= Nested intervals extract a convergent subsequence
{title2=$|I_j|=(b-a)/2^j$}
Begin with a <closed interval> $I_0=[a,b]$ containing the entire <bounded sequence>. Split the interval at its midpoint. At least one of the two closed halves contains infinitely many indexed terms; choose such a half and repeat. This gives <nested intervals> $I_j$ with lengths $(b-a)/2^j$. The <nested interval theorem>, using <completeness of the real numbers>, supplies their unique common point $\ell$.
Choose $n_j>n_{j-1}$ with $x_{n_j}\in I_j$. Infinitely many indexed terms in $I_j$ guarantee that the next index can be chosen. Both $x_{n_j}$ and $\ell$ belong to $I_j$, so
$$
|x_{n_j}-\ell|\leq\frac{b-a}{2^j}\longrightarrow0.
$$
This is a short existence proof, not necessarily a finite algorithm for an arbitrary input <sequence>: deciding which half contains infinitely many future terms may require mathematical knowledge about that <sequence>.
\Image[/nested-interval-selection-bounded-subsequence.png]
{title=<Nested intervals> isolate a <subsequential limit> while the selected indices keep increasing}
{description=For $x_n=(-1)^n+1/(n+1)$, the illustrated halves contain infinitely many even-indexed terms and narrow around $1$. Each row marks one selected term inside its interval; the table gives the original index. The right plot shows the selected error below the interval-length bound. The exact selections follow the displayed model, rather than deciding an arbitrary infinite <sequence> by sampling.}
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