Double Pointed Topology is not T0

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Theorem

Let $T_1 = \left({S, \tau_S}\right)$ be a topological space.

Let $D = \left({A, \left\{ {\varnothing, A}\right\} }\right)$ be the indiscrete topology on an arbitrary doubleton $A = \left\{{a, b}\right\}$.

Let $T = \left({T_1 \times D, \tau}\right)$ be the double pointed topological space on $T_1$.


Then $T$ is not a $T_0$ (Kolmogorov) space.


Proof 1

By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$.


Let $x \in S$, and consider the point $\tuple {x, a} \in S \times A$.

Then:

$\forall U \in \tau: \tuple {x, a} \in U \implies \tuple {x, b} \in U$
$\forall U \in \tau: \tuple {x, b} \in U \implies \tuple {x, a} \in U$

as $T_2$ is an indiscrete space.


Hence the result, by definition of $T_0$ (Kolmogorov) space.

$\blacksquare$


Proof 2

By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$.

By definition, $D$ is the indiscrete space on a doubleton.


Aiming for a contradiction, suppose $T$ is a $T_0$ (Kolmogorov) space.

Then from Product Space is $T_0$ iff Factor Spaces are $T_0$ it follows that $D$ is also a $T_0$ (Kolmogorov) space.

But from Indiscrete Non-Singleton Space is not $T_0$, $D$ is not a $T_0$ (Kolmogorov) space.


The result follows by Proof by Contradiction.

$\blacksquare$