Excluded Set Topology is Topology

Theorem
Let $T = \struct {S, \tau_{\bar H} }$ be an excluded set space.

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

Proof
We have by definition that $S \in \tau_{\bar H}$.

Also, as $H \cap \O = \O$, we have that $\O \in \tau_{\bar H}$.

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

By definition:
 * $H \cap U_1 = \O$

and:
 * $H \cap U_2 = \O$

and so by definition of set intersection:
 * $H \cap \paren {U_1 \cap U_2} = \O$

So:
 * $U_1 \cap U_2 \in \tau_{\bar H}$

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

We have that:
 * $\forall U \in \UU: H \cap U = \O$

Hence from Subset of Union:
 * $H \cap \bigcup \UU = \O$

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