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Appendix 1: Solutions to Selected Exercises 341
2
2
To see that y is unique, suppose that x z = x − z. Then z(x + 1) = x,
2
2
and since x + 1 = 0, we can divide both sides by x + 1 to conclude that
2
z = x/(x + 1) = y.
4. Suppose x = 0. Let y = 1/x. Now let z be an arbitrary real number. Then
zy = z(1/x) = z/x, as required.
To see that y is unique, suppose that y is a number with the property that

∀z ∈ R(zy = z/x). Then in particular, taking z = 1, we have y = 1/x,

so y = y.

6. (a) Let A =∅ ∈ P (U). Then clearly for any B ∈ P (U), A ∪ B =
∅ ∪ B = B.

To see that A is unique, suppose that A ∈ P (U) and for all
B ∈ P (U), A ∪ B = B. Then in particular, taking B = ∅ , we can

conclude that A ∪ ∅ = ∅. But clearly A ∪ ∅ = A ,sowehave A =
∅ = A.
(b) Hint: Let A = U.
11. Existence: We are given that for every G ⊆ F, ∪G ∈ F, so in particular,
since F ⊆ F, ∪F ∈ F. Let A =∪F. Now suppose B ∈ F. Then by ex-
ercise 8 of Section 3.3, B ⊆∪F = A, as required.
Uniqueness: Suppose that A 1 ∈ F, A 2 ∈ F, ∀B ∈ F(B ⊆ A 1 ), and
∀B ∈ F(B ⊆ A 2 ). Applying this last fact with B = A 1 we can conclude
that A 1 ⊆ A 2 , and similarly the previous fact implies that A 2 ⊆ A 1 . Thus
A 1 = A 2 .

Section 3.7
1. Hint: Comparing (b) to exercise 16 of Section 3.3 may give you an idea of
what to use for A.
4. Suppose P (∪ i∈I A i ) ⊆∪ i∈I P (A i ). Clearly ∪ i∈I A i ⊆∪ i∈I A i , so
∪ i∈I A i ∈ P (∪ i∈I A i ) and therefore ∪ i∈I A i ∈∪ i∈I P (A i ). By the defini-
tion of the union of a family, this means that there is some i ∈ I such that
∪ i∈I A i ⊆ A i . Now let j ∈ I be arbitrary. Then it is not hard to see that
A j ⊆∪ i∈I A i ,so A j ⊆ A i .
7. Suppose that lim x→c f (x) = L > 0. Let = L. Then by the definition of
limit, there is some δ> 0 such that for all x,if0 < | x − c| <δ then
| f (x) − L| < = L. But if | f (x) − L| < L then −L < f (x) − L < L,
so 0 < f (x) < 2L. Therefore, if 0 < | x − c| <δ then f (x) > 0.
9. The proof is correct.