CHAPTER 1
Solution to Problem 1.1.
We have
A
=
{
2
,
4
,
6
}
,B
=
{
4
,
5
,
6
}
,
so
A
∪
B
=
{
2
,
4
,
5
,
6
}
,and
(
A
∪
B
)
c
=
{
1
,
3
}
.
On the other hand,
A
c
∩
B
c
=
{
1
,
3
,
5
}∩{
1
,
2
,
3
}
=
{
1
,
3
}
.
Similarly, we have
A
∩
B
=
{
4
,
6
}
(
A
∩
B
)
c
=
{
1
,
2
,
3
,
5
}
.
On the other hand,
A
c
∪
B
c
=
{
1
,
3
,
5
}∪{
1
,
2
,
3
}
=
{
1
,
2
,
3
,
5
}
.
Solution to Problem 1.2.
(a) By using a Venn diagram it can be seen that for any
sets
S
and
T
,wehave
S
=(
S
∩
T
)
∪
(
S
∩
T
c
)
.
(Alternatively, argue that any
x
must belong to either
T
or to
T
c
,so
x
belongs to
S
if and only if it belongs to
S
∩
T
or to
S
∩
T
c
.) Apply this equality with
S
=
A
c
and
T
=
B
, to obtain the Frst relation
A
c
A
c
∩
B
)
∪
(
A
c
∩
B
c
)
.
Interchange the roles of
A
and
B
to obtain the second relation.
(b) By De Morgan’s law, we have
(
A
∩
B
)
c
=
A
c
∪
B
c
,
and by using the equalities of part (a), we obtain
(
A
∩
B
)
c
=
(
(
A
c
∩
B
)
∪
(
A
c
∩
B
c
)
)
∪
(
(
A
∩
B
c
)
∪
(
A
c
∩
B
c
)
)
A
c
∩
B
)
∪
(
A
c
∩
B
c
)
∪
(
A
∩
B
c
)
.
(c) We have
A
=
{
1
,
3
,
5
}
and
B
=
{
1
,
2
,
3
}
A
∩
B
=
{
1
,
3
}
. Therefore,
(
A
∩
B
)
c
=
{
2
,
4
,
5
,
6
}
,
2