Subset of Indiscrete Space is Compact

Theorem

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

Let $H \subseteq S$.

$H$ is compact in $T$.

Proof

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 $T$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $3$