c University of Bristol 2014 Further Topics in Analysis: Solutions 2 We assume that all sets in questions 2 and 3 are disjoint. 1. Prove that the following sets are countable. a) The set f2; 4; 6; 8; 10;::: g of positive even numbers. b) The set f2; 3; 5; 7; 11;::: g of prime numbers. c) The set f(x; y) 2 R × Rjx 2 Q; y 2 Qg, where Q is the set of rational numbers. 2 d) The set of spheres on the plane R whose centers have integer coordinates and whose radii are rational. Solution. a) Let X be a set of positive even numbers. Define f : N ! X by f(n) = 2n. The inverse function f −1 : X ! N is f −1(m) = m=2. By Theorem A1.7.61 f is a bijection. b) Obviously P ⊂ N. It is known that the set of primes is infinite. By Lemma 3.10 any subset of a countable set is finite or countable. Therefore P is countable. c) In Example 3.19 we proved that Q is countable. By Lemma 3.17 the cartesian product of two countable sets is countable. Therefore Q × Q is countable. d) Let S be the set of spheres on the plane R2 whose centers have integer coordinates and whose radii are rational. It is clear that every sphere in S is uniquely described by a pair (n; m) 2 N × N which is the centre of the sphere and a positive rational number r 2 Q+ which is the radius of the sphere. Hence there is a one{to{one correspondence between the set S and the set (N × N) × Q+. We know that N×N is countable (Example 3.8) and Q+ ⊂ Q is countable (Example 3.19). Since the cartesian product of two countable sets is countable (Lemma 3.17) we conclude that (N × N) × Q+ is countable. Therefore S is countable. 2. a) Prove that the union of two finite sets is finite; b) Prove that the Cartesian product of two finite sets is finite. Solution. a) Let f : X ! f1; 2;:::;Ng and g : Y ! f1; 2;:::;Mg are bijections. Define the function h : X [ Y ! f1; 2;:::;N + Mg by the formula f(z) if z 2 X; h(z) := g(z) + N if z 2 Y: The inverse h−1 : f1; 2;:::;N + Mg ! X [ Y is f −1(k) if 1 ≤ k ≤ N; h(k) := g−1(k − N) if N + 1 ≤ k ≤ N + M: By Theorem A1.7.6 h is a bijection. So X [ Y is finite and has N + M elements. b) Let f : f1; 2;:::;Ng ! X and g : f1; 2;:::;Mg ! Y are bijections. Then X and Y can be arranged in finite lists X = fx1; x2; : : : ; xN g;Y = fy1; y2; : : : ; yM g: 1Ax.y.z refers to Lecture Notes in Analysis by Vitali Liskevich. You can download them on the Web: http://www.maths.bris.ac.uk/ ∼ majm/analysis1/ Write elements of X × Y = fx1; x2; : : : ; xN g × fy1; y2; : : : ; yM g as a rectangular array (figure on the left) and then read them off along the horizontal lines (figure on the right): (x1; y1)(x1; y2) ::: (x1; yM ) 1 2 :::M (x ; y )(x ; y ) ::: (x ; y ) M + 1 M + 2 ::: 2M 2 1 2 2 2 M h(xn;ym) . −−−−−−! . (xN ; y1)(xN ; y2) ::: (xN ; yM ) (N −1)M + 1 (N −1)M + 2 :::NM In this process every element of X × Y is counted and none of them is counted twice. This means that we established a bijection between h : X × Y ! f1; 2;:::;NMg. So the cartesian product X × Y is finite and has NM elements. Such bijection h : X × Y ! f1; 2;:::;NMg can be represented analytically by the formula h(xn; ym) := (n − 1)M + m. 3. a) Prove that the union of a finite and a countable set is countable. b) Prove that the union of a finite and an infinite set is infinite. c) Prove that the union of a finite number of countable sets is countable. d) Prove that the cartesian product of a finite number of countable sets is countable. e) Prove that the cartesian product of a finite and a countable set is countable. Solution. a) Let f : X ! f1; 2;:::;Ng and g : Y ! N are bijections. Define h : X [ Y ! N by the formula f(z) if z 2 X; h(k) := g(z) + N if z 2 Y: The inverse h−1 : N ! X [ Y is f −1(k) if 1 ≤ k ≤ N; h(k) := g−1(k − N) if k ≥ N + 1: By Theorem A1.7.6 h is a bijection. So X [ Y is countable. b) Let X be a finite set and Y be an infinite set. Then Y ⊂ (X [ Y ), that is X [ Y contains an infinite subset. Therefore X [ Y is infinite. c) Let X1;X2;:::;XN be countable sets. First we prove first that X1 [X2 is countable. Let N1 and N2 be two "copies" of the set of positive natural numbers N. Let f1 : N1 ! X1 and f2 : N2 ! X2 are bijections. Define h : N1 [ N2 ! X1 [ X2 by the formula f (k) if x 2 ; h(k) := 1 N1 f2(k) if x 2 N2: −1 The function h is a bijection with inverse h : X1 [ X2 ! N1 [ N2 given by −1 −1 f1 (y) if y 2 X1; h (y) := −1 f2 (y) if y 2 X2: Now let ' : N ! N1 × N1 be a bijection (Example 3.9). Then the composition h ◦ ' : N ! X1 [ X2 is a bijection. Therefore X1 [ X2 is countable. Now we proceed by induction. If k = 1 then the set X1 is countable; Assume that X1 [ X2 [···[ Xk (k 2 N, 1 < k < N) is countable. Then X1 [ X2 [···[ Xk [ Xk+1 = (X1 [ X2 [···[ Xk) [ Xk+1 is countable. 2 We conclude that X1 [ X2 [···[ XN is countable. d) Let X1;X2;:::;XN are countable sets. Define Yk := X1 × X2 × · · · × Xk (k = 1;:::;N). Thus YN := X1 × X2 × · · · × XN We need to prove that YN is countable. By induction: If k = 1 then the set Y1 = X1 is countable; Assume that Yk (k 2 N, 1 ≤ k < N) is countable; Then Yk+1 = (X1 × · · · × Xk) × Xk+1 = Yk × Xk+1 where Yk and Xk+1 are both countable. By Lemma 1.17 the cartesian product of countable sets is countable. So Yk+1 is countable. We conclude that X1 × X2 × · · · × XN is countable. e) Let f : f1; 2;:::;Ng ! X and g : f1; 2; : : : ; m; : : : g ! Y are bijections. Then X and Y can be arranged in the lists: X = fx1; x2; : : : ; xN g;Y = fy1; y2; : : : ; ym;::: g: Write elements of X × Y as a rectangular array (figure on the left) and then read them off along the vertical lines (figure on the right): (x1; y1)(x1; y2) ::: (x1; ym) ::: 1 N + 1 ::: (m−1)N + 1 ::: (x ; y )(x ; y ) ::: (x ; y ) ::: 2 N + 2 ::: (m−1)N + 2 ::: 2 1 2 2 2 m h(xn;ym) . −−−−−−! . (xN ; y1)(xN ; y2) ::: (xN ; ym) ::: N 2N : : : mN : : : In this process every element of X × Y is counted and none of them is counted twice. This means that we established a bijection between X × Y and N. So X [ Y is countable. Such bijection h : X × Y ! N can be represented analytically by the formula h(xn; ym) := (m − 1)N + n. 4. Let X be a countable set. Prove the following statements. a) If there exists an injection f : Y ! X then the set Y is at most countable. b) If there exists a surjection f : X ! Y then the set Y is at most countable. Solution. a) Let f : Y ! X be an injection. Then Im(f) is a subset of X. Since X is countable by Lemma 3.11 we conclude that Im(f) is at most countable (finite or countable). But (compare the proof of Theorem A1.7.6) the function f : Y ! Im(f) is a bijection. Therefore Y is at most countable. b) Let f : X ! Y be a surjection. Then (8y 2 Y )[f −1(y) 6= ;]. For each y 2 Y select one −1 (and only one!) element g(y) 2 f (y). Then g : Y ! X is a function. Since y1 6= y2 implies −1 −1 f (y1) \ f (y1) = ; we conclude that g : Y ! X is an injection. By Q4, b) above we conclude that Y is at most countable. 5. Let functions f : X ! Y and g : Y ! Z are bijections. Prove that the composition g ◦ f : X ! Z is a bijection. Solution. By Theorem A1.7.6 a function is a bijection if and only if there exists an inverse function. Let f −1 : Y ! X and g−1 : Z ! Y are inverse functions. Then f −1 ◦ g−1 : Z ! X is an inverse function to f ◦ g : X ! Z. Hence f ◦ g : X ! Z is a bijection. 3.