KindiakMath

The Beauty of Bounded Sequences

Sequences can converge, but they may not. That remains true with bounded sequences. Nevertheless, we do gain some convenient convergence properties from bounded sequences that are, in general, not guaranteed when sequences are not bounded.

Monotone Convergence Theorem. Let \{x_n\} be a sequence that is bounded above (resp. bounded below) and non-decreasing (resp. non-increasing). Then \{x_n\} converges.

Proof. Assume \{x_n\} is bounded above and non-decreasing. Since \{x_n\} is bounded above, its supremum M := \sup \{x_n\} exists. Fix \epsilon > 0. By definition, there exists N \in \mathbb N such that x_N > M - \epsilon. Then, for n > N,

M - \epsilon < x_N \leq x_n \leq M < M+ \epsilon \quad \Rightarrow \quad |x_n - M| <\epsilon.

Therefore, x_n \to M, so that \{x_n\} converges. Now, if \{x_n\} is bounded below, then \{-x_n\} is bounded above. Thus, -x_n \to L for some L \in \mathbb R, so that x_n \to -L. Thus, \{x_n\} converges.

Bolzano-Weierstrass Theorem. Let \{x_n\} be a bounded sequence. Then there exists an increasing sequence \{n_k\} such that the subsequence \{x_{n_k}\} converges.

Proof. We first remark that for nonempty sets K, L that are bounded above (resp. bounded below), K \subseteq L implies \sup K \leq \sup L (resp. \inf L \leq \inf K). Since \{x_n\} is bounded, |x_n| \leq M for some M > 0.

Call a number x_n a peak term if x_n = \sup\{x_k : k \geq n\}.

Suppose \{ x_n \} has finitely many peak terms with the last peak term x_N. Define n_1 := N+1. Define inductively n_{k+1} as follows: for n_k > N, there exists n_{k+1} > n_k such that x_{n_k} < x_{n_{k+1}} \leq M. Therefore, we have defined a non-decreasing subsequence \{x_{n_k}\} that is bounded above by M. By the monotone convergence theorem, \{x_{n_k}\} converges, and we are done.

Now suppose \{x_n\} has infinitely many peak terms x_{n_1},x_{n_2},\dots. By definition, \{x_{n_k}\} is non-increasing. Since \{ x_n \} is bounded below by -M, so is \{x_{n_k}\}. By the monotone convergence theorem again, \{x_{n_k}\} converges, and we are done.

Corollary 1. Let \{x_n\} be a bounded sequence in [a, b]. Then \{x_n\} contains a subsequence that converges in [a, b].

Proof. By the Bolzano-Weierstrass theorem, let \{x_{n_k}\} be a convergent subsequence of \{x_n\} with limit x. We have

\displaystyle a \leq x_{n_k} \leq b \quad \Rightarrow a \leq \underbrace{\lim_{k \to \infty} x_{n_k}}_x \leq b,

so that x \in [a, b], as required.

Bounded sequences help us prove the celebrated continuity theorems in classic calculus. We do that next time.

—Joel Kindiak, 20 Dec 24, 1200H

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