Subset of Indiscrete Space is Sequentially Compact

Theorem

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

Let $H \subseteq S$.


$H$ is sequentially compact in $T$.


Proof

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 $T$.

$\blacksquare$


Sources

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