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Solution to Problem 4.1.5.15.
By uniform continuity, we have that for
every
²>
0, there exists
δ>
0 such that
a<x<y<a
+
δ
=
⇒
f
(
x
)

f
(
y
)

<².
Choose
²
n
=1
/n
, and let
δ
n
be a corresponding
δ
.
We may take
δ
0
n
=
min(
δ
n
,
1
/n
), and choose
x
n
=
a
+
δ
n
/
2.
Then
n, m
≥
N
=
⇒
x
n
,x
m
∈
(
a, a
+
δ
n
)=
f
(
x
n
)

f
(
x
m
)

<
1
n
,
so the sequence
n
7→
f
(
x
n
) is Cauchy, hence convergent. Call the limit
A
, and
do a similar construction near
b
to ﬁnd a number
B
.
Then the function
g
:[
a, b
]
→
R
biven by
g
(
x
A
if
x
=
a
f
(
x
)i
f
a<x<b
B
if
x
=
b
is a continuous map which coincides with
f
on (
a, b
). To see that it is continuous
at
x
, we need to show that for any
0 there exists
0 such that if
a
≤
x<ea
+
δ
then
f
(
x
)

A

<²
, and we have veriﬁed this above. The
construction at
b
is analogous.
Solution to Problem 4.2.4.5.
This is immediate. If

f
(
x
)

f
(
y
)
≤
M
1

x

y

and

g
(
x
)

g
(
y
)
M
2

x

y

,
then

(
f
+
g
)(
x
)

(
f
+
g
)(
y
)

=
ﬂ
ﬂ
ﬂ
‡
f
(
x
)

f
(
y
)
·
+
‡
g
(
x
)

g
(
y
)
·ﬂ
ﬂ
ﬂ
≤
(
M
1
+
M
2
)

x

y

.
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 Fall '07
 HUBBARD
 Continuity

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