Excluded Point Topology is Topology

Theorem

Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.

Then $\tau_{\bar p}$ is a topology on $S$, and $T$ is a topological space.

We have by definition that $S \in \tau_{\bar p}$, and as $p \notin \O$ we have that $\O \in \tau_{\bar p}$.

Now let $U_1, U_2 \in \tau_{\bar p}$.

By definition $p \notin U_1$ and $p \notin U_2$.

By definition of set intersection:

$p \notin U_1 \cap U_2$

So $U_1 \cap U_2 \in \tau_{\bar p}$.

Now let $\UU \subseteq \tau_{\bar p}$.

We have that $\forall U \in \UU: p \notin U$.

Hence from Subset of Union $p \notin \bigcup \UU$.

So all the properties are fulfilled for $\tau_{\bar p}$ to be a topology on $S$.

$\blacksquare$

1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology