Excluded Set Topology is Topology

Theorem
Let $T = \left({S, \tau_{\bar H}}\right)$ 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 $H \cap \varnothing = \varnothing$, we have that $\varnothing \in \tau_{\bar H}$.

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

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

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

and so by definition of set intersection:
 * $H \cap \left({U_1 \cap U_2}\right) = \varnothing$

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

Now let $\mathcal U \subseteq \tau_{\bar H}$.

We have that:
 * $\forall U \in \mathcal U: H \cap U = \varnothing$

Hence from Subset of Union:
 * $H \cap \bigcup \mathcal U = \varnothing$

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