Countable Complement Space is not Separable

Theorem

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

Then $T$ is not a separable space.

Proof

Let $U$ be a countable subset of $S$.

By the definition of $T$, $U$ is closed.

As $U$ is countable but $S$ is uncountable:

$U \subsetneq S$

and so:

$U \ne S$

From Closed Set Equals its Closure:

$U^- = U$

where $U^-$ is the closure of $U$.

Thus:

$U^- \ne S$

So by definition, $U$ is not everywhere dense in $T$.

As $U$ is arbitrary, there is no countable set which is everywhere dense in $T$.

The result follows by definition of separable 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: $3$