To prove the Bolzano-Weierstrass theorem, we will first need two lemmas.

Lemma 1.

All bounded monotone sequences converge.

proof.

Let $(s_{n})$ be a bounded, nondecreasing sequence. Let $S$ denote the set $\{s_{n}:n\in\mathbb{N}\}$. Then let $b=\sup S$ (the supremum of $S$.)

Choose some $\epsilon>0$. Then there is a corresponding $N$ such that $s_{N}>b-\epsilon$. Since $(s_{n})$ is nondecreasing, for all $n>N$, $s_{n}>b-\epsilon$. But $(s_{n})$ is bounded, so we have $b-\epsilon<s_{n}\leq b$. But this implies $|s_{n}-b|<\epsilon$, so $\lim s_{n}=b$. $\square$

(The proof for nonincreasing sequences is analogous.)

Lemma 2.

Every sequence has a monotonic subsequence.

proof.

First a definition: call the $n$th term of a sequence dominant if it is greater than every term following it.

For the proof, note that a sequence $(s_{n})$ may have finitely many or infinitely many dominant terms.

First we suppose that $(s_{n})$ has infinitely many dominant terms. Form a subsequence $(s_{{n_{k}}})$ solely of dominant terms of $(s_{n})$. Then $s_{{n_{{k+1}}}}<s_{{n_{k}}}$ $k$ by definition of “dominant”, hence $(s_{{n_{k}}})$ is a decreasing (monotone) subsequence of ($s_{n}$).

For the second case, assume that our sequence $(s_{n})$ has only finitely many dominant terms. Select $n_{1}$ such that $n_{1}$ is beyond the last dominant term. But since $n_{1}$ is not dominant, there must be some $m>n_{1}$ such that $s_{m}>s_{{n_{1}}}$. Select this $m$ and call it $n_{2}$. However, $n_{2}$ is still not dominant, so there must be an $n_{3}>n_{2}$ with $s_{{n_{3}}}>s_{{n_{2}}}$, and so on, inductively. The resulting sequence

is monotonic (nondecreasing). $\square$

proof of Bolzano-Weierstrass.

The proof of the Bolzano-Weierstrass theorem is now simple: let $(s_{n})$ be a bounded sequence. By Lemma 2 it has a monotonic subsequence. By Lemma 1, the subsequence converges. $\square$