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Cardinality Equivalent Sets Definition: We Say That Two Sets Are Equivalent (Sometimes Called Equipotent), Denoted by a ~ B Iff There Exists a Bijection F: a → B

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Cardinality Equivalent Sets Definition: We Say That Two Sets Are Equivalent (Sometimes Called Equipotent), Denoted by a ~ B Iff There Exists a Bijection F: a → B Cardinality Equivalent Sets Definition: We say that two sets are equivalent (sometimes called equipotent), denoted by A ~ B iff there exists a bijection f: A → B. This is an equivalence relation on the class of all sets: A ~ A for all sets A. (I : A → A is a bijection for all sets A) A If A ~ B then B ~ A. (If f:A → B is a bijection, f -1:B → A is also) If A ~ B and B ~ C then A ~ C. (If f:A → B is a bijection and g:B → C is a bijection, then g⊙f: A → C is a bijection) The equivalence classes under this relation are called cardinalities. Finite Sets We define some special sets of natural numbers: ℕ ℕ ℕ ℕ = {1} = {1,2} = {1,2,3}, ..., = {1,2,...,m} 1 2 3 m These sets are sometimes called initial segments. Definition: A set A is finite iff A = ∅ or A ~ ℕ for some m ∈ℕ. m A set is infinite iff it is not finite. We say that ∅ is of cardinality 0. If A ~ ℕ we say that A is of cardinality m. m ℕ This makes sense since A and are in the same equivalence m class, i.e., "are of the same cardinality". Finite Cardinalities When two finite sets are of the same cardinality, say of cardinality k, then by definition, there is a bijection between them, and from each of them onto ℕ . k Since a bijection sets up a one-to-one pairing of the elements in the domain and codomain, it is easy to see that all the sets of cardinality k, must have the same number of elements, namely k. Indeed, for any set that has k elements we can set up a bijection between that set and ℕ . So, for finite sets, all the sets in the k same cardinality have the same number of elements. This is why we often refer to a cardinality as a cardinal number. ℕ Since the definition of an infinite set is a "negation", we should expect that most proofs about them will use contradiction methods. Theorem: ℕ is an infinite set. Pf: BWOC assume that ℕ is a finite set. ℕ ℕ Then there exists a bijection f: → for some k. k Let n = f(1) + f(2) + ... + f(k) + 1. Then n is a natural number (being the sum of natural numbers) and n > f(i) for any i. So n is in the codomain of f, but since it can not equal any f(i), it is not in the Rng(f). So, f is not onto →← Denumerable Sets Definition: A set is denumerable iff it is of the same cardinality as ℕ. (Also known as countably infinite.) The cardinality of the denumerable sets is denoted ℵ which is read 0 as "aleph naught" or "aleph null". (ℵ is the first letter of the Hebrew alphabet.) One may be tempted to say, in analogy with finite sets, that all denumerable sets have the same number of elements, or all denumerable sets have ℵ elements. As our next example will 0 show, you probably should avoid this temptation. ℕ~2ℕ Let 2ℕ denote the set of all even natural numbers. Theorem: 2ℕ is denumerable. Pf: To prove this we must find a bijection f: ℕ → 2ℕ. Our candidate will be f(x) = 2x with domain ℕ. This f is one-to-one: Suppose f(x) = f(y). Then 2x = 2y, so x=y. This f is onto: Let y be an arbitrary element of the codomain 2ℕ Since y is an even natural number, y = 2k for some k in ℕ. Thus, f(k) = 2k = y, and f is onto. f is a bijection, and so, ℕ~2ℕ i.e., 2ℕ is denumerable. So, the number of even natural numbers is the same as the number of all natural numbers ????? So, what gives? On the one hand, our intuition tells us that this last statement can't be correct. Every even natural number is in ℕ, but ℕ also has odd natural numbers ... so ℕ must have more elements than 2ℕ ! But, the bijection between these sets, pairs up the elements exactly, so there are just as many elements in one set as in the other! Something has to give ... these are contradictory results. Realize that intuition is built up from experience. Our direct experience with sets is limited to finite sets ... so our intuition is usually ok for these. But we have no direct experience with infinite sets (for instance, you've never counted the elements in one), and our intuition leads us astray when it comes to infinite sets! Denumerable + 1 = Denumerable Theorem: If A is denumerable and x ∉ A then A ∪ {x} is denumerable. Pf: Since A is denumerable there exists a bijection g: ℕ → A. Now define f: ℕ → A ∪ {x} by: f(1) = x f(2) = g(1) f(3) = g(2) .... and in general f(n) = g(n-1) for n > 1. f is one-to-one since no two images of f can be the same (no two images of g are the same, and none of them is x) f is onto because g is onto A and x is certainly in Rng(f). So, f is a bijection. Listable Sets? This is a Cherowitzo special definition – you will not find this anywhere in the literature. A denumerable set is one whose elements can written in a list (an infinite list) where all the elements appear somewhere and no element appears twice. If you can create such a list of elements of the set, then you can define a function whose arguments are the elements of the set and whose values are the positions in the list where the elements appear. This function is a bijection between the set and ℕ ... thus proving that the set is denumerable. Thus, you can prove that a set is denumerable by creating this list. So, maybe, denumerable sets should be called listable sets. Union of Denumerable Sets Theorem: If A and B are disjoint denumerable sets then A ∪ B is denumerable. Pf: Since A is denumerable we can list its elements as: a , a , a , ... 1 2 3 Since B is denumerable we can also list its elements as: b , b , b , .... 1 2 3 We can now form a list of the elements of A ∪ B this way: a , b , a , b , a , b , ..... 1 1 2 2 3 3 Clearly, every element of A ∪ B appears exactly once on this list, so A ∪ B is denumerable. More formally, we can define the bijection f from A ∪ B onto ℕ by: − = = 2k 1 when x a k f x { = } . 2k when x bk ℤ is denumerable We can apply the theorems we have just proved to obtain this result. First, notice that -ℕ is denumerable. (Consider f(x) = -x ). Since 0 ∉ ℕ we have that A = ℕ ∪ {0} is denumerable. Then ℤ = ℕ ∪ {0} ∪ -ℕ = A ∪ -ℕ is denumerable. ℚ is denumerable From the last chapter we know that ℕ×ℕ is denumerable since the function f: ℕ×ℕℕ given by f(n,m) = 2n-1(2m – 1) is a bijection. Theorem: If A and B are denumerable sets, then A×B is denumerable. Pf: Since A and B are denumerable, there exist bijections f:Aℕ and g:Bℕ. Now consider the function h:A×Bℕ×ℕ given by h(a,b) = (f(a), g(b)). We show that h is a bijection. Suppose that h(a,b) = h(c,d) ⇒ (f(a),g(b)) = (f(c),g(d)) ⇒ f(a) = f(c) and g(b) = g(d). Since f is an injection a = c, and since g is an injection b = d. Now, let (c,d) be an arbitrary element of ℕ×ℕ. Since f is onto, there is an a ∈ ℕ with f(a) = c, and since g is onto there is a b ∈ ℕ with g(b) = d. So, h(a,b) = (f(a),g(b)) = (c,d), so h is a bijection. The composition of h with f above shows that A×B is denumerable. ℚ is denumerable Since ℤ is denumerable, the last theorem shows that ℤ×ℤ is denumerable. Theorem: Any infinite subset of a denumerable set is denumerable. Pf: Since the original set is denumerable, its elements can be put into a list. By passing through this list and removing any element which is not in the subset, we obtain a list of the elements of the subset. By identifying each fraction p/q with the ordered pair (p,q) in ℤ×ℤ we see that the set of fractions is denumerable. By identifying each rational number with the fraction in reduced form that represents it, we see that ℚ is denumerable. Countable Sets Definition: A countable set is a set which is either finite or denumerable. In most theorems involving denumerable sets the term denumerable can be replaced by countable. Proofs involve extending the proofs for denumerable sets by checking the cases when one or more of the sets involved are finite. Thus: Theorem: If A is countable and x ∉ A then A ∪ {x} is countable. Theorem: If A and B are disjoint countable sets then A ∪ B is countable. Theorem: If A and B are countable sets, then A×B is countable. Theorem: Any subset of a countable set is countable. Countable Union Theorem: Let A be a countable family of countable sets. Then ∪A is countable. This theorem while plausible can not be proved from the generally accepted axioms of set theory (Zermelo-Fraenkel axioms).
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