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Let $X,Y$ be topological spaces and equip $C(X,Y)$ with the point-open topology (generalized the topology of pointwise convergence) generated by the following sub-base: $$ V(x,U)\triangleq \left\{ f\in C(X,Y):\, f(x)\subseteq U \right\}, $$ where $x \in X$ and $U$ is open in $Y$. If $Y$ is Hausdorff and $X$ is $T_1$, then is the point-open topology Hausdorff also?

ABIM
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  • The point-open topology agrees with the topology of pointwise convergence. You do not need $X$ to be $T_1$ to show that If $Y$, then the point-open topology is Hausdorff. – Paul Frost Dec 04 '19 at 09:25
  • @PaulFrost interesting; I just happened to notice the same thing in a comment below. This is a nice coincidence :) – ABIM Dec 04 '19 at 10:30

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If $f \neq g$ then there exist $x$ such that $f(x) \neq g(x)$. There exist disjoint open sets $U$ and $W$ containing $f(x)$ and $g(x)$ respectively. Now $V(x,U)$ and $V(x,W)$ are disjoint open sets containing $f$ and $g$.

Kavi Rama Murthy
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  • Thanks you Kabo; so the proof is the same as in the Compact-open topology then. – ABIM Dec 04 '19 at 09:21
  • @AIM_BLB that topology is finer, so also Hausdorff. You use singletons in the compact- open case. – Henno Brandsma Dec 04 '19 at 09:43
  • @HennoBrandsma I think it is only finer if $X$ is also Hausdorff, since for example if $X$ is equipped with a topology which is not $T_1$ then no singelton ${x}$ for $x \in X$ is closed (and therefore is not compact). – ABIM Dec 04 '19 at 10:29
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    @AIM_BLB no, it’s always finer as finite sets are compact in any space. – Henno Brandsma Dec 04 '19 at 10:34
  • Ah, you're right; very silly of me. – ABIM Dec 04 '19 at 10:42