Double Pointed Countable Complement Topology fulfils no Separation Axioms

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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 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 if and only if $T$ is a $T_3$ space
$T \times D$ is a $T_4$ space if and only if $T$ is a $T_4$ space
$T \times D$ is a $T_5$ space if and only if $T$ is a $T_5$ space


But from Countable 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.

$\blacksquare$


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