Closure of Open Ball in Metric Space

Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M$.

Let $y \in \map \cl {\map {B_\epsilon} x}$, where $\cl$ denotes the closure of $\map {B_\epsilon} x$.

Then $\map d {x, y} \le \epsilon$.

Proof

Suppose $\map d {x, y} > \epsilon$.

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

Hence:

$y \notin \map \cl {\map {B_\epsilon} x}$

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.7$: Definitions: Proposition $3.7.23$