Hence,
l

< a
k
≥
n
< l
+
so

a
k
≥
n

l

<
. Since
k
≥
n
≥
N
, we can say for
n
≥
N
,

a
n

l

<
.
(
⇐
) Assume that lim
a
n
=
l
.
Then given
>
0, there exists an
N
∈
N
such that
n
≥
N
implies

a
n

l

<
/
2.
This means for every
a
n
≥
N
,
l

/
2
< a
n
≥
N
< l
+
/
2.
Now, since
l

/
2 is a lower bound for
a
n
≥
N
, it follows by the definition of greatest lower bound that
l

< l

/
2
≤
inf
{
a
n
:
n
≥
N
}
.
Similarly, sup
{
a
n
:
n
≥
N
} ≤
l
+
/
2
< l
+
.
Thus,
l

<
inf
{
a
n
:
n
≥
N
} ≤
a
n
≥
N
≤
sup
{
a
n
:
n
≥
N
}
< l
+ . Thus,

inf
{
a
n
:
n
≥
N
} 
l

<
and

sup
{
a
n
:
n
≥
N
} 
l

<
. Hence lim inf
a
n
= lim sup
a
n
=
l
.
2.5.1 Give an example or argue that one does not exist.
(a) A sequence that has a subsequence that is bounded but contains no subsequence that converges.
Impossible.
Let
a
n
be a sequence and let (
a
n
k
) be a bounded subsequence of
a
n
.
Then by
the the BolzanoWeierstrass Theorem, there is some subsequence, call it (
a
n
k
i
), of (
a
n
k
) that
converges. But (
a
n
k
i
) is also a subsequence of (
a
n
), and it converges.
(b) A sequence that does not contain 0 or 1, as a term but contains subsequences converging to each
of these values.
If (
a
n
) = (1
/
2
,
1

1
/
2
,
1
/
3
,
1

1
/
3
,
1
/
4
,
1

1
/
4
· · ·
), then the subsequence of odd indexed terms
(1
/
2
,
1
/
3
,
1
/
4
,
· · ·
) converges to 0 and the subsequence of even indexed terms (1

1
/
2
,
1

1
/
3
,
1

1
/
4
,
· · ·
) converges to 1.
(c) A sequence that contains subsequences converging to every point in the infinite set
A
=
{
1
,
1
/
2
,
1
/
3
,
1
/
4
,
1
/
5
,
· · ·
).
Let (
a
n
) = (1
,
1
/
2
,
1
,
1
/
2
,
1
/
3
,
1
,
1
/
2
,
1
/
3
,
1
/
4
,
1
,
1
/
2
,
1
/
3
,
1
/
4
,
1
/
5
,
· · ·
).
Then this sequence con
tains a subsequence (
a
n
k
)
k
∈
N
such that (
a
n
k
) = (
p, p, p, p,
· · ·
) where
p
is any fixed element of
A
.
Since (
a
n
k
) converges to
p
and
p
was an arbitrary element of
A
, we see that given any element of
A
, we can find a subsequence that converges to it.
(d) A sequence that contains subsequences converging to every point in the infinite set
{
1
,
1
/
2
,
1
/
3
,
1
/
4
,
1
/
5
,
· · · }
,
and no subsequences converging to points outside of this set.
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 Spring '11
 DanRomik
 Calculus, lim, Supremum, Limit of a sequence, Limit superior and limit inferior, subsequence