Open Extension of Double Pointed Countable Complement Topology is T4 Space
Theorem
Let $T = \struct {S, \tau_S}$ be a countable complement topology on an uncountable set $S$.
Let $D = \struct {\set {0, 1}, \tau_D}$ be the indiscrete topology on two points.
Let $T \times D$ be the double pointed topology on $T$.
Let $\paren {T \times D}^*_{\bar p}$ be the open extension topology on $S \times \set {0, 1} \cup \set p$ where $p \notin S \times \set {0, 1}$.
Then $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space, and no other separation axioms are fulfilled.
That is, $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space or $T_5$ space.
Proof
From Open Extension Topology is T4 we have that $\paren {T \times D}^*_{\bar p}$ is a $T_4$ space.
From Double Pointed Countable Complement Topology fulfils no Separation Axioms, we have that $T \times D$ is not a $T_0$ space or a $T_5$ space.
From Condition for Open Extension Space to be $T_0$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_0$ space.
From Condition for Open Extension Space to be $T_5$ Space, it follows that $\paren {T \times D}^*_{\bar p}$ is not a $T_5$ space.
From Open Extension Topology is not $T_1$, we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space.
From Open Extension Topology is not $T_3$ we have that $\paren {T \times D}^*_{\bar p}$ is not a $T_3$ space.
Finally, from $T_2$ Space is $T_1$ Space, as $\paren {T \times D}^*_{\bar p}$ is not a $T_1$ space it is not a $T_2$ space.
$\blacksquare$
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: $8$