Subset of Indiscrete Space is Compact

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Let $T = \left({S, \left\{{\varnothing, S}\right\}}\right)$ be an indiscrete topological space.

Let $H \subseteq S$.

$H$ is compact in $T$.


The subspace $T_H = \left({H, \left\{{\varnothing, S \cap H}\right\}}\right)$ is trivially also an indiscrete space.

The only open cover of $T_H$ is $\left\{{H}\right\}$ itself.

The only subcover of $H$ is, trivially, also $\left\{{H}\right\}$, which is finite.

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