A Convergent Subsequence Always Hides Inside

Every bounded sequence hides a convergent subsequence. Halve the interval and follow the half holding infinitely many terms.

Examples & Exercises

Find two convergent subsequences of $a_n = (-1)^n$ and compute their limits.

a_{2k} \to 1, \qquad a_{2k+1} \to -1

Bisection on the demo sequence (4 steps) — the same "halve and follow" idea.

Show proof

Taking even indices, $a_{2k} = (-1)^{2k} = 1$ for every $k$ , so $(a_{2k})$ is the constant sequence $1, 1, 1, \ldots$ , converging to $1$ .
Taking odd indices, $a_{2k+1} = (-1)^{2k+1} = -1$ for every $k$ , so $(a_{2k+1})$ converges to $-1$ .
Both $(2k)$ and $(2k+1)$ are strictly increasing index sequences, so these are genuine subsequences. The bounded sequence $(-1)^n$ was hiding more than one convergent subsequence, with different limits.

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

|\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

$|\sin n| \leq 1$ for every integer $n$ , so $(\sin n)$ is a bounded sequence.
By the Bolzano–Weierstrass theorem, a convergent subsequence $(\sin(n_k))$ exists (its limit is some value in $[-1,1]$ ).
The theorem only guarantees existence; it does not say which subsequence. As a numerical illustration (not a proof), terms pile up near $1$ : $\sin(35489) \approx 0.9999990$ and $\sin(36199) \approx 0.9999991$ .

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

n_k \geq k \to \infty

The demo stays trapped in

[0,1]

, which is why bisection works. Without boundedness this picture cannot even be drawn.

Show proof

Let $(n_k)$ be any subsequence of $(n)$ . Its indices are strictly increasing, so $n_k \geq k$ by induction: $n_1 \geq 1$ , and $n_{k+1} > n_k \geq k$ gives $n_{k+1} \geq k+1$ .
Hence $n_k \geq k \to \infty$ : every subsequence is unbounded above, hence divergent.
A convergent sequence is bounded, so the unbounded $(n_k)$ cannot converge. Thus $(n)$ has no convergent subsequence at all — dropping the boundedness hypothesis makes the theorem false.