Math 315 Section 1 — Fall 2006 — Midterm Exam 2 — Solutions
1. Give a simple example of each of the following, or argue that such a request is impossible.
(a) A Cauchy sequence with a divergent subsequence.
This is impossible. Every subsequence of a convergent sequence (i.e. a Cauchy sequence)
converges to the same limit as the original sequence. [See Theorem 2.5.1]
(b) An unbounded sequence containing a subsequence that is Cauchy.
Let
a
n
= 1
/n
is
n
is odd and
a
n
=
n
is
n
is even. That is,
(
a
1
, a
2
, a
3
, a
4
, . . .
) = (
1
1
,
2
,
1
3
,
4
,
1
5
,
6
, . . .
)
.
This sequence is unbounded because the subsequence
(
a
2
k
)
is unbounded, but the subse
quence
(
a
1
, a
3
, a
5
, a
7
, . . .
)
is a Cauchy sequence.
(c) A continuous function
f
: (0
,
1)
→
R
and a Cauchy sequence (
x
n
) such that
(
f
(
x
n
)
)
is
not a Cauchy sequence.
Let
f
(
x
) = 1
/x
and
(
x
n
) = (1
/n
)
. Then
f
(
x
)
is continuous on
(0
,
1)
and the sequence
(
x
n
)
is Cauchy, but
f
(
x
n
) =
n.
So, the sequence
(
f
(
x
n
)
)
is not a Cauchy sequence.
(d) A continuous function
f
: [0
,
1]
→
R
and a Cauchy sequence (
x
n
) such that
(
f
(
x
n
)
)
is
not a Cauchy sequence.
This is impossible. Continuous functions on closed sets map Cauchy sequences to Cauchy
sequences. [See Theorem 4.3.2(iv).]
(e) A function
f
that is diﬀerentiable on the interval [0
,
1] such that
f
0
(0) =

1 and
f
0
(1) = 1,
but
f
0
(
x
) is never equal to 0 for any
x
∈
[0
,
1].
This is impossible. By Darboux’s Theorem, since
f
0
(0) =

1
<
0
< f
0
(1) = 1
, there must
exist some
c
∈
(0
,
1)
where
f
0
(
c
) = 0
.
1