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 $S \in \tau_{\bar H}$, and 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 $H \cap \left({U_1 \cap U_2}\right) = \varnothing$ by definition of set intersection.

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 properties are fulfilled for $\tau_{\bar H}$ to be a topology on $S$.