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 $x$.

$\blacksquare$


Sources