Interior via balls - Knowlpedia

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

a\in \operatorname{int}(E)\quad\Longleftrightarrow\quad \exists\,\delta>0 \text{ such that } B(a;\delta)\subset E.

Remarks

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 $a\in\operatorname{int}(E)$ .