Subset of Metric Space contains Limits of Sequences iff Closed

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Theorem

Let $M = \left({A, d}\right)$ be a metric space.

Let $H \subseteq A$.


Then $H$ is closed in $M$ iff:

for each sequence $\left\langle{a_n}\right\rangle$ 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:

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

and:

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

If the set $\left\{ {a_1, a_2, \ldots}\right\}$ 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 $\left\{ {a_1, a_2, \ldots}\right\}$ is finite, then for some $N \in \N$:

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

Since:

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

Then:

$\forall n > N: d \left({a_n, a}\right) = 0$

Thus:

$a_n = a$

and so:

$a \in H$

$\Box$


Sufficient Condition

Let $H$ be a set such that:

for each sequence $\left\langle{a_n}\right\rangle$ such that $\displaystyle \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$.

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

Then $b$ is the limit of a convergent sequence of points of $H$.

By hypothesis, $b \in H$.

Thus $H$ is a closed set by definition.

$\blacksquare$


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