Subset of Indiscrete Space is Compact

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $H \subseteq S$.


$H$ is compact in $T$.


Proof

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 $T$.

$\blacksquare$


Sources