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Counting and Sizes of Sets Learning Goals Counting and Sizes of Sets Learning goals: students determine how to tell when two sets are the same \size" and start looking at different sizes of sets. Two sets are called similar, written A ∼ B if there is a one-to-one and onto function whose domain is A and whose image is B. The function is a one-to-one correspondence and we call such functions bijections. Clearly the inverse of such a function is also such a function, so if A ∼ B then B ∼ A. Also, A ∼ A using the identity function. Other words often used for similar are equinumerous, equipollent, and equipotent. It shouldn't be too hard to prove that if A ∼ B and B ∼ C then A ∼ C. So similarity possesses the qualities of reflexivity, symmetry, and transitivity. We call such relationships equivalence relations. Equivalence relations seek to classify some kind of object. In this case, we are classifying sets by size, and all sets that a similar to A are the same size as A. The fancy word for \size" is cardinality. If n is a positive integer, a set A will have n elements in it if and only if A ∼ f1; 2; 3; : : : ; ng. (This actually requires some proof! We need to show that different values of n give sets that are not similar.) We also say that A has cardinal number n. It should be easy to prove that if f1; 2; : : : ; mg ∼ f1; 2; : : : ; ng then m = n. It should also be clear that if A ∼ ; then A = ;. The empty set will have cardinal number zero. All the sets in this paragraph are called finite. A set that is not finite is called infinite. Instead of actually trying to find the cardinal number of a set, it is often easier to tell if a set is finite or not by: Theorem. Every infinite set is similar to a proper subset of itself; no finite set is similar to a proper subset of itself. This theorem is actually usually used as the definition of finite and infinite. A set A is called countably infinite if it is similar to the set of positive integers. In this way, we have a one-to-one correspondence between A and the positive integers, so that we could \list" the elements of A in order: a1; a2;:::. That is, the set could be written as a sequence. A set is countable if it is either finite or countably infinite. A countably infinite subset is said to have cardinality @0. Another word for countable is denumerable. Theorem. Every subset of a countable set is countable. Proof. Set A be a countable set and B a subset. If B is finite we are done. Otherwise, since A is countable there is a bijective function f : Z+ ! A. Now define an increasing function k on Z+ as follows. Among all positive integers, there will be a non-empty subset of them for which f(z) 2 B. By well-ordering, define k(1) to be this least positive integer. Now assume k(1); k(2); : : : ; k(n) have been defined. Since B is not finite there are elements of B not among f(k(1)); f(k(2)); : : : ; f(k(n)). Let k(n + 1) be the smallest integer t so that f(t) is a member of B and not one of these earlier elements. The composition f ◦k is a one-to-one correspondence of B with Z. Since we have countable sets, there ought to be uncountable (or nondenumerable) sets. Here's one: 1 Theorem. The real numbers R are uncountable. Proof. Consider any one-to-one function from the positive integers to the reals|basically a list of real numbers a(1); a(2);:::. Now each real number has a perfectly good base-10 expansion (+=−)(z):(d1d2d3 :::) where we will ignore the sign and the whole part z and just look at the fractional part. Define a new number x to be the number 0:f1f2f3 ::: where fi = 1 unless di = 1 in which case we set fi = 2. By checking, it can easily be seen that x is not a(1) because the first decimal place is wrong. Nor is x = a(2) because the second decimal place is wrong, and so forth. Thus x can't be anywhere on the list, so our one-to-one function is not onto because x is not in the image. (Note: we don't have to worry about whether we wrote a real number as, say, 0:3999 ::: versus 0:4000 ::: because x can't have an infinite string of nines or zeroes|it has none at all!) Thus no one-to-one correspondence exists from the positive integers to the reals, so the reals are not countable. Note that this proof relies on the ability to write any real number as a decimal expansion. This is not hard to prove (we'll do so later, once we know what the real numbers actually are!). The Algebra of Sets You are undoubtedly familiar with the idea of union and intersection. The union of two sets, A [ B is the set of all things in either A or B (or both), while the intersection A \ B is the set of things only in both. We can extend the idea of union and intersection to any arbitrary (even uncountable) collection of sets. For example, if x is a positive real number and Ax is the interval (−1 − x; 1 + x), then the intersection of all of these is the closed interval [−1; 1]. T That is, x2R+ (−1 − x; 1 + x) = [−1; 1]. If A and B are sets, the complement of A relative to B, denoted B − A is the set of all b's that are not in A: B − A = fx : x 2 B; x 62 Ag. Note that this splits B into two separate parts: the part in A and the part not in A. That is, B = (B \ A) [ (B − A), while (B \ A) \ (B − A) = ;. Because of this relationship, we have DeMorgan's laws. If An is a collection of sets, then: \ [ [ \ B − An = (B − An);B − An = (B − An) and this extends to uncountable collections of A's as well. More on Countability One nifty thing about countability is that it is hard to become uncountable! For example, the set of all integers is countable, though clearly \larger" than just the positive integers. A simple one-to-one correspondence is f(x) = −x=2 is x is even and equals (x − 1)=2 is x is odd. Not even the rational are uncountable. For every rational can be written as p=q for some integer p and some positive integer q that is relatively prime to p. Then f(p=q) = 2p3q if p is positive and 2p5q if p is negative (and, say, 7 if p = 0) gives a one-to-one correspondence between Q and some subset of the positive integers. We know subsets of countable sets are countable, so Q is also countable. 2 It should even be clear that Z × Z is countable by an extension of the above argument. Strange that the real numbers should be so much larger than the rationals! We will find that some real numbers are not rational, in fact there are infinitely many of them between any two rationals. But we will also find infinitely many rationals between any two reals. How can there be more reals than rationals?? We can even put together countable sets and they stay countable. First, a lemma: 1 Theorem. Let fAngn=1 be a countable collection of sets. Define B1 = A1 and Bn = An − Sn−1 S1 S1 k=1 Ak when n > 1. Then the intersection of any two Bi is empty, but n=1 An = n=1 Bn. This theorem is saying that we can construct the B's so that, upto any point, they have all the same elements as the A's but are completely non-redundant. Bascially, each B only adds the things in An but not in any other A. Si Sj−1 Sj−1 Proof. If i < j then Bi ⊆ Ai ⊆ k=1 Ak ⊆ k=1 Ak. But Bj = Aj − k=1 Ak so Bi \ Bj = ; as needed. The equality of the unions is a simple induction on the construction. Theorem. A countable union of countable sets is countable. Proof. In light of the lemma, we only need to prove the fact for disjoint sets. So assume the sets are disjoint, and that the set An has been listed An = fan1; an2;:::g (which may be finite, or even empty). Let x be any element in the union of all these. Then x is in exactly one of these sets, say x 2 Ap. Exactly one, because it must be in at least one to be in the union, and can only be in at most one because the sets are disjoint. Since x 2 Ap we have x = apq for some q. Thus we have a one-to-one correspondence between the union and a subset of Z × Z. But we know subsets of countable sets are countable! This allows another proof that the rationals are countable: they are the union of the obviously countable sets An = fk=n; k 2 Zg for n = 1; 2;:::. The set of polynomials with integer coefficients is also countable. For the linear polyno- mials are simply the countable union over all n of the countable sets of polynomials nx + b, unioned with the same set of functions, but with negative leading coefficients, union the in- tegers for the constant polynomials.
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