Metric Closure and Topological Closure of Subset are Equivalent

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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 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 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:

$H^- = \map \cl H$

$\blacksquare$