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The Bolzano-Weierstrass Property and Compactness
We know that not all sequences converge. In fact, the ones that do converge are just the “very good” ones. But even “very good” sequences may not converge. Cauchy sequences, for example, are very good; they’re not really any different than convergent sequences: a Cauchy sequence actually does converge in “good” spaces (i.e., complete spaces), and fails to converge only if the point that “should be” its limit is not in the space — i.e., it fails to converge because the space is not good (is incomplete), not because the sequence is bad.
Now let’s ask about a weaker property of sequences: which sequences have cluster points? Equivalently, when does a sequence have a subsequence that converges? Consider the alter- nating real sequence { 1 , − 1 , 1 , − 1 , 1 ,.. .}. The sequence certainly doesn’t converge, but it has subsequences that do, such as { 1 , 1 , 1 ,.. .}. We now study an easy-to-check condition that guarantees that a sequence in R or Rn^ has a convergent subsequence: every bounded sequence in Rn^ has a Cauchy subsequence, a subsequence that therefore converges in Rn. That’s the content of the Bolzano-Weierstrass Theorem. (We’re assuming throughout this section that Rn^ is endowed with a norm; we’ve already seen that which norm we use in Rn has no effect on convergence — i.e., on which sequences converge.)
The Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a convergent subsequence.
Proof: Let {xn} be a bounded sequence and without loss of generality assume that every term of the sequence lies in the interval [0, 1]. Divide [0, 1] into two intervals, [0, 12 ] and [^12 , 1]. (Note: this is not a partition of [0, 1].) At least one of the halves contains infinitely many terms of {xn}; denote that interval by I 1 , which has length 12 , and let {xn′ } be the subsequence of {xn} consisting of every term that lies in I 1.
Now divide I 1 into two halves, each of length 14 = (^12 )^2 , at least one of which contains infinitely many terms of the (sub)sequence {xn′^ }, and denote that half by I 2. Let {xn′′^ } be the subsequence of {xn′ } consisting of all of the terms that lie in I 2. Continuing in this way, we construct a sequence of nested intervals I 1 ⊇ I 2 ⊇... , where the length of In is (^12 )n, and each interval contains an infinite number of terms of the original sequence {xn}. Finally, we construct a subsequence {zn} of {xn} made up of one term from each interval In. This subsequence is clearly Cauchy: ∀N: m, n > N ⇒ |zm − zn| < (^12 )N^. Therefore the subsequence {zn} converges, according to the Cauchy-sequence version of the Completeness Axiom. ‖
The Bolzano-Weierstrass Theorem is true in Rn^ as well:
The Bolzano-Weierstrass Theorem: Every bounded sequence in Rn^ has a convergent subsequence. Proof: Let {xm} be a bounded sequence in Rn. (We use superscripts to denote the terms of the sequence, because we’re going to use subscripts to denote the components of points in Rn.) The sequence {xm 1 } of first components of the terms of {xm} is a bounded real sequence, which has a convergent subsequence {xm 1 k}, according to the B-W Theorem in R. Let {xmk^ } be the corresponding subsequence of {xm}. Then the sequence {xm 2 k} of second components of {xmk^ } is a bounded sequence of real numbers, so it too has a convergent subsequence, and we again have a corresponding subsequence of {xmk^ } (and therefore of {xm}), in which the sequences of first and second components both converge. Continuing for n iterations, we end up with a subsequence {zm} of {xm} in which the sequences of first, second,... , nth components all converge, and therefore the subsequence {zm} itself converges in Rn. ‖
It’s elementary to show that the following form of the B-W Theorem is equivalent to the one we’ve just proved:
The Bolzano-Weierstrass Theorem: Every sequence in a closed and bounded set S in Rn^ has a convergent subsequence (which converges to a point in S). Proof: Every sequence in a closed and bounded subset is bounded, so it has a convergent subsequence, which converges to a point in the set, because the set is closed. ‖
Conversely, every bounded sequence is in a closed and bounded set, so it has a convergent subsequence.
Subsets of Rn^ that are both closed and bounded are so important that we give them their own name: a closed and bounded subset of Rn^ is said to be compact. And in any metric space, the sets in which all bounded sequences have convergent subsequences are so important that we give that property of sets its own name as well:
Definition: A set S in a metric space has the Bolzano-Weierstrass Property if every sequence in S has a convergent subsequence — i.e., has a subsequence that converges to a point in S.
The B-W Theorem states that closed and bounded (i.e., compact) sets in Rn^ have the B-W Property. We can also prove the converse of the B-W Theorem, that any set in Rn^ with the B-W Property is closed and bounded (compact). See Theorem 29.6 of Simon & Blume.
Now we can easily prove the Weierstrass Theorem; in fact, we can prove the following gen- eralized form of the Weierstrass Theorem, which says that continuous functions preserve compactness.
Theorem: Let f : X → Y. If X is compact and f is continuous, then f (X) is compact. Proof: We must show that f (X) has the B-W Property. Let {yn} be a sequence in f (X); we must show that {yn} has a convergent subsequence. For each n ∈ N, let xn ∈ X be such that f (xn) = yn (which we can do because yn ∈ f (X)). Since X is compact, {xn} has a subsequence {xnk } that converges to some x ∈ X. Since f is continuous, {f (xnk )} converges to f (x) ∈ Y. Since x ∈ X, we have f (x) ∈ f (X). ‖
Corollary (The Weierstrass Theorem): A continuous real-valued function on a compact subset S of a metric space attains a maximum and a minimum on S. Proof: f (S) is a compact subset of R, i.e., a closed and bounded subset of R. Since f (S) is a bounded subset of R, it has both a least upper bound M and a greatest lower bound m; and since f (S) is closed, it contains m and M. Therefore m = min f (S) and M = max f (S). ‖
Exercise: In the example on the preceding page, (X, d) is an infinite discrete metric space. Which subsets of X are compact? Which subsets are closed and bounded? Which subsets are open? Let X = Z, the set of all integers, and let f : Z → R be the real-valued function f (x) = x. Does f attain a maximum or a minimum on Z? Is f continuous? On which subsets of Z does f attain a maximum or a minimum?
Note: You may also see a definition that says a compact set is one that has the Heine-Borel Property — every open cover has a finite subcover. Just as “closed and bounded” didn’t get us what we wanted when we went from Rn^ to a metric space, the B-W Property doesn’t get us what we want when we go to a topological space (a space where open and closed sets are the defining concepts but the space may not have a metric structure).
