Open Extension of Double Pointed Countable Complement Topology is T4 Space

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Theorem

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

Let $D = \left({\left\{{0, 1}\right\}, \vartheta}\right)$ be the indiscrete topology on two points.

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

Let $\left({T \times D}\right)^*_{\bar p}$ be the open extension topology on $S \times \left\{{0, 1}\right\} \cup \left\{{p}\right\}$ where $p \notin S \times \left\{{0, 1}\right\}$.


Then $\left({T \times D}\right)^*_{\bar p}$ is a $T_4$ space, and no other separation axioms are fulfilled.

That is, $\left({T \times D}\right)^*_{\bar p}$ 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 $\left({T \times D}\right)^*_{\bar p}$ 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 $\left({T \times D}\right)^*_{\bar p}$ is not a $T_0$ space.

From Condition for Open Extension Space to be $T_5$ Space, it follows that $\left({T \times D}\right)^*_{\bar p}$ is not a $T_5$ space.


From Open Extension Topology is not $T_1$, we have that $\left({T \times D}\right)^*_{\bar p}$ is not a $T_1$ space.

From Open Extension Topology is not $T_3$ we have that $\left({T \times D}\right)^*_{\bar p}$ is not a $T_3$ space.


Finally, from $T_2$ Space is $T_1$ Space, as $\left({T \times D}\right)^*_{\bar p}$ is not a $T_1$ space it is not a $T_2$ space.

$\blacksquare$


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