# Subset of Metric Space contains Limits of Sequences iff Closed

## Theorem

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

Let $H \subseteq A$.

Then $H$ is closed in $M$ if and only if:

for each sequence $\sequence {a_n}$ of points of $H$ that converges to a point $a \in A$, it follows that $a \in H$.

## Proof

### Necessary Condition

Let $H$ be closed in $M$.

Suppose that:

$\ds \lim_{n \mathop \to \infty} a_n = a$

and:

$\forall n \in \N_{>0}: a_n \in H$

If the set $\set {a_1, a_2, \ldots}$ is infinite then every neighborhood of $a$ contains infinitely many points of $H$.

Thus $a$ is a limit point of $H$.

So by definition of closed set, $a \in H$.

On the other hand, if $\set {a_1, a_2, \ldots}$ is finite, then for some $N \in \N$:

$n, m > N \implies a_n = a_m$

Since:

$\ds \lim_{n \mathop \to \infty} a_n = a$

Then:

$\forall n > N: \map d {a_n, a} = 0$

Thus:

$a_n = a$

and so:

$a \in H$

$\Box$

### Sufficient Condition

Let $H$ be a set such that:

for each sequence $\sequence {a_n}$ such that $\ds \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$.

Let $b$ be a limit point of $H$.

By Definition of Limit Point (Metric Space), $b$ is the limit of a convergent sequence of points of $H$.

By hypothesis, $b \in H$.

Thus $H$ is a closed set by definition.

$\blacksquare$