B. Cardinal Arithmetic, Exams of Construction

Download B. Cardinal Arithmetic and more Exams Construction in PDF only on Docsity! B. Cardinal Arithmetic In this Appendix we discuss cardinal arithmetic. We assume the Axiom of Choice is true. Definitions. Two sets A and B are said to have the same cardinality, if there exists a bijective map A → B. It is clear that this defines an equivalence relation on the class1 of all sets. A cardinal number is thought as an equivalence class of sets. In other words, if we write a cardinal number as a, it is understood that a consists of all sets of a given cardinality. So when we write cardA = a we understand that A belongs to this class, and for another set B we write cardB = a, exactly when B has the same cardinality as A. In this case we write cardB = card, A. Notations. The cardinality of the empty set ∅ is zero. More generally the cardinality of a finite set is equal to its number of elements. The cardinality of the set N, of all natural numbers, is denoted by ℵ0. Definition. Let a and b be cardinal numbers. We write a ≤ b if there exist sets A ⊂ B with cardA = a and cardB = b. This is equivalent to the fact that, for any sets A and B, with cardA = a and cardB = b, one of the following equivalent conditions holds: • there exists an injective function f : A→ B; • there exists a surjective function g : B → A. For two cardinal numbers a and b, we use the notation a < b to indicate that a ≤ b and a 6= b. Theorem B.1 (Cantor-Bernstein). Suppose two cardinal numbers a and b sat- isfy a ≤ b and b ≤ a. Then a = b. Proof. Fix two sets A and B with cardA = a and cardB = b, so there exist injective functions f : A → B and g : B → A. We shall construct a bijective function h : A→ B. Define the sets A0 = Ar g(B) and B0 = Ar f(A). Then define recursively the sequences (An)n≥0 and (Bn)n≥0 by An = g(Bn−1) and Bn = f(An−1), ∀n ≥ 1. Claim 1: One has Am ∩An = Bm ∩Bn, ∀m > n ≥ 0. Let us first observe that the case when n = 0 is trivial, since we have the inclusions Am = g(Bm−1) ⊂ g(B) = ArA0 and Bm = f(Am−1) ⊂ f(A) = B rB0. Next we prove the desired property by induction on m. The case m = 1 is clear (this forces n = 0). Suppose the statement is true for m = k, and let us prove it for m = k+1. Start with some n < k+1. If n = 0, we are done, by the above discussion. Assume first n ≥ 1. Since f and g are injective we have Ak+1 ∩An = g(Bk) ∩ g(Bn−1) = g(Bk ∩Bn−1) = ∅, Bk+1 ∩Bn = f(Ak) ∩ f(An−1) = f(Ak ∩An−1) = ∅, 1 The term class is used, because there is no such thing as the “set of all sets.” 2001 2002 APPENDICES and we are done. Put C = Arn≥0 An and D = B r ⋃ n≥0Bn. Claim 2: One has the equality f(C) = D. First we prove the inclusion f(C) ⊂ D. Start with some point c ∈ C, but assume f(c) 6∈ D. This means that there exists some n ≥ 0 such that f(c) ∈ Bn. Since f(c) ∈ f(A) = BrB0, we must have n ≥ 1. But then we get f(c) ∈ Bn = f(An−1), and the injectivity of f will force c ∈ An−1, which is impossible. Second, we prove that D ⊂ f(C). Start with some d ∈ D. First of all, since D ⊂ B r B0 = f(A), there exists some c ∈ A with d = f(c). If c 6∈ C, then there exists some n ≥ 0, such that c ∈ An, and then we would get d = f(c) ∈ f(An) = Bn+1, which is impossible. We now begin constructing the desired bijection. First we define φ : ⋃ n≥0Bn → B by φ(b) = { b if b ∈ Bn and n is odd (f ◦ g)(b) if b ∈ Bn and n is even Claim 3: The map φ defines a bijection φ : ⋃ n≥0 Bn → ⋃ n≥1 Bn. It is clear that, since φ ∣∣ Bn is injective, the map φ is injective. Notice also that, if n ≥ 0 is even, then φ(Bn) = f ( g(Bn) ) = f(An+1) = Bn+2. When n ≥ 0 is odd we have φ(Bn) = Bn, so we have indeed the equality φ ( ⋃ n≥0 Bn ) = ⋃ n≥1 Bn. Now we define ψ : ⋃ n≥0An → B by ψ = φ−1 ◦ f . Clearly ψ is injective, and ψ ( ⋃ n≥0 An ) = φ−1 ( ⋃ n≥0 f(An) ) = φ−1 ( ⋃ n≥0 Bn+1 ) = φ−1 ( ⋃ n≥1 Bn ) = ⋃ n≥0 Bn, so ψ defines a bijection ψ : ⋃ n≥0 An → ⋃ n≥0 Bn. We then combine ψ with the bijection f : C → D, i.e. we define the map h : A→ B by h(x) = { ψ(x) if x ∈ ⋃ n≥0An f(x) if x ∈ Ar ⋃ n≥0An = C. Clearly h is injective, and h(B) = ψ ( ⋃ n≥0 An ) ∪ f(C) = ( ⋃ n≥0 Bn ) ∪D = B, so h is indeed bijective.  Theorem B.2 (Total ordering for cardinal numbers). Let a and b be cardinal numbers. Then one has either a ≤ b, or b ≤ a. Proof. Choose two sets A and B with cardA = a and cardB = b. In order to prove the theorem, it suffices to construct either an injective function f : A → B, or an injective function f : B → A. B. Cardinal Arithmetic 2005 Proof. (i). Let a be an infinite cardinal number, and let A be an infinite set A, with cardA = a. Since for every finite subset F ⊂ A, there exists some x ∈ Ar F , one to construct a sequence (xn)n∈N ⊂ A, with xm 6= xn, ∀m > n ≥ 1. Then the subset B = {xn : n ∈ N} has cardB = ℵ0, so the inclusion B ⊂ A gives the desired inequality. (ii). Consider the sets A0 = {n ∈ N : n, even} and A1 = {n ∈ N : n, odd}. Then clearly cardA0 = cardA1 = ℵ0, and the equality A0 ∪A1 = N gives ℵ0 + ℵ0 = cardA0 + cardA1 = card(A0 ∪A1) = card N = ℵ0. (iii). Take the set P = N × N, so that ℵ0 · ℵ0 = cardP . It is obvious that cardP ≥ ℵ0. To prove the other inequality, we define a surjection φ : N → P as follows. For each n ≥ 1 we take sn = n(n− 1)/2, we set Bn = {m ∈ N : sn < m ≤ sn+1}, and we define φn : Bn → P by φ(m) = (n+ sn −m,m− sn + 1), ∀m ∈ Bn. Notice that (1) φn(Bn) = {(p, q) ∈ N× N : p+ q = n+ 1}. Notice also that ⋃ n≥1Bn = N, and Bj ∩ Bk = ∅, ∀ j > k ≥ 1, so there exists a (unique) function φ : N → P , such that φ ∣∣ Bn = φn, for all n ≥ 1. By (1) it is clear that φ is surjective.  Theorem B.3. Let a and b be cardinal numbers, with 1 ≤ b ≤ a, and a infinite. Then: (i) a + b = a; (ii) a · b = a. Proof. It is clear that a ≤ a + b ≤ a + a, a ≤ a · b ≤ a · a, so in order to prove the theorem, we can assume that a = b. (i). Fix some set A with cardA = a. Use Zorn Lemma, to find a maximal non-empty family {Ai : i ∈ I} of subsets of A with (a) cardAi = ℵ0, for all i, j ∈ I; (b) Ai ∩Aj = ∅, for all i, j ∈ I with i 6= j. If we put B = A r ( ⋃ i∈I Ai ) , then by maximality it follows that B is finite. In particular, if we take i0 ∈ I then obviously card(Ai0 ∪ B) = ℵ0, so if we replace Ai0 with Ai0 ∪ B, we will still have the above properties (a) and (b), but also A = ⋃ i∈I Ai. This proves that a = cardA = ℵ0 · d, where d = card I. In other words, we have a = card(N× I). Consider then the sets C0 = {n ∈ N : n even} and C1 = {n ∈ N : n odd}, 2006 APPENDICES so that (C0 × I) ∪ (C1 × I) = I × N, and (C0 × I) ∩ (C1 × I) = ∅. In particular, we get a = card(C0 × I) + card(C1 × I) = = (cardC0) · (card I) + (cardC1) · (card I) = = ℵ0 · d + ℵ0 · d = a + a. (ii). Fix A a set with cardA = a. We are going to employ Zorn Lemma to find a bijection A→ A×A. Define X = { (D, f) : D ⊂ A, f : D → D ×D bijective } . Equip X with the following order (D, f) ≺ (D′, f ′) ⇐⇒ { D ⊂ D′ f = f ′ ∣∣ D Notice that X is non-empty, since we can find at leas one set D ⊂ A with cardD = ℵ0. We now check that X satisfies the hypothesis of Zorn Lemma. Let T ={ (Di, fi) : i ∈ I } be a totally ordered subset of X. It is fairly clear that if one takes D = ⋃ i∈I and one defines f : D → D ×D as the unique function with f ∣∣ Di = fi, ∀ i ∈ I, then f is injective, and f(D) = ⋃ i∈I f(Di) = ⋃ i∈I fi(Di) = ⋃ i∈I (Di ×Di) = D ×D, so the pair (D, f) indeed belongs to X, and is an upper bound for T. Use Zorn Lemma to produce a maximal element (D, f) ∈ X. Notice that, if we take d = cardD, then by construction we have (2) d · d = d. We would like to prove that D = A. In general this is not the case (for example, when A = N, every (D, f) ∈ X, with N rD finite, is automatically maximal). We notice however that all we need to show is the equality (3) d = a. We prove this equality by contradiction. We know that we already have d ≤ a. Suppose d < a. Put G = ArD notice that d + cardG = a. Since d < a, by (i) we see that we must have the equality cardG = a. Then there exists a subset E ⊂ G with cardE = d. Consider the set P = (E × E) ∪ (E ×D) ∪ (D × E). Since E ∩D = ∅, the three sets above are pairwise disjoint, so using (2) combined again with part (i), we get cardP = card(E × E) + card(E ×D) + card(D × E) = = d · d + d · d + d · d = d + d + d = d = cardE. This means that there exists a bijection g : E × P , which combined with the fact that E ∩D = P ∩ (D×D) = ∅, will produce a bijection h : D∪E → P ∪ (D×D), such that h ∣∣ D = f and h ∣∣ E = g. Since we have P ∪ (D×D) = (D ∪E)× (D ∪E), the pair (D ∪ E, h) ∈ X will contradict the maximality of (D, f).  B. Cardinal Arithmetic 2007 Corollary B.1. If a is an infinite cardinal number, and if b is a cardinal number with 2 ≤ b ≤ 2a, then ba = 2a. Proof. We have 2a ≤ ba ≤ (2a)a = 2a·a = 2a, and the desired equality follows from the Cantor-Bernstein Theorem.  Corollary B.2. Let a be an infinite cardinal number, let A be a set with cardA = a, and define Pfin(A) = {F ∈ P(A) : F finite}. Then cardPfin(A) = a. Proof. First of all, the map A 3 a 7−→ {a} ∈ Pfin(A) is injective, so a ≤ cardPfin(A). We now prove the other inequality. For every integer n ≥ 1, let An denote the n-fold cartesian product. We treat the sequence A1, A2, . . . as pairwise disjoint. For every n ≥ 1 we define the map φn : An → Pfin(A), by φ(a1, . . . , an) = {a1, . . . , an}, and we define the map φ : ⋃∞ n=1A n → Pfin(A) as the unique map such that φ ∣∣ An = φn, ∀n ≥ 1. Notice now that, since cardAn = an = a, ∀n ≥ 1, it follows that card ( ∞⋃ n=1 An ) = ℵ0 · a = a, which gives card(Rangeφ) ≤ a. But it is clear that {∅} ∪ Rangeφ = Pfin(A), and the fact that Pfin(A) is infinite, proves that cardPfin(A) = card(Rangeφ) ≤ a.  We conclude with a result on the cardinal number c = card R. Proposition B.2. (i) For two real numbers a < b, one has card(a, b) = card[a, b) = card(a, b] = card[a, b] = c. (ii) c = 2ℵ0 .