P1: PBU/OVY
P2: PBU/OVY
QC: PBU/OVY
T1: PBU
JWDD027-11
JWDD027-Salas-v1
December 2, 2006
16:49
SECTION 11.1
589
27.
Let
c
= lub
S
. Since
b
∈
S,
b
≤
c.
Since
b
is an upper bound for
S,
c
≤
b.
Thus,
b
=
c
.
28.
S
consists of a single element, equal to lub
S
.
29.
(a) Suppose that
K
is an upper bound for
S
and
k
is a lower bound. Let
t
be any element of
T
. Then
t
∈
S
which implies that
k
≤
t
≤
K
.Thus
K
is an upper bound for
T
and
k
is a lower bound, and
T
is bounded.
(b) Let
a
= glb
S.
Then
a
≤
t
for all
t
∈
T.
Therefore,
a
≤
glb
Similarly, if
b
= lub
S,
then
t
≤
b
for all
t
∈
T,
so lub
T
≤
b.
It now follows that glb
S
≤
glb
T
≤
lub
T
≤
lub
S.
30.
(a) Let
S
=
{
r
:
r<
√
2
,r
rational
}
.
(b) Let
T
=
{
t
:
t<
2
,t
irrational
}
.
31.
Let
c
be a positive number and let
S
=
{
c,
2
c,
3
c, .
..
}
. Choose any positive number
M
and consider the
positive number
M/c
. Since the set of positive integers is not bounded above, there exists a positive
integer
k
such that
k
≥
. This implies that
kc
≥
M
. Since
∈
S
, it follows that
S
is not
bounded above.
32.
(a) If
S
is a set of negative numbers, then 0 is an upper bound for
S
. It follows that
α
= lub
S
≤
0.
(b) If
T
is a set of positive numbers, then 0 is a lower bound for
T
. It follows that
β
= glb
T
≥
0.
33.
(a) See Exercise 75 in Section 1.2.
(b) Suppose
x
2
0
>
2. Choose a positive integer
n
such that
2
x
0
n
−
1
n
2
<x
2
0
−
2
.
Then,
2
x
0
n
−
1
n
2
2
0
−
2=
⇒
2
2
0
−
2
x
0
n
+
1
n
2
=
±
x
0
−
1
n
²
2
(c) If
x
2
0
<
2, then choose a positive integer
n
such that
2
x
0
n
+
1
n
2
<
2
−
x
2
0
.
Then
x
2
0
+
2
x
0
n
+
1
n
2
<
⇒
(
x
0
+
1
n
)
2
<
2
34.
Assume that there are only Fnitely many primes,
p
1
,p
2
,
···
n
and let
Q
=
p
1
·
p
2
p
n
+1.