Interior is Union of Elements of Basis
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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:
- $V^\circ = \bigcup \left\{ {G \in B: G \subseteq V}\right\}$
$\blacksquare$
Sources
- Mizar article YELLOW_8:11