Proof:
Assume that (
s
n
) is unbounded above (for the case of bounded below, we can use
the similar argument). We can find
s
n
1
from (
s
n
) such that
s
n
1
≥
1
.
Secondly we can find
s
n
2
from (
s
n
) such that
s
n
2
≥
2
,
with n
2
> n
1
.
Next we can find
s
n
3
from (
s
n
) such that
s
n
3
≥
3
,
with n
3
> n
2
.
Repeating this process, we obtain a subsequence
s
n
k
→
+
∞
.
Limit Superior and Limit Inferior
Recall that we have notions of sup
S
and inf
S
for any subset
S
⊂
R
. We want to consider the case where
S
is a sequence.
Definition
Let (
s
n
) be a bounded sequence. A number
α
is a
subsequential limit
of (
s
n
)
if there is a subsequence (
s
n
k
) of (
s
n
) such that
s
n
k
→
α
[Examples and remarks]
1. Let
s
n
=
1+(

1)
n
+1
2
and (
s
n
) = (1
,
0
,
1
,
0
,
1
,
0
, ...
).
Then 0 and 1 are subsequencial
limits.
2. Let
s
n
= (

1)
n n
+1
n
and (
s
n
) = (

1
2
,
3
2
,

4
3
,
5
4
,

6
5
, ...
). Then

1 and 1 are sunsequencial
limits.
3. Let
s
n
=

2 +
1
n
,
n
= 3
k

2
,
1
n
,
n
= 3
k

1
,
1 +
1
n
,
n
= 3
k.
Then

2
,
0 and 1 are subsequenctial limits.
4. Let
s
n
=
sin
nπ
6
, and
(
s
n
) =
1
2
,
√
3
2
,
1
,
√
3
2
,
1
2
,
0
,

1
2
,

√
3
2
,

1
,

√
3
2
,

1
2
,
1
2
, ...
.
Then
1
2
,
√
3
2
,
1
,
0
,

1
2
,

√
3
2
,

1
.
are subsequenctial limits.
81