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368 Appendix 1: Solutions to Selected Exercises
17. Following the hint, we recursively define partial orders R n , for n ∈ N,so
that R = R 0 ⊆ R 1 ⊆ R 2 ⊆ ... and

∗
∀i ∈ I n ∀ j ∈ Z ((a i , a j ) ∈ R n ∨ (a j , a i ) ∈ R n ). ( )
+
Let R 0 = R. Given R n , to define R n+1 we apply exercise 2 of Section 6.2,
with B ={a i | i ∈ I n }. Finally, let T =∪ n∈N R n . Clearly T is reflexive,
because every R n is. To see that T is transitive, suppose that (a, b) ∈ T
and (b, c) ∈ T . Then for some natural numbers m and n, (a, b) ∈ R m
and (b, c) ∈ R n .If m ≤ n then R m ⊆ R n , and therefore (a, b) ∈ R n and
(b, c) ∈ R n . Since R n is transitive, it follows that (a, c) ∈ R n ⊆ T . A sim-
ilar argument shows that if n < m then (a, c) ∈ T ,so T is transitive. The
proof that T is antisymmetric is similar. Finally, to see that T is a total
order, suppose x ∈ A and y ∈ A. Since we have numbered the elements of
A, we know that for some positive integers m and n, x = a m and y = a n .
∗
But then by ( ) we know that either (a m , a n )or(a n , a m ) is an element of
R n , and therefore also an element of T.
20. (a) We follow the hint.
Base case: n = 0. Suppose A and B are finite sets and |B|= 0. Then
B = ∅,so A × B = ∅ and |A × B|= 0 =|A|· 0.
Inductionstep:Letnbeanarbitrarynaturalnumber,andsupposethat
for all finite sets A and B,if |B|= n then A × B is finite and |A × B|=
|A|· n. Now suppose A and B are finite sets and |B|= n + 1. Choose

an element b ∈ B, and let B = B \{b}, a set with n elements. Then

A × B = A × (B ∪{b}) = (A × B ) ∪ (A ×{b}), and since b /∈ B ,

A × B and A ×{b} are disjoint. By inductive hypothesis, A × B

is finite and |A × B |=|A|· n. Also, it is not hard to see that A ∼
A ×{b}–just match up each x ∈ A with (x, b) ∈ A ×{b}–so A ×{b}
is finite and |A ×{b}| = |A|. By Theorem 7.1.7, it follows that A × B

is finite and |A × B|=|A × B |+|A ×{b}| = |A|· n +|A|=|A|·
(n + 1).
(b) To order a meal, you name an element of A × B, where A ={steak,
chicken, pork chops, shrimp, spaghetti} and B ={ice cream, cake,
pie}. So the number of meals is |A × B|=|A|·|B|= 5 · 3 = 15.
22. (a) Base case: n = 0. If |A|= 0 then A = ∅,so F ={∅}, and |F|=
1 = 0!.
Induction step: Suppose n is a natural number, and the desired
conclusion holds for n. Now let A be a set with n + 1 elements,
and let F ={ f | f is a one-to-one, onto function from I n+1 to A}.
Let g : I n+1 → A be a one-to-one, onto function. For each i ∈ I n+1 ,
let A i = A \{g(i)}, a set with n elements, and let F i ={ f | f is a