## Math 3000 Homework Answers #21

From Smith, Eggen, & St. Andre, A Transition to Advanced Mathematics, 5th Ed.

Answers to * problems are given in the back of the book and will not be reproduced here.

(pg. 213 : 3, 5, 7, 8, 12, 16, 21 )

(b) Consider the function f: A A × {x} given by f(a) = (a,x) for all aA. We show that f is a bijection. If (b, x)A × {x}, then bA and we have f(b) = (b, x). Thus, f is onto A × {x}. If f(a) = f(b) then (a,x) = (b,x) and so, a = b. Thus, f is one-to-one. We have then that f is a bijection, so A is equivalent to A × {x}.
5. (b) not finite.
(d) not finite.
(f) finite (in fact empty)
(g) finite
(h) finite
(i) not finite
(j) not finite
(k) finite
(l) finite
(m) not finite.
7. (b) Let A be infinite and AB. If B is finite, it would have an infinite subset, namely A. This contradicts Theorem 5.5, every subset of a finite set is finite. Therefore, B must be infinite.
8. (a) Define the function f: Nkx NmNkm by f(a,b) = (a-1)m + b. We will show that this function is a bijection and thus show that Nkx Nm is finite because it is equivalent to Nkm. Suppose f(a,b) = f(c,d), then (a-1)m + b = (c-1)m + d. This can be rewritten as (a-c)m = d - b. Since both d and b are m, and both are at least 1, there is only one way that d-b can be a multiple of m, and that is that d = b. Thus we have (a-1)m = (c-1)m which implies that a = c. So, we have (a,b) = (c,d) and f is one-to-one. Now, let r be any element of Nkm. By the division algorithm we can write r = sm + t with t < m. If t = 0, we can write r = (s-1)m + m. Notice that with this convention, the largest the coefficient of m can be is k-1. Now, f(s+1, t) = sm + t = r, so the function is onto, and thus a bijection.
(b) Since A is finite, there exists a bijection f: A Nk and since B is finite there exists a bijection g: BNm. Define the function h: A x B Nkx Nm by h(a,b) = (f(a), g(b)). We show that h is a bijection. Suppose h(a,b) = h(c,d). Then (f(a), g(b)) = (f(c), g(d)). But this means that f(a) = f(c) and g(b) = g(d). Since f is one-to-one, we have a = c. Since g is one-to-one we have b = d, so (a,b) = (c,d) and h is one-to-one. Now, let (r, s) be any element of Nkx Nm. Since f is onto, there exists an a so that f(a) = r. Since g is onto, there exists a b so that g(b) = s, therefore, h(a,b) = (f(a), g(b)) = (r, s). So, h is onto and therefore a bijection. Now, by composing h with the bijection of part (a) we get a bijection from A x B onto Nkm, and so, A x B is finite.

12. Let A be a finite set and B an infinite set. If B - A is finite, then (B - A) A is finite, being the union of two finite sets. However, B (B - A) A, which is a contradiction, since all subsets of a finite set are finite. Therefore, B - A must be an infinite set.
16. Let n = 2, then r = 1 since r < n. Any function from {1} to {1,2} can not be onto, clearly. So assume that the statement is true for n = k. Consider a function f: NrNk+1. Suppose that f is onto. Then, there exists an a in Nr such that f(a) = k+1. If a is not r, let f(r) = m and define a new function f':NrNk+1 by f'(a) = m, f'(r) = k+1 and f'(x) = f(x) otherwise. The functions f and f' have the same image. Now, restrict f (or f') to the set Nr - {r}, which is Nr-1. The image of this restriction is Nk. By the induction hypothesis, this restricted function is not onto. So, there exists an element y in Nk which has no pre-image under the restriction. But this this element would also have no pre-image under the original function f either, so f is not onto. Therefore, the statement is true by PMI.
21. (a) C . This proof does not correctly construct the bijection needed to show that this is true.
(c) F. The claim is false. If B is the empty set, then A x B will be finite, but A could be any size set.