Double Pointed Countable Complement Topology is Weakly Countably Compact
Jump to navigation
Jump to search
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: $7$