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Math 523H Homework #2
Sections 1.2, 1.3
1.2 #1. Let
S
be the open interval
(1
,
2)
and let
T
be the closed interval
[

2
,
2]
.
Describe the following sets:
(a)
S
∪
T
= [

2
,
2] =
T
(b)
S
∩
T
= (1
,
2) =
S
(c)
R
\
S
= (
∞
,
1]
∪
[2
,
∞
)
(d)
T
\
S
= [

2
,
1]
∪ {
2
}
(e)
R
\
(
T
\
S
) = (
∞
,

2)
∪
(1
,
2)
∪
(2
,
∞
)
1.2 #2. Describe the following sets of real numbers:
(a)
{
x
∈
R
x
2
>
2
}
=
{
x
∈
R
x <

√
2 or
x >
√
2
}
(b)
{
x
∈
R

x
 ≤
3
}
=
{
x
∈
R

3
≤
x
≤
3
}
(c)
{
x
∈
R

x

2
 ≤
3
}
=
{
x
∈
R

1
≤
x
≤
5
}
(d)
{
x
∈
Q

x
 ≤
1
}
=
{
x
∈
Q

1
≤
x
≤
1
}
1.2 #3. If
A
,
B
, and
C
are sets, prove that
A
∩
(
B
∪
C
) = (
A
∩
B
)
∪
(
A
∩
C
)
.
Proof.
Let
x
∈
A
∩
(
B
∪
C
). Then
x
∈
A
and
x
∈
B
∪
C
. So
x
is always in
A
and it may be in
B
or
C
or both.
•
Case 1:
x
∈
B
but
x /
∈
C
. Since
x
∈
A
,
x
∈
A
∩
B
.
•
Case 2:
x
∈
C
but
x /
∈
B
. Since
x
∈
A
,
x
∈
A
∩
C
.
•
Case 3:
x
∈
B
and
x
∈
C
. Then
x
∈
A
∩
B
and
x
∈
A
∩
C
.
In all cases,
x
∈
A
∩
B
or
x
∈
A
∩
C
. So
x
∈
(
A
∩
B
)
∪
(
A
∩
C
). Therefore,
A
∩
(
B
∪
C
)
⊆
(
A
∩
B
)
∪
(
A
∩
C
).
Now let
x
∈
(
A
∩
B
)
∪
(
A
∩
C
). Then
x
∈
A
and
x
∈
B
and
x /
∈
C
,
x
∈
A
and
x
∈
C
and
x /
∈
B
,
or
x
∈
A
and
x
∈
B
and
x
∈
C
.
•
Case 1:
x
∈
A
and
x
∈
B
and
x /
∈
C
. Since
x
∈
B
,
x
∈
B
∪
C
. So
x
∈
A
∩
(
B
∪
C
).
•
Case 2:
x
∈
A
and
x
∈
C
and
x /
∈
B
. Since
x
∈
C
,
x
∈
B
∪
C
. So
x
∈
A
∩
(
B
∪
C
).
•
Case 3:
x
∈
A
and
x
∈
B
and
x
∈
C
. It is clearly the case then that
x
∈
A
∩
(
B
∪
C
).
In all cases,
x
∈
A
and
x
∈
B
∪
C
. So
x
∈
A
∩
(
B
∪
C
). Therefore,
A
∩
(
B
∪
C
)
⊇
(
A
∩
B
)
∪
(
A
∩
C
).
Since
A
∩
(
B
∪
C
)
⊆
(
A
∩
B
)
∪
(
A
∩
C
) and
A
∩
(
B
∪
C
)
⊇
(
A
∩
B
)
∪
(
A
∩
C
),
A
∩
(
B
∪
C
) =
(
A
∩
B
)
∪
(
A
∩
C
)
±
1.2 #5a. Describe the Cartesian product
S
×
T
where
S
= [0
,
1]
and
T
=
{
x
x
≥
0
}
S
×
T
=
{
(
s,t
)
0
≤
s
≤
1 and
t
≥
0
}
1.2 #5d. Describe the Cartesian product
S
×
T
where
S
=
N
and
T
=
R
.
S
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This note was uploaded on 12/07/2010 for the course MATH 523 taught by Professor Nahmod during the Fall '10 term at UMass (Amherst).
 Fall '10
 Nahmod
 Math, Sets

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