Included Set Topology is Topology

Theorem
Let $T = \struct {S, \tau_H}$ be an included set space.

Then $\tau_H$ is a topology on $S$, and $T$ is a topological space.

Let $U_1, U_2 \in \tau_H$.

By definition of included set topology:
 * $H \subseteq U_1$ and $H \subseteq U_2$

From Intersection is Largest Subset:
 * $H \subseteq U_1 \cap U_2$

By definition of included set topology:
 * $U_1 \cap U_2 \in \tau_H$

Let $\UU \subseteq \tau_H$.

By definition of included set topology:
 * $\forall U \in \UU: H \subseteq U$

Hence from Subset of Union:
 * $H \subseteq \bigcup \UU$

By definition of included set topology:
 * $S \in \tau_H$

So all the properties are fulfilled for $\tau_H$ to be a topology on $S$.