Open Ball is Open Set/Metric Space

Theorem
Let $M = \left({A, d}\right)$ be a metric space.

Let $x \in A$.

Let $\epsilon \in \R_{>0}$.

Let $B_\epsilon \left({x}\right)$ be an open $\epsilon$-ball of $x$ in $M$.

Then $B_\epsilon \left({x}\right)$ is an open set of $M$.

Proof
Let $y \in B_\epsilon \left({x}\right)$.

From Open Ball of Point Inside Open Ball, there exists $\delta \in \R_{>0}$ such that $B_\delta \left({y}\right) \subseteq B_\epsilon \left({x}\right)$

The result follows from the definition of open set.