Open Extension Topology is Topology

Theorem
Let $T = \struct {S, \tau}$ be a topological space.

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

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

Proof
By definition:


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

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

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

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 $\UU \subseteq \tau^*_{\bar p}$.

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

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

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

So $\tau^*_p$ is a topology on $S \cup \set p$.