MAT 300, Mathematical Structures
Dr. L. Mantini
Problem Set 6 Solutions
Spring 2010
4.1
: 3, 4, 5, 6, 7, 10, 12;
6.1
: 2, 5, 6, 9, 12
4.1.3 Describe the truth sets of the following statements as subsets of
R
2
(pictures omitted).
(a)
y
=
x
2

x

2: region is graph of the parabola opening up with
x
intercepts at (2
,
0)
and (

1
,
0)
(b)
y < x
: region is the open halfplane below the line
y
=
x
(c) either
y
=
x
2

x

2 or
y
= 3
x

2: region is union of given parabola and line
(d)
y < x
, and either
y
=
x
2

x

2 or
y
= 3
x

2: region is the portion of the graph from
(c) that lies strictly below the line
y
=
x
4.1.5 Prove parts 2 and 3 of Theorem 4.1.3.
Solution
: We prove part 2 as a twopart proof showing both set containments sepa
rately, and we prove part 3 using a string of equivalent statements.
2. Prove that
A
×
(
B
∪
C
) = (
A
×
B
)
∪
(
A
×
C
).
Proof
: (
⊆
) Let (
x, y
)
∈
A
×
(
B
∪
C
). Then
x
∈
A
and
y
∈
B
∪
C
. So either
y
∈
B
or
y
∈
C
. If
y
∈
B
, then (
x, y
)
∈
A
×
B
. If
y
∈
C
, then (
x, y
)
∈
A
×
C
. In either case,
(
x, y
)
∈
(
A
×
B
)
∪
(
A
×
C
).
(
⊇
) Now let (
x, y
)
∈
(
A
×
B
)
∪
(
A
×
C
). So either (
x, y
)
∈
A
×
B
or (
x, y
)
∈
A
×
C
.
If (
x, y
)
∈
A
×
B
, then
x
∈
A
and
y
∈
B
⊆
B
∪
C
, so (
x, y
)
∈
A
×
(
B
∪
C
).
If
(
x, y
)
∈
A
×
C
, then
x
∈
A
and
y
∈
C
⊆
B
∪
C
, so (
x, y
)
∈
A
×
(
B
∪
C
). In either
case, (
x, y
)
∈
A
×
(
B
∪
C
), which was what we had to prove.
3. Prove that (
A
×
B
)
∩
(
C
×
D
) = (
A
∩
C
)
×
(
B
∩
D
).
Proof
: Let
x
and
y
be arbitrary. Then
(
x, y
)
∈
(
A
×
B
)
∩
(
C
×
D
)
⇔
(
x, y
)
∈
A
×
B
∧
(
x, y
)
∈
C
×
D
⇔
x
∈
A
∧
y
∈
B
∧
x
∈
C
∧
y
∈
D
⇔
(
x
∈
A
∧
x
∈
C
)
∧
(
y
∈
B
∧
y
∈
D
)
⇔
x
∈
A
∩
C
∧
y
∈
B
∩
D
⇔
(
x, y
)
∈
(
A
∩
C
)
×
(
B
∩
D
)
.
4.1.6 What’s wrong with the proof given in the book that for any sets
A
,
B
,
C
, and
D
,
(
A
∪
C
)
×
(
B
∪
D
)
⊆
(
A
×
B
)
∪
(
C
×
D
)?