Math 3210-1
Worksheet 1
Not to be turned in.
Set Operations
1. Fill in the blanks in the proof of the following theorem.
Theorem. Let A and B be subsets of a universal set U. Then
A ∩ (U \ B) = A \ B.
Proof. According to our definition of equality of sets we must show that
[A ∩ (U \ B)] ⊆ [A \ B] and [A \ B] ⊆ [A ∩ (U \ B)].
Equivalently,
x ∈ A ∩ (U \ B) iff
x ∈ A \ B.
Let us start by showing that x ∈ A ∩ (U \ B) implies x ∈ A \ B. If x ∈ A ∩ (U \ B), then
x ∈ A and x ∈
, by the definition of intersection. But x ∈ U \ B means
that x ∈ U and
. Since x ∈ A and x ∈
/ B, we have x ∈
, as
required. Thus A ∩ (U \ B) ⊆ A \ B.
⊆
. If
, then
Conversely, we mush show that
x ∈ A and x ∈
/ B. Since A ⊆ U, we have x ∈
. Thus x ∈ U and x ∈
/ B,
so
. But then
and x ∈ U \ B, so x ∈ A ∩ (U \ B). Hence
A \ B ⊆ A ∩ (U \ B).
2. Use a Venn diagram to illustrate A \ B
3. Fill in the blanks in the proof of the following theorem.
Theorem. Let A, B and C be subsets of a universal set U. Then,
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
,
Proof. We begin by showing that A∪(B ∩C) ⊆ (A∪B)∩(A∪C). If x ∈
then either x ∈ A or x ∈ B ∩ C. If x ∈ A, then certainly x ∈ A ∪ B and x ∈ A ∪ C.
Thus x ∈
. On the other hand, if
, then x ∈ B and x ∈ C.
But this implies that x ∈ A ∪ B and
, so x ∈ (A ∪ B) ∩ (A ∪ C). Hence
A ∪ (B ∩ C) ⊆ (A ∪ B) ∩ (A ∪ C).
and
. There are
Conversely, if y ∈ (A ∪ B) ∩ (A ∪ C), then
two cases to consider: when y ∈ A and when y ∈
/ A. If y ∈ A, then y ∈ A ∪ (B ∩ C)
and this part is done. On the other hand, if
, then since y ∈ A ∪ B, we
must have y ∈ B. Similarly, since y ∈ A ∪ C and y ∈
/ A, we have
. Thus
, and this implies that y ∈ A ∪ (B ∩ C). Hence (A ∪ B) ∩ (A ∪ C) ⊆
A ∪ (B ∩ C).
4. Fill in the blanks in the proof of the following theorem.
Theorem. Let A, B and C be subsets of a universal set U. Then,
A \ (B ∪ C) = (A \ B) ∩ (A \ C).
Proof. We wish to prove that A \ (B ∪ C) = (A \ B) ∩ (A \ C). To this end, let
x ∈ A\(B∪C). Then
and
. Since x ∈
/ B∪C,
and x ∈
/ C (for if it were in either B or C then it would be in their union). Thus x ∈ A
and x ∈
/ B and x ∈
/ C. Hence x ∈ A\B and x ∈ A\C, which implies that
.
We conclude that A \ (B ∪ C) ⊆ (A \ B) ∩ (A \ C).
Conversely, suppose that x ∈
x∈
and x ∈
/
blank and x ∈
/
. Hence
so x ∈
. Then x ∈ A \ B and x ∈ A \ C. But then
⊆
. This implies that x ∈
/ (B ∪ C),
as desired.