Homework 2
H´
ector Guillermo Cu´
ellar R´
ıos
February 2, 2006
12.12 Let
D
be a nonempty set and suppose that
f
:
D
→
R
and
g
:
D
→
R
.
Define the function
f
+
g
:
D
→
R
by (
f
+
g
)(
x
) =
f
(
x
) +
g
(
x
).
(a) If
f
(
D
) and
g
(
D
) are bounded above, then prove that (
f
+
g
)(
D
) is
bounded above and sup[(
f
+
g
)(
D
)]
≤
sup
f
(
D
) + sup
g
(
D
).
Proof.
Let sup
f
(
D
) =
m
and sup
g
(
D
) =
n
. We know that
∀
x
∈
D,
f
(
x
)
≤
m
and
g
(
x
)
≤
n
, then
f
(
x
) +
g
(
x
)
≤
m
+
n
and since
f
(
x
) +
g
(
x
) = (
f
+
g
)(
x
) (
f
+
g
)(
x
)
≤
m
+
n
thus (
f
+
g
)(
x
) is
bounded above by m+n and by the completeness axiom (
f
+
g
)(
x
) has
a least upper bound, let’s call it
d
, since
d
is the smallest of the upper
bounds,
d
≤
m
+
n
.
(b) Find an example to show that a strict inequality in part (a) may occur.
Let
D
=
{
x
: 0
≤
x
≤
10
}
,
f
(
x
) =
x
and
g
(
x
) =

x
. sup
f
(
x
) = 10
and sup
g
(
x
) = 0 and sup [(
f
+
g
)(
x
)] = 0 and 0
≤
10 + 0.
(c) State and prove the analog of part (a) for infima.
If
f
(
D
) and
g
(
D
) are bounded below, then prove that (
f
+
g
)(
D
) is
bounded below and inf[(
f
+
g
)(
D
)]
≥
inf
f
(
D
) + inf
g
(
D
).
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 Fall '08
 Akhmedov,A
 Empty set, Supremum, Order theory, upper bound, Closed set, 13.6 second