Double Pointed Topology is not T0/Proof 1

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Theorem

Let $T_1 = \struct {S, \tau_S}$ be a topological space.

Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.

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


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


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 $D$ is an indiscrete space.


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

$\blacksquare$