Indiscrete Space is T3
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Theorem
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space .
Then $T$ is a $T_3$ space.
Proof
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.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $10$