Countable Complement Space is not Weakly Countably Compact
Theorem
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Then $T$ is not 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 $H \subseteq S$ be a countable set.
From Limit Points of Countable Complement Space, it contains all its limit points.
Let $x \in H$.
$\relcomp S {\relcomp S H \cup \set x} = H \setminus \set x$ is countable.
Then $\relcomp S H \cup \set x$ is an open neighborhood of $x$.
However $H \cap \paren {\paren {\relcomp S H \cup \set x} \setminus \set x} = H \cap \relcomp S H = \O$.
So $x$ is not a limit point of $H$.
Since $x$ is arbitrary, there are no limit points in $H$.
This shows that $T$ is not 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: $20$. Countable Complement Topology: $7$