Excluded Point Space is T0/Proof 1
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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
Let $T$ be a trivial space.
That is, let $S = \set p$.
Then the result holds vacuously, as there are no two distinct points in $T$.
Now suppose $T$ is not trivial.
Then $\exists x \in S: x \ne p$.
Now we have that $\set x \subseteq T$ is open in $T$ such that $p \notin \set x$ but $x \in \set x$.
Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.
Then we have that (for example) $\set x \subseteq T$ is open in $T$ such that $x \in \set x$ but $y \notin \set x$
Hence the result.
$\blacksquare$