Section 2.4
#4)
(a) We prove that sup(
aS
) =
a
sup
S
. The proof for inf(
aS
) is similar. Let
x
∈
aS
. Then
x/a
∈
S
.
So sup
S
≥
x/a
since sup
S
is an upper bound for
S
.
Therefore,
a
sup
S
≥
x
, and
a
sup
S
is
an upper bound for
aS
.
Now let
v
be any upper bound for
aS
.
If
y
∈
S
, then
v
≥
ay
.
So
v/a
≥
y
. Thus,
v/a
is an upper bound for
S
, and so
v/a
≥
sup
S
. Therefore,
v
≥
a
sup
S
, and so
a
sup
S
= sup(
aS
) as desired.
(b) We prove that inf(
bS
) =
b
sup
S
. The proof for sup(
bS
) is similar. Let
x
∈
bS
. Then
x/b
∈
S
. So
sup
S
≥
x/b
since sup
S
is an upper bound for
S
. Therefore,
b
sup
S
≤
x
, and
b
sup
S
is a lower
bound for
bS
. Now let
v
be any lower bound for
bS
. If
y
∈
S
, then
v
≤
by
. So
v/b
≥
y
. Thus,
v/b
is an upper bound for
S
, and so
v/b
≥
sup
S
. Therefore,
v
≤
b
sup
S
, and so
a
sup
S
= inf(
aS
) as
desired.
#6)
We prove that sup(
A
+
B
) = sup
A
+ sup
B
. The proof of inf(
A
+
B
) is similar. Let
x
∈
A
+
B
. Then
x
=
a
+
b
for some
a
∈
A
and some
b
∈
B
. Therefore,
x
=
a
+
b
≤
sup
A
+ sup
B
since sup
A
and
sup
B
are upper bounds for
A
and
B
respectively. Thus, sup
A
+ sup
B
is an upper bound for
A
+
B
.
To prove it is the least upper bound, we show that for all
>
0, there exists
x
∈
A
+
B
such that
sup
A
+ sup
B

< x
≤
sup
A
+ sup
B
. Let
>
0. Choose
a
∈
A
such that sup
A

2
< a
≤
sup
A
,
and choose
b
∈
B
such that sup
B

2
< b
≤
sup
B
. Then
a
+
b
∈
A
+
B
and sup
A
+ sup
B

<
a
+
b
≤
sup
A
+ sup
B
as desired.
#7)
We will prove the relation sup
{
f
(
x
) +
g
(
x
) :
x
∈
X
} ≤
sup
{
f
(
x
) :
x
∈
X
}
+ sup
{
g
(
x
) :
x
∈
X
}
. The
corresponding relation for inf
{
f
(
x
) +
g
(
x
)
}
is proved similarly. It suffices to prove the right hand side
is an upper bound for
{
f
(
x
) +
g
(
x
) :
x
∈
X
}
, for then it must be at least as large as the supremum of
the set. Let
x
0
∈
X
be arbitrary. Then
f
(
x
0
) +
g
(
x
0
)
≤
sup
{
f
(
x
) :
x
∈
X
}
+ sup
{
g
(
x
) :
x
∈
X
}
, and
the right hand side is an upper bound as desired.
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 Spring '08
 JUNGE
 Supremum, Order theory, LHS, sup

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