Countable Complement Space is not Sigma-Compact

Theorem

Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.

Then $T$ is not a $\sigma$-compact space.

Proof

From Compact Sets in Countable Complement Space, the only compact sets in $T$ are finite.

A countable union of finite sets can not be an uncountable set.

Hence the result by definition of $\sigma$-compact space.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $20$. Countable Complement Topology: $2$