Page 167 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Existence and Uniqueness Proofs 153
name, say b, for something such that b ∈ A and b ∈ B. Similarly, by the second
given we can let c be something such that c ∈ A and c ∈ C. At this point the
third given is redundant. We already know that there’s something in A, because
in fact we already know that b ∈ A and c ∈ A. We may as well skip to the last
given, which says that if we ever come across two objects that are elements of
A, we can conclude that they are equal. But as we have just observed, we know
that b ∈ A and c ∈ A! We can therefore conclude that b = c. Since b ∈ B and
b = c ∈ C, we have found something that is an element of both B and C,as
required to prove the goal.

Solution
Theorem. Suppose A, B, and C are sets, A and B are not disjoint, A and C are
not disjoint, and A has exactly one element. Then B and C are not disjoint.
Proof. Since A and B are not disjoint, we can let b be something such that
b ∈ A and b ∈ B. Similarly, since A and C are not disjoint, there is some object
c such that c ∈ A and c ∈ C. Since A has only one element, we must have
b = c. Thus b = c ∈ B ∩ C and therefore B and C are not disjoint.

Exercises

1. Prove that for every real number x there is a unique real number y such
∗
2
that x y = x − y.
2. Prove that there is a unique real number x such that for every real number
y, xy + x − 4 = 4y.
3. Prove that for every real number x,if x = 0 and x = 1 then there is a
unique real number y such that y/x = y − x.
4. Prove that for every real number x,if x = 0 then there is a unique real
∗
number y such that for every real number z, zy = z/x.
5. Recall that if F is a family of sets, then ∪F ={x |∃A(A ∈ F ∧ x ∈ A)}.
Suppose we define a new set ∪!F by the formula ∪!F ={x |∃!A(A ∈
F ∧ x ∈ A)}.
(a) Prove that for any family of sets F, ∪!F ⊆∪F.
(b) A family of sets F is said to be pairwise disjoint if every pair of
distinct elements of F are disjoint; that is, ∀A ∈ F∀B ∈ F(A =
B → A ∩ B =∅). Prove that for any family of sets F, ∪!F =∪F
iff F is pairwise disjoint.
p ∗
d 6. Let U be any set.
(a) Prove that there is a unique A ∈ P (U) such that for every B ∈
P (U), A ∪ B = B.