Math 295. Solutions to Homework 3 (1) Let S be a set. A function f : S ! S is called an involution provided that f ◦ f (s) = s for all s 2 S. Show: If g : S ! S is an involution, then g is bijective. We first show that g is surjective. Choose s 2 S. We need to produce and s0 2 S so that g(s0) = s. Let s0 = g(s). Then g(s0) = g(g(s)) = s: Now for injectivity. Suppose s; s0 2 S with g(s) = g(s0). We need to show that s = s0. Well, s = g(g(s)) = g(g(s0)) = s0: (2) Suppose a; b 2 R with a < b. Show that there exist c; d 2 R with c 2 Q and d 62 Q so that a < c < d < b. 1 First we prove that c exists. By the archimedian property of R, there is a natural number n such that n < (b−a). Thus multiplying by n, we have bn − an > 1. By a theorem proved in class, we can find a natural number m such that an < m < bn. Dividing by n, we see that we can take c = m . n p It remains to find d 2 R n Q such that c < d < b. We know that 2 is irrational, and if q 2 Q is not zero, then p p p −1 q · 2 is not rational. Since 2 > 0, we have 2 > 0 and p p c= 2 < b= 2: Thus, from what we just proved, there exists q 2 Q so that p p c= 2 < q < b= 2: p It follows that c < q 2 < b. (3) Show that every nonempty bounded above subset of N has a maximal element. That is, if ; 6= S ⊂ N is bounded above, then there is an element t 2 S so that t ≥ s for all s 2 S. Let S ⊂ N be a bounded above, nonempty subset of N. Since S is bounded above and nonempty, it has a supremum, call is σ. It will be enough to show that σ belongs to S. Since the set of natural numbers is unbounded, the set T := fn 2 N j n ≥ σg is nonempty. By well-ordering, T has a minimal element, call it j. Thus, since σ is positive, we have j−1 < σ ≤ j. We claim that σ = j. Since any element s of S is both a natural number and less than or equal to σ, we conclude, from the fact that there are no natural numbers strictly between (j − 1) and j, that either s ≤ (j − 1) or s = j. Thus, we have two cases: either there is an s for which s = j, or, for all s 2 S we have s ≤ (j − 1). In the former case, we must have j = s ≤ σ ≤ j; so σ = s 2 S. In the latter case, we conclude that for all s 2 S we have s ≤ (j − 1) < σ ≤ j: But this means that σ is not the least upper bound, a contradiction. (4) Fun with maps. (a) Suppose A is a set and f : N ! A is an injection. Show that there is a bijection between A and a proper subset of A. (That is, there is a subset A0 of A and a map g : A ! A0 such that A0 6= A and g is bijective.) For all n 2 N, define an 2 A by an := f(n). Let =(f) denote the image of f in A. Thus, =(f) = 0 fa1; a2;:::g. Let A = fa2; a3; a4;:::g [ (A n =(f)). 0 0 We claim that a1 62 A , so A is a proper subset of A. Since a1 = f(1), we have a1 62 An=(f). Consequently, 0 if a1 2 A , then it must be in the set fa2; a3;:::g: Thus, there is an n 2 N with n > 1 so that a1 = an = f(n). Since f is injective, the only way an can be a1 0 is if n = 1. Thus, a1 62 A . Since every a 2 A belongs to exactly one of =(f) or A n =(f), we can define the function g : A ! A0 by ( a if a 2 =(f) and a = a , g(a) = n+1 n a if a 2 A n =(f). We need to show that g is bijective. We first show that g is surjective. Choose a0 2 A0. If a0 2 [A n =(f)] ⊂ A0, then g(a0) = a0. Otherwise, if 0 0 a 2 fa2; a3; a4;:::g, then a = an for some n 2 N with n > 1. Since n > 1, we have (n − 1) 2 N and 0 g(a(n−1)) = an = a . Finally, we show that g is injective. Suppose a; a¯ 2 A and g(a) = g(¯a). We want to conclude that a =a ¯. Suppose first that g(a) 2 [A n =(f)] ⊂ A0. In this case, we have a = g(a) = g(¯a) =a: ¯ Now suppose that g(a) 62 [An=(f)]. We must have g(a) 2 fa2; a3;:::g. Thus, since g(¯a) = g(a), we have g(¯a) 2 fa2; a3;:::g as well. Thus, there are n; m 2 N for which a = an and a¯ = am. We have f(n + 1) = g(a) = g(¯a) = f(m+ 1). Consequently, since f is injective, we conclude that n + 1 = m + 1, or n = m. Hence a = an = am =a ¯. (b) For n 2 N, let Nn := fk 2 N: k ≤ ng. Suppose `; m 2 N. Show that N` = Nm if and only if ` = m. Serious hint: One direction is easy; for the other direction, use induction on max(`; m). If n = m, then Nn = Nm. Now suppose that Nn = Nm. Both Nn and Nm are nonempty bounded above subset of N. Thus, from problem number 2, they have maximal elements. In the first case, the maximal element is n and in the second case the maximal element is m. We have shown in class that supremums are unique (when they exist). Since a maximal element is a supremum, we conclude that n is the unique maximal element of Nn = Nm. Since m is the unique maximal element of Nm = Nn, we conclude that n = m. (c) Suppose n 2 N. Show that if S is a nonempty subset of Nn, then there is a bijective map from S to N` for some ` ≤ n. Hint: use induction. Let P (n) be the statement: If S is a nonempty subset of Nn, then there is a bijective map from S to N` for some ` ≤ n. Define T ⊂ N by T := fn 2 N: P (n) is trueg: We need to show that T is an inductive subset of N. Since the only nonempty subset of N1 is N1 itself, we have 1 2 T . Suppose n 2 T . Let S ⊂ Nn+1 be nonempty. If (n + 1) 62 S, then S ⊂ Nn and since P (n) is true, we conclude that there is an ` ≤ n < (n + 1) 0 and a bijective map from S to N`. If (n + 1) 2 S, then consider the set S = S n fn + 1g. We note that 0 0 S is a subset of Nn, so there is an ` ≤ n < (n + 1) and a bijective map f : S ! N`. Define the function g : S ! N`+1 by ( f(s) if s 2 S0 g(s) = ` + 1 if s = n + 1 Since f is bijective, the function g is clearly bijective. (d) Compare your answers to parts (4a) and (4c). We see in these two problems the difference between finite sets and infinite sets. A finite set can not be bijectively mapped to a proper subset of itself while an infinite set can be bijectively mapped to a proper subset of itself. (5) What is a finite set anyway? (a) A set A is said to be finite if there is an n 2 N and a bijective function f : Nn ! A. Conclude from the previous problem that n is uniquely determined by A. We say that A has n elements. Suppose n; m 2 N and we have bijective maps fn : Nn ! A and fm : Nm ! A. We want to show n = m. We begin by establishing the general fact: if B and C are sets and g : B ! C is a bijection, then g−1 is a bijective function from C to B. Since g is injective, we know that g−1 : g(B) ! B is a function. Since g is surjective, we have g(B) = C. Thus, g−1 : C ! B. We now show that g−1 bijective. For surjective: −1 Choose b 2 B. Note that the image of g(b) under g is b. For injective: Suppose c1 and c2 are in C and −1 −1 −1 −1 −1 g (c1) = g (c2). We then have c1 = g(g (c1)) = g(g (c2)) = c2; so g is injective. We now show that if h: Nm ! Nn is a bijection, then m = n. We proceed by induction on max (n; m). More specifically, let S := f` 2 N: ` ≥ max (m; n) implies that if h: Nn ! Nm is a bijection, then n = mg: We shall show that S is inductive. If max (n; m) = 1, then n = m = 1; so 1 2 S. Suppose ` 2 S. We need to show that ` + 1 2 S. Suppose g : Nm ! Nn a bijection and max (m; n) = ` + 1.