Limit Point of Subset of Metric Space is at Zero Distance

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

Let $H \subseteq A$ be an arbitrary subset of $A$.

Let $x \in A$ be a limit point of $H$.

Let $\map d {x, H}$ denote the distance between $x$ and $H$:
 * $\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$

Then:
 * $\map d {x, H} = 0$

Proof
Let $x$ be a limit point of $H$.

$\map d {x, H} \ne 0$.

By definition of metric, that means:
 * $\map d {x, H} > 0$

Then:
 * $\exists \epsilon \in \R_{>0}: \forall y \in H: \map d {x, y} > \epsilon$

That is:
 * $\forall y \in H: y \notin \map {B_\epsilon} x \setminus \set x$

where $\map {B_\epsilon} x \setminus \set x$ denotes the deleted $\epsilon$-neighborhood of $x$.

That is:
 * $\paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$

Hence, by definition, $x$ is not a limit point of $H$.

From this contradiction it follows that it cannot be the case that $\map d {x, H} \ne 0$

That is:
 * $\map d {x, H} = 0$

Warning
The converse is not necessarily true.

It may be the case that $H$ has no limit points.

The corresponding result is in fact:

If $\map d {x, H} = 0$, then either:
 * $x$ is a limit point of $H$

or:
 * $x \in H$

This is demonstrated in Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero.