upper bounds (respectively) for the set
A
n
=
{
a
n
, a
n
+1
, a
n
+2
, . . .
}
for each
n
≥
N
.
Since
y
n
= sup
A
n
, it now follows that
α

< y
n
< α
+
for all
n
≥
N
.
Since
(
y
n
)
→
y
, the OLT now implies that
α

≤
y
≤
α
+
. Since
is arbitrary, we have
that
y
=
α
by Theorem 1.2.6. The argument showing that
lim
z
n
=
α
is
very
similar.
Exercise 2.5.1
(a)
Impossible. If the subsequence is bounded, then it must have a convergent subsequence
by BW Theorem.
(b)
Define the sequences
x
n
= 1
/
(
n
+ 1)
and
y
n
= 1 +
x
n
. Then,
(
x
n
)
→
0
and
(
y
n
)
→
1
.
Moreover,
x
n
, y
n
/
∈ {
0
,
1
}
for all
n
∈
N
. Thus, the sequence
(
x
1
, y
1
, x
2
, y
2
, . . .
)
satisfies
the request.
(c)
The following sequence satisfies the request:
1
,
1
,
1
2
,
1
,
1
2
,
1
3
,
1
,
1
2
,
1
3
,
1
4
,
1
,
1
2
,
1
3
,
1
4
,
1
5
,
1
,
1
2
,
1
3
,
1
4
,
1
5
,
1
6
,
1
, . . .
.
(d)
This request is impossible because there must also be a subsequence that converges to
zero. To see why this is so, let
(
a
n
)
be a sequence that has subsequences that converge
to every number in the set
A
=
{
1
/n
:
n
∈
N
}
. We will construct a subsequence
(
a
n
k
)
of
(
a
n
)
that converges to zero in the following way:
•
Since
(
a
n
)
has a subsequence converging to 1,
∃
n
1
∈
N
such that

a
n
1

1

<
1
.
•
Since
(
a
n
)
has a subsequence converging to
1
/
2
,
∃
n
2
∈
N
such that

a
n
2

1
/
2

<
1
/
2
.
•
Since
(
a
n
)
has a subsequence converging to
1
/
3
,
∃
n
3
∈
N
such that

a
n
3

1
/
3

<
1
/
3
.
•
.
.
.
•
Since
(
a
n
)
has a subsequence converging to
1
/k
,
∃
n
k
∈
N
such that

a
n
k

1
/k

<
1
/k
.
Continuing in this fashion gives us the subsequence
(
a
n
k
)
such that

a
n
k

1
/k

<
1
/k
for each
k
∈
N
. To show that
(
a
n
k
)
→
0
, let
>
0
be arbitrary. Choose
N
∈
N
such
that
N >
2
/
. Then, for any
k
≥
N
,

a
n
k

=
a
n
k

1
k
+
1
k
≤
a
n
k

1
k
+
1
k
<
2
+
2
=
.
Therefore,
(
a
n
k
)
→
0
.
Solutions
Math 3320: Homework #4
Page 4 of 4
Exercise 2.5.5
Assume
(
a
n
)
is a bounded sequence with the property that every convergent subsequence
of
(
a
n
)
converges to the same limit
a
∈
R
. For a contradiction, assume that
(
a
n
)
does not
converge to
a
. Then, there exists an
0
>
0
with the following property:
For every
N
∈
N
, there exists
n
≥
N
such that

a

a
n
 ≥
0
.
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 Spring '15
 AmarjitNijjar
 Math, lim, Supremum, Limit of a sequence, Limit superior and limit inferior, Xn, subsequence