Subset of Metric Space contains Limits of Sequences iff Closed
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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.
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.8$