**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 *)

3. (a) Hint: Use contradiction.

(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:

(b) Since A is finite, there exists a bijection f: A

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:

21. (a)

(c)