Interior is Union of Elements of Basis

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

Let $B$ be a basis of $T$.

Let $V$ be a subset of $S$.

Then $V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$

where $V^\circ$ denotes the interior of $V$.

Proof
By definition of interior:
 * $\left\{ {G \in B: G \subseteq V}\right\} = \left\{ {G \in B: G \subseteq V^\circ}\right\}$

and
 * $V^\circ$ is open.

Thus by Open Set is Union of Elements of Basis:
 * $V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$