Balzano-Weierstrass Theorem

Scratch work:

Here we split each interval, and choose the one that has infinitely-many terms. Here we have it that:

We can pick an element $b_1\in I_1$, then the next element $b_2\in I_2$, and so on.

More formally, pick an $n_1\in \mathbb{N}$ such that $a_{n_1}\in I_1$, $n_2\in \mathbb{N}$ where $a_{n_2}\in I_2$ and $n_2>n_1$. Continue with $n_3\in \mathbb{N}$ such that $a_{n_3}\in I_3$ and $n_3>n_2$. Repeat as needed. We get that, by the Nested Interval Property, that
Proof

Let $(a_n)$ be a bounded sequence so that $\mathrm{\exists }M>0$ where $|a_n|\le M$ for all $n\in \mathbb{N}$. Bisect the closed interval $[-M,M]$ into the two closed intervals $[-M,0]$ and $[0,M]$ (the midpoint is in both halves). Now it must be that at least one of these closed intervals contains an infinite number of the terms in the sequence $(a_n)$. Select a half for which this is the case and label that interval as $I_1$. Then let $a_{n_1}$ be some term in the sequence $(a_n)$ satisfying $a_{n_1}\in I_1$.

Next we bisect $I_1$ into closed intervals of equal length and let $I_2$ be again the half that contains infinitely many terms from $(a_n)$. Select an $a_{n_2}$ from the original sequence with $n_2>n_1$ and $a_{n_2}\in I_2$.

In general if we construct the closed interval $I_k$ by taking a half of $I_{k-1}$ containing an infinite number of terms of $(a_n)$ and then select $n_k>n_{k-1}>\cdots >n_2>n_1$ so that $a_{n_k}\in I_k$.

We want to argue that $(a_{n_k})$ is a convergent subsequence. But we need a candidate for the limit. Via the Nested Interval Property since:

Then $\mathrm{\exists }x\in \mathbb{R}$ where $x\in I_k$ for every $k$. It just remains to show that $(a_{n_k})\to x$. Let $\epsilon >0$. Via 25 Subsequences and the Bolzano-Weierstrass Theorem Practice#2.5.3 and the Limit Laws (Algebraic Limit Theorem) then
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