Double Pointed Topology is not T0/Proof 1

Proof
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.