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Math 295. Solutions to Homework 3 (1) Let be a . A f : S → S is called an provided that f ◦ f (s) = s for all s ∈ S. Show: If g : S → S is an involution, then g is bijective.

We first show that g is surjective. Choose s ∈ S. We need to produce and s0 ∈ S so that g(s0) = s. Let s0 = g(s). Then g(s0) = g(g(s)) = s. Now for injectivity. Suppose s, s0 ∈ 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 ∈ R with a < b. Show that there exist c, d ∈ R with c ∈ Q and d 6∈ Q so that a < c < d < b.

1 First we prove that c exists. By the archimedian of R, there is a natural n such that n < (b−a). Thus multiplying by n, we have bn − an > 1. By a proved in , we can find a m such that an < m < bn. Dividing by n, we see that we can take c = m . n √ It remains to find d ∈ R \ Q such that c < d < b. We know that 2 is irrational, and if q ∈ Q is not zero, then √ √ √ −1 q · 2 is not rational. Since 2 > 0, we have 2 > 0 and √ √ c/ 2 < b/ 2.

Thus, from what we just proved, there exists q ∈ Q so that √ √ c/ 2 < q < b/ 2. √ It follows that c < q 2 < b. (3) Show that every nonempty bounded above of N has a maximal . That is, if ∅ 6= S ⊂ N is bounded above, then there is an element t ∈ S so that t ≥ s for all s ∈ 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 is unbounded, the set

T := {n ∈ N | n ≥ σ} 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 ∈ S we have s ≤ (j − 1). In the former case, we must have j = s ≤ σ ≤ j; so σ = s ∈ S. In the latter case, we conclude that for all s ∈ S we have s ≤ (j − 1) < σ ≤ j. But this means that σ is not the least upper bound, a . (4) Fun with maps. (a) Suppose A is a set and f : N → A is an injection. Show that there is a between A and a proper subset of A. (That is, there is a subset A0 of A and a g : A → A0 such that A0 6= A and g is bijective.)

For all n ∈ N, define an ∈ A by an := f(n). Let =(f) denote the of f in A. Thus, =(f) = 0 {a1, a2,...}. Let A = {a2, a3, a4,...} ∪ (A \ =(f)). 0 0 We claim that a1 6∈ A , so A is a proper subset of A. Since a1 = f(1), we have a1 6∈ A\=(f). Consequently, 0 if a1 ∈ A , then it must be in the set {a2, a3,...}.

Thus, there is an n ∈ 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 6∈ A . Since every a ∈ A belongs to exactly one of =(f) or A \ =(f), we can define the function g : A → A0 by ( a if a ∈ =(f) and a = a , g(a) = n+1 n a if a ∈ A \ =(f). We need to show that g is bijective. We first show that g is surjective. Choose a0 ∈ A0. If a0 ∈ [A \ =(f)] ⊂ A0, then g(a0) = a0. Otherwise, if 0 0 a ∈ {a2, a3, a4,...}, then a = an for some n ∈ N with n > 1. Since n > 1, we have (n − 1) ∈ N and 0 g(a(n−1)) = an = a . Finally, we show that g is injective. Suppose a, a¯ ∈ A and g(a) = g(¯a). We want to conclude that a =a ¯. Suppose first that g(a) ∈ [A \ =(f)] ⊂ A0. In this case, we have a = g(a) = g(¯a) =a. ¯ Now suppose that g(a) 6∈ [A\=(f)]. We must have g(a) ∈ {a2, a3,...}. Thus, since g(¯a) = g(a), we have g(¯a) ∈ {a2, a3,...} as well. Thus, there are n, m ∈ 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 ∈ N, let Nn := {k ∈ N: k ≤ n}. Suppose `, m ∈ N. Show that N` = Nm ` = 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 ∈ 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 := {n ∈ N: P (n) is true}. 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 ∈ T . Suppose n ∈ T . Let S ⊂ Nn+1 be nonempty. If (n + 1) 6∈ 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) ∈ S, then consider the set S = S \{n + 1}. 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 ∈ 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 ∈ 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 ∈ 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 ∈ 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 := {` ∈ N: ` ≥ max (m, n) implies that if h: Nn → Nm is a bijection, then n = m}. We shall show that S is inductive. If max (n, m) = 1, then n = m = 1; so 1 ∈ S. Suppose ` ∈ S. We need to show that ` + 1 ∈ S. Suppose g : Nm → Nn a bijection and max (m, n) = ` + 1. We have three cases: (a) n = m = ` + 1, (b) n < ` + 1 = m, or (c) m < ` + 1 = n. In case (a) we are finished. We take up case (b): If n < ` + 1, then n ≤ ` and so by induction, n = m, a contradiction. In case (c): If m < ` + 1, then m ≤ ` −1 and g : Nn → Nn is a bijection. Thus, n = m, a contradiction. (b) Show that every finite subset of an ordered field has a maximal element.

Let F be an ordered field. Define the subset R of N by R := {` ∈ N: if A is a finite subsetof F with ` elements, then A has a maximal element} We need to show that R is inductive. Since a set with only one element (such a set is called a ) obviously has a maximal element, we know that 1 ∈ R. Suppose ` ∈ R. Let A be a subset of F with ` + 1 elements. Choose a ∈ A. The set A \{a} has ` elements (why?). Hence A \{a} has a maximal element, call it b. If a > b, then a is the maximal element of A. If b > a, then b is the maximal element of A. (c) Show that every finite subset of an ordered field has a minimal element.

In the above, replace the word “maximal” with the word “minimal”. (6) Suppose A, B, and C are sets. Suppose f : A → B and g : B → C. Under what conditions on f and g will g ◦ f be (a) injective?

We let =(f) denote the image of f in B. In for g ◦ f to be injective, we will require f to be injective, and the restriction of g to =(f) must be injective as well. Suppose these two conditions are met and a1, a2 ∈ A. Since the restriction to =(f) of g is injective, we have g(f(a1)) = g(f(a2)) if and only if f(a1) = f(a2). Since f is injective, this means that a1 = a2. (b) surjective?

We need that the restriction of g to =(f) is surjective. Suppose that the restriction of g to =(f) is surjective. Suppose c ∈ C. Since the restriction of g to =(f) is surjective, there is an a ∈ A so that g(f(a)) = c. Hence, g ◦ f is surjective. (c) bijective?

In order for g ◦ f to be bijective, we need that it is both surjective and injective. Hence, from the above, we require: (a) f must be injective, and (b) the restriction of g to =(f) must be both surjective and injective. Thus, f must be injective, and the restriction of g to =(f) must be bijective. Suppose f −1 and g−1 are functions. Is the inverse of g ◦ f a function?

Yes. It is f −1 ◦ g−1. Book Problems Chapter 1: 20. We have

|(x + y) − (x0 + y0)| = |(x − x0) + (y − y0)|

≤ |x − x0| + |y − y0| < ε/2 + ε/2 = ε We also have |(x − y) − (x0 − y0)| = |(x − x0) + (y0 − y)|

≤ |x − x0| + |y0 − y| < ε/2 + ε/2 = ε Chapter 4: 14 (i) - (vii)

(i) Graph is shifted up by c. (ii) Graph is shifted left by c. (iii) Let us assume c > 0. In this case, the graph is stretched vertically by a factor of c. (iv) Let us assume c > 0. In this case, the graph is stretched horizontally by a factor of c. (v) Not sure how to describe without pictures, which I can’t type. (vi) The new graph is produced as follows: Retain the right half of the graph. Reflect, through the y-axis, the right half of the original graph. This becomes the left half of the new graph. (vii) The new graph is produced as follows: Retain the top half of the graph. Reflect, through the x-axis, the bottom half of the original graph. Combine the old top half with this reflection of the bottom half. Chapter 12: 2,3,4 (solutions in book)