The Timpson Lyceum · Real Analysis (2020)

Lecture 8 (Real Analysis 2020)

Overview

This lecture introduces subsequences of a sequence and develops their core properties before using them to define sequential compactness of a set. We prove that every convergent sequence passes its limit on to all of its subsequences, that every sequence (no matter how erratic) contains a monotone subsequence, and that every bounded sequence therefore has a convergent subsequence — the Bolzano–Weierstrass theorem. We close by defining sequentially compact sets and showing, with examples, that exactly the closed bounded intervals $[a,b]$ are sequentially compact. These tools are the foundation for the convergence and compactness arguments that dominate the rest of real analysis.

Key Concepts

Subsequence. Given a sequence $(a_n)$ and a strictly increasing sequence of natural numbers $n_1 < n_2 < n_3 < \cdots$ , the sequence $(a_{n_k})$ is a subsequence of $(a_n)$ . Intuitively, you scan left to right through $(a_n)$ and keep or discard each term; you may never repeat a term or reorder them. The strict increase of the indices is exactly what enforces this "keep order, no repeats" rule. A subsequence still has infinitely many terms.
- Any sequence is a subsequence of itself (take $n_k = k$ ). Also, the shifted sequence $(a_{n+1})$ is a subsequence of $(a_n)$ — it is just $(a_n)$ with the first term dropped.
- Index fact: for any increasing index sequence, $n_k \ge k$ for every $k$ (the $k$ -th chosen index is at least $k$ , since indices start at $1$ and strictly increase).
Subsequences of a convergent sequence. If $\lim_{n\to\infty} a_n = a$ , then every subsequence $(a_{n_k})$ also converges to $a$ . Consequently a sequence with two subsequences tending to different limits cannot converge.
Peak index. An index $m$ is a peak index of $(a_n)$ if every later term is no larger: $a_m \ge a_n \quad \text{for all } n \ge m.$ A peak index is a property of the index, not of the value at it. A sequence may have zero peak indices (e.g. $1,2,3,4,\dots$ ), finitely many, or infinitely many (e.g. $\tfrac1n$ , where every index is a peak; or $(-1)^n$ , where every even index is a peak).
Monotone Subsequence Theorem. Every sequence contains a monotone subsequence. (Proof sketch: if there are infinitely many peak indices, the subsequence on those indices is monotonically decreasing; if there are only finitely many, then past the last peak no index is a peak, so each chosen term can be exceeded by a later one, building a monotonically increasing subsequence.)
Bolzano–Weierstrass Theorem. Every bounded sequence has a convergent subsequence. (A bounded sequence has a monotone subsequence; that subsequence is also bounded; and every bounded monotone sequence converges.)
Sequential compactness. A set $S \subseteq \mathbb{R}$ is sequentially compact if every sequence $(a_n)$ with all terms in $S$ has a subsequence converging to a point of $S$ . This is a property of the set, not of any one sequence.
- The half-open interval $(0,2]$ is not sequentially compact: $\tfrac1n$ lies in it, but every subsequence converges to $0 \notin (0,2]$ .
- The unbounded interval $[0,\infty)$ is not sequentially compact: the sequence $1,2,3,\dots$ lies in it but is unbounded, so it has no convergent subsequence at all.
- Closed bounded intervals are sequentially compact. Every $[a,b]$ is sequentially compact, since a sequence in $[a,b]$ is bounded (has a convergent subsequence by Bolzano–Weierstrass), and because $[a,b]$ is closed, the subsequence's limit lies back in $[a,b]$ .

Worked Example

Prove that the recursive sequence $a_1 = 1,\ a_{n+1} = \dfrac{1 + a_n}{2 + a_n}$ converges, and find its limit.

Step 1 — Show it converges. Every term is positive (sum and quotient of positives), so the sequence is bounded below by $0$ . To see it is monotonically decreasing, examine consecutive differences. Algebra gives the key identity:

\begin{aligned} a_{n+2} - a_{n+1} &= \frac{1+a_{n+1}}{2+a_{n+1}} - \frac{1+a_n}{2+a_n} \\ &= \frac{a_{n+1} - a_n}{\,(2+a_{n+1})(2+a_n)\,}. \end{aligned}

The denominator is positive, so $a_{n+2}-a_{n+1}$ has the same sign as $a_{n+1}-a_n$ . Since $a_2 - a_1 = \tfrac{2}{3} - 1 = -\tfrac13 < 0$ , induction forces every difference to be negative: the sequence is monotonically decreasing. Bounded below and decreasing $\Rightarrow$ it converges to some limit $a$ .

Step 2 — Find the limit. Since $a_{n+1}$ is a subsequence (shift) of $a_n$ , both sides have the same limit $a$ . Passing to the limit in the recursion and solving:

\begin{aligned} a &= \frac{1 + a}{2 + a} \\ a(2 + a) &= 1 + a \\ a^2 + a - 1 &= 0 \\ a &= \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}. \end{aligned}

The terms are bounded below by $0$ , so the limit cannot be negative. Discarding the negative root, $a = \frac{-1 + \sqrt{5}}{2}.$

Summary

A subsequence $(a_{n_k})$ is built from a strictly increasing index sequence ( $n_1 < n_2 < \cdots$ ), preserving order with no repeats; it satisfies $n_k \ge k$ and still has infinitely many terms.
If $(a_n) \to a$ , then every subsequence converges to the same $a$ ; differing subsequential limits prove divergence.
A peak index $m$ satisfies $a_m \ge a_n$ for all $n \ge m$ ; splitting on "infinitely vs. finitely many peaks" yields a monotone subsequence inside any sequence.
Bolzano–Weierstrass: every bounded sequence has a convergent subsequence (bounded $\to$ monotone subsequence $\to$ bounded monotone $\to$ convergent).
A set is sequentially compact when every sequence in it has a subsequence converging to a point of the set; closed bounded intervals $[a,b]$ qualify, while $(0,2]$ (not closed) and $[0,\infty)$ (not bounded) do not.