Math 350
Advanced Calculus I
Spring 2008
Homework Solutions 3
# 1:
Supplementary Problem A
Let
S
⊂
R
be a nonempty and bounded set. Let
m
= sup
S
.
Prove: If
m
6∈
S
, then
S
has no maximum. This result can be used in Problem 12.3.
Solution
We prove the contrapositive: If
S
has a maximum
x
, then
x
has to be sup
S
. That is,
sup
S
∈
S
. Let
x
= max
S
. Then for any
s
∈
S
,
s
≤
x
. Hence
x
is an upper bound of
S
. For
any
y
∈
IR
such that
y < x
, since
x
∈
S
and
x > y
,
y
is not an upper bound of
S
. Hence by the
deﬁnition of supremum,
x
= sup
S
. Since
x
∈
S
, we have sup
S
∈
S
.
# 2:
Supplementary Problem B
Indicate the logical error in the following proof:
“For
S
= (0
,
4), since all
x
such that
x >
4 are upper bounds of
S
, sup
S
= 4.”
Solution
The reasoning misses half of the proof. It only proves that sup
S
≤
4 (as the least upper
bound), but does not prove that sup
S
= 4. That is, it does not show for any
x
such that
x <
4,
x
is not a upper bound.
12.3 For each subset of
IR
, give its supremum and its maximum, if they exist. For each party, you
must verify your results. That is, if you are claiming that a number
a
is the supremum, you must
show that it is using the deﬁnition.
(d) (0
,
4)
Solution
sup
S
= 4, No maximum
We want to prove that sup
S
= 4. To show this, we need to show
(1) 4 is an upper bound for
S
:
Suppose
s
is an arbitrary element of
S
. Then 0
< s <
4, so
s
≤
4, as required. Therefore 4 is an upper
bound for
S
.
(2) For all
m
∈
IR
, if
m <
4, then
m
is not an upper bound for
S
:
Suppose
m
is an arbitrary real number and
m <
4. We need to show that
m
is not an upper bound
for
S
. To show this, we need to show there exists an element
s
∈
S
such that
s > m
.
In the case that
m
≤
0, choose
s
= 1, then clearly
s
= 1
∈
S
and
s
= 1
> m
.
In the case that
m >
0, choose
s
to be the midpoint of
m
and 4, so
s
=
m
+ 4
2
=
m
+
4

m
2
.
Then by the ﬁrst part of the equation, 0
< s <
4, so
s
∈
S
. By the second part of the equation,
since 4

m >
0, we have
s > m
.
Therefore
m
is not an upper bound for
S
. Thus 4 is the least upper bound of
S
.
Since sup
S
= 4 and 4
6∈
S
, by Problem #1,
S
has no maximum.
(f)
'
1

1
n

n
∈
IN
“
Solution
sup
S
= 1, No maximum
We want to prove that sup
S
= 1. To show this, we need to show
(1) 1 is an upper bound for
S
:
1
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View Full DocumentSuppose
s
is an arbitrary element of
S
. Then
s
= 1

1
n
for some
n
∈
IN
. Since 1
/n >
0, then
s <
1,
as required. Therefore 1 is an upper bound for
S
.
(2) For all
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 Spring '08
 QIAN
 Calculus, Interior, Supremum, Order theory, upper bound, General topology, Boundary

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