaystyle f(x_1)=y_1,f(x_2)=y_2$, respectively. It follows that $\displaystyle Y$ is Hausdorff.

EDIT: The latex software is acting up, but you can quote my post to see the code.
• Apr 18th 2011, 09:24 PM
Drexel28
Quote:

Originally Posted by tom007
Hello. I need help with this:
Let $\displaystyle f:X\rightarrow Y$ be open surjection such that graph {(x,f(x)):x in X} is closed subset of $\displaystyle X\times Y$. 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 Y:(x,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.
• Apr 19th 2011, 08:51 AM
tom007
wow.. guys.. thanks a lot! :)