Condition for Point being in Closure/Metric Space

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

Let $H \subseteq S$.

Let $\map \cl H$ denote the closure of $H$ in $M$.

Let $x \in S$.

Then $x \in \map \cl H$ :
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$.

Proof
By definition of closure of $H$ in $M$:


 * $\map \cl H = H^i \cup H'$

where:
 * $H^i$ denotes the set of isolated points of $H$
 * $H'$ denotes the set of limit points of $H$.

Sufficient Condition
Let $x \in \map \cl H$.

Suppose $x \in H^i$.

By definition of isolated point:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set a$

But also by definition of isolated point, we do at least have that:
 * $x \in H$

and from Center is Element of Open Ball, we have that:
 * $\forall \epsilon \in \R_{>0}: x \in \map {B_\epsilon} x$

and so:
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

Suppose $x \in H'$:

By definition of limit point:
 * $\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$

and so:
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

So in both cases:
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

Necessary Condition
Let $x$ be such that:
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \O$

To show that $x \in \map \cl H$, it is sufficient to show that either:
 * $x \in H^i$

or that:
 * $x \in H'$

There are $2$ cases to consider.

$(1): \quad$ Suppose that:
 * $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$

Then $x$ is by definition an isolated point of $H$.

That is:
 * $x \in H^i$

$(2): \quad$ Suppose that:
 * $\forall \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H \ne \set x$

Let $\epsilon$ be arbitrary.

As $\map {B_\epsilon} x \cap H \ne \O$, it must follow that:
 * $\exists y \in \map {B_\epsilon} x \cap H: y \ne x$

and so:
 * $y \in \paren {\map {B_\epsilon} x \setminus \set x} \cap H$

Hence:
 * $\paren {\map {B_\epsilon} x \setminus \set x} \cap H \ne \O$

Then $x$ is by definition a limit point of $H$.

That is:
 * $x \in H'$

Hence the result.