Subset of Indiscrete Space is Compact

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

Let $H \subseteq S$.

$H$ is compact in $T$.


The subspace $T_H = \struct {H, \set {\O, S \cap H} }$ is trivially also an indiscrete space.

The only open cover of $T_H$ is $\set H$ itself.

The only subcover of $H$ is, trivially, also $\set H$, which is finite.

So $H$ is (trivially) compact in $T$.