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370 Appendix 1: Solutions to Selected Exercises
Finally, notice that there are three kinds of elements of P n+1 : those that are
elements of P n , the set {n + 1}, and sets of the form S ∪{n + 1}, where
|S|+1
S ∈ P n . It follows that the last formula above is just (−1) |A S |,
S∈P n+1
as required.

Section 7.2

1. (a) By Theorem 7.1.6, Q is countable. If R \ Q were countable then, by
Theorem 7.2.1, Q ∪ (R \ Q) = R would be countable, contradicting
Theorem 7.2.6. Thus, R \ Q must be uncountable.
√
+
(b) Let A ={ 2 + n | n ∈ Z }. It is not hard to see that A and Q are dis-
√
joint, since 2 is irrational, and A is denumerable. Now apply Theo-
rems 7.1.6 and 7.2.1 to conclude that A ∪ Q is denumerable, and there-
fore A ∪ Q ∼ A. Finally, observe that R = (R \ (A ∪ Q)) ∪ (A ∪ Q)
and R \ Q = (R \ (A ∪ Q)) ∪ A, and apply part 2 of Theorem 7.1.2.
4. Suppose that A ∼ P (A). Then there is a one-to-one, onto function f :
A → P (A). Let X ={a ∈ A | a /∈ f (a)}∈ P (A). Since f is onto, there
must be some a ∈ A such that f (a) = X. But then according to the defi-
nition of X, a ∈ X iff a /∈ f (a), so X = f (a), which is a contradiction.
7. Hint: Define f : P (A) × P (B) → P (A ∪ B) by the formula f (X, Y) =
X ∪ Y, and prove that f is one-to-one and onto.
9. Hint: First note that if F = ∅ then g can be any function. If F = ∅,
then since F is countable, we can write its elements in a list: F =
{ f 1 , f 2 ,...}. Now define g : Z → R by the formula g(n) = max{| f 1 (n)|,
+
| f 2 (n)|,..., | f n (n)|}.

Section 7.3
1. (a) The function i A : A → A is one-to-one.
(b) Suppose A B and B C. Then there are one-to-one functions f :
A → B and g : B → C. By part 1 of Theorem 5.2.5, g ◦ f : A → C
is one-to-one, so A C.
5. Let g : A → B and h : C → D be one-to-one functions.
(a) Since A = ∅, we can choose some a 0 ∈ A. Notice that g −1 :
Ran(g) → A. Now define j : B → A as follows:

−1
g (b)if b ∈ Ran g,
j(b) =
a 0 otherwise.
We let the reader verify that j is onto.