Further Topics in Analysis: Solutions 2

Home , Cartesian product, Union (set theory)

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 {2, 4, 6, 8, 10,... } of positive even numbers. b) The set {2, 3, 5, 7, 11,... } of prime numbers. c) The set {(x, y) ∈ R × R|x ∈ Q, y ∈ Q}, 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) ∈ N × N which is the centre of the sphere and a positive rational number r ∈ 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 → {1, 2,...,N} and g : Y → {1, 2,...,M} are bijections. Define the function h : X ∪ Y → {1, 2,...,N + M} by the formula f(z) if z ∈ X, h(z) := g(z) + N if z ∈ Y.

The inverse h−1 : {1, 2,...,N + M} → 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 : {1, 2,...,N} → X and g : {1, 2,...,M} → Y are bijections. Then X and Y can be arranged in ﬁnite lists

X = {x1, x2, . . . , xN },Y = {y1, y2, . . . , yM }.

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 = {x1, x2, . . . , xN } × {y1, y2, . . . , yM } 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 → {1, 2,...,NM}. So the cartesian product X × Y is ﬁnite and has NM elements. Such bijection h : X × Y → {1, 2,...,NM} 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 → {1, 2,...,N} and g : Y → N are bijections. Define h : X ∪ Y → N by the formula f(z) if z ∈ X, h(k) := g(z) + N if z ∈ 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 ∈ , h(k) := 1 N1 f2(k) if x ∈ N2.

−1 The function h is a bijection with inverse h : X1 ∪ X2 → N1 ∪ N2 given by −1 −1 f1 (y) if y ∈ X1, h (y) := −1 f2 (y) if y ∈ X2.

Now let ϕ : N → N1 × N1 be a bijection (Example 3.9). Then the composition h ◦ ϕ : 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 ∈ 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 ∈ 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 : {1, 2,...,N} → X and g : {1, 2, . . . , m, . . . } → Y are bijections. Then X and Y can be arranged in the lists:

X = {x1, x2, . . . , xN },Y = {y1, y2, . . . , ym,... }.

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 (∀y ∈ Y )[f −1(y) 6= ∅]. For each y ∈ Y select one −1 (and only one!) element g(y) ∈ 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.