Metric Closure and Topological Closure of Subset are Equivalent

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

Let $T = \struct{A, \tau}$ be the topological space with the topology induced by $d$.

Let $H \subseteq A$.

Then:
 * the metric closure of $H$ in $M$ equals the topological closure of $H$ in $T$

Proof
Let $H^i$ be the set of isolated points of $H$ in $M$.

From Leigh.Samphier/Sandbox/Isolated Point in Metric Space iff Isolated Point in Topological Space:
 * $H^i$ equals the set of isolated points of $H$ in the topological space $T$.

Let $H'$ be the set of limit points of $H$ in $M$.

From Leigh.Samphier/Sandbox/Limit Point in Metric Space iff Limit Point in Topological Space:
 * $H'$ equals the set of limit points of $H$ in the topological space $T$.

Let $H^-$ denote the closure of $H$ in the metric space $M$.

By definition of the closure of $H$ in the metric space $M$
 * $H^- = H' \cup H^i$

Let $\map \cl H$ denote the closure of the closure of $H$ in the topological space $T$.

By definition of the closure of $H$ in the topological space $T$
 * $\map \cl H = H' \cup H^i$

Thus:
 * $H^- = \map \cl H$