Double Pointed Finite Complement Topology fulfils no Separation Axioms

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

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

Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.

Proof
From Separation Axioms on Double Pointed Topology, we have that:
 * $T \times D$ is not a $T_0$ space.


 * $T \times D$ is not a $T_1$ space.


 * $T \times D$ is not a $T_2$ space.

Also from Separation Axioms on Double Pointed Topology, we have that:


 * $T \times D$ is a $T_3$ space $T$ is a $T_3$ space


 * $T \times D$ is a $T_4$ space $T$ is a $T_4$ space


 * $T \times D$ is a $T_5$ space $T$ is a $T_5$ space

But from Finite Complement Space is not $T_3$, $T_4$ or $T_5$:


 * $T$ is not a $T_3$ space


 * $T$ is not a $T_4$ space


 * $T$ is not a $T_5$ space

Hence the result.