proof of Bolzano-Weierstrass Theorem


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

Lemma 1.

All bounded monotone sequences converge.

proof.

Let (sn) be a bounded, nondecreasing sequence. Let S denote the set {sn:n}. Then let b=supS (the supremum of S.)

Choose some ϵ>0. Then there is a corresponding N such that sN>b-ϵ. Since (sn) is nondecreasing, for all n>N, sn>b-ϵ. But (sn) is bounded, so we have b-ϵ<snb. But this implies |sn-b|<ϵ, so limsn=b.

(The proof for nonincreasing sequences is analogous.)

Lemma 2.

Every sequence has a monotonicPlanetmathPlanetmath subsequence.

proof.

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

For the proof, note that a sequence (sn) may have finitely many or infinitely many dominant terms.

First we suppose that (sn) has infinitely many dominant terms. Form a subsequence (snk) solely of dominant terms of (sn). Then snk+1<snk k by definition of “dominant”, hence (snk) is a decreasing (monotone) subsequence of (sn).

For the second case, assume that our sequence (sn) has only finitely many dominant terms. Select n1 such that n1 is beyond the last dominant term. But since n1 is not dominant, there must be some m>n1 such that sm>sn1. Select this m and call it n2. However, n2 is still not dominant, so there must be an n3>n2 with sn3>sn2, and so on, inductively. The resulting sequence

s1,s2,s3,

is monotonic (nondecreasing).

proof of Bolzano-Weierstrass.

The proof of the Bolzano-Weierstrass theorem is now simple: let (sn) be a bounded sequence. By Lemma 2 it has a monotonic subsequence. By Lemma 1, the subsequence converges.

Title proof of Bolzano-Weierstrass Theorem
Canonical name ProofOfBolzanoWeierstrassTheorem
Date of creation 2013-03-22 12:22:26
Last modified on 2013-03-22 12:22:26
Owner akrowne (2)
Last modified by akrowne (2)
Numerical id 5
Author akrowne (2)
Entry type Proof
Classification msc 40A05
Classification msc 26A06