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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 class^1 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 card A = a we understand that A belongs to this class, and for another set B we write card B = a, exactly when B has the same cardinality as A. In this case we write card B = 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 ℵ 0.
Definition. Let a and b be cardinal numbers. We write a ≤ b if there exist sets A ⊂ B with card A = a and card B = b. This is equivalent to the fact that, for any sets A and B, with card A = a and card B = 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 card A = a and card B = 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
A 0 = A r g(B) and B 0 = A r 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) = A r A 0 and Bm = f (Am− 1 ) ⊂ f (A) = B r B 0. 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 = A rn≥ 0 An and D = B r
n≥ 0 Bn. 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) = B rB 0 , 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 B 0 = 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≥ 0 Bn^ → 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≥ 0 An^ →^ 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+
= 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≥ 0 An f (x) if x ∈ A r
n≥ 0 An^ =^ 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 card A = a and card B = 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.
2004 APPENDICES
- a + 0 = a,
- a · b = b · a,
- (a · b) · d = a · (b · d),
- a · 1 = a,
- a · (b + d) = (a · b) + (a · d),
- (a · b)d^ = (ad) · (bd),
- ab+d^ = (ab) · (ad),
- (ab)d^ = (ab·d,
for all cardinal numbers a, b, d ≥ 1.
Remark B.2. The order relation ≤ is compatible with all the operations, in the sense that, if a 1 , a 2 , b 1 , and b 2 are cardinal numbers with a 1 ≤ a 2 and b 1 ≤ b 2 , then
- a 1 + b 1 ≤ a 2 + b 2 ,
- a 1 · b 1 ≤ a 2 · b 2 ,
- ab 1 1 ≤ ab 22.
Proposition B.1. Let a ≥ 1 be a cardinal number. (i) If A is a set with card A = a, and if we define
P(A) = {B : B subset of A},
then 2 a^ = card P(A). (ii) a < 2 a.
Proof. (i). Put
P = { 0 , 1 }A^ =
f : f function from A to { 0 , 1 }
so that 2a^ = card P. We need to define a bijection φ : P → P(A). We take
φ(f ) = {a ∈ A : f (a) = 1}, ∀ f ∈ P.
It is clear that, since a function f : A → { 0 , 1 } is completely determined by the set {a ∈ A : f (a) = 1}, the map φ is indeed bijective. (ii). The map A 3 a 7 −→ {a} ∈ P(A) is clearly injective. This prove the inequality a ≤ 2 a. We now prove that a 6 = 2a, by contradiction. Assume there is a bijection θ : A → P(A). Define the set
B = {a ∈ A : a 6 ∈ θ(a)},
and choose b ∈ A such that B = θ(b). If b ∈ B, then by construction we get b 6 ∈ θ(b) = B, which is impossible. If b 6 ∈ B, we have b 6 ∈ θ(b), which forces b ∈ B, again an impossibility.
We now discuss the properties of these operations, when infinite cardinal num- bers are used.
Lemma B.1 (Properties of ℵ 0 ). (i) For any infinite cardinal number a, one has the inequality ℵ 0 ≤ a. (ii) ℵ 0 + ℵ 0 = ℵ 0 ; (iii) ℵ 0 · ℵ 0 = ℵ 0 ;
B. Cardinal Arithmetic 2005
Proof. (i). Let a be an infinite cardinal number, and let A be an infinite set A, with card A = a. Since for every finite subset F ⊂ A, there exists some x ∈ A r 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 card B = ℵ 0 , so the inclusion B ⊂ A gives the desired inequality. (ii). Consider the sets A 0 = {n ∈ N : n, even} and A 1 = {n ∈ N : n, odd}.
Then clearly card A 0 = card A 1 = ℵ 0 , and the equality A 0 ∪ A 1 = N gives
ℵ 0 + ℵ 0 = card A 0 + card A 1 = card(A 0 ∪ A 1 ) = card N = ℵ 0.
(iii). Take the set P = N × N, so that ℵ 0 · ℵ 0 = card P. It is obvious that card P ≥ ℵ 0. 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≥ 1 Bn^ =^ 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 card A = a. Use Zorn Lemma, to find a maximal non-empty family {Ai : i ∈ I} of subsets of A with
(a) card Ai = ℵ 0 , 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 i 0 ∈ I then obviously card(Ai 0 ∪ B) = ℵ 0 , so if we replace Ai 0 with Ai 0 ∪ B, we will still have the above properties (a) and (b), but also A =
i∈I Ai.^ This proves that^ a^ = card^ A^ =^ ℵ^0 ·^ d, where^ d^ = card^ I.^ In other words, we have a = card(N × I). Consider then the sets
C 0 = {n ∈ N : n even} and C 1 = {n ∈ N : n odd},
B. Cardinal Arithmetic 2007
Corollary B.1. If a is an infinite cardinal number, and if b is a cardinal number with 2 ≤ b ≤ 2 a, then
ba^ = 2a.
Proof. We have 2 a^ ≤ 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 card A = a, and define
Pfin(A) = {F ∈ P(A) : F finite}.
Then card Pfin(A) = a.
Proof. First of all, the map A 3 a 7 −→ {a} ∈ Pfin(A) is injective, so a ≤ card Pfin(A). We now prove the other inequality. For every integer n ≥ 1, let An^ denote the n-fold cartesian product. We treat the sequence A^1 , A^2 ,... as pairwise disjoint. For every n ≥ 1 we define the map
φn : An^ → Pfin(A),
by
φ(a 1 ,... , an) = {a 1 ,... , an},
and we define the map φ :
n=1 A n (^) → Pfin(A) as the unique map such that
φ
An^ =^ φn,^ ∀^ n^ ≥^ 1. Notice now that, since card An^ = an^ = a, ∀ n ≥ 1 ,
it follows that
card
n=
An
= ℵ 0 · a = a,
which gives
card(Range φ) ≤ a.
But it is clear that
{∅} ∪ Range φ = Pfin(A),
and the fact that Pfin(A) is infinite, proves that
card Pfin(A) = card(Range φ) ≤ 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 = 2ℵ^0.
2008 APPENDICES
Proof. (i). It is clear that, since (a, b) is infinite, we have card[a, b] = 2 + card(a, b) = card(a, b).
The inclusions (a, b) ⊂ [a, b) ⊂ [a, b] and (a, b) ⊂ (a, b] ⊂ [a, b], combined with the Cantor-Bernstein Theorem, immediately give
card[a, b) = card(a, b] = card(a, b).
Finally, the bijection
(a, b) 3 t 7 −→ tan
π(2t − a − b) 2(b − a)
∈ R
shows that card(a, b) = c. (ii). The proof of this result uses a certain construction, which is useful for many other purposes. Therefore we choose to work in full generality. Consider the set T = { 0 , 1 }ℵ^0 =
a = (αn)n∈N : αn ∈ { 0 , 1 }, ∀ n ∈ N
so 2ℵ^0 = card P. For any real number r ≥ 2, we define the map φr : T → [0, 1] by
φ(a) = (r − 1)
∑^ ∞
n=
αn rn^
, ∀ a = (αn)n∈N ∈ T.
The maps φr , r ≥ 2 are “almost” injective. To clarify this, we define the set
T 0 =
a = (αn)n∈N ∈ T : the set {n ∈ N : αn = 0} is infinite
Note that
T r T 0 =
(αn)n∈N ∈ T : there exists N ∈ N, such that αn = 1, ∀ n ≥ N
Clearly φ is surjective. In fact φ is “almost” bijective.
Claim 1: Fix r ≥ 2. For elements a = (αn)n∈N, b = (βn)n∈N ∈ T 0 , the following are equivalent (∗) φr (a) > φr (b); (∗∗) there exists k ∈ N, such that alphak > βk, and αj = βj , for all j ∈ N with j < k.
We first prove the implication (∗∗) ⇒ (∗). If a, b ∈ T 0 satisfiy (∗∗), then
(4) φr (a) − φr (b) = r − 1 rk^
∑^ ∞
n=k+
αn − βn rn^
r − 1 rk^
− (r − 1)
∑^ ∞
n=k+
βn 2 n^
Notice now that there are infinitely many indices n ≥ k + 1 such that βn = 0. This gives the fact that ∑∞
n=k+
βn rn^
∑^ ∞
n=k+
rn^
(r − 1)rk^
so if we go back to (4) we get
φr (a) − φr (b) ≥
r − 1 rk^
− (r − 1)
∑^ ∞
n=k+
βn rn^
r − 1 rk^
rk^
r − 2 rk^
so in particular we get φr (a) > φr (b. Conversely, if φr (a) > φr (b), we choose k = min{n ∈ N : αn 6 = βn}.
