Excluded Point Space is T0

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Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.


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


Proof 1

Let $T$ be a trivial space.

That is, let $S = \left\{{p}\right\}$.

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 $\left\{{x}\right\} \subseteq T$ is open in $T$ such that $p \notin \left\{{x}\right\}$ but $x \in \left\{{x}\right\}$.


Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$.

Then we have that (for example) $\left\{{x}\right\} \subseteq T$ is open in $T$ such that $x \in \left\{{x}\right\}$ but $y \notin \left\{{x}\right\}$


Hence the result.

$\blacksquare$


Proof 2

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.

$\blacksquare$


Sources