Condition for Closure of Open Ball to be Closed Ball of Same Radius

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Theorem

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


The following statements are equivalent:

$(1) \quad$ for all $x \in X$ and $r > 0$ we have $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$, where $\cl$ denotes closure
$(2) \quad$ for each $\epsilon > 0$ and $x, y \in X$ with $x \ne y$, there exists $z \in X$ with $\map d {z, y} < \epsilon$ and $\map d {x, z} < \map d {x, y}$.


Proof

$(1)$ implies $(2)$

Suppose that $(1)$ holds.

Let $\epsilon > 0$.

Let $x, y \in X$ have $x \ne y$.

We aim to show that there exists $z \in X$ with $\map d {z, y} < \epsilon$ and $\map d {x, z} < \map d {x, y}$.

Let $r = \map d {x, y}$.

From $(1)$, we have $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$.

So $y \in \map \cl {\map {B_r} x}$.

Hence by the definition of topological closure, for each $\epsilon > 0$ we have:

$\map {B_\epsilon} y \cap \map {B_r} x \ne \O$

Let $z \in \map {B_\epsilon} y \cap \map {B_r} x$.

Then $\map d {y, z} < \epsilon$ and $\map d {x, z} < r = \map d {x, y}$.

$\Box$


$(2)$ implies $(1)$

Suppose that $(2)$ holds.

Let $x \in X$ and $r > 0$.

From Closure of Open Ball in Metric Space, we have:

$\map \cl {\map {B_r} x} \subseteq \map { {B_r}^-} x$

It remains to show that:

$\map { {B_r}^-} x \subseteq \map \cl {\map {B_r} x}$

Since $\map {B_r} x \subseteq \map \cl {\map {B_r} x}$, we only need to show that if $y \in X$ has $\map d {x, y} = r$, then $y \in \map \cl {\map {B_r} x}$.

Let $y \in X$ have $\map d {x, y} = r$.

Let $\epsilon > 0$.

Then there exists $z \in X$ such that $\map d {x, z} < \map d {x, y} = r$ and $\map d {z, y} < \epsilon$.

So we have $z \in \map {B_\epsilon} y$ and $z \in \map {B_r} x$.

So $\map {B_\epsilon} y \cap \map {B_r} x \ne \O$ for each $\epsilon > 0$.

Hence by the definition of closure, we have $y \in \map \cl {\map {B_r} x}$.

Hence we obtain $\map \cl {\map {B_r} x} = \map { {B_r}^-} x$.

$\blacksquare$


Sources

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