Interior via balls

Proposition. Let $(X,d)$ be a metric space, let $a\in X$, and let $E\subset X$. Then

Context. This equivalence connects the “union of all open subsets” definition of interior to the ball-based definition of openness in metric spaces.

Proof sketch.

• If $a\in\operatorname{int}(E)$, then $a$ lies in some open set $G\subset E$, so by openness there is a ball around $a$ contained in $G\subset E$.
• If $B(a;\delta)\subset E$, then $B(a;\delta)$ is open and contained in $E$, hence lies in the union defining $\operatorname{int}(E)$, so $a\in\operatorname{int}(E)$.