Closure of Open Ball in Metric Space

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Theorem

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

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


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

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


Proof

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

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

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

$\blacksquare$


Sources