Page 55 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Operations on Sets 41
diagram that the sets A ∩ B and A \ B do not overlap, and therefore they will
always be disjoint for any sets A and B. Another way to see this would be to
write out what it means to say that x ∈ (A ∩ B) ∩ (A \ B):

x ∈ (A ∩ B) ∩ (A \ B) means (x ∈ A ∧ x ∈ B) ∧ (x ∈ A ∧ x ∈ B),
which is equivalent to x ∈ A ∧ (x ∈ B ∧ x ∈ B).

But this last statement is clearly a contradiction, so the statement x ∈ (A ∩
B) ∩ (A \ B) will always be false, no matter what x is. In other words, nothing
can be an element of (A ∩ B) ∩ (A \ B), so it must be the case that (A ∩ B) ∩
(A \ B) = ∅. Therefore, A ∩ B and A \ B are disjoint.
The next theorem gives another example of a general fact about set oper-
ations. The proof of this theorem illustrates that the principles of deductive
reasoning we have been studying are actually used in mathematical proofs.

Theorem 1.4.7. For any sets A and B, (A ∪ B) \ B ⊆ A.
Proof. We must show that if something is an element of (A ∪ B) \ B, then it
must also be an element of A, so suppose that x ∈ (A ∪ B) \ B. This means
that x ∈ A ∪ B and x ∈ B, or in other words x ∈ A ∨ x ∈ B and x ∈ B. But
notice that these statements have the logical form P ∨ Q and ¬Q, and this
is precisely the form of the premises of our very first example of a deductive
argument in Section 1.1! As we saw in that example, from these premises we
can conclude that x ∈ A must be true. Thus, anything that is an element of
(A ∪ B) \ B must also be an element of A,so(A ∪ B) \ B ⊆ A.

You might think that such a careful application of logical laws is not needed
to understand why Theorem 1.4.7 is correct. The set (A ∪ B) \ B could be
thought of as the result of starting with the set A, adding in the elements of
B, and then removing them again. Common sense suggests that the result will
just be the original set A; in other words, it appears that (A ∪ B) \ B = A.
However, as you are asked to show in exercise 9, this conclusion is incorrect.
This illustrates that in mathematics, you must not allow imprecise reasoning
to lead you to jump to conclusions. Applying laws of logic carefully, as we did
in our proof of Theorem 1.4.7, may help you to avoid jumping to unwarranted
conclusions.

Exercises

∗
1. Let A ={1, 3, 12, 35}, B ={3, 7, 12, 20}, and C ={x | x is a prime
number}. List the elements of the following sets. Are any of the sets