proof of Bolzano-Weierstrass Theorem

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 $b=supS$ (the supremum of $S$.)

Choose some $ϵ>0$. Then there is a corresponding $N$ such that $s_N>b-ϵ$. Since $(s_n)$ is nondecreasing, for all $n>N$, $s_n>b-ϵ$. But $(s_n)$ is bounded, so we have . But this implies , so $lim{s}_n=b$. $\mathrm{\square }$

(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 $k$ 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

$$s_1,s_2,s_3,\mathrm{\dots }$$

is monotonic (nondecreasing). $\mathrm{\square }$

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. $\mathrm{\square }$