1. 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)) $\displaystyle \in$ X x Y: x $\displaystyle \in$ 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

2. Originally Posted by Sogan
Hello,
I'm not sure, why i need compactness of Y.
You need $\displaystyle Y$ to be compact, not Hausdorff.

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

Then, for every $\displaystyle F\subset Y$ closed,

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

is closed.

Regards.

3. Originally Posted by FernandoRevilla
You need $\displaystyle Y$ to be compact, not Hausdorff.

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

Then, for every $\displaystyle F\subset Y$ closed,

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

is closed.

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

If $\displaystyle C\subseteq Y$ is closed then $\displaystyle X\times C$ is closed in $\displaystyle X\times Y$ with the product topology. Now, note then $\displaystyle \Gamma_f\cap\left(X\times C\right)$ is a closed subset of the subspace $\displaystyle \Gamma_f$, but by continuity this implies that $\displaystyle \Gamma_f\cap\left(X\times C\right)$ is compact. Thus, since the canonical injection onto $\displaystyle X$, $\displaystyle \pi_X$ is continuous we may conclude that $\displaystyle \pi_X\left(\Gamma_f\cap\left(X\times \right)\right)\subseteq X$ is compact, and since $\displaystyle X$ is Hausdorff it follows that $\displaystyle \pi_X\left(\Gamma_f\left(X\times C\right)\right)\subseteq X$ is closed. But, using the above poster's correct formula we note that $\displaystyle f^{-1}\left(C\right)=\pi_X\left(\Gamma_f\cap\left(X\ti mes C\right)\right)$ it follows that the preimage under $\displaystyle f$ of every closed $\displaystyle C\subseteq Y$ is closed and thus $\displaystyle f$ continuous.

4. Originally Posted by Drexel28
You're obviously right, but the reasoning is very (intentionally?) vague.

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

Regards.