Page 49 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Operations on Sets 35
Because the word or is always interpreted as inclusive or in mathematics,
anything that is an element of either A or B, or both, will be an element of
A ∪ B. Thus, we can think of A ∪ B as the set resulting from throwing all the
elements of A and B together into one set. A \ B is the set you would get if you
started with the set A and removed from it any elements that were also in B.

Example 1.4.2. Suppose A ={1, 2, 3, 4, 5} and B ={2, 4, 6, 8, 10}. List the
elements of the following sets:

1. A ∩ B. 4. (A ∪ B) \ (A ∩ B).
2. A ∪ B. 5. (A \ B) ∪ (B \ A).
3. A \ B.

Solutions
1. A ∩ B ={2, 4}.
2. A ∪ B ={1, 2, 3, 4, 5, 6, 8, 10}.
3. A \ B ={1, 3, 5}.
4. We have just computed A ∪ B and A ∩ B in solutions 1 and 2, so all we
need to do is start with the set A ∪ B from solution 2 and remove from it
any elements that are also in A ∩ B. The answer is (A ∪ B) \ (A ∩ B) =
{1, 3, 5, 6, 8, 10}.
5. We already have the elements of A \ B listed in solution 3, and B \ A =
{6, 8, 10}. Thus, their union is (A \ B) ∪ (B \ A) ={1, 3, 5, 6, 8, 10}.Isit
just a coincidence that this is the same as the answer to part 4?

Example 1.4.3. Suppose A ={x | x is a man} and B ={x | x has brown hair}.
What are A ∩ B, A ∪ B, and A \ B?

Solution

By definition, A ∩ B ={x | x ∈ A and x ∈ B}. As we saw in the last section,
the definitions of A and B tell us that x ∈ A means the same thing as “x is a
man,” and x ∈ B means the same thing as “x has brown hair.” Plugging this
into the definition of A ∩ B, we find that

A ∩ B ={x | x is a man and x has brown hair}.
Similar reasoning shows that

A ∪ B ={x | either x is a man or x has brown hair}
and
A \ B ={x | x is a man and x does not have brown hair}.