Subset of Metric Space is Subset of its Closure

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

Let $H \subseteq A$ be a subset of $A$.

Then:
 * $H \subseteq H^-$

where $H^-$ denotes the closure of $H$.

Proof
By definition of closure:


 * $H^- = H' \cup H^i$

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

Let $a \in H$.

If $a$ is a limit point of $H$ then $a \in H'$.

Suppose $a \notin H'$.

Then by definition of limit point:
 * $\neg \forall \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} \ne \O$

That is:
 * $\exists \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} = \O$

and so as $\map d {a, a} = 0$:
 * $\exists \epsilon \in \R_{>0}: \set {x \in A: \map d {x, a} < \epsilon} = \set a$

Thus, by definition, $a$ is an isolated point of $H$.

So $a \in H'$ or $a \in H^i$.

Hence by definition of set union:
 * $a \in H \subseteq H^-$

By definition of subset:
 * $H \subseteq H \subseteq H^-$

and hence the result by definition of closure.