Theorem 04 every unbounded sequence has a monotone

Theorem 0.4 Every unbounded sequence has a monotone subsequence that diverges either
to + ∞ or to -∞ .
80

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

Let ( s
n
) be a bounded sequence. Let
S = { α | α is a subsequential limit of ( s
n
) } .
Then
1. S 6 = ∅ .
2. S is a bounded set.
By Theorem 0.4, every bounded sequence has a convergent subsequence. Then S is not
an empty set. Since ( s
n
) is bounded, so is S .
Definition Let ( s
n
) be a bounded sequence and let S be the set of its all subsequential
limits.
1. The limit superior of ( s
n
) (denoted by lim sup s
n
) is
lim sup s
n
= sup S.
2. The limit inferior of ( s
n
) (denoted by lim inf s
n
) is
lim inf s
n
= inf S.
[Examples and remarks]
1. In some books, we use the notation: lim = lim sup and lim = lim inf.
2. lim inf s
n
≤ lim sup s
n
.
3. If ( s
n
) is convergent (i.e., s
n
→ s ), since every its subsequence converges to the same
limit s , we see S = { s } . In this case, lim inf s
n
= lim sup s
n
.
if ( s
n
) is not convergent, then we must have lim inf s
n
< lim sup s
n
. In this case, we
say that ( s
n
) oscillates .