Page 353 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 339
Section 3.5

1. Suppose x ∈ A ∩ (B ∪ C). Then x ∈ A, and either x ∈ B or x ∈ C.
Case 1. x ∈ B. Then since x ∈ A, x ∈ A ∩ B,so x ∈ (A ∩ B) ∪ C.
Case 2. x ∈ C. Then clearly x ∈ (A ∩ B) ∪ C.
Since x was arbitrary, we can conclude that A ∩ (B ∪ C) ⊆
(A ∩ B) ∪ C.
4. Suppose x ∈ A. We now consider two cases:
Case 1. x ∈ C. Then x ∈ A ∩ C, so since A ∩ C ⊆ B ∩ C, x ∈ B ∩ C,
and therefore x ∈ B.
Case 2. x /∈ C. Since x ∈ A, x ∈ A ∪ C, so since A ∪ C ⊆ B ∪ C,
x ∈ B ∪ C. But x /∈ C, so we must have x ∈ B.
Thus, x ∈ B, and since x was arbitrary, A ⊆ B.
7. Hint: Assume x ∈ P (A) ∪ P (B), which means that either x ∈ P (A)or
x ∈ P (B). Treat these as two separate cases. In case 1, assume x ∈ P (A),
which means x ⊆ A, and prove x ∈ P (A ∪ B), which means x ⊆ A ∪ B.
Case 2 is similar.
11. Let x be an arbitrary real number.
(←) Suppose |x − 4| > 2.
Case 1. x − 4 ≥ 0. Then |x − 4|= x − 4, so we have x − 4 > 2, and
therefore x > 6.Addingxtobothsidesgivesus2x > 6 + x,so2x − 6 > x.
Since x > 6, this implies that 2x − 6 is positive, so |2x−6|= 2x − 6> x.
Case 2. x − 4 < 0. Then |x − 4|= 4 − x,sowehave4 − x > 2, and
therefore x < 2. Therefore 3x < 6, and subtracting 2x from both sides
we get x < 6 − 2x. Also, from x < 2 we get 2x < 4, so 2x − 6 < −2.
Therefore 2x − 6isnegative,so |2x − 6|= 6 − 2x > x.
(→) Hint: Imitate the “←” direction, using the cases 2x − 6 ≥ 0 and
2x − 6 < 0.
15. (a) Suppose x ∈∪(F ∪ G). Then we can choose some A ∈ F ∪ G such
that x ∈ A. Since A ∈ F ∪ G, either A ∈ F or A ∈ G.If A ∈ F then,
since x ∈ A, it follows that x ∈∪F. Similarly, if A ∈ G then x ∈∪G.
Thus either x ∈∪F or x ∈∪G,so x ∈ (∪F) ∪ (∪G).
Nowsupposethat x ∈ (∪F) ∪ (∪G).Theneither x ∈∪F or x ∈∪G.
If x ∈∪F, then we can choose some A ∈ F such that x ∈ A. Since
A ∈ F, A ∈ F ∪ G, so since x ∈ A, it follows that x ∈∪(F ∪ G). A
similar argument shows that if x ∈∪G then x ∈∪(F ∪ G).
(b) The theorem is: ∩(F ∪ G) = (∩F) ∩ (∩G).
19. (→) Suppose that A B and C are disjoint. Let x be an arbitrary element of
A ∩ C. Then x ∈ A and x ∈ C.If x /∈ B, then since x ∈ A, x ∈ A \ B, and
therefore x ∈ A B. But also x ∈ C, so this contradicts our assumption