A Convergent Subsequence Always Hides Inside

Every bounded sequence hides a convergent subsequence inside it. This scene finds that subsequence by halving the interval.

The trick is to keep halving the interval. Always follow the half that holds infinitely many terms. Each halving makes the interval narrower. The points trapped inside pile onto a single value. Stringing those points together gives a convergent subsequence.

For example, thirty 🧑 classmates scatter across a playground. Split the playground in half and one side still holds a crowd of them. Keep halving that side and you corner a single spot.

(an) bounded⇒∃ (ank) convergent(a_n)\ \text{bounded}\Rightarrow\exists\ (a_{n_k})\ \text{convergent}

Part of these learning paths

Examples & Exercises

Find two convergent subsequences of an=(−1)na_n = (-1)^n and compute their limits.

a2k→1,a2k+1→−1a_{2k} \to 1, \qquad a_{2k+1} \to -1
Bisection on the demo sequence (4 steps) — the same "halve and follow" idea.
Show proof

  1. Taking even indices, a2k=(−1)2k=1a_{2k} = (-1)^{2k} = 1 for every kk, so (a2k)(a_{2k}) is the constant sequence 1,1,1,…1, 1, 1, \ldots, converging to 11.

  2. Taking odd indices, a2k+1=(−1)2k+1=−1a_{2k+1} = (-1)^{2k+1} = -1 for every kk, so (a2k+1)(a_{2k+1}) converges to −1-1.

  3. Both (2k)(2k) and (2k+1)(2k+1) are strictly increasing index sequences, so these are genuine subsequences. The bounded sequence (−1)n(-1)^n was hiding more than one convergent subsequence, with different limits.

Show that (sin⁡n)(\sin n) is bounded, and conclude via Bolzano–Weierstrass that a convergent subsequence exists.

∣sin⁡n∣≤1 ⟹ ∃ n1<n2<⋯: lim⁡k→∞sin⁡(nk)∈[−1,1]|\sin n| \leq 1 \ \Longrightarrow\ \exists\, n_1 < n_2 < \cdots:\ \lim_{k \to \infty} \sin(n_k) \in [-1,1]
Bisection on the demo sequence (6 steps) — boundedness is the key hypothesis.
Show proof

  1. ∣sin⁡n∣≤1|\sin n| \leq 1 for every integer nn, so (sin⁡n)(\sin n) is a bounded sequence.

  2. By the Bolzano–Weierstrass theorem, a convergent subsequence (sin⁡(nk))(\sin(n_k)) exists (its limit is some value in [−1,1][-1,1]).

  3. The theorem only guarantees existence; it does not say which subsequence. As a numerical illustration (not a proof), terms pile up near 11: sin⁡(35489)≈0.9999990\sin(35489) \approx 0.9999990 and sin⁡(36199)≈0.9999991\sin(36199) \approx 0.9999991.

Prove that an=na_n = n has no convergent subsequence — showing why the boundedness hypothesis is essential.

nk≥k→∞n_k \geq k \to \infty
The demo stays trapped in [0,1][0,1], which is why bisection works. Without boundedness this picture cannot even be drawn.
Show proof

  1. Let (nk)(n_k) be any subsequence of (n)(n). Its indices are strictly increasing, so nk≥kn_k \geq k by induction: n1≥1n_1 \geq 1, and nk+1>nk≥kn_{k+1} > n_k \geq k gives nk+1≥k+1n_{k+1} \geq k+1.

  2. Hence nk≥k→∞n_k \geq k \to \infty: every subsequence is unbounded above, hence divergent.

  3. A convergent sequence is bounded, so the unbounded (nk)(n_k) cannot converge. Thus (n)(n) has no convergent subsequence at all — dropping the boundedness hypothesis makes the theorem false.

3. Sequences