# Particular Point Space is T0

## Theorem

Let $T = \left({S, \tau_p}\right)$ be a particular point space.

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

## Proof 1

Let $T$ be a trivial space, so that $S = \left\{{p}\right\}$.

Then the result holds vacuously -- 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 $\left\{{p}\right\} \subseteq T$ is open in $T$ such that $p \in \left\{{p}\right\}$ but $x \notin \left\{{p}\right\}$.

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

Then we have that $\left\{{p, x}\right\} \subseteq T$ is open in $T$ such that $x \in \left\{{p, x}\right\}$ but $y \notin \left\{{p, x}\right\}$.

Hence the result.

$\blacksquare$

## Proof 2

We have:

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

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

$\blacksquare$

## Sources

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{II}: \ 8 - 10: \ 4$