Closure of Neighborhood in Metric Space

Theorem

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

Let $N_\epsilon \left({x}\right)$ be an $\epsilon$-neighborhood in $M = \left({A, d}\right)$.

Let $y \in \operatorname{cl} \left({N_\epsilon \left({x}\right)}\right)$, where $\operatorname{cl}$ denotes the closure of $N_\epsilon \left({x}\right)$.

Then $d \left({x, y}\right) \le \epsilon$.

Proof

Suppose $d \left({x, y}\right) > \epsilon$.

Then $N_{d \left({x, y}\right) - \epsilon} \left({y}\right)$ is an open set containing $y$ and not meeting $N_\epsilon \left({x}\right)$.

Hence $y \notin \operatorname{cl} \left({N_\epsilon \left({x}\right)}\right)$.

$\blacksquare$