0.2 Sets
1.
a. A = {x 
x = 5k, 
, k ≥ 1}
x = 5k, , k ≥ 1}
2.
a. {1, 3, 5, 7, . . . }
c. { − 1, 1, 3}
e. { } = ∅ is the empty set by Example 3.
3.
a. Not equal: −1 
A but −1 ∉ B.
A but −1 ∉ B.
c. Equal to {a, l, o, y}.
e. Not equal: 0 A but 0 ∉ B.
A but 0 ∉ B.
g. Equal to { − 1, 0, 1}.
4.
a. ∅, {2}
c. {1}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}
5.
a. True. B ⊆ C means each element of B (in particular A) is an element of C.
c. False. For example, A = {1}, B = C = {{1}, 2}.
6.
a. Clearly A ∩ B ⊆ A and A ∩ B ⊆ B; If X ⊆ A and X ⊆ B, then x X implies x A and x B, that is x A ∩ B. Thus X ⊆ A ∩ B.
X implies x A and x B, that is x A ∩ B. Thus X ⊆ A ∩ B.
7. If x A ∪ (B1 ∩ B2 ∩ . . . ∩ Bn), then x A or x Bi for all i. Thus x A ∪ Bi for all i, that is x (A ∪ B1) ∩ (A ∪ B2) ∩ . . . ∩ (A ∪ Bn). Thus
A ∪ (B1 ∩ B2 ∩ . . . ∩ Bn), then x A or x Bi for all i. Thus x A ∪ Bi for all i, that is x (A ∪ B1) ∩ (A ∪ B2) ∩ . . . ∩ (A ∪ Bn). Thus
and the reverse argument proves equality. The other formula is proved similarly.
9. A = {1, 2}, B = {1, 3}, C = {2, 3}.
10.
a. Let A × B = B × A, and fix a A and b B (since these sets are nonempty). If x A, then (x, b) A × B = B × A. This implies x B; so A ⊆ B. Similarly B ⊆ A.
A and b B (since these sets are nonempty). If x A, then (x, b) A × B = B × A. This implies x B; so A ⊆ B. Similarly B ⊆ A.
c. If x A ∩ B, then x A and x B, so (x, x) A × B. If (x, x) A × B, then x A and x B, so x A ∩ B.
A ∩ B, then x A and x B, so (x, x) A × B. If (x, x) A × B, then x A and x B, so x A ∩ B.
11.
a. (x, y) A × (B ∩ C)
A × (B ∩ C)
c. (x, y) (A ∩ B) × (A′ ∩ B′)
(A ∩ B) × (A′ ∩ B′)
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