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 $S^*_p = S \cup \set p$.

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:

$\ds U_1, U_2$

$\in$

$\ds \tau$

$\ds \leadsto \ \ $

$\ds U_1 \cap U_2$

$\in$

$\ds \tau$

$\ds \leadsto \ \ $

$\ds U_1 \cap U_2$

$\in$

$\ds \tau^*_{\bar p}$

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

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

So:

$\ds\bigcup \UU \in \tau$

and so:

$\ds \bigcup \UU \in \tau^*_p$

If $S^*_p \in \UU$ then:

$\ds \bigcup \UU = S^*_p \in \tau^*_p$

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

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $16$. Open Extension Topology: $8$