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Solutions homework 4.
Page 108 Problem 4:
Let (
x
n
) be a Cauchy sequence and let
± >
0. Then there exists
δ >
0 such that

x

y

< δ
implies

f
(
x
)

f
(
y
)

< ±
. For this
δ
there exist
N
such that

x
n

x
m

< δ
for all
n,m
≥
N
. Hence

f
(
x
n
)

f
(
x
m
)

< ±
for all
n,m
≥
N
, i.e., (
f
(
x
n
)) is
a Cauchy sequence.
Page 108 Problem 5.
a. If
f
and
g
are uniformly continuous, then so is
f
+
g
. Proof: Let
± >
0 then there
exists
δ
1
>
0 such that

x

y

< δ
1
implies

f
(
x
)

f
(
y
)

<
±
2
. Similarly there
exists
δ
2
>
0 such that

x

y

< δ
2
implies

g
(
x
)

g
(
y
)

<
±
2
. let
δ
= min
{
δ
1
,δ
2
}
.
Then
δ >
0 and

x

y

< δ
implies

f
(
x
) +
g
(
x
)

f
(
y
)

g
(
y
)

< ±
. For the
proof that

f

is uniformly continuous, let
δ >
0 such that

x

y

< δ
implies

f
(
x
)

f
(
y
)

< ±
. Then the inequality

f
(
x
)
 
f
(
y
)
 ≤ 
f
(
x
)

f
(
y
)

implies that

x

y

< δ
implies

f
(
x

)

f
(
y
)

< ±
, i.e.

f

is uniformly continuous. Now sup(
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This note was uploaded on 02/05/2012 for the course MATH 555 taught by Professor Staff during the Spring '07 term at South Carolina.
 Spring '07
 Staff

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