Explanation of the Bolzano-Weierstrass theorem proof

[BifSlamkovich]

Homework Statement

Every bounded sequence has a convergent subsequence.

Homework Equations

Suppose that closed intervals I_0 $\supset$ I_1$\supset$ ... $\supset$ I_m and natural numbers n$_{1}$ < n$_{2}$ <
... < n$_{m}$ have been chosen such that for each 0 $\leq$ k $\leq$ m,
(2)
|I$_{k}$| = $b-a/2^{k}$, x$_{n}_{k}$$\in$I$_{k}$n and x$_{n}$ $\in$ I$_{k}$ for infinitely many n.

The Attempt at a Solution

 

[lippyka]

Let a_1, a_2, ... be a bounded sequence. For each natural number k, defineI_k = [a_m-2^k(a_m-a_1), a_m+2^k(a_m-a_1)]. Since the sequence is bounded, we have that b-a/2^k <= b-a < \infty for all k, where b and a are the upper and lower bounds of the sequence respectively. Thus, the intervals I_k are well-defined.By the Bolzano-Weierstrass theorem, there exists a convergent subsequence of the bounded sequence a_1, a_2, ... . Let n_1, n_2, ... n_m be the indices of the subsequence. For each 0 \leq k \leq m, we have that x_{n_k} \in I_k and x_{n} \in I_k for infinitely many n. Thus, the conditions of (2) are satisfied and we can conclude that the bounded sequence has a convergent subsequence.