Originally Posted by

**Sogan** I have a solution, but i'm not sure whether it is correct:

"<="

We want to show f is proper <=> f is closed.

Let V be a closed subset of X.

f(V) is closed in Y <=> Y-f(V) is open in Y.

so let y $\displaystyle \in$ Y-f(V) a arbitrary point. We know Y is Hausdorff, hence for every point v $\displaystyle \in$ f(V) we have a open set $\displaystyle U_v$ of v and a open set U of y, s.t.

$\displaystyle U_v \bigcap $ U= $\displaystyle \emptyset$

(*)hence it exist a nbh U of y , s.t. U $\displaystyle \bigcap$ f(V) = $\displaystyle \emptyset$

hence f(V) is closed.

I'm not sure whether this prove is correct.

I mean, i havnÄt use locally compactness of Y. So i think this can't be correct.

the part with (*) above isn't correct, or? how can i involve locally compactness in that point, such that my proof is correct?

Regards