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.

Proof
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 $S$.