Math 3210-1
Handout 1
THEOREM 5: Let A and B be subsets of a universal set U . Then A ∩ (U r B) = A r B.
Proof
According to our definition of equality of sets, we must show that
[A ∩ (U r B)] ⊆ [A r B] and [A r B] ⊆ [A ∩ (U r B)]
or equivalently,
x ∈ A ∩ (U r B) iff x ∈ A r B.
Let us begin by showing that x ∈ A ∩ (U r B) implies that x ∈ A r B. If x ∈ A ∩ (U r B), then
x ∈ A and x ∈
, by the definition of intersection. But x ∈ U r B means that x ∈ U
and
. Since x ∈ A and x ∈
/ B, we have x ∈
, as required. Thus
A ∩ (U r B) ⊆ A r B.
⊆
. If
, then
Conversely, we must show that
. Thus x ∈ U and x ∈
/ B, so
x ∈ A and x ∈
/ B. Since A ⊆ U , we have x ∈
. But then
and x ∈ U r B, so x ∈ A ∩ (U r B). Hence A r B ⊆
A ∩ (U r B).
˜
THEOREM 6: Let A, B, and C be subsets of a universal set U . Then the following statements are
true.
(a) A ∪ (U r A) = U .
(b) A ∩ (U r A) = ∅.
(c) U r (U r A) = A.
(d) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
(e) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
(f) A r (B ∪ C) = (A r B) ∩ (A r C).
(g) A r (B ∩ C) = (A r B) ∪ (A r C).
Proof We will start with proving part (d).
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 two
Conversely, if y ∈ (A ∪ B) ∩ (A ∪ C), then
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
. Thus
, and this implies that y ∈ A ∪ (B ∩ C).
and y ∈
/ A, we have
Hence (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C).
˜
Proof We will now prove part (f).
We wish to prove that A r (B ∪ C) = (A r B) ∩ (A r C). To this end, let x ∈ A r (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 r B and
x ∈ A r C, which implies that
. We conclude that A r (B ∪ C) ⊆ (A r B) ∩ (A r C).
Conversely, suppose that x ∈
. Then x ∈ A r B and x ∈ A r C. But then x ∈
and x ∈
/
and x ∈
/
. This implies that x ∈
/ (B ∪C),
so x ∈
. Hence
⊆
as desired.
˜