Subset of Metric Space is Subset of its Closure

Theorem
Let $M = \left({A, d}\right)$ 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:
 * $\neg \forall \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} \ne \varnothing$

That is:
 * $\exists \epsilon \in \R_{>0}: \left\{{x \in A: 0 < d \left({x, a}\right) < \epsilon}\right\} = \varnothing$

and so as $d \left({a, a}\right) = 0$:
 * $\exists \epsilon \in \R_{>0}: \left\{{x \in A: d \left({x, a}\right) < \epsilon}\right\} = \left\{{a}\right\}$

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.