Page 352 - HOW TO PROVE IT: A Structured Approach, Second Edition

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338 Appendix 1: Solutions to Selected Exercises
arbitrary, this shows that x ⊆ A,so x ∈ P (A). A similar argument shows
that x ⊆ B, and therefore x ∈ P (B). Thus, x ∈ P (A) ∩ P (B).
Now suppose that x ∈ P (A) ∩ P (B). Then x ∈ P (A) and x ∈ P (B),
so x ⊆ A and x ⊆ B. Suppose that y ∈ x. Then since x ⊆ A and x ⊆ B,
y ∈ A and y ∈ B,so y ∈ A ∩ B. Thus, x ⊆ A ∩ B,so x ∈ P (A ∩ B).
9. Suppose that x and y are odd. Then we can choose integers j and k such that
x = 2 j + 1 and y = 2k + 1. Therefore xy = (2 j + 1)(2k + 1) = 4 jk +
2 j + 2k + 1 = 2(2 jk + j + k) + 1. Since 2 jk + j + k is an integer, it
follows that xy is odd.
12. Hint: Let x ∈ R be arbitrary, and prove both directions of the biconditional
separately. For the “→” direction, use existential instantiation and proof
by contradiction. For the “←” direction, assume that x = 1 and then solve
the equation x + y = xy for y in order to decide what value to choose
for y.
15. Suppose that ∪F and ∩G are not disjoint. Then we can choose some x such
that x ∈∪F and x ∈∩G. Since x ∈∪F, we can choose some A ∈ F such
that x ∈ A. Since we are given that every element of F is disjoint from
some element of G, there must be some B ∈ G such that A ∩ B = ∅.
Since x ∈ A, it follows that x /∈ B. But we also have x ∈∩G and B ∈ G,
from which it follows that x ∈ B, which is a contradiction. Thus, ∪F and
∩G must be disjoint.
17. (a) Suppose x ∈∪(F ∩ G). Then we can choose some A ∈ F ∩ G such
that x ∈ A. Since x ∈ A and A ∈ F, x ∈∪F, and similarly since
x ∈ A and A ∈ G, x ∈∪G. Therefore, x ∈ (∪F) ∩ (∪G). Since x was
arbitrary, this shows that ∪(F ∩ G) ⊆ (∪F) ∩ (∪G).
(b) The sentence “Thus, we can choose a set A such that A ∈ F, A ∈ G,
and x ∈ A” is incorrect. (Why?)
(c) One example is F ={{1}, {2}}, G ={{1}, {1, 2}}.
21. Suppose that ∪F ⊆∪G. Then there is some x ∈∪F such that x /∈∪G.
Since x ∈∪F, we can choose some A ∈ F such that x ∈ A. Now let
B ∈ G be arbitrary. If A ⊆ B, then since x ∈ A, x ∈ B. But then since
x ∈ B and B ∈ G, x ∈∪G, which we already know is false. Therefore
A ⊆ B. Since B was arbitrary, this shows that for all B ∈ G, A ⊆ B.
Thus, we have shown that there is some A ∈ F such that for all B ∈ G,
A ⊆ B.
23. (a) Suppose x ∈∪ i∈I (A i \ B i ). Then we can choose some i ∈ I such
that x ∈ A i \ B i , which means x ∈ A i and x /∈ B i . Since x ∈ A i ,
x ∈∪ i∈I A i , and since x /∈ B i , x /∈∩ i∈I B i . Thus, x ∈ (∪ i∈I A i ) \
(∩ i∈I B i ).
(b) One example is I ={1, 2}, A 1 = B 1 ={1}, A 2 = B 2 ={2}.