Prove that if $Y$ is hausdorff then $Y^X$ is it too when it is equipped with open-compact topology.

Question

So in this question is asked to show that if $Y$ is an Hausdorff space then for any other space $X$ the subset of continuous function $\mathcal C(X,Y)$ of $Y^X$ is Hausdorff but I think that really just $Y^X$ is it too: indeed it seem to me that Brian Scott's answer do not use continuity, right? So is really $Y^X$ Hausdorff when $Y$ is it? Could someone explain if Brian Scott's answer use continuity?

Answer 1

You can define the compact-open topology on the whole set $Y^X$, simply by using the family $\{M(C,U):C$ compact, $U$ open$\}$ as a subbase, where $M(C,U)=\{f:f[C]\subseteq U\}$. It contains the product topology, which has $\{M(\{x\},U):x\in X$, $U$ open$\}$ as a subbase, so it is Hausdorff if $Y$ is. For stronger properties, like regularity and complete regularity, you need to work in $C(X,Y)$ because there you use compactness of $f[C]$ all the time.