# proof of Bolzano-Weierstrass Theorem

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. But this implies $|s_{n}-b|<\epsilon$, so $\lim s_{n}=b$. $\square$

(The proof for nonincreasing sequences is analogous.)

Lemma 2.

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}} $k$ 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

 $s_{1},s_{2},s_{3},\ldots$

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. $\square$

Title proof of Bolzano-Weierstrass Theorem ProofOfBolzanoWeierstrassTheorem 2013-03-22 12:22:26 2013-03-22 12:22:26 akrowne (2) akrowne (2) 5 akrowne (2) Proof msc 40A05 msc 26A06