Math 3361-Modern Algebra Lecture 08 9/26/ Cardinality

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1 Math 336-Modern Algebra Lecture 08 9/26/4. Cardinality I started talking about cardinality last time, and you did some stuff with it in the Homework, so let s continue. I said that two sets have the same cardinality, if there is a one-to-one correspondence between them. Let me give an alternate approach here that we ll have an easier time using. Note that we re talking about sets now, not just groups. Basic Principle. Given two sets A and B, exactly one of the following is true: () A and B are the same size, (2) A is larger than B, or (3) B is larger than A. Basic Principle 2. If there is no onto function f : A B, then B must be larger than A. Basic Principle 3. If there is an onto function f : A B, then B cannot be larger than A. Let s just operate as if these are fundamental truths. From Homework 07, therefore, we know that that there is an onto function f : N Z, so Z is not larger than N. It is easy to show that N is not larger than Z. Therefore, they must be the same size. Basic Principle 4. With regards to infinite sets, twice as big, is not really bigger at all. Twice as big is the same size. Let s throw this in too. Basic Principle 5. If A B, then A can t be larger than B. It could be the same size, however. 2. The Rationals are Countable Any set that is the same size (formally, have the same cardinality) as the natural numbers, N, is said to be countable. Some people also include finite sets with the infinite countable sets. That s fine. Sets that are larger than N are called uncountable. I want to show today that the rationals are countable, and the reals are uncountable. In some sense, the rationals are quite a large set. For example, between any two real numbers, there are infinitely many rational numbers. It seems, therefore, that there must be as many rationals as reals. That s not true. At least from our point of view. We ll see this later. First, let s prove that Q is countable. Actually, I m going to prove that Q + is countable, and from what we ve seen, if that s true, then Q must be countable also. Consider the following infinite array of numbers.

2 2, 2, 2, 3, 2 2, 3, () 4, 2 3, 3 2, 4, 5, 2 4, 3 3, 4 2, 5, 6, 2 5, 3 4, 4 3, 5 2, 6, 7, 2 6, 3 5, 4 4, 5 3, 6 2, 7,. For any fraction, I can add the numerator and denominator to get a positive integer. I ve gathered all the fractions with the same sum in rows. For example, the third row has all the fractions with a sum of 4, and there are only three of these. All of Q + is in this array with many repeats. In particular, all the fractions with a given denominator line up along a diagonal. I can now put these all into a single line of numbers. (2), 2, 2, 3, 2 2, 3, 4, 2 3, 3 2, 4, 5, 2 4, 3 3, 4 2, 5, 6, 2 5, 3 4, 4 3, 5 2, 6, 7, 2 6, 3 5, 4 4, 5 3, 6 2, 7,... Now, you can define a function f : N Q + such that f() =, f(2) = 2, etc., where f(n) is simply the n-th thing in the list. Since all the positive rationals are in the list at least once, f must be onto, and Q + cannot be bigger than N. It follows that Q + and Q must be countable.. Is there any positive rational that is not listed in (2) more than once? 3. The Reals are Uncountable Now, I m going to show you that the reals are uncountable. As you may have gathered, showing a set is countable essentially means stuffing all of it into a sequence. You can t do that with the reals. What I ll do is show that the interval (0, ) is uncountable, and since this is a subset of R, we must conclude that R is uncountable too. Let s suppose that we could put all the real numbers between 0 and into a list, and then we ll show we really couldn t have.

3 3 OK. So suppose the following is a list containing all the real numbers between 0 and. Since every real number has a decimal expansion, possibly infinite, we ll list them out that way. a = (3) a 2 = a 3 = a 4 = a 5 = a 6 = a 7 = Note that I put 3 and 4 sixth and seventh on the list. Another list might put them somewhere else, but they would have to be somewhere. I claim that there is no way that all the numbers in (0, ) can be in this list. I ll describe how to find one that s missing. We re going to construct the decimal expansion of a number x that does not belong to the list. The first digit of a = is, so if I choose x to be x = , then definitely x a. The second digit of a 2 = is 5, so if I choose x to be x = , then definitely x a 2. In general, I will choose the n-th digit of x to be different from the n-th digit of a n. Carrying this out to infinity, x is not equal to anything in the list. That is, the list is missing x. The one thing we have to be careful about is choosing the digit 9. The decimal , for example, is actually equal to. That s an analysis issue, so I won t talk about it more. It s easy to avoid choosing 9 s, so we re OK. Note that there are always at least eight choices for each digit of x, so there really are a lot of numbers missing. Not surprisingly, any such list will leave out an uncountable number of reals. In any case, (0, ) and R are uncountable. 4. The Continuum Hypothesis When we get within sight of the boundaries of mathematics, I like to point that out. We ve just seen that R is bigger than N. Cantor wondered about something that is almost too obvious to notice. Is there a set that is smaller than R and bigger than N? We can state this question as a conjecture. (Conjecture) The Continuum Hypothesis. The cardinality of R is the smallest uncountable infinity. It is relatively easy to show that there are sets larger than R (the set of all f : R R, for example). It turns out, however, that all the basic facts that we assume about mathematics are consistent with this conjecture being true and also with it being false. In other words, The Continuum Hypothesis is not provable and not disprovable. This is really odd. For example, we could just assume that there is a subset of R that is uncountable, but smaller than R. We would never run into a contradiction. We could also just assume that there isn t such a set. Again, no contradictions. This is why we re careful about proving things. There is a lot of obvious stuff that isn t obvious at all.

4 4 5. Homework 08 Note that an isomorphism is a one-to-one onto function, so isomorphic groups must be the same size. For infinite groups, its actually possible for a group to be isomorphic to a proper subset of itself. Consider the following subset of Z. (4) 3Z = { 3z z Z } = {..., 6, 3, 0, 3, 6,...}. These are the multiples of 3 in the integers. I could just as easily talk about sets 2Z or 7Z.. Consider the function f : Z 3Z such that f(z) = 3z. Is this an isomorphism? 2. Consider the function f : Z 3Z such that f(z) = 3z. Is this an isomorphism? 3. Consider the function f : Z 3Z such that f(z) = 3z + 3. Is this an isomorphism? 4. Are Z and 3Z the same size (explicitly, do they have the same cardinality)? 5. Is it possible that the groups Q +, and R +, are isomorphic? (a) yes (b) no, they do not have the same cardinality (c) no, they have the same cardinality, but no function can preserve the operation (d) no, a proper subset of a set is never isomorphic to that set it is not known whether they are or not It s time for a multiple choice proof. I claim that if f : G H is an isomorphism, then f(e) = e. 6. In the statement f(e) = e, in which group is the second e? In We know there is only one identity in a group. In fact, our proof showed that if x a = a for even one a, x has to be the identity. So, let h be any old element in H. 7. I claim that there is a g G such that f(g) = h. Why is this true? (a) because f is one-to-one (b) because f is onto (c) because f preserves the operation (d) well, like most things Howard says, it s actually not true (cont.)

5 5 8. It follows that f(e) h = h, because... (Remember that we don t want to use what we re trying to prove in the proof.) (a) f(e) h = f(e) f(g) = f(e g) = f(g) = h (b) f(e) h = e h = h (c) f(e) h = f(e h) = f(h) = h (d) f(e) f(e) h = f(e) h OK. This completes the proof, and isomorphisms take the identity to the identity. Isomorphisms also take inverses to inverses. Let s look at that. 9. Let a G. Which of the following statements best describes f(a )? (a) The element that f maps the inverse of a to. (b) The inverse of the element that f maps a to. (c) The inverse image of the element a. (d) One over the element f(a). 0. Let a G. Which of the following statements best describes ( f(a))? (a) The element that f maps the inverse of a to. (b) The inverse of the element that f maps a to. (c) The inverse image of the element a. (d) One over the element f(a). The proof of the fact that isomorphisms take inverses to inverses is pretty simple. We already have proven that inverses are unique. In fact, our proof showed that for any element a G, any element that acts at all like the inverse of a is, in fact, the inverse of a. Consider the following string of equalities (5) f(a) f(a ) = f(a a ) = f(e) = e. It follows f(a) and f(a ) are inverses of each other. Bye.

One-to-one functions and onto functions

MA 3362 Lecture 7 - One-to-one and Onto Wednesday, October 22, 2008. Objectives: Formalize definitions of one-to-one and onto One-to-one functions and onto functions At the level of set theory, there are