# Thread: continuous function and graph

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)) $\in$ 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

2. Originally Posted by Sogan
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.

3. Originally Posted by FernandoRevilla
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.

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.