The cardinal comparison of sets

(B) The cardinal comparison of sets I think we can agree that there is some kind of fundamental difference between finite sets and infinite sets. For a finite set we can count its members and so give it a size 0, 1, 2, 3,... a unique natural number. In this way we see how two finite sets can be equivalent if we don t worry about which members it has. Two finite sets are equivalent, in this sense, if they have the same number of members. It is also obvious that all infinite sets are somehow equivalent. This was an accepted fact 150 years ago (and it still is by many non-mathematicians). For several years Cantor tried to set up an enumeration of R, a bijection N R. Around 1875, after some correspondence with Dedekind and so more thought, Cantor showed that N and R are not equivalent. There are different sizes of infinite sets. After that all hell broke out, although this time I don t think the Pope got involved. In this document we look at the way sets, finite and infinite, can be given a size. Contents 1 The comparison relations............................... 1 1.1 Exercises: The comparison relations..................... 5 1.2 Problems: The comparison relations..................... 6 2 The diagonal argument................................. 8 2.1 Exercises: The diagonal argument...................... 11 2.2 Problems: The diagonal argument...................... 11 3 The Cantor-Schröder-Bernstein theorem....................... 12 3.1 Exercises: The CSB theorem......................... 16 3.2 Problems: The CSB theorem......................... 16 1 The comparison relations A set A is something with elements or members. It also has a family of subsets a A X A and the whole family of these is the power set PA of A. Often a set carries more gadgets to form an algebraic structure, or a geometric structure, or a topological structure, or several other kinds of structures. Here we are concerned solely with the set, its elements, and its subsets. We wish to not distinguish between sets that are in some sense equivalent. How can we make that precise? Each finite set has a given size. It is equivalent to precisely one of the sets {0} {0, 1} {0, 1, 2} {0,..., n} 1

using the natural numbers n N. Beyond these we have the infinite sets, and as we will see shortly there are different sizes of such sets. To make that precise we set up some notation and terminology. 1.1 DEFINITION. For two sets A, B we write A B A B if there is at least one function f : A B which is an injection a bijection respectively. These are the cardinal comparison cardinal equivalence relations, respectively. Observe that if A, B are finite sets then the following hold. A B A has fewer members than B (or the same number of members) A B A has exactly the same number of members as B We wish to extend this idea to infinite sets. In due course we will write A B for A B A = B for A B but for the time being we stick to the notation given above. The following result is a simple consequence of the behaviour of functions. The proof is left as an exercise. 1.2 LEMMA. Let A, B, C be arbitrary sets. The following hold. (i) A B = A B (ii) A A (iii) A B = B A (iv) A B C = A C (v) A B C = A C Observe that there seems to be one property missing, namely that A B A = A B for arbitrary A, B. This is true be the proof is not immediate. We return to this later in Section 3. What use is this notion of cardinal equivalence? Intuitively it is seems obvious that any two infinite sets are equivalent. Let s look at some examples of this. 2

1.3 EXAMPLES. (a) Consider the set N of natural numbers. 0, 1, 2, 3, 4,..., n,... We use this as a template against which we can measure other infinite sets. (b) Consider the set Z of integers. We can match this against N as follows. 0, 1, 1, 2, 2, 3, 3, 4, 4,... This shows that Z N. A simple calculation show that this enumeration is produced by the function ζ : N Z given by ζ(n) = ( 1) n+1 (n + 1)/2 for each n N. Here is the integer part function. (c) Consider Z Z. We can think of this set as the points in the cartesian plane that have integer components. Then the path (0, 0). shows that Z Z N. There are many other paths through Z Z and some of these can be described by a formula. Although, at the moment I can t think of one. Have a look at Problems 1.7, 1.8, 1.9. (d) Consider the set Q of rational numbers. Each such number has a representation n/d in lowest common form. That is with n Z, d N and where n and d have no common factors (which ensures d 0). Thus each rational can be viewed as sitting on some point in the diagram of (c). There will be some points that don t have a rational friend. However, this shows that Q N. (e) There is a neater enumeration of the rationals. Consider the function f : N N given by f(0) = 1 f(2m + 1) = f(m) f(2m + 2) = f(m) + f(m + 1) for each m N. It doesn t take long to see that for each n N the two consecutive outputs f(n), f(n + 1) are relatively prime. With quite a bit more work it can be shown that N Q n f(n)/f(n + 1) is a bijection. We may look at the details of this later. (f) What about the set of algebraic numbers? We can convince ourselves that this set is equivalent to N, but a proper proof needs some background developments. 3

An algebraic number is a (real) root of a polynomial a 0 x n + a 1 x n 1 + + a n 2 x 2 + a n 1 x + a n with integer coefficients and a 0 0. Each such polynomial has just finite many roots (in fact, precisely n root if we allow multiple roots). Since each such polynomial is determined by its finite list of coefficients, there is an enumeration of all such polynomials. This leads to an enumeration of all algebraic numbers. (g) What about the set R of real numbers? Cantor showed that N R, that is R can not be enumerated as an N-sequence. Later, in Section 3, we will see a quite simple proof of this. Notice that by part (f) there must be a lot of transcendental numbers, a real (or complex) number that is not algebraic. However, it wasn t until around 1850 that an explicit example of a transcendental was produced by Liouville. This was 0 (1/10)n! which, if you write this down as a decimal expansion, you will see what is going on. It was Hermite in 1873 who first proved that e is transcendental, and Lindemann in 1882 proved that π is transcendental. There are still many numbers, such as e + π for which we do not know if they are algebraic or transcendental. Around 500 BC one of the Pythagoreans, called Hippasus, produced the first known non-rational number, namely 2. This caused a bit of a splash. The Pythagoreans, lead by Pythagoras, were a pseudo-religious group who investigated the nature of the world, and mathematics in particular. One of their dogmas was that the world is rational, and the comparison of any two like things could be measured by a rational number. When Hippasus proved that 2 is not rational this gave the Pythagoreans a bit of a quandary. They threw Hippasus off a ship into the sea to drown him, and so prevent the awkward result becoming more well-known. This sets up the problem we look at in this document. We know how to measure the size of a finite set, using the natural numbers. How can we measure the size of an infinite set? To help with this make use of a bit of notation. 1.4 DEFINITION. We say a set A has cardinality ℵ 0 if N A. We sometime write A = ℵ 0 to indicate this. We say a set A has cardinality ℶ 1, or 2 ℵ 0, if PN A. We sometime write A = ℶ 1 to indicate this. More generally, we say two sets A, B have the same cardinality and write A = B if A B, that is there is a bijection between them. Here ℵ 0 is a particular cardinal number, the first in an very long increasing chain ℵ 0 < ℵ 1 <... < ℵ n <... < ℵ ω <... of cardinal numbers. We may look at cardinal numbers in more detail later. There is also another chain of cardinals ℶ 0 = ℵ 0 < ℶ 1 = 2 ℶ 0 < ℶ 2 = 2 ℶ 1 <... 4

which corresponds to taking powers sets. As indicated, it is known that ℵ 0 < ℶ 1, and we will prove this in Section 2. It is known that ℶ 1 must fit somewhere in the ℵ-series, but it is not known where. It can be proved that ℶ 1 ℵ ω, but it could be anywhere strictly between ℵ 0 and ℵ ω. It could also be beyond ℵ ω. Later we may look at some of the basics of cardinal arithmetic, and we may be able to show ℶ 1 ℵ ω. I should tell you that the symbol ℵ is the first letter of the Hebrew alphabet. It is now standard notation. In fact, the first four letters ℵ (aleph), ℶ (beth), ℸ (daleth), ג (gimel) of the Hebrew alphabet are now standard AMS LaTeX symbols. However, we don t yet use Cyrillic subscripts, very often. If you ask me I can tell you a story of why the letter ℵ was used. However, I don t know if it is true. 1.1 Exercises: The comparison relations 1.1 Prove the five parts of Lemma 1.2 by exhibiting appropriate functions. 1.2 Recall that for a set A a characteristic function on A is a function χ : A 2 where 2 = {0, 1}. Recall also that for each sets A, B we let [A, B] = [A B] be the set of all functions from A to B. Both notations are useful at different times. (a) Show there is a bijection PA [A, 2] for arbitrary A. (b) Show that if A is finite with m members, that PA has 2 m members. 1.3 (a) For arbitrary sets A, X, Y show the following. (i) PA [A, 2] (ii) If X Y then [X, A] [Y, A] and [A, X] [A, Y ] (iii) [X, [Y, A]] [X Y, A] (b) Show that both N N N and [N, PN] PN 1.4 (a) Show that for each pair A, B of countable sets the cartesian product A B is countable. (b) Show that for each countable set A, each of the sets A 0, A 1, A 2,..., A n,... is countable. Here for each n N the power A n is the set of all n-tuples of elements of A. (b) Show that for each countable set A, the set A of all finite sequences from A is countable. 1.5 Consider Example 1.3(e). Start to write down the suggested enumeration, and see if you can sort out a picture of what is happening. 5

1.2 Problems: The comparison relations 1.6 Consider a collection X of subsets of N. (a) Suppose X is pairwise disjoint, that is for X, Y X. Show that X is countable. (b) We say X is almost disjoint if X = Y or X Y = X = Y or X Y is finite for X, Y X. Show that there is such a collection in bijection with S = (0, ) the strictly positive reals. To do that you might want to consider the odd numbers for m N, x S. 1.7 Consider the function where, of course, x, y N. (a) Show the following 2 mx + 1 f : N N Q given by f(x, y) = (x + y)2 + 3x + y 2 (i) f(0, z + 1) = f(z, 0) + 1 (ii) f(x + 1, y) = f(x, 1 + y) + 1 (iii) f(x, y) = f(0, x + y) + x (iv) f(z, 0) = f(0, z) + z for all x, y, z N. (b) Show that (i) f(z, 0) N (ii) f(x, y) N for all x, y, z N. (c) Show that each m N is a value of f. (d) Show that f is injective, that is for each m N there are unique x, y N with m = f(x, y). 6

1.8 Consider the integral points N N in the upper right hand quadrant of the cartesian plane R R. Suppose we enumerate these by taking longer and longer downward diagonals, as illustrated.. 10 6 11 3 7 1 4 0 2 Each point (x, y) N 2 is given a position f(x, y). Thus 12 8 5 f(0, 0) = 0, f(0, 1) = 1, f(1, 0) = 2, f(0, 2) = 3, f(1, 1) = 4, f(2, 0) = 5, f(3, 0) = 6,... are the first few positions. Show that f(x, y) = (x + y)2 + 3x + y 2 gives the position of a general point (x, y). 1.9 (a) Consider the function f : N N N given by f(x, y) = max{x 2, y 2 } + x + (x y) { x y if y x x y = 0 if x y gives the last component. Write out an explicit description of this function for the two cases. (b) Consider the integral points N N in the upper right hand quadrant of the cartesian plane R R. Suppose we enumerate these by taking larger and larger square, as illustrated.. 16 17 9 9 10 11 12 4 5 6 13 1 2 7 14 0 3 8 15 Each point (x, y) N 2 is given a position f(x, y). Thus f(0, 0) = 0, f(0, 1) = 1, f(1, 1) = 2, f(1, 0) = 3, f(0, 2) = 4, f(1, 2) = 5, f(2, 2) = 6,... are the first few positions. Show that f is the function of part (a). 7

2 The diagonal argument The main aim of this section is to show that the set R of reals is not countable, that is there is no bijection between N and R. This along with several related results is given in Theorem 2.4. Before that we develop some preliminary material for arbitrary sets. Each set A has two associate sets PA [A, 2] = [A 2] its power set and the set of characteristic functions on A. Recall that, as on the right hand side, we use alternative notions for the set of functions from A to 2. This can be useful when there are several other functions around. By Exercise 1.2(a) or Exercise 1.3(a)(i) we know that these two associated sets PA and [A, 2] are in bijective correspondence. That ensures that the analysis we carry out here for one of them is easily transferred to the other. However, it is instructive to see both versions done independently. 2.1 LEMMA. Let A be any set. Then that is there is at least one injection A PA A [A, 2] A f PA A φ [A, 2] from A to the associated set. Proof. (f) The following function will do. A a f PA {a} (φ) Consider any a A and let φ(a) : A 2 be the function given by for arbitrary x A. We have φ(a)(x) = { 1 if x = a 0 if x a φ(a) : A 2 and hence φ : A [A, 2] by releasing a. Thus it suffices to show that φ is injective. To this end consider a, b A with φ(a) = φ(b). Thus a = b is required. But φ(b)(a) = φ(a)(a) = 1 so that b = a, as required. Next we come to the crucial result. This proof of this is quite simple, but it has had an enormous impact on certain areas of Mathematics (and has influenced many people who don t know what they are talking about). Again the result is given in two parts, and each part can be deduced from the other. But again it is instructive to see both versions. 8

2.2 THEOREM. Let A be any set and consider any pair of functions. A Then neither is surjective. f PA A φ [A, 2] Proof. (f) Given the function f we must produce a subset D A which is not an output of f. For each x A we know that f(x) A, so the construction x D x / f(x) makes sense to produce some D A. By way of contradiction suppose D is an output of f, that is D = f(a) for some a A. Then a D a / f(a) = D a / D which can not be, and so gives the contradiction. (φ) Given the function φ we must produce a function : A 2 which is not an output of φ. For each x A we know that φ(x) : A 2, so we may also evaluate φ(x) at any y A to produce φ(x)(y) 2 = {0, 1}. Thus we may set (x) = 1 φ(x)(x) to produce : A 2. By way of contradiction suppose is an output of φ, that is = φ(a) for some a A. Then which is the contradiction. (a) = φ(a)(a) 1 φ(a)(a) = (a) As mentioned at the beginning of this section, for each set A the two sets PA and [A, 2] have the same cardinality. We now have the following crucial result. 2.3 COROLLARY. For each set A, the cardinal equivalent sets PA and [A, 2] do not have the same cardinality as A. Given two sets A, B we write A < B if there is at least one injection from A to B, but there is no bijection between the two sets. We say B has strictly larger cardinality than A. We have just seen that for each set A. But now A < PA A, PA, P 2 A, P 3 A,... is a chain of larger and larger sets. When does this end? It doesn t! When A = ℵ 0, that is when A is countably infinite, This chain produces the chain of cardinals ℵ 0 = ℶ 0 < ℶ 1 < ℶ 2 < ℶ 3 <... which goes on for ever, and further than you think. As yet we haven t determined the cardinality of the set R of real numbers. Let s do that now, after a fashion. 9

2.4 THEOREM. Each non-trivial interval of R has cardinality ℶ 1 = 2 ℵ 0. Proof. Let I be any non-trivial interval of R. We first observe the following. [0, 3/2] I R PQ [Q 2] [N 2] The first comparison is obtained by an appropriate linear function. inclusion. For the third we send each real r to the rational interval The second is an (, r) Q and observe that this is an injection. The fourth is an instance of the remarks at the beginning of this section. The fifth holds since Q N. Observe that this is a place where the notation [A B] is useful, so we don t get [A, B] confused with a real interval. Given these comparisons it is now sufficient to produce an injection f : [N 2] [0, 3/2] and this is where the choice of 3/2 becomes useful. For each f : N 2 consider the following series. x f = {f(r)3 r r < ω} This is a series of positive terms and, since each f(r) {0, 1}, we have 0 {3 r r < ω} = 3/2 to show that the series converges. Thus we have a function [N 2] [0, 3/2] x x f and it suffices to check that this is injective. Consider f, g [N 2] with x f = x g. We require f = g. To this end, for each r N let a r = f(r) g(r) so that a r { 1, 0, 1}, and we require a r = 0. Consider the following series. {ar 3 r r < ω} This has value x f x g which is 0. Since a r 1 the series is absolutely convergent. This gives a 0 = {a r 3 r 0 < r < ω} { a r 3 r 0 < r < ω} {3 r 0 < r < ω} = 1/2 so that to give a 0 = 0. a 0 { 1, 0, 1} 1/2 a 0 1/2 10

A repeat of this argument now gives a n = 0 by induction on n. Thus f = g, as required. The crucial results of this and the previous section is that Q is countable, but R is not. We have Q = ℵ 0 R = ℶ 1 where we know that ℵ 0 < ℵ 1 ℶ 1. The standard axioms of Axiomatic Set Theory tell us very little about the size of ℶ 1. Some people, such as Gödel, believe that ℶ 1 = ℵ 1, and it seems that some people now believe that ℶ 1 = ℵ 2. If that is not the case then what is the value of ℶ 1. As we move up the possible values ℵ 1, ℵ 1, ℵ 3,..., ℵ 42,... these seem to get less and less likely. If ℶ 1 ℵ 1 then the next most likely value seems to be ℵ ω. But there is a quite simple proof that this can not be the value. What is going on? Is someone having us on? 2.1 Exercises: The diagonal argument 2.1 Consider the proof of Lemma 2.1. How are the two functions f and φ related? 2.2 Consider the (missing) proof of the last part of Theorem 2.4. Using an induction on n, or otherwise, fill in this missing argument. 2.3 This continues Exercise 1.4 (a) Show that the set of all real algebraic numbers is countable. (b) Show that N (0, 1) and that (0, 1) is not countable. (c) Show that there is at least one transcendental number. (d) Can you exhibit a transcendental number and prove that it is transcendental. 2.2 Problems: The diagonal argument 2.4 Let A be the set of all monotone (increasing) ω-sequences of natural numbers. For we say if a = (a n n < ω) b = (b n n < ω) a is dominated by b or b dominates a a n /b n 0 as n holds. Let f : N A be an arbitrary function. (a) Let d n = max{f(0) n, f(1) n,..., f(n) n } for each n N. Show that d = (d m n < ω) is a member of A. 11

(b) Show that for arbitrary m N and a A, if b = f(m) dominates a then we have 2a n < b n d n for almost all n, that is for all sufficiently large n. (c) Show that no sequence f(m) dominates d. (d) Show that there is no countable B A such that each a A is dominated by some b B. 2.5 Calculate the cardinality of each of the following. (i) The set of real sequence, that is the set of ω-sequences of real numbers. (ii) The set of convergent real sequences. (iii) The set of real sequences with a convergent series. (iv) The set of continuous real functions R R. (v) Any set of pairwise disjoint non-trivial real intervals. (vi) The set of discontinuities of any monotone function R R. 2.6 Let {A n n N} be an increasing chain of sets with union A. A 0 A 1 A n A = {A n n N} Suppose also that for each n N there is no surjective function A n A. Now let φ : A [N A] be an arbitrary function. For each n N let to obtain a family of subsets of A. X n = {φ(a)(n) a A n } (i) Show that each X n is a proper subset of A. (ii) Show that if f φ[a] then f(n) X n for at least one n N. (iii) Show that φ is not surjective. 2.7 By making use of Problem 2.6 show that the cardinality of PN can not be ℵ λ for any countable limit ordinal λ. If you are not sure what such a λ is, then take the particular case λ = ω. 3 The Cantor-Schröder-Bernstein theorem We have two comparisons between sets A B A B given by the existence of an injection a bijection respectively. These are designed to capture the informal notions A is no bigger than B A has the same size as B 12

where it is the number of elements we are trying to measure. It ought to be the case that A B and B A = A B but this is by no means obvious. In this section we prove that this implication does hold. The result has various names attached to it, usually referring to two of the three people in the title of this section. In Levy Basic set theory page 85 it is stated that the result was proved by Dedekind in 1887, conjectured by Cantor in 1895, and proved by Bernstein in 1898. Those dates look a bit dodgy. In Temple 100 years of mathematics page 32 it is stated that the result was proved independently by Schröder in 1896 and Bernstein in 1905. I have been told by others that Cantor suggested the idea, Schröder wrote the words, and Bernstein set it to music. But I m not sure I believe that one. The aim of this section is to prove the following. 3.1 THEOREM. Consider a pair of injections f A B g between two sets going in opposite directions. Then there is a bijection h : A B between the sets. Furthermore, such a bijection h can be built from parts of the two given injections f, g. Proof. We partition each set into two parts in a way that the restrictions of f, g become bijections. A = A + A B = B + B A + f + B A g+ B + A = g[b + ] B = f[a + ] f + = f A+ g + = g B + Let h + = f + and let h be the inverse of g +. We thus have two bijections h + : A + B h : A B + between disjoint sets. We let h = h + h to obtain the required bijection between the given sets. Of course, this proof is not complete. How do we know that such a partition exists? In fact, that is probably the more interesting part of the result. In the remainder of this section we show how such a partition can be produced. 13

We are given a pair of injections f, g between two sets A, B and we wish to find partitions such that each of the two restrictions is a bijection. One way to achieve that is to use a simple fixed point construction. However, here it is a fixed set construction. We work on the power set PA of A and combine the two given functions f and g into one function H : PA PA on subsets of A. For each subset X A we put H(X) = A g[b f[x]] to produce another subset of A. Observe how H produces the output H(X). For the input X A we take the image of X across f to obtain a subset f[x] B on the other side. We take the complement of that in B to obtain B f[x]. We pass that back to A using g to obtain g[b f[x]] A. We take the complement of that in A to obtain H(X). We want to find some X A which is fixed by H, that is H(X) = X. In general the input X and the output H(X) have no relationship with each other. However, H does track variations of X. 3.2 LEMMA. The function H is monotone, that is for U, X A. U X = H(U) H(X) Proof. Consider U, X A with U X. Then so that by taking complements in B. But now f[u] f[x] B f[x] B f[u] g[b f[x]] g[b f[u]] so that by taking complements in A. H(U) H(X) To find a fixed set we look at those subsets that are inflated by H. 3.3 DEFINITION. Let U be the collections of those sets U A with U H(U). 14

Trivially, we have U. In some circumstances this could be the only member of U. The trick is to look for a maximal member of U. To do that we let L = U so that any maximal member of U is a subset of L. In fact, we get more than that. 3.4 LEMMA. We have L H(L), so that L U. Proof. Consider any U U. Then U L, so that since H is monotone. But now to give the required result. U H(U) H(L) L = U H(L) This shows that L is a special member of U. It is the unique maximum member of U. It also has another special relationship with H. 3.5 LEMMA. We have H(L) = L. Proof. We already know that L H(L), so it suffices to verify the converse inclusion. But this known inclusion gives H(L) H ( H(L) ) since H is monotone. This ensures that H(L) U, and hence H(L) L by the definition of L. This show that H has at least one fixed set. To obtain the required partition we can use any fixed set. 3.6 DEFINITION. Let A + be any fixed set of H. We put to obtain partitions of A and B. A = A A = B = f[a + ] B + = B B We could try to carry out this construction starting with any subset A + A. However, it is the fixed set property that gives us the conditions we want. 3.7 LEMMA. Let A + be any fixed set of H. Then holds. f[a + ] = B g[b + ] = A Proof. The left hand equality is given by the construction. We must verify the right hand equality. But so that as required. A + = H(A + ) = A g[b f[a + ]] = A g[b B ] = A g[b + ] A = A A + = g[b + ] This completes the proof of Theorem 3.1. 15

3.1 Exercises: The CSB theorem 3.1 Consider the first proof of the CSB result. There may be many fixed sets of the function H. (a) Show that the example produced L is the largest fixed set, that is if F is any fixed set then F L. (b) Show there is a smallest fixed set. 3.2 This exercise produces a generalization of the constructions and results of Exercise 3.1. Let (P, ) be any linearly ordered set. Suppose this linear set is complete, that is for each X P the supremum X and the infimum X exist. Recall that the supremum is just the least upper bound, and the infimum is the greatest lower bound. For instance, the real interval [0.1] is an example of such a set. Let h : P P a monotone function on P, that is x y = h(x) h(y) for all x, y P. An element a P is a fixed point (or a fixed element) of h if h(a) = a. (a) Show that h has a largest and a smallest fixed point. (b) Find examples where h has precisely 1,2,3,4,... fixed points. (c) Using P = [0, 1], the real interval, how many fixed points do you think h can have (provided h is not the identity function). 3.2 Problems: The CSB theorem 3.3 (a) Suppose A, B, C are three sets with A B C, A B = and where A, B have cardinality ℵ 0. Show that C A = C. (b) Show that if A R is countably infinite, then R A = ℶ 1. (c) What is the cardinality of the set I of irrationals? 16