Indiscrete Space is T3

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Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space .

Then $T$ is a $T_3$ space.


Let $F \subseteq S$ be a closed set in $S$.

Let $y \in S$ such that $y \notin F$.

The only way this can happen is if $F = \O$.

So there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

That is, $U = \O$ and $V = S$.

Hence (trivially) the result.