Subset of Indiscrete Space is Sequentially Compact

From ProofWiki
Jump to navigation Jump to search


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

Let $H \subseteq S$.

$H$ is sequentially compact in $T$.


From Sequence in Indiscrete Space converges to Every Point, every sequence in $T$ converges to every point of $S$.

So every infinite sequence has a subsequence which converges to every point in $S$.

Hence $H$ is (trivially) sequentially compact in $T$.