Double Pointed Countable Complement Topology is Weakly Countably Compact

Theorem

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

Let $T \times D$ be the double pointed topology on $T$.

Then $T \times D$ is weakly countably compact.

Proof

By definition, $T$ is weakly countably compact if and only if every infinite subset of $S$ has a limit point in $S$.

Let $D = \set {0, 1}$.

Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$.

Then its twin $\tuple {p, 1}$ is a limit point of $A$.

Hence the result by definition of weakly countably compact.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $21$. Double Pointed Countable Complement Topology: $7$