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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?

Antonio Maria Di Mauro
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    Can you remind us what $Y^X$ is? – Dietrich Burde Dec 25 '22 at 12:19
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    The notation $Y^{X}$ for two sets $X$ and $Y$ usually denotes the set of functions from $X$ to $Y$. (The motivation for this notation is that for finite sets $\vert Y^{X}\vert=\vert Y\vert^{\vert X\vert}$). – G. Blaickner Dec 25 '22 at 12:43
  • Well, the open-compact topology on $C(X,Y)$ uses the fact that elements of this set are continuous functions (by definition). Since the answer by Brian Scott uses the subbasis of this specific topology, does this not necessarily uses continuity? – G. Blaickner Dec 25 '22 at 12:46
  • @G.Blaickner Well really open compact topology is defined in $Y^X$ and Brian Scott use subbasis elements of $Y^X$ as you can see here where is defined open-compact topology: is this false? – Antonio Maria Di Mauro Dec 25 '22 at 13:08
  • For a locally compact space $X$, the compact-open topology $C_\mathrm{co}(X,-)$ is uniquely defined by being the right adjoint to $-\times X$ (therefore being the internal hom). I currently don't know how, but maybe this could be useful for an elegant answer, since when using this adjunction, the space $Y$ (from which the seperation axiom is inherited) will be isolated on the right side: $C(A,C_\mathrm{co}(X,Y))\cong C(A\times X,Y)$ – Samuel Adrian Antz Dec 26 '22 at 03:46

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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.

hartkp
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