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)) X x Y: x 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
You're obviously right, but the reasoning is very (intentionally?) vague.
If is closed then is closed in with the product topology. Now, note then is a closed subset of the subspace , but by continuity this implies that is compact. Thus, since the canonical injection onto , is continuous we may conclude that is compact, and since is Hausdorff it follows that is closed. But, using the above poster's correct formula we note that it follows that the preimage under of every closed is closed and thus continuous.