Math 3345-Real Analysis — Lecture 19 10/26/05 1. Counting Measures

Math 3345-Real Analysis — Lecture 19 10/26/05

1. Counting Measures and Cardinality

Derivatives and integrals depend heavily on the notion of length. The essence of the relevant concepts depends on the concept of measure. Let’s start with a simple measure. Question. How big is the set { 1, 3, 4, 7 }?

Well, this set has four elements. What else is there to say? We can measure the size of finite sets by counting how many elements there are in the set. This is known as a counting measure, and if you can measure sets, then it’s possible to talk about derivatives and integrals. Let’s see how that works. Consider the function f : Z → R defined by f(x)=x2. Note that this function is only defined on Z. What’s the derivative of this function? For functions defined on R, the derivative is a rate of change. For example, if f 0(2) = 5, then the slope of the tangent line is m =5atx = 2, and y is increasing by 5 as x increases by 1. In this case, what should f 0(2) be? Well, f(2) = 4 and f(3) = 9, so as x changes from 2 → 3, y changes from 4 → 9. In other words, y compared to x is changing at a rate of 5 to 1. f 0(2) must be 5. Let’s define this in general. The derivative of a function f : Z → R at x is f(x +1)− f(x) (1) f 0(x)= . 1 Here we get f 0(0) = 1 f 0(1) = 3 (2) f 0(2) = 5 f 0(3) = 7 It looks like (3) f 0(x)=2x +1. This may look odd, but we’ve got a different kind of derivative. In regular differentiation and integration, we have the fundamental theorem of calculus. It looks like b (4) Z f 0(x) dx = f(b) − f(a). a Do we have something similar with functions defined on Z? Let’s take a look. How should 2 (5) Z f 0(x) dx 0 be calculated? With the Riemann integeral, we multiplied function values times little lengths (like ∆x)in the domain. Then we added everything up. We don’t have little lengths in our domain, but we have the counting measure, which is the same sort of thing. Each point in the domain counts as 1. Therefore, 2 (6) Z f 0(x) dx = f 0(0) · 1+f 0(1) · 1+f 0(2) · 1=1+3+5=9. 0 In other words, with a counting measure, we just add up the function values. Note that the 9 looks familiar. 2 (7) Z f 0(x) dx =9=f(3) − f(0). 0 1 2

Does this make sense? Of course. The function f increases by 1 as x goes from 0 to 1. It increases by 3 as x goes from 1 to 2. It increases by 5 as x goes from 2 to 3. There is a total increase of 1 + 3 + 5 = 9. We also know that f increases from 0 to 9 as x goes from 0 to 3.

7 0 1. Compute R2 f (x) dx.

2. Compute f(8) − f(2). Is the real fundamental theorem of calculus this simple? Yes. Well, actually no. But the basic idea holds.

2. Cardinality

Counting the number of elements in a set measures a set’s cardinality. One way of describing this is as follows. Consider the set { 1, 3, 4, 7 }. We count the elements one, two, three, four. In other words, we form a one-to-one correspondence

1 → 1 2 → 3 (8) 3 → 4 4 → 7

A ﬁnite set will have the same cardinality as one of the following sets:

{}= ∅ { 1 } { 1, 2 } { } (9) 1, 2, 3 { 1, 2, 3, 4 } { 1, 2, 3, 4, 5 } . .

This corresponds to saying how many elements are in the set. An inﬁnite set could have the same cardinality as the natural numbers N = { 1, 2, 3, 4, 5,...}. In this case, we’ll say the set is countably inﬁnite, or just countable. The integers are countable. An indication of this is the correspondence below.

1 → 0 2 → 1 3 →−1 4 → 2 (10) 5 →−2 6 → 3 . .

Essentially, this indicates how all the integers will ﬁt into a sequence. 3

The rationals are also countable. The correspondence below puts all the positive rationals into a sequence.

1 1 → 1 1 2 → 2 2 3 → 1 1 4 → (11) 3 2 5 → 2 3 6 → 1 1 7 → 4 . .

Note that this takes all the fractions whose denominator and numerator add up to two, then all the fractions whose denominator and numerator add up to three, then four, then ﬁve, etc. Can you see how every positive fraction must be in this list? Actually, there are lot’s of repetitions, so each positive fraction is in the list at least once. It should be intuitively apparent that we can skip over the repetitions to get a one-to-one correspondence. We could also play the same trick used with the integers to get the negative rationals into the list. The reals are not countable. The reals between 0 and 1 are not countable. To see this, suppose that we could put all the reals between 0 and 1 into a list. Each of these numbers has a decimal expansion, and it might look something like this. We’ll also make sure that we don’t use the repeating 9’s representation for any number.

1 → 0.458796542145544 ... 2 → 0.578945457856215 ... 3 → 0.145872525463245 ... 4 → 0.333333333333333 ... (12) 5 → 0.668877445587445 ... 6 → 0.000014587544587 ... 7 → 0.999948754745555 ... . .

From here, it’s easy to construct the decimal expansion of a number not on the list. Since the first digit of the first number is 4, I’ll make sure that the first digit of my number isn’t a 4. How about 7? The second digit of the second number is 7, so I’ll pick 2 for the second digit of my number. The third digit of the third number is 5, so I’ll pick 8 for the third digit of my number. Now, I could get messed up, if I pick all 0’s or all 9’s, so I’ll never pick a 0 or a 9. So I’ve got

(13) x =0.728 ... and x is not on the list. I can do this for any list, so it must be impossible to get all the real numbers into a sequence. The reals are not countable. We’ll say that the reals are uncountable. 4

3. Summary of cardinality

We can specifically count the number of elements in a finite set. There are countably infinite sets, like Q and Z, and there are uncountable sets, like R and I. It is relatively easy to show that there are sets, such as the set of all real-valued functions, that have cardinalities larger than R’s. In fact, given a set, it’s “easy” to construct a set with a larger cardinality. So, we’ve got all the finite cardinalities, the countably infinite cardinality, and an infinite number of uncountable cardinalities. We could ask the question, is there a cardinality between countably infinite and the cardinality of R? This question is known to be unanswerable. We can adopt the following Continuum Hypothesis, which we generally do, or we can assume that it’s false. Either way, there is no contradiction with the normal “facts” of mathematics. The Continuum Hypothesis. The cardinality of the real numbers is the smallest uncountable cardinality.

Homework 19 starts on the next page. 5

4. Homework 19

1. The correspondence between N and Q+ that I have above has repetitions. If you were to remove the repetitions, what would be the 20th fraction in the list?

2. Your list from problem 1 could generate a list for the negative rationals. We could then get all of Q into a list by starting with 0, taking the first thing in the “positive” list, then the first thing in the “negative” list, then the second thing in the “positive” list, etc. What’s the fifteenth thing on this list?

We can’t talk about uncountable cardinalities with lists. To see a one-to-one correspondence with the reals, consider the graph of the tangent function. One brach of this function sets up a one-to-one correspondence − π π between 2 , 2 and R. Once you buy that pile of canned goods, you have to accept that any open interval has the same cardinality as the real numbers. To work with cardinalities, it’s convenient to nail down some intuitive axioms. For a set A, we will denote the cardinality of A by | A |.

3. Would you be willing to accept Axiom I? If A ⊂ B, then | A |≤|B |.

4. Would you be willing to accept Axiom II? If A ⊂ B and A =6 B (i.e., A is a proper subset of B), then the cardinality of A could be the same as B’s.

5. Would you be willing to accept Axiom III? Given two sets A and B, then exactly one of the following is true: | A | < | B |, | A | = | B |,or| A | > | B |.

6. What axiom says that | I |≤|R |? If I were countable, then we would have R as the union of two countable sets. If this were true, then both Q and I could be put into sequences, and then alternating terms from these two sequences would put all of R into a sequence. We know this can’t happen. I is not countable.

7. By Axiom III, problem 6, and the , | I | = | R |.

We just saw that countable plus countable has to be countable. What must countable plus uncountable be? In particular, suppose | A | = | R | and | B | = | N |. What’s the cardinality of A ∪ B? For simplicity, let’s assume that A and B are disjoint.

8. Must it be true that A = R?

9. Since | A | = | (0, 1)| and | B | = | Z |, do you agree that | A ∪ B | = | (0, 1)∪ Z |? By Axiom I, | A ∪ B | = | (0, 1)∪ Z |≥|(0, 1)| = | R |. Also by Axiom I, | A ∪ B | = | (0, 1)∪ Z |≤|R |.

10. Since ≤ is the same as >6 , and ≥ is the same as <6 , which Axiom says that | A ∪ B | = | R |? Bye.