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Homework answers, week of Oct. 7
#1, p. 75 Because
f
is continuous on
[
a
,
b
]
, the extremevalue theorem implies that
there exist real numbers
m
and
M
such that
m
≤
f
(
x
)
≤
M
for all
x
∈
[
a
,
b
]
,
and, moreover, that there exist numbers
x
m
,
x
M
∈
[
a
,
b
]
such that
f
(
x
m
) =
m
and
f
(
x
M
) =
M
. Since
f
is an increasing function,
f
(
x
)
<
f
(
y
)
if and only
if
x
<
y
. Consequently, if
x
M
<
b
, then
M
=
f
(
x
M
)
<
f
(
b
)
, a contradiction.
Therefore,
M
=
f
(
b
)
and
x
M
=
b
. Similarly, if
x
m
>
a
, then
m
=
f
(
x
m
)
>
f
(
a
)
, which contradicts the deﬁnition of
m
. Hence
m
=
f
(
a
)
and
x
m
=
a
.
Therefore,
f
([
a
,
b
]) = [
f
(
a
)
,
f
(
b
)]
.
#2, p. 75 Let
M
=
sup
f
([
a
,
b
])
. Then there exists a sequence
{
y
n
} ⊂
f
([
a
,
b
])
such
that
{
y
n
} →
M
. Consequently, there exists a sequence
{
x
n
} ⊂
[
a
,
b
]
such
that
f
(
x
n
) =
y
n
for every
n
. The BolzanoWeierstrauss theorem implies that
there is a subsequence
{
x
n
k
}
that converges to a point
p
∈
[
a
,
b
]
. Since
f
is
continuous,
{
f
(
x
n
k
)
} →
f
(
p
)
. Furthermore, since
f
(
x
n
k
) =
y
n
k
for every
k
,
{
f
(
x
n
k
)
} →
M
. The uniqueness theorem for limits implies that
f
(
p
) =
M
.
Hence
f
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This note was uploaded on 04/02/2009 for the course MAT 371 taught by Professor Thieme during the Spring '07 term at ASU.
 Spring '07
 thieme
 Real Numbers

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