Open Extension Topology is Topology

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $\tau^*_{\bar p}$ be the open extension topology of $\tau$.

Then $\tau^*_{\bar p}$ is a topology on $S^*_p = S \cup \left\{{p}\right\}$.

Proof
By definition:


 * $\tau^*_{\bar p} = \left\{{U: U \in \tau}\right\} \cup \left\{{S^*_p}\right\}$

We have that $S^*_p \in \tau^*_{\bar p}$ by definition.

We also have that $\varnothing \in \tau$ so $S\varnothing \in \tau^*_{\bar p}$.

Now let $U_1, U_2 \in \tau^*_{\bar p}$.

Then $U_1, U_2 \in \tau \implies U_1 \cap U_2 \in \tau \implies U_1 \cap U_2 \in \tau^*_{\bar p}$.

Finally consider $\mathcal U \subseteq \tau^*_{\bar p}$.

Assuming $S^*_p \notin \mathcal U$ we have that $\mathcal U \subseteq \tau$.

So:
 * $\displaystyle \bigcup \mathcal U \in \tau$ and so $\displaystyle \bigcup \mathcal U \in \tau^*_p$.

If $S^*_p \in \mathcal U$ then $\displaystyle \bigcup \mathcal U = S^*_p \in tau^*_p$.

So $\tau^*_p$ is a topology on $S \cup \left\{{p}\right\}$.