Every lingering sequence has a convergent subsequence

Statement

Let [ilmath](X,d)[/ilmath] be a metric space, then:

• [math]\forall(x_n)_{n=1}^\infty\subseteq X\left[\left(\exists x\in X\ \forall\epsilon>0[\vert B_\epsilon(x)\cap(x_n)_{n=1}^\infty\vert=\aleph_0]\right)\implies\left(\exists(k_n)_{n=1}^\infty\subseteq\mathbb{N}\left[(\forall n\in\mathbb{N}[k_n<k_{n+1}])\implies\left(\exists x'\in X\left[\lim_{n\rightarrow\infty}(x_{k_n})=x'\right]\right)\right]\right)\right][/math]

This is just a verbose way of expressing the statement that:

• Given a sequence [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] if it is a lingering sequence then it has a subsequence that converges

Proof

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TODO: Write proof

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Proof outline:

1. Take [ilmath]k_1[/ilmath] to be the index of any point of the sequence in [ilmath]B_1(x)[/ilmath]
2. Take [ilmath]k_2[/ilmath] to be any index AFTER [ilmath]k_1[/ilmath] of the sequence in the ball [ilmath]B_\frac{1}{2}(x)[/ilmath]
3. ...
4. Show the sequence [ilmath](x_{k_n})_{n=1}^\infty[/ilmath] converges to [ilmath]x[/ilmath]

We have exhibited a convergent subsequence, we're done.

See also

• Compactness
• Lingering sequence