Excluded Point Space is T0

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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 1

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.

$\blacksquare$


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

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