Graphics Reference
In-Depth Information
Deduce that if
x
and
y
,
then 0
y
x
1/
n
and
f
(
x
)
f
(
y
)
n
. So thefunction is
not uniformly continuous.
(v) If you attempted to extend
f
to a function
g
: [0, 1]
R such
that
f
(
x
)
g
(
x
) for rational
x
, is there any value that you
could takefor
g
(1/
2) which would makethefunction
g
continuous at thepoint 1/
2?
Q
R
47 Thefunction
f
: [0, 1]
is uniformly continuous.
(i) If (
a
) is a Cauchy sequence of rational numbers in [0, 1],
show that (
f
(
a
)) is a Cauchy sequence.
(ii) If (
a
) are rational Cauchy sequences in [0, 1] with the
samereal limit, provethat (
f
(
a
) and (
b
)) and (
f
(
b
)) havethesame
limit.
(iii) Is there a well-defined function
g
: [0, 1]
R such that
g
(
x
)
a
is a
Cauchy sequence of rational numbers with an irrational limit
a
?
(iv) Finally, show that
g
is uniformly continuous on [0, 1]. Given
0, choose
such that
x
y
implies
f
(
x
)
f
(
y
)
, for rational
x
and
y
.Let
a
and
b
bereal
numbers in [0, 1] such that
a
b
, and let (
a
f
(
x
) for rational
x
and (
f
(
a
g
(
a
), when (
a
))
)
)
a
and
(
b
b
be Cauchy sequences of rational numbers wholly
within theintrval [
a
,
b
]. Usetheinequality
)
g
(
a
)
g
(
b
)
g
(
a
)
f
(
a
)
f
(
a
)
f
(
b
)
f
(
b
)
g
(
b
)
to complete the proof that the function
f
may be extended to
a continuous function on thewholeintrval [0, 1].
The Intermediate Value Theorem and the Maximum
—
minimum
Theorem, like the theorem that a continuous function on a closed
interval is uniformly continuous, can only be proved using the
completeness of the real numbers. However, uniform continuity, like
continuity, can be defined whether the domain of the function is
complete or not. What we have shown in qns 46 and 47 is that uniform
continuity is the necessary and suMcient condition that a continuous
function defined on a dense subset of a closed interval may be extended
to a continuous function on thewholeintrval. As a final optional
exercise we indicate how this theorem can be used to define exponents.
