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Revision of How to use the Bolzano-Weierstrass theorem from Sat, 24/01/2009 - 17:27

Quick description

The Bolzano-Weierstrass theorem asserts that every bounded sequence of real numbers has a convergent subsequence. More generally, it states that if X is a closed bounded subset of \mathbb{R}^n then every sequence in X has a subsequence that converges to a point in X. This article is not so much about the statement, or its proof, but about how to use it in applications. As we shall see, there are certain signs to look out for: if you come across a statement of a certain form (to be explained in the article), then the Bolzano-Weierstrass theorem may well be helpful.

one of the signs to look out for is any statement that has the form "For every \delta>0 there exists a\in X such that P(a,\delta)," where X is some bounded set and P is some property of pairs (a,\delta) of real numbers such that if \delta_1<\delta_2 then P(a,\delta_1) implies P(a,\delta_2).

Example 1: every continuous function on a closed bounded interval is bounded

Let f be a continuous function defined on the closed interval [u,v]. A well-known theorem says that f is bounded. There are various proofs, but one easy one uses the Bolzano-Weierstrass theorem.

The purpose of this article is to show that the proof using the Bolzano-Weierstrass theorem is not just easy to follow, but easy to spot in the first place. However, this is not quite so obvious, since the theorem makes no mention of sequences, so let us interrupt the discussion of this example and talk about how to convert statements that do not involve sequences into statements that do.

General discussion: how to produce sequences

Suppose you have a statement like "For every \delta>0 there exists a\in X such that |f(a)-c|\leq\delta." Then in particular you know that for every n\in\mathbb{N} there exists a\in X such that |f(a)-c|\leq 1/n. If for each n you choose such an a and call it a_n, then you get a sequence (a_n)_{n=1}^\infty such that |f(a_n)-c|\leq 1/n for every n.

So far, this applies to any statement of the form "For every \delta>0 there exists a\in X such that P(a,\delta)," where P(a,\delta) is some statement that involves a and \delta. It gives us a sequence (a_n)_{n=1}^\infty such that P(a_n,1/n) holds for every n\in\mathbb{N}. However, the resulting statement may not be equivalent to the statement we started with. For instance, the statement "For every \delta>0 there exists a\in X such that f(a)=\delta" is not equivalent to the statement "There is a sequence (a_n) such that f(a_n)=1/n for every n."

Suppose, however, that we knew that the statement P(a,\delta) was such that if \delta_1<\delta_2 then P(a,\delta_1) implies P(a,\delta_2). (An example of such a statement is "|f(a)-c|<\delta.") Then suppose that we have a sequence (a_n) such that P(a_n,1/n) for every n. The Archimedean property of the real numbers tells us that for every \delta>0 we can find n\in\mathbb{N} such that 1/n<\delta. But then P(a_n,1/n) holds, which implies that P(a_n,\delta) holds, by our assumption about the property P. This implies that there exists a such that P(a,\delta) holds.

Similar reasoning shows that if P is a property such that for every a, if C_1>C_2 then P(C_1,a) implies P(C_2,a), then the statement "For every C there exists a\in X such that P(a,C)" is equivalent to the statement "There is a sequence (a_n) in X such that P(a_n,n) for every n\in\mathbb{N}."

Example 1 continued

We are told that f is continuous. This is the statement that for every x\in [a,b] and every \epsilon>0 there exists \delta>0 such that if |y-x|<\delta then |f(x)-f(y)|<\epsilon. Unfortunately, because "for every x" is involved in this statement, and because what is guaranteed to exist is \delta rather than an element of [a,b], we don't get very far if we apply the above procedure to this statement.

Does that mean we cannot apply the Bolzano-Weierstrass theorem? Not at all, because we also have the option of looking at the contrapositive of the statement we are trying to prove. That is, it would be enough to prove that if f is unbounded then f cannot be continuous. So now we have a different hypothesis to look at.

That hypothesis is "For every C there exists a\in [u,v] such that |f(a)|>C." If C_1>C_2 and |f(a)|>C_1, then |f(a)|>C_2, so this time we are in exactly the situation we want in order to generate a sequence.