Equivalence of Definitions of Closed Set in Metric Space

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

Let $H \subseteq A$.

Proof
Let $H'$ denote the set of limit points of $H$.

Definition 1 implies Definition 2
Let $H$ be closed in $M$ by definition 1.

Then by definition $A \setminus H$ is open in $M$.

Let $b \in M: b \in A \setminus H$.

Then by definition of open set:
 * $\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \subseteq A \setminus H$

where $\map {B_\delta} {b; d}$ denotes the open $\delta$-ball of $b$.

From Intersection with Complement is Empty iff Subset:
 * $\map {B_\delta} {b; d} \cap H = \O$

So by definition of limit point:
 * $b \notin H'$

Hence:
 * $A \setminus H \cap H' = \O$

By Intersection with Complement is Empty iff Subset:
 * $H' \subseteq H$

Thus $H$ contains all its limit points.

That is, $H$ be closed in $M$ by definition 2.

Definition 2 implies Definition 1
Let $H$ be closed in $M$ by definition 2.

That is:
 * $H' \subseteq H$

Then by Set Complement inverts Subsets:
 * $A \setminus H \subseteq A \setminus H'$

Let $b \in A \setminus H$.

Then by definition of relative complement:
 * $b \notin H'$

Then by definition of limit point:
 * $\exists \delta \in \R_{>0}: \map {B_\delta} {b; d} \cap H = \O$

and so:
 * $\map {B_\delta} {b; d} \subseteq A \setminus H$

As $b$ is arbitrary, it follows that $A \setminus H$ is open in $M$.

Hence by definition, $H$ is closed in $M$ by definition 1.

Also see

 * Complement of Closed Set in Complex Plane is Open
 * Complement of Open Set in Complex Plane is Closed