A bounded sequence need not converge, but it cannot avoid having some convergent piece. This theorem is the engine behind the extreme value theorem, the convergence of Cauchy sequences, and compactness. Its proof combines subsequences with the monotone convergence theorem (through nested intervals).
Key idea
Bolzano–Weierstrass theorem. Every bounded sequence of real numbers has a convergent subsequence.
Why it works
Say every term lies in $I_0 = [a, b]$. Cut $I_0$ in half. At least one half contains $a_n$ for infinitely many $n$; call it $I_1$ (if both do, take the left one). Repeat: $I_2 \subseteq I_1$ is a half of $I_1$ holding infinitely many terms, and so on. The interval $I_k$ has length $\frac{b - a}{2^{k}}$. Now choose indices: $n_1$ with $a_{n_1} \in I_1$, then $n_2 > n_1$ with $a_{n_2} \in I_2$ (possible, since $I_2$ holds infinitely many terms), and so on. The left ends of the $I_k$ increase and are bounded, so they converge to some $c$ (monotone convergence), and the right ends converge to $c$ too, since the lengths go to $0$. Both $a_{n_k}$ and $c$ lie in $I_k$, so $|a_{n_k} - c| \le \frac{b - a}{2^{k}} \to 0$. $\blacksquare$
Worked examples
Worked example: Running the bisection
Let $a_n = \frac{1}{3} + \frac{1}{10n}$ for odd $n$ and $a_n = \frac{4}{5} + \frac{1}{10n}$ for even $n$, all in $[0, 1]$. Keeping the left half whenever it holds infinitely many terms, find $I_3$.
- The terms pile up at $\frac{1}{3}$ and $\frac{4}{5}$ (from above). A closed half holds infinitely many terms exactly when one of these values lies in it.
- $[0, \frac{1}{2}]$ contains $\frac{1}{3}$: keep it. Then $[0, \frac{1}{4}]$ does not contain $\frac{1}{3}$ or $\frac{4}{5}$, so keep $[\frac{1}{4}, \frac{1}{2}]$.
- Then $[\frac{1}{4}, \frac{3}{8}]$ contains $\frac{1}{3}$: keep it.
Answer: $I_3 = \left[\frac{1}{4}, \frac{3}{8}\right]$, closing in on $\frac{1}{3}$.
Worked example: What the theorem promises
Does $(-1)^{n}$ converge? Does it have a convergent subsequence?
- It is bounded and diverges.
- The constant subsequence $a_{2k} = 1$ converges.
Answer: No; yes. The theorem promises a convergent subsequence, never convergence of the whole sequence.
Common mistake
Keeping the half with more of the early terms. What matters is a half with infinitely many terms; any finite number of terms can be ignored. In the example, the first terms of the sequence sit near $0.43$ and $0.85$, but where they are does not decide anything. Also, the theorem needs boundedness: $a_n = n$ has no convergent subsequence.