Excluded Point Space is T0/Proof 2

Theorem

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.

Then $T$ is a $T_0$ (Kolmogorov) space.

Proof

We have:

Excluded Point Topology is Open Extension Topology of Discrete Topology

Discrete Space satisfies all Separation Properties (including being a $T_0$ space)

Then by Condition for Open Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its open extension.

$\blacksquare$