5 Set Operations, Functions, and Counting

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1 5 Set Operations, Functions, and Counting Let N denote the positive integers, N 0 := N {0} be the non-negative integers and Z = N 0 ( N) the positive and negative integers including 0, Q the rational numbers, R the real numbers, and C the complex numbers. We will also use F to stand for either of the fields R or C. 5.1 Set Operations and Functions Notation 5.1 Given two sets X and Y, let Y X denote the collection of all functions f : X Y. If X = N, we will say that f Y N is a sequence with values in Y and often write f n for f (n) and express f as {f n } n=1. If X = {1, 2,..., N}, we will write Y N in place of Y {1,2,...,N} and denote f Y N by f = (f 1, f 2,..., f N ) where f n = f(n). Notation 5.2 More generally if {X α : α A} is a collection of non-empty sets, let X A = X α and π α : X A X α be the canonical projection map defined α A by π α (x) = x α. If If X α = X for some fixed space X, then we will write as X A rather than X A. X α α A Recall that an element x X A is a choice function, i.e. an assignment x α := x(α) X α for each α A. The axiom of choice states that X A provided that X α for each α A. Notation 5.3 Given a set X, let 2 X denote the power set of X the collection of all subsets of X including the empty set. The reason for writing the power set of X as 2 X is that if we think of 2 meaning {0, 1}, then an element of a 2 X = {0, 1} X is completely determined by the set A := {x X : a(x) = 1} X. In this way elements in {0, 1} X are in one to one correspondence with subsets of X. For A 2 X let A c := X \ A = {x X : x / A} and more generally if A, B X let B \ A := {x B : x / A} = A B c. We also define the symmetric difference of A and B by A B := (B \ A) (A \ B). As usual if {A α } α I is an indexed collection of subsets of X we define the union and the intersection of this collection by α I A α := {x X : α I x A α } and α I A α := {x X : x A α α I }. Example 5.4. Let A, B, and C be subsets of X. Then A (B C) = [A B] [A C]. Indeed, x A (B C) x A and x B C x A and x B or x A and x C x A B or x A C x [A B] [A C]. Notation 5.5 We will also write α I A α for α I A α in the case that {A α } α I are pairwise disjoint, i.e. A α A β = if α β. Notice that is closely related to and is closely related to. For example let {A n } n=1 be a sequence of subsets from X and define {A n i.o.} := {x X : # {n : x A n } = } and {A n a.a.} := {x X : x A n for all n sufficiently large}. (One should read {A n i.o.} as A n infinitely often and {A n a.a.} as A n almost always.) Then x {A n i.o.} iff and this may be expressed as Similarly, x {A n a.a.} iff N N n N x A n {A n i.o.} = N=1 n N A n.

2 34 5 Set Operations, Functions, and Counting which may be written as N N n N, x A n {A n a.a.} = N=1 n N A n. We end this section with some notation which will be used frequently in the sequel. Definition 5.6. If f : X Y is a function and B Y, then If A X we also write, f 1 (B) := {x X : f (x) B}. f (A) := {f (x) : x A} Y. Example 5.7. If f : X Y is a function and B Y, then f 1 (B c ) = [ f 1 (B) ] c or to be more precise, To prove this notice that f 1 (Y \ B) = X \ f 1 (B). x f 1 (B c ) f (x) B c f (x) / B x / f 1 (B) x [ f 1 (B) ] c. On the other hand, if A X it is not necessarily true that f (A c ) = [f (A)] c. For example, suppose that f : {1, 2} {1, 2} is the defined by f (1) = f (2) = 1 and A = {1}. Then f (A) = f (A c ) = {1} where [f (A)] c = {1} c = {2}. Notation 5.8 If f : X Y is a function and E 2 Y If G 2 X, let let f 1 E := f 1 (E) := {f 1 (E) E E}. f G := {A 2 Y f 1 (A) G}. Definition 5.9. Let E 2 X be a collection of sets, A X, i A : A X be the inclusion map (i A (x) = x for all x A) and E A = i 1 A (E) = {A E : E E} Exercises Let f : X Y be a function and {A i } i I be an indexed family of subsets of Y, verify the following assertions. Exercise 5.1. ( i I A i ) c = i I A c i. Exercise 5.2. Suppose that B Y, show that B \ ( i I A i ) = i I (B \ A i ). Exercise 5.3. f 1 ( i I A i ) = i I f 1 (A i ). Exercise 5.4. f 1 ( i I A i ) = i I f 1 (A i ). Exercise 5.5. Find a counterexample which shows that f(c D) = f(c) f(d) need not hold. 5.2 Cardinality In this section, X and Y be sets. Definition 5.10 (Cardinality). We say card (X) card (Y ) if there exists an injective map, f : X Y and card (Y ) card (X) if there exists a surjective map g : Y X. We say card (X) = card (Y ) (also denoted as X Y ) if there exists a bijections, f : X Y. Proposition If X and Y are sets, then card (X) card (Y ) iff card (Y ) card (X). Proof. If f : X Y is an injective map, define g : Y X by g f(x) = f 1 and g Y \f(x) = x 0 X chosen arbitrarily. Then g : Y X is surjective. If g : Y X is a surjective map, then Y x := g 1 ({x}) for all x X and so by the axiom of choice there exists f x X Y x. Thus f : X Y such that f (x) Y x for all x. As the {Y x } x X are pairwise disjoint, it follows that f is injective. Theorem 5.12 (Schröder-Bernstein Theorem). If card (X) card (Y ) and card (Y ) card (X), then card (X) = card (Y ). Stated more explicitly; if there exists injective maps f : X Y and g : Y X, then there exists a bijective map, h : X Y. Proof. See the Appendix, Theorem B.10. Exercise 5.6. If X = X 1 X 2 with X 1 X 2 =, Y = Y 1 Y 2 with Y 1 Y 2 =, and X i Y i for i = 1, 2, then X Y. This exercise generalizes to an arbitrary number of factors. Page: 34 job: unanal macro: svmonob.cls date/time: 2-Nov-2012/8:26

3 5.3 Finite Sets Notation 5.13 (Integer Intervals) For n N we let J n := {1, 2,..., n} := {k N : k n}. Definition We say a non-empty set, X, is finite if card (X) = card (J n ) for some n N. We will also write # (X) = n 1 to indicate that card (X) = card (J n ). [It is shown in Theorem 5.17 below that # (X) is well defined, i.e. it is not possible for card (X) = card (J n ) and card (X) = card (J m ) unless m = n.] Lemma Suppose n N and k J n+1, then card (J n+1 \ {k}) = card (J n ). Proof. Let f : J n J n+1 \ {k} be defined by { x if x < k f (x) = x + 1 if x k Then f is the desired bijection. 2 Alternatively. If n = 1, then J 2 = {1, 2} and either J 2 \ {k} = J 1 or J 2 \{k} = {2}, either way card (J 2 \ {k}) = card (J 1 ). Now suppose that result holds for a given n N and k J n+2. If k = {n + 2} we have J n+2 \{k} = J n+1 so card (J n+2 \ {k}) = card (J n+1 ) while if k J n+1 J n+2, then J n+2 \ {k} = (J n+1 \ {k}) {n + 2} J n {n + 2} J n {n + 1} = J n+1. Lemma If m, n N with n > m, then every map, f : J n J m, is not injective. Proof. If f : J n J m were injective, then f Jm+1 : J m+1 J m would be injective as well. Therefore it suffices to show there is no injective map, f : J m+1 J m for all m N. We prove this last assertion by induction on m. The case m = 1 is trivial as J 1 = {1} so the only function, f : J 2 J 1 is the function, f (1) = 1 = f (2) which is not injective. Now suppose m 1 and there were an injective map, f : J m+2 J m+1. Letting k := f (m + 2) we would have, f Jm+1 : J m+1 J m+1 \{k} J m, which would produce and injective map from J m+1 to J m. However this contradicts the induction hypothesis and thus completes the proof. 1 You should read # (X) = n, as X is a set with n elements. 2 The inverse map, g : J n+1 \ {k} J n, to f is given by g (y) = { y if y < k y 1 if y > k. 5.3 Finite Sets 35 Theorem If m, n N, then card (J m ) card (J n ) iff m n. Moreover, card (J n ) = card (J m ) iff m = n and hence card (J m ) < card (J n ) iff m < n. Proof. As J m J n if m n and J m = J n if m = n, it is only the forward implications that have any real content. If card (J m ) card (J n ), there exists an injective map, g : J m J n. According to Lemma 5.16 this can only happen if m n. If card (J n ) = card (J m ), then card (J n ) card (J m ) and card (J m ) card (J n ) and hence m n and n m, i.e. m = n. Proposition If X is a finite set with # (X) = n and S is a non-empty subset of X, then S is a finite set and # (S) = k n. Moreover if # (S) = n, then S = X. Proof. It suffices to assume that X = J n and S J n. We now give two proofs of the result. Proof 1. Let S 1 = S and f (1) := min S 1. If S 2 := S 1 \ {f (1)} is not empty, let f (2) := min S 2 2. We then continue this construction inductively. So if f (k) = min S k k has been constructed, then we define S k+1 := S k \ {f (k)}. If S k+1 we define f (k + 1) := min S k+1 k + 1. As f (k) k for all k that f is defined, this process has to stop after at most n steps. Say it stops at k so that S k+1 =. Then f : J k S is a bijection and therefore S is finite and # (S) = k n. Moreover, the only way that k = n is if f (k) = k at each step of the construction so that f : J n S is the identity map in this case, i.e. S = J n. Proof 2. We prove this by induction on n. When n = 1 the only no-empty subset of S of J 1 is J 1 itself. Thus # (S) = 1 and S = J 1. Now suppose that the result hold for some n N and let S J n+1. If n + 1 / S, then S J n and by the induction hypothesis we know # (S) = k n < n + 1. So now suppose that n + 1 S and let S := S \ {n + 1} J n. Then by the induction hypothesis, S is a finite set and # (S ) = k n, i.e. there exists a bijection, f : J k S and S = J n is k = n. Therefore f : J k+1 S given by f = f on J k and f (k + 1) = n+1 is a bijections from J k+1 to S. Therefore # (S) = k +1 n+1 with equality iff S = J n which happens iff S = J n+1. Proposition If f : J n J n is a map, then the following are equivalent, 1. f is injective, 2. f is surjective, 3. f is bijective. Proof. If n = 1, the only map f : J 1 J 1 is f (1) = 1. So in this case there is nothing to prove. So now suppose the proposition holds for level n and f : J n+1 J n+1 is a given map. Page: 35 job: unanal macro: svmonob.cls date/time: 2-Nov-2012/8:26

4 36 5 Set Operations, Functions, and Counting If f : J n+1 J n+1 is an injective map and f (J n+1 ) is a proper subset of J n+1, then card (J n+1 ) < card (f (J n+1 )) = card (J n+1 ) which is absurd. Thus f is injective implies f is surjective. Conversely suppose that f : J n+1 J n+1 is surjective. Let g : J n+1 J n+1 be a right inverse, i.e. f g = id, which is necessarily injective, see the proof of Proposition By the pervious paragraph we know that g is necessarily surjective and therefore f = g 1 is a bijection. Theorem A subset S N is finite iff S is bounded. Moreover if # (S) = n N then the sup (S) n with equality iff S = J n. Proof. If S is bounded then S J n for some n N and hence S is a finite set by Proposition Also observe that if # (S) = n = sup (S), then S J n and # (S) = n = # (J n ). Thus it follows from Proposition 5.18 that S = J n. Conversely suppose that S N is a finite set and let n = # (S). We will now complete the proof by induction. If n = 1 we have S J 1 and therefore S = {k} for some k N. In particular sup S = k 1 with equality iff S = J 1. Suppose the truth of the statement for some n N and let S N be a set with # (S) = n + 1. If we choose a point, k S, we have by Lemma 5.15 that # (S \ {k}) = n. Hence by the induction hypothesis, sup (S \ {k}) n with equality iff S \ {k} = J n. If sup (S \ {k}) > n then sup (S) sup (S \ {k}) n + 1 as desired. If sup (S \ {k}) = n then S \ {k} = J n therefore S k > n. Hence it follows that sup (S) = k n + 1. Corollary Suppose S is a non-empty subset of N. Then S is an unbounded subset of N iff card (J n ) card (S) for all n N. Proof. If S is bounded we know card (S) = card (J k ) for some k N which would violate the hypothesis that card (J n ) card (S) for all n N. Conversely if card (S) card (J n ) for some n N, then there exists and injective map, f : S J n. Therefore card (S) = card (f (S)) = card (J k ) for some k n. So S is finite and hence bounded in N by Theorem Exercise 5.7. Suppose that m, n N, show J m+n = J m (m + J n ) and (m + J n ) J m =. Use this to conclude if X is a disjoint union of two nonempty finite sets, X 1 and X 2, then # (X) = # (X 1 ) + # (X 2 ). Exercise 5.8. Suppose that m, n N, show J m J n J mn. Use this to conclude if X and Y are two non-empty sets, then # (X Y ) = # (X) # (Y ). 5.4 Countable and Uncountable Sets Definition 5.22 (Countability). A set X is said to be countable if X = or if there exists a surjective map, f : N X. Otherwise X is said to be uncountable. Remark From Proposition 5.11, it follows that X is countable iff there exists an injective map, g : X N. This may be succinctly stated as; X is countable iff card (X) card (N). From a practical point of view as set X is countable iff the elements of X may be arranged into a linear list, X = {x 1, x 2, x 3,... }. Example The integers, Z, are countable. In fact N Z, for example define f : N Z by (f (1), f (2), f (3), f (4), f (5), f (6), f (7),... ) := (0, 1, 1, 2, 2, 3, 3,... ). Lemma If S N is an unbounded set, then card (S) = card (N). Proof. The main idea is that any subset, S N, may be given as an finite or infinite list written in increasing order, i.e. S = {n 1, n 2, n 3,... } with n 1 < n 2 < n 3 <.... If the list is finite, say S = {n 1,..., n k }, then n k is an upper bound for S. So S will be unbounded iff only if the list is infinite in which case f : N S defined by f (k) = n k defines a bijection. Formal proof. Define f : N S via, let S 1 := S and f (1) := min S 1, S 2 := S 1 \ {f (1)} and f (2) := min S 2, S 3 := S 2 \ {f (2)} and f (3) := min S 3. In more detail, let T denote those n N such that there exists f : J n S and {S k S} n k=1 satisfying, S 1 = S, f (k) = min S k and S k+1 = S k \ {f (k)} for 1 k < n. If n T, we may define S n+1 := S n \ {f (n)} and f (n + 1) := min S n+1 in order to show n + 1 T. Thus T = N and we have constructed an injective map, f : N S. Moreover k N S k N \ J n for all n and therefore k N S k =. Thus it follows that f is a bijection. End of Lecture 14, 10/31/2012. The following theorem summarizes most of what we need to know about counting and countability. Theorem The following properties hold; 1. N N is countable and in fact N N N, i.e. there exists a bijective map, h, from N to N N. Page: 36 job: unanal macro: svmonob.cls date/time: 2-Nov-2012/8:26

5 5.4 Countable and Uncountable Sets If X and Y are countable, then X Y is countable. 3. If {X n } n N are countable sets then X := n=1x n is a countable set. 4. If X is countable, then either there exists n N such that X J n or X N. 5. If S N and S J n for some n N then S is bounded. 6. If X is a set and card J n card X for all n N then card N card X. 7. If A X is a subset of a countable set X then A is countable. Proof. We take each item in turn. 1. Put the elements of N N into an array of the form (1, 1) (1, 2) (1, 3)... (2, 1) (2, 2) (2, 3)... (3, 1) (3, 2) (3, 3) and then count these elements by counting the sets {(i, j) : i + j = k} one at a time. For example let h (1) = (1, 1), h(2) = (2, 1), h (3) = (1, 2), h(4) = (3, 1), h(5) = (2, 2), h(6) = (1, 3) and so on. In other words we put N N into the following list form, N N = {(1, 1), (2, 1), (1, 2), (3, 1), (2, 2), (1, 3), (4, 1),..., (1, 4),... }. 2. If f : N X and g : N Y are surjective functions, then the function (f g) h : N X Y is surjective where (f g) (m, n) := (f (m), g(n)) for all (m, n) N N. 3. By assumption there exists surjective maps, f n : N X n, for each n N. Let h (n) := (a (n), b (n)) be the bijection constructed for item 1. Then f : N X defined by f (n) := f a(n) (b (n)) is a surjective map. 4. To see this let f : N X be a surjective map and let g (x) := min f 1 ({x}) for all x X. Then g : X N is an injective map. Let S := g (X), then g : X S N is a bijection. So it remains to show S N or S J n for some n N. If S is unbounded, then S N as we have already seen. So it suffices to consider the case where S is bounded. If S is bounded by 1 then S = {1} = J 1 and we are done. Now assume the result is true if S is bounded by n N and now suppose that S is bounded by n + 1. If n + 1 / S, then S is bounded by n and so by induction, S J k for some k n < n + 1. If n + 1 S, then from above, S \ {n + 1} J k for some k n, i.e. there exists a bijection, f : J k S \ {n + 1}. We then extend f to J k+1 by setting f (k + 1) := n + 1 which shows J k+1 S. 5. We again prove this by induction on n. If n = 1, then S = {m} for some m N which is bounded. So suppose for some n N, every subset S N with S J n is bounded. Now suppose that S N with S J n+1. Then f (J n ) J n and hence f (J n ) is bounded in N. Then max f (J n ) {f (n + 1)} is an upper bound for S. This completes the inductive argument. 6. For each n N there exists and injection, f n : J n X. By replacing X by X 0 := n N f n (J n ) we may assume that X = n N f n (J n ). Thus there exists a surjective map, f : N X by item 3. Let g : X N be defined by g (x) := min f 1 ({x}) for all x X and let S := g (X). To finish the proof we need only show that S is unbounded. If S were bounded, then we would find k N such that J k S X. However this is impossible since card J n card X = card J k would imply n k even though n can be chosen arbitrarily in N. 7. If g : X N is an injective map then g A : A N is an injective map and therefore A is countable. Lemma If X is a countable set which contains Y X with Y N, then X N. Proof. By assumption there is an injective map, g : X N and a bijective map, f : N Y. It then follows that g f : N N is injective from which it follows that g (X) is unbounded. [Indeed, (g f) (J n ) g (X) for all n implies card (J n ) card (g (X)) for all n which implies g (X) is unbounded by Corollary 5.21.] Therefore X g (X) N by Lemma Corollary We have card (Q) = card (N) and in fact, for any a < b in R, card (Q (a, b)) = card (N). Proof. First off Q is a countable since Q may be expressed as a countable union of countable sets; Q = { n } m N m : n Z. From this it follows that Q (a, b) is countable for all a < b in R. As these sets are not finite, they must have the cardinality of N. Theorem 5.29 (Uncountabilty results). If X is an infinite set and Y is a set with at least two elements, then Y X is uncountable. In particular 2 X is uncountable for any infinite set X. Proof. Let us begin by showing 2 N = {0, 1} N is uncountable. For sake of contradiction suppose f : N {0, 1} N is a surjection and write f (n) as (f 1 (n), f 2 (n), f 3 (n),... ). Now define a {0, 1} N by a n := 1 f n (n). By construction f n (n) a n for all n and so a / f (N). This contradicts the assumption that f is surjective and shows 2 N is uncountable. For the general Page: 37 job: unanal macro: svmonob.cls date/time: 2-Nov-2012/8:26

6 38 5 Set Operations, Functions, and Counting case, since Y0 X Y X for any subset Y 0 Y, if Y0 X is uncountable then so is Y X. In this way we may assume Y 0 is a two point set which may as well be Y 0 = {0, 1}. Moreover, since X is an infinite set we may find an injective map x : N X and use this to set up an injection, i : 2 N 2 X by setting i (A) := {x n : n N} X for all A N. If 2 X were countable we could find a surjective map f : 2 X N in which case f i : 2 N N would be surjective as well. However this is impossible since we have already seen that 2 N is uncountable. Corollary The set (0, 1) := {a R : 0 < a < 1} is uncountable while Q (0, 1) is countable. More generally, for any a < b in R, card (Q (a, b)) = card (N) while card (Q c (a, b)) > card (N). Proof. From Section 3.4, the set {0, 1, 2..., 8} N can be mapped injectively into (0, 1) and therefore it follows from Theorem 5.29 that (0, 1) is uncountable. For each m N, let A m := { n m : n N with n < m}. Since Q (0, 1) = m=1x m and # (X m ) < for all m, another application of Theorem 5.26 shows Q (0, 1) is countable. The fact that these results hold for any other finite interval follows from the fact that f : (0, 1) (a, b) defined by f (t) := a + t (b a) is a bijection. Definition We say a non-empty set X is infinite if X is not a finite set. Example Any unbounded subset, S N, is an infinite set according to Theorem Theorem Let X be a non-empty set. The following are equivalent; 1. X is an infinite set, 2. card (J n ) card (X) for all n N, 3. card (N) card (X), 4. card (X \ {x}) = card (X) for some (or all) x X. Proof. 1. = 2. Suppose that X is an infinite set. We show by induction that card (J n ) card (X) for all n N. Since X is not empty, there exists x X and we may define f : J 1 X by f (1) = x in order to learn card (J 1 ) card (X). Suppose we have shown card (J n ) card (X) for some n N, i.e. there exists and injective map, f : J n X. If f (J n ) = X it would follows that card (X) = card (J n ) and would violated the assumption that X is not a finite set. Thus there exists x X \ f (J n ) and we may define f : J n+1 X by f Jn = f and f (n + 1) = x. Then f : J n+1 X is injective and hence card (J n+1 ) card (X) This is the content of Theorem B = 4. Let x 1 X and f : N X be an injective map such that f (1) = x 1. We now define a bijections, ψ : X X \ {x 1 } by { x if x / f (N) ψ (x) = f (i + 1) if x = f (i) f (N). 4. = 1. We will prove the contrapositive. If X is a finite and x X, we have seen that card (X \ {x}) < card (X), namely # (X \ {x}) = # (X) 1. The next two theorems summarizes the properties of cardinalities that have been proven above. Theorem 5.34 (Cardinality/Counting Summary I). Given a non-empty set X, then one and only one of the following statements holds; 1. There exists a unique n N such that card X = card J n. 2. card X = card N, 3. card X > card N. Cases 2. or 3. hold iff card J n card X for all n N which happens iff card N card X. If X satisfies case 1. we say X is a finite set. If X satisfies case 2 we say X is a countably infinite set and if X satisfies case 3. we say X is an uncountably infinite set. Theorem 5.35 (Cardinality/Counting Summary II). Let X and Y be sets and S be a subset of N. 1. If S N is an unbounded set, then card (S) = card (N). 2. If S N is a bounded set then card (S) = card (J n ) for some n N. 3. If {X k } k=1 are subsets of X such that card (X k) card (N), then card ( k=1 X k) card (N). 4. If X and Y are sets such that card (X) card (N) and card (Y ) card (N), then card (X Y ) card (N). 5. card (Q) = card (N) = card (Z). 6. For any a < b in R, card (Q (a, b)) = card (N) while card (Q c (a, b)) > card (N) Exercises Exercise 5.9. Show that Q n is countable for all n N. Exercise Let Q [t] be the set of polynomial functions, p, such that p has rationale coefficients. That is p Q [t] iff there exists n N 0 and a k Q for 0 k n such that n p (t) = a k t k for all t R. (5.1) Show Q [t] is a countable set. k=0 Page: 38 job: unanal macro: svmonob.cls date/time: 2-Nov-2012/8:26

7 Definition 5.36 (Algebraic Numbers). A real number, x R, is called algebraic number, if there is a non-zero polynomial p Q [t] such that p (x) = 0. [That is to say, x R is algebraic if it is the root of a non-zero polynomial with coefficients from Q.] Note that for all q Q, p (t) := t q satisfies p (q) = 0. Hence all rational numbers are algebraic. But there are many more algebraic numbers, for example y 1/n is algebraic for all y 0 and n N. Exercise Show that the set of algebraic numbers is countable. [Hint: any polynomial of degree n has at most n real roots.] In particular, most irrational numbers are not algebraic numbers, i.e. there is still any uncountable number of non-algebraic numbers.

ADVANCED CALCULUS - MTH433 LECTURE 4 - FINITE AND INFINITE SETS

ADVANCED CALCULUS - MTH433 LECTURE 4 - FINITE AND INFINITE SETS 1. Cardinal number of a set The cardinal number (or simply cardinal) of a set is a generalization of the concept of the number of elements