Open Ball is Open Set

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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.

$\blacksquare$


Sources