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The Arithmetic of Cardinal Numbers

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The Arithmetic of Cardinal Numbers The arithmetic of cardinal numbers We wish to extend the familiar operations of addition, multiplication and expo- nentiation of the natural numbers so as to apply to all cardinal numbers. We start with addition and multiplication, then go on consider the unary exponen- tiation function assigning 2 n to each number n and finally the general binary exponentiation operation assigning mn to numbers n, m . Addition and Multiplication Lemma 2 For finite sets A, B , (i) If A ∩ B = ∅ then A ∪ B is finite and |A ∪ B| = |A| + |B|, (ii) A × B is finite and |A × B| = |A| · | B|. Proof see Example 1.9. For sets A, B let A + B = ( {0} × A) ∪ ({1} × B). We call this the disjoint union of A and B. Lemma 3 If A ∼ C and B ∼ D then (i) A + B ∼ C + D, (ii) A × B ∼ C × D. Proof Let f : A ∼ C, g : B ∼ D. Then h : A + B → C + D and e : A × B → C × D defined by h(0 , a ) = (0 , f (a)) for a ∈ A, h (1 , b ) = (1 , g (b)) for b ∈ B e(a, b ) = ( f(a), g (b)) for a ∈ A, b ∈ B are bijections so the result follows. Definition 10 For cardinal numbers n, m let (i) n + m = |A + B| for some/any sets A, B such that n = |A| and m = |B|, (ii) n · m = |A × B| for some/any sets A, B such that n = |A| and m = |B|, Note that the previous lemma ensures that these definitions make sense. Proposition 21 For cardinal numbers n, m and any sets A, B such that A ∩ B = ∅ and n = |A| and m = |B| we have n + m = |A ∪ B|. Proof For disjoint sets A, B we have the obvious bijection h : A + B ∼ A ∪ B defined by h(0 , a ) = a for a ∈ A and h(1 , b ) = b for b ∈ B. 21 Proposition 22 Let n, p, m, k be cardinal numbers such that n ≤ p and m ≤ k. Then (i) n + m ≤ p + k, (ii) n · m ≤ p · k. Proof Let A, B, C, D be some sets with n = |A|, m = |B|, p = |C|, k = |D| and let f : A ¹ C, g : B ¹ D. The function h and e defined as in the proof of Lemma 3 are injections from A + B to C + D and from A × B to C × D respectively, so the result follows. Proposition 23 If m, n, k are cardinal numbers then (i) (m + n) + k = m + ( n + k), (ii) m + n = n + m, (iii) m · (n + k) = m · n + m · k, (iv) (m · n) · k = m · (n · k), (v) m · n = n · m. Proof We will prove (i), leaving the rest as exercise. Let A, B, C be some pairwise disjoint sets with m = |A|, n = |B|, k = |C|. By Proposition 21 we have m + n = |A ∪ B|, n + k = |B ∪ C| so since ( A ∪ B) ∩ C = ∅ and A ∩ (B ∪ C) = ∅, we only need to show that ( A ∪ B) ∪ C ∼ A ∪ (B ∪ C) which is clearly true since the sets are equal. Unary Exponentiation Lemma 4 For each finite set A, the set P ow (A) is finite and |P ow (A)| = 2 |A|. Proof see Example 1.9. Lemma 5 If A ∼ B then P ow (A) ∼ P ow (B). Proof Let f : A ∼ B. Then h : P ow (A) → P ow (B) defined by h(X) = {f(x) | x ∈ X} for X ∈ P ow (A) is easily checked to be a bijection. Definition 11 For each cardinal number n let 2n = |P ow (A)| for some/any set A such that n = |A|. 22 Proposition 24 For cardinal numbers n, m (i) If n ≤ m then 2n ≤ 2m, (ii) 2n+m = 2 n · 2m, (iii) n < 2n, (iv) c = 2 ℵ0 . Proof (i) is an exercise. (ii) Let A and B be disjoint sets with cardinalities n and m respectively. The mapping h : P ow (A) × P ow (B) → P ow (A ∪ B) defined by h(X, Y ) = X ∪ Y is a bijection, hence 2 n · 2m = 2 n+m. (iii) is the Cantor’s theorem. (iv) is the Proposition 19. Binary Exponentiation Recall that for sets A, B we defined AB = {f | f : B→A}. Lemma 6 If A, B are finite sets then so is AB and |AB| = |A||B|. Proof See Example 3.1. Lemma 7 If A ∼ C and B ∼ D then AB ∼ CD. Proof Let f : A ∼ C, g : B ∼ D. Then h : AB → CD defined by h(e) = f ◦ (e ◦ g−1) for e ∈ AB is a bijection. (Draw a picture.) Definition 12 For cardinal numbers n, m let mn = |AB| for some/any sets A, B such that m = |A| and n = |B|. Note that for any set A, A∅ contains exactly one element (the function that does not assign anything to anything), hence n0 = 1 for every cardinal number n. If A is non-empty then ∅A is empty so for n 6= 0 we have 0 n = 0. The following result ensures that binary exponentiation agrees with unary ex- ponentiation. Proposition 25 For any set A, P ow (A) ∼ { 0, 1}A. 23 Proof The function that assigns Z ⊆ A its characteristic function χZ : A → { 0, 1} 1 if a ∈ Z χZ (a) = ½ 0 if a∈ / Z is a bijection. Proposition 26 For cardinal numbers n, m, p, k , (i) If 0 < n and m ≤ k then nm ≤ nk, (ii) If n ≤ p then nm ≤ p m. Proof Exercise. Proposition 27 If m, n, k are cardinal numbers then (i) mn+k = mn · mk, (ii) mn·k = ( mn)k, (iii) (m · n)k = mk · nk. Proof Let A, B, C be sets with |A| = m, |B| = n, |C| = k and B ∩ C = ∅. (i) The function h : AB∪C → (AB) × (AC ) which assigns e ∈ AB∪C the pair hf, g i with f being the restriction of e to B and g being the restriction of e to C is a bijection. (ii) The function h : AB×C → (AB)C which assigns e ∈ AB×C the function f ∈ (AB)C satisfying [f(c)]( b) = e(b, c ) for c ∈ C and b ∈ B is a bijection. (iii) The function h : ( A × B)C → (AC ) × (BC ) which assigns e ∈ (A × B)C the pair hf, g i such that e(c) = hf(c), g (c)i for c ∈ C is a bijection. 24 Ordinal Numbers Cardinal numbers generalize the natural numbers in their capacity of measuring sizes of sets, telling us how many elements a set has. However, natural numbers are also used for counting some things one by one, like putting elements of a set in order or labeling steps in some process. The distinction is perhaps subtle because in the case of finite sets, whichever way we order the elements of a set one after another, we get a finite sequence of the same length, and this length is equal to the number of elements in the finite set. Cantor was led to consider this role of natural numbers early on, when he introduced the notion of the derived set for a given subset of R: the derived set of a set A ⊆ R is the set of its limit points, that is, A without its isolated points. Cantor was interested in what happens when we start with a set and repeat the operation of taking the derived set. There are sets A such that if we keep taking the derived set we obtain a sequence A′, (A′)′,... of ever smaller sets such that the intersection of all of them again has isolated points so we may take the derived set of that again and so on .... For steps in this process Cantor needed a sequence of ordinal numbers which start just as the natural numbers but continue: 0, 1, 2, 3,...,ω, ω +1 , ω +2 ,...,ω + ω, ω ·2+1 ,...,ω ·3, ω ·3+1 ,...,...,ω ·ω,... ω·2 | {z } Cantor called numbers like these ordinal numbers 1. Note that when consid- ering e.g. the set {0, 1, 2,...,ω,ω + 1 ,...,ω + ω} we find that it is equinu- merous with N, although counting one by one up to ω + ω is ‘much longer’ than counting up to ω. Cantor developed the arithmetic of both cardinal and ordinal numbers and they are quite different . Both cardinal and ordinal addi- tion/multiplication/exponentiation generalize the corresponding operations on natural numbers and the same notation (+ , · , something something ) is used for both 2. We will not talk about the arithmetic of ordinal numbers until much later on in the course. 1We will define them precisely later. 2So the meaning has to be understood from the context. 25 Paradoxes We have developed some naive set theory, largely as it was first developed, as a framework accomodating previously known and studied mathematical struc- tures like natural, rational and real numbers. A framework which allowed treat- ing these collections of numbers along with their arbitrary subcollections and collections of their subcollections as a single sort of objects, namely sets. There was a considerable resistance to this process which we have already men- tioned, notably from Leopold Kronecker, who opposed even the publication of Cantor’s papers and his obtaining a mathematical position at Gottingen or Berlin University. On the other hand the freedom afforded by this new frame- work appealed to many. In 1900 David Hilbert declared the set theory to be a genuine part of mathematics, asserting that ”no one can drive us out of the paradise created for us by Cantor” and making the proof or refutation of the Continuum Hypothesis the first on the list of his famous problems.
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