Compact Subset of Compact Space is not necessarily Closed

Theorem

A compact subset of a compact space is not necessarily closed.

Proof

Let $S$ be a set containing more than one element.

Let $\tau = \set {S, \O}$ be the indiscrete topology on $S$.

Let $x \in S$.

Let $H = S \setminus \set x$.

Then $H$ is a proper subset of $S$.

Then from Subset of Indiscrete Space is Compact and Sequentially Compact, the subspace induced by $\tau$ on $H$ is a compact subspace of $\struct {S, \tau}$.

But from Closed Sets in Indiscrete Topology, the only closed subsets of $\struct {S, \tau}$ are $S$ and $\O$.

Thus $H$ is a compact subspace of $\struct {S, \tau}$ which is not closed in $S$.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction