Closure of Open Ball may not equal Closed Ball of Same Radius

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

Let $x \in A$.

Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball of $x$ of radius $\epsilon$ for some $\epsilon \in \R_{>0}$.

Let $\map { {B_\epsilon}^-} x$ be the closed $\epsilon$-ball of $x$ of radius $\epsilon$.

Then it is not necessarily the case that:

$\map \cl {\map {B_\epsilon} x} = \map { {B_\epsilon}^-} x$

where $\cl$ denotes closure.


Proof by Counterexample:

Let $M = \struct {A, d}$ be the standard discrete metric space on a set $A$.

From Closure of Open $1$-Ball in Standard Discrete Metric Space we have that:

$\map \cl {\map {B_1} x} = \set x$


\(\ds \map { {B_1}^-} x\) \(=\) \(\ds \set {y \in A: \map d {x, y} \le 1}\) Definition of Closed Ball of Metric Space
\(\ds \) \(=\) \(\ds A\)

Hence the result.