Closure of Subset of Metric Space by Convergent Sequence
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Lemma
Let $M$ be a metric space.
Let $C \subseteq M$.
Let $x \in M$.
Let $\map \cl C$ denote the closure of $C$ in $T$.
Then $x \in \map \cl C$ if and only if there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.
Proof
Necessary Condition
Suppose there exists a sequence $\sequence {x_n}$ in $C$ which converges to $x$.
Let $\epsilon > 0$.
Then by definition:
- $\exists N \in \N: \forall n > N: x_n \in \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$ in $M$.
Since $\forall n: x_n \in C$, it follows that:
- $\forall \epsilon > 0: \map {B_\epsilon} x \cap C \ne \O$
Hence $x \in \map \cl C$.
$\Box$
Sufficient Condition
Now suppose $x \in \map \cl C$.
By definition of closure:
- $\forall n \in \N: \exists x_n \in C \cap \map {B_{1 / n} } x$
Thus clearly $\sequence {x_n}$ converges to $x$.
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$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $7.2$: Sequential compactness: Lemma $7.2.2$