[Understanding Analysis by Stephen Abbott] - [Chapter 2 Sequences and Series] - [2.5 Subsequences and the Bolzano-Weierstrass Theorem]

Definition 2.5.1 Let \((a_n)\) be a sequence of real numbers, and let \(n_1 < n_2 < n_3 < n_4 < n_5 < ...\) be an increasing sequence of natural numbers. Then the sequence $$ (a_{n_1}, a_{n_2}, a_{n_3}, a_{n_4}, a_{n_5},...) $$ is called a subsequence of \((a_n)\) and is denoted by \((a_{n_k})\), where \(k \in N\) indexes the subsequence.

Theorem 2.5.2 Subsequences of a convergent sequence converge to the same limit as the original sequence.

Proof. Assume \((a_n) \to a\), and let \(a_{n_k}\) be a subsequence. Given \(\epsilon > 0\), there exists \(N\) such that \(|a_n - a| < \epsilon\) whenever \(n \geqslant N\). Because \(n_k \geqslant k\) for all \(k\), the same \(N\) will suffice for the subsequence; that is, \(|a_{n_k} - a| < \epsilon\) whenever \(k \geqslant N\)

Theorem 2.5.5 Bolzano-Weierstrass Theorem. Every bounded sequence contains a convergent subsequence

Proof. Let \((a_n)\) be a bounded sequence so that there exists \(M > 0\) satisfying \(|a_n| \leqslant M\) for all \(n \in M\). Bisect the closed interval \([-M, M]\) into the two closed intervals \([-M, 0]\) and \([0, M]\). (THe midpoint is included 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 the 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 a half that again contains an infinite number of terms of the original sequence. Because there are an infinite number of terms from \((a_n)\) to choose from, we can select an \(a_{n_2}\) from the original sequence with \(n_2 > n_1\) and \(a_{n_2} \in I_2\). In general, 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 \(n_k > n_{k - 1} > \cdot \cdot \cdot > n_2 > n_1\) so that \(a_{n_k} \in I_l\)

We want to argue that \(a_{n_k}\) is a convergent subsequence, but we need a candidate for the limit. The sets $$ I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdot \cdot \cdot $$ form a nested sequence of closed intervals, and by the Nested Interval Property there exists at least one point \(x \in R\) contained in every \(I_k\). This provides us with the candidate we were looking for. It just remains to show that \(a_{n_k} \to x\)

Let \(\epsilon > 0\). By construction, the length of \(I_k\) is \(M(1/2)^{k - 1}\) which converges to zero. Choose \(N\) so that \(k \geqslant N\) implies that the length of \(I_k\) is less than \(\epsilon\). Because \(x\) and \(a_{n_k}\) are both in \(I_k\), it follows that \(|a_{n_k} - x| < \epsilon\)