Subset of Metric Space is Closed iff contains all Zero Distance Points

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

Let $H \subseteq A$.

Then $H$ is closed in $M$ :
 * $\forall x \in A: \map d {x, H} = 0 \implies x \in H$

where $\map d {x, H}$ denotes the distance between $x$ and $H$.

Necessary Condition
Let $H$ be closed in $M$.

Let $x \in A: \map d {x, H} = 0 \implies x \in H$.

By Existence of Sequence in Subset of Metric Space whose Limit is Infimum, there exists a sequence $\sequence {a_n}$ of points of $H$, such that:
 * $\ds \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$

Thus, every neighborhood of $x$ contains points of $H$.

If some $a_n \in H$, then $x \in H$.

Otherwise, each $a_n$ is different from $x$.

Then $x$ is the limit of $\sequence {a_n}$, and hence of $H$.

Thus $x \in H$, by definition of closed set.

Sufficient Condition
Suppose $H$ is such that:
 * $\forall x \in A: \map d {x, H} = 0 \implies x \in H$

From Point at Zero Distance from Subset of Metric Space is Limit Point or Element:
 * $\map d {x, H} = 0$

either:
 * $x$ is a limit point of $H$
 * $x \in H$

Let $x$ be limit point of $H$.

Then by hypothesis $x \in H$.

That is:
 * $H$ contains all its limit points

So by definition, $H$ is a closed set of $M$.

Also see

 * Point at Distance Zero from Closed Set is Element, which presents this result as:


 * If $H$ is closed in $M$, then $\map d {x, H} = 0$ $x \in H$.