Every lingering sequence has a convergent subsequence

Statement

Let $(X,d)$ be a metric space, then:

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

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

• Given a sequence $(x_n)_{n=1}^\infty\subseteq X$ 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 $k_1$ to be the index of any point of the sequence in $B_1(x)$
2. Take $k_2$ to be any index AFTER $k_1$ of the sequence in the ball $B_\frac{1}{2}(x)$
3. ...
4. Show the sequence $(x_{k_n})_{n=1}^\infty$ converges to $x$

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

See also

• Compactness
• Lingering sequence