# Included Set Topology is Topology

## Theorem

Let $T = \left({S, \tau_H}\right)$ be an included set space.

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

## Proof

We have by definition that $\varnothing \in \tau_H$, and as $H \subseteq S$ we have that $S \in \tau_H$.

Now let $U_1, U_2 \in \tau_H$.

By definition $H \subseteq U_1$ and $H \subseteq U_2$.

From Intersection is Largest Subset it follows that $H \subseteq \left({U_1 \cap U_2}\right)$.

So $U_1 \cap U_2 \in \tau_H$.

Now let $\mathcal U \subseteq \tau_H$.

We have by definition that $\forall U \in \mathcal U: H \subseteq U$.

Hence from Subset of Union $H \subseteq \bigcup \mathcal U$.

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

$\blacksquare$