Excluded Point Space is T0/Proof 2
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Theorem
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $T$ is a $T_0$ (Kolmogorov) space.
Proof
We have:
- 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.
$\blacksquare$