Bolzano-Weierstrass Theorem

This theorem is (I’ve read) a classic in analysis. It can be stated (for the case of one dimension) as follows:

Let a and b be numbers such that a < b. Then every sequence in the interval [a,b] has a subsequence that converges to a point in [a,b].

Before stating the proof, we consider a key lemma: Every sequence has a monotone subsequence.

Proof of lemma: Consider a sequence $\{a_{n}\}$ . We can a natural number m a peak index for the sequence provided that $\{a_{n}\} \leq \{a_{m}\}$ for all integers $n \geq m$ . Now, we can have two possibilities: (1) there are finitely (including 0) many peak indices or (2) there are an infinite number of peak indices. Assume (1), then we can choose an N such that for all n greater than N there are no peak indices. But that means, by definition of a peak index, that choosing some $a_{n}$ for n greater than N, we can always find another element $a_{n'}$ that is larger. And we can do this indefinitely, so we can build a string of numbers, each greater than the last. This gives us the desired subsequence (increasing monotonically). If, instead, we assume (2), then we have an infinite number of peak indices, each of which (again, by definition) is smaller than the previous index. So we have a monotonically decreasing subsequence. $\Box$

Another lemma: Every bounded sequence has a convergent subsequence. Well, we know by the preceding lemma that every sequence has a monotone subsequence, and we know from the monotone convergence theorem that every bounded monotone (sub)sequence converges, so the lemma is proved.

Ok, now to the proof of the main theorem.

Proof of BW Theorem: Let $\{x_{n}\}$ be a sequence in [a,b]. Then $\{x_{n}\}$ is bounded. By the preceding lemma, there is a convergent subsequence of $\{x_{n}\}$ . Now, since $\{x_{n}\}$ is a sequence in [a,b], the limit is also in [a,b] (by another result in Fitzpatrick which I won’t state or prove here).

That is it. Pretty simple, really. I’ll admit the concept of a peak index wasn’t completely obvious to me at first. I mean, it is straightforward the way it is defined, but I wanted to come up with an intuitive picture (a graph of a function, for example) of why it would have to be true. I don’t really have that picture. But I’ll keep working on it and see if I can’t come up with something later on.

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This entry was posted on January 8, 2010 at 11:28 pm and is filed under Basics: Advanced Calc, Math Notes. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Bolzano-Weierstrass Theorem”

Anonymous Says:
January 19, 2010 at 2:54 pm | Reply
Don’t get too hung up on developing an intuitive picture of a graph–that’ll bite you in the ass over and over again. (That’s what I loved about undergrad and graduate real analysis!) Once the naive intuition you bring from calculus has taken enough beatings, the correct intuition will develop on its own.
- jessebstone Says:
  January 19, 2010 at 4:35 pm | Reply
  Thanks for the advice. I guess I can see why that sort of intuition I was hoping for may not get me very far, but how some other sort of intuition (which I have yet to develop) could take its place.

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