## Saturday, September 09, 2006

### Bolzano-Weierstrass Theorem

The Bolzano-Weierstrass Theorem is a very important theorem in the realm of analysis. It was first proved by Bernhard Bolzano but it became well known with the proof by Karl Weierstrass who did not know about Bolzano's proof. In light of this history, the proof gets its current name. I use it in my proof for the Cauchy Criterion which also an important theorem in analysis.

It says:

every bounded sequence has a convergent subsequence

To make heads or tails of this, it is necessary to offer some definitions of bounded sequence and convergent subsequence. Since this is one of the subtleties of analysis, I will attempt to provide a complete formal definition. Interspersed, I will provide some comments that will make the ideas clear even if you wish to skip some of the formalisms.

Let's start with sequence. This one is a tougher nut to crack, then it might seem. From one perspective, a "sequence" is just a set of numbers but from another perspective, this definition in itself is too limiting. For example, what about a sequence of points on a line?

To apply to all these domains, we will use an abstraction called a metric space and to define this, we must first define a metric.

Definition 1: metric

A metric is a binary function d(a,b) on a set X such that for any a,b,c ∈ X:

(a) d(a.b) is a real number.

(b) d(a,b) ≥ 0

(c) d(a,b) = 0 if and only if a = b

(d) d(a,b) = d(b,a)

(e) d(a,c) ≤ d(a,b) + d(b,c)

In other words, a metric is a distance function. For real numbers, the standard metric is an absolute subtraction. The distance between any two numbers x,y is abs(x - y).

Definition 2: Metric Space

A metric space is a pairing of a set X with a metric, that is, a distance function. Any element in X can be thought of as a point in the metric space.

From this perspective, all the number systems (integers, reals, rationals) are examples of metric spaces as are geometric coordinates (lines and planes).

Definition 3: Sequence

A sequence is a mapping from the set of natural numbers to the points of a given metric space.

In other words, there is a first element, a second element, a third element, etc.

For example:

the set of even numbers = {2, 4, 6, ...}

Definition 4: Bounded Sequence

A sequence an is bounded if there exists some number k such that abs(an) ≤ k for all elements in the sequence.

By this definition, a bounded sequence has both a least upper bound and a greatest lower bound.

OK, good. To continue on with formal definitions, we now need a definition of strictly increasing.

Definition 5: Strictly Increasing

A function f(x) is said to be strictly increasing if x is less than y implies f(x) is less than f(y).

I now use the definition of strictly increasing in the next definition.

Definition 6: Subsequence

If (an) is sequence, then a subsequence is (anr) where each nr is a strictly increasing sequence of natural numbers.

The main idea here is that if a sequence is a mapping of the set of natural numbers to points in a metric space. Then a subsequence is a mapping of a set of natural numbers to a subset of natural numbers to points in a metric space.

Here's an example. Let's say we have the sequence of natural numbers = (1,2,3,4,...)

In this case, we have a mapping from the set of natural numbers to the set of natural numbers. The first element is 1, the second element is 2, and so on. Remember, the set of natural numbers is a metric space, that is, they are a set combined with a distance function for any two elements from that set.

Now, let's say we have the sequence of odd numbers which is subsequence of the natural numbers so that we have (1,3,5,7,9,..)

This is a mapping from the natural numbers to the natural numbers to a metric space in the sense that the first element of the odd numbers, 1, maps to the first element of the sequence of integers which maps to 1. The second element of the odd numbers, 3, maps to the third element of the sequence of integers which maps to 3.

The main idea here is that a subsequence maintains the order of the sequence but deletes some of the elements. So, all subsequences are sequences in themselves but they also have a mapping to the elements of another sequence.

Definition 7: Convergent Sequence

A sequence (an) in the metric space (X,d(a,b)) is said to be convergent if there exists a point x ∈ X such that:

for every real number ε greater than 0, there exists a natural number N such that if n greater than N, d(an,x) is less than ε.

The notation for this is:

lim (sn) = x.

In other words, if we are talking about a sequence of real numbers where the distance function d(a,b) is just subtraction and the metric space is just the set of real numbers, then this says that for any convergent sequence, there is a limit for this sequence (which is a real number) such that we can pick a point in the sequence where all elements of the remaining elements of the sequence are arbitrarily close to the limit. In other words, the sequence is said to converge to this limit.

For the proof below, we also need:

Definition 8: Monotone Sequence

A sequence sn is monotonic if it meets one of the four following conditions:

(1) Monotonically Increasing: i greater than j → si greater than sj

(2) Monotonically Decreasing: i greater than j → si is less than sj

(3) Monotonically Nondecreasing: i greater than j → si ≥ sj

(4) Monotonically Nonincreasing: i greater than j → si ≤ sj

The main idea behind a monotone sequence is that it moves in a consistent direction. It is either gradually increasing, gradually decreasing, or just staying the same without ever regressing.

Now, with these definitions aside, we are ready to start on the proof of the Bolzano-Weierstrass Theorem. We will also need some lemmas to begin.

Lemma 1: All bounded monotone sequences converge.

Proof:

(1) Assume that S is a bounded nondecreasing sequence = s1, ..., sn. (We will be able to make an analogous argument for bounded nonincreasing sequences so we only need to prove this one case)

(2) Let b = sup S

NOTE: sup S means the supremum for the sequence S which is the least upper bound for the sequence S. We know that this exists since S is bounded.

For nonincreasing sequences, we would let b = inf(S) where inf(S) is the infimum for the sequence S which is the greatest lower bound for the sequence S.

(3) Let ε be any positive number.

(4) Then, there is some N such that sN is greater than b - ε

We know that this exists otherwise b would not be the least upper bound since b - ε is less than b. We can make an analogous argument using the greatest lower bound for a bounded nonincreasing sequence.

(5) Since S is nondecreasing, for all n ≥ N, sn is greater than b - ε. [See Definition of monotone sequence above]

(6) S is also bounded by b so we have:

b is greater than sn which is greater than b - ε

(7) But this implies that abs(sn - b) is less than ε since:

if we subtract b from all sides then we get:

0 is greater than sn - b which is greater than

Since ε is greater than 0, we have:

sn - b is between and .

(8) So, by the definition of a convergent sequence, we have:

lim(sn) = b.

QED

Lemma 2: Every sequence has a monotonic subsequence

Proof:

(1) Let us call a term si in a sequence S dominant if it is greater than all the terms that follow it.

(2) It is clear that each sequence either has a finite number of dominant terms or an infinite number of dominant terms.

(3) Let us start by handling the case where a sequence S has an infinite number of dominant terms.

(4) We can now form a subsequence that only includes these dominant terms. [See definition of subsequence above]

(5) Then for every Snk that makes up the subsequence, Snk is greater than Snk+1 which means that the subsequence is monotonic decreasing sequence. [See definition of monotone sequence above]

(6) Now, let's handle the case where there is a finite number of dominant terms.

(7) Let's label the first term after the last dominant term (since there are only a finite number of them), n1. If there aren't any dominant terms, then let's label the first term in the sequence n1.

(8) Now since n1 is not dominant, there must be another term in the sequence which is greater than n1. We can call this next term n2. But n2 is not dominant so there must be an n3 and so on.

(9) We now have constructed a nondecreasing monotone subsequence. [See definition of monotone sequence above]

QED

Theorem: Bolzano-Weierstrass Theorem

Every bounded sequence has a convergent subsequence

Proof:

(1) Let (sn) be a bounded sequence.

(2) By Lemma 2 above, we know that it has a monotonic subsequence.

(3) By Lemma 1 above, the monotonic subsequence converges.

QED

References

### Cauchy's Criterion

Cauchy's criterion is a well known criteria for when an infinite sum converges, that is, has a finite limit.

Augustin Cauchy was not the first to come up with the criteria. Leonhard Euler, for example, used a similiar criteria. It may be that the significance of the criteria became appreciated in the context of Cauchy's great work on the foundations of calculus.

Definition 1: Cauchy Sequence

A sequence si is Cauchy Sequence if and only if given any positive number ε, there exists an integer N such that if m,n are greater than N, then absolute(sm - sn) is less than ε

In other words, elements of the sequence get arbitrarily close to one another.

I will need a few properties of absolute inequalities:

Lemma 1: absolute(a - b) ≤ absolute(a) + absolute(b)

Proof:

(1) Case I: a - b is nonnegative

So abs(a-b) = a - b

If b ≥ 0, then a - b ≤ a ≤ abs(a) + abs(b)

If b is less than 0, then a - b = a + abs(b) ≤ abs(a) + abs(b)

(2) Case II: a - b is negative

So abs(a-b) = -(a-b) = b - a = abs(b - a)

Using step #1, we know that abs(b - a) ≤ abs(b) + abs(a) = abs(a) + abs(b)

So that:

abs(a - b) ≤ abs(a) + abs(b)

QED

Lemma 2: abs(a) - abs(b) ≤ abs(a - b)

Proof:

(1) Case I: a - b is nonnegative so that abs(a - b) = a - b

If a ≥ 0, then abs(a - b) = a - b ≥ abs(a) - abs(b)

NOTE: It is = except for the case where b is negative.

If a is less than 0, then abs(b) is greater than abs(a) and abs(a) - abs(b) must be a negative number.

(2) Case II: a - b is negative and abs(a - b) = -(a - b) = b - a

If a is ≥ 0, then abs(b) is greater than a and abs(a) - abs(b) is a negative number.

If a is less than 0 and b is less than 0, then b - a = b + abs(a) = abs(a) - abs(b) so that abs(a - b) = abs(a) - abs(b)

If a is less than 0 and b ≥ 0, then b - a = b + abs(a) = abs(a) + abs(b) ≥ abs(a) - abs(b).

QED

Here are some properties of Cauchy Sequences which I will use below:

Lemma 3: Any convergent sequences is a Cauchy Sequence

Proof:

(1) Let ai be a convergent sequence (that is, as ai gets larger, it approaches a limit) so that it's limit = L. [See definition 7, here for definition of a convergent sequence]

(2) So from the above definition, we know for any positive number ε, there exists a positive number N such that:

if n is greater than N, then absolute(an - L) is less than ε

(3) So, for a value (1/2)ε, there exists an integer N such that if n is greater than N, absolute(an - L) is less than (1/2)ε

(3) So let's assume that we have two integers m,n both greater than N.

(4) This means that in both cases absolute(am - L) is less than (1/2)ε and absolute(an - L) is less than (1/2)ε

(5) This gives us:

absolute(am - an) = absolute([am - L] - [an - L]) ≤ absolute(am - L) + absolute(an - L) [See Lemma 1 above]

(6) Finally,

absolute(am - L) - absolute(an - L) is less than (1/2)ε + (1/2)ε = ε

QED

Lemma 4: A Cauchy Sequence has a bound

Proof:

(1) Let (ai) be a Cauchy Sequence.

(2) Then, for any positive number ε greater than 0, there is an integer N such that:

for any integer m,n ≥ N, abs(an - am) is less than ε [See Definition of a Cauchy Sequence above]

(3) So that if we ε = 1 (since ε can be any positive number), we have:

abs(an) - abs(am) ≤ abs(an - am) less than 1 for all n,m ≥ N. [See Lemma 2 above]

(4) Let m = N (since m can be any integer ≥ N), then we have:

abs(an) - abs(aN) is less than 1 which means that:

abs(an) is less than abs(aN) + 1 for n ≥ N.

(5) Now, let n = N (since n can be any integer ≥ N), then we have:

abs(aN) - abs(am) is less than 1 which means that:

abs(am) is greater than abs(aN) - 1 for all n ≥ N.

(6) So, for all n, we have:

abs(an) is less than max { abs(a1), ..., abs(aN-1), abs(aN) + 1 }

and

abs(an) is greater than min { abs(an1), ..., abs(aN-1), abs(aN) - 1 }

(7) This shows that for all finite subsets of the sequence, there exists a bound for ai where upper bound = max { abs(a1), ..., abs(aN-1), abs(aN) + 1 } and a lower bound = min { abs(a1, ..., abs(aN-1), abs(aN)-1 }

QED

Lemma 5: If a Cauchy Sequence has a subsequence convergent to b, then the Cauchy sequence itself converges to b.

Proof:

(1) Let an be a Cauchy sequence with the subsequence ain convergent to b.

(2) By the definition of convergence, we know that for a positive number ε/2, there exists an integer M such that for all n ≥ M abs(ain - b) ≤ ε/2.

(3) By the definition of a Cauchy sequence, we know that there exists an integer n0 such that for all m,n ≥ n0, abs(an - am) ≤ ε/2.

(4) Now, if iM (this is the start of the subsequence that converges) is less than n0, we can always find a M' which is greater than M such that iM' ≥ n0.

We can assume this since we are assuming an infinite subsequence.

(5) So for all n ≥ n0, we have:

abs(an - b) = abs(an - aiM' + aiM' - b) ≤ abs(an - aiM') + abs(aiM' - b) [See Lemma 1 above]

(6) Now, abs(an - aiM') + abs(aiM' - b) is less than ε/2 + ε/2 = ε

We know that abs(an - aiM') is less than ε/2 from the definition of a Cauchy sequence.

We know that abs(aiM') is less than ε/2 from the definition of the convergent sequence.

(7) Now putting this all together gives us that:

abs(an - b) ≤ ε

Which by definition (see Definition 7, here) means that lim(an) = b.

QED

Lemma 6: Every real Cauchy sequence is convergent.

Proof:

(1) By Lemma 4 above, every Cauchy sequence is bounded.

(2) So, by the Bolzano-Weierstrass Theorem (see Theorem, here), every Cauchy sequence has a convergent subsequence.

(3) So, by Lemma 5 above, every Cauchy sequence is convergent.

QED

Lemma 7: A sequence of reals converges if and only if it is a Cauchy sequence

Proof:

(1) By Lemma 6 above, we know that a Cauchy sequence is convergent.

(2) By Lemma 3 above, we know that a convergent sequence is a Cauchy sequence.

QED

Here is the Criterion:

Theorem: Cauchy's Criterion

A series ai is convergent (that is, has a finite limit) if and only if for every positive number ε, there exists a positive integer N such that:

for all n greater than N and p ≥ 1:

absolute(an+1 + an+2 + ... + an+p) is less than ε

Proof:

(1) Let sn = ∑ ai

The assumption here is that i ranges from 0 to n.

(2) sn converges if and only if it is a Cauchy Sequence. [See Lemma 7 above]

(3) Assume than sn is a Cauchy Sequence.

(4) Then, for every positive number ε, there exists a number N such that for all integers n,m greater than N, absolute(sm - sn) is less than ε [See Definition 1 above]

(5) Let's assume that m is greater than n.

At this point, we've made no assumption about m or n and this is consistent with our assumption in step #3.

(6) We know that there exists an integer p ≥ 1 such that m = n + p

(7) Based on the definition of sn, we can see that:

absolute(sm - sn) = absolute(sn+p - sn) = absolute(an+1 + an+2 + ... + an+p)

(8) Now, using step #6, we can see that ∑ ai is convergent if and only if the conditions of the given apply.

QED

References