Case 2: suppose there are only finitely many dominant terms, and we only focus
on this case in the following.
Select
n
1
, such that
s
n
1
is larger than all dominant terms. Then
(11.1)
For any
N
≥
n
1
, there exists
m > N
, such that
s
m
≥
s
N
.
We are going to use “inductive construction” to find a monotone sequence.
(1) Applying (11.1) with
N
=
n
1
gives
n
2
> n
1
, such that
s
n
2
≥
s
n
1
;
(2) Suppose
n
1
,
· · ·
, n
k
have been selected such that
n
1
< n
2
<
· · ·
< n
k
,
and
s
n
1
≤
s
n
2
≤ · · · ≤
s
n
k
.
Applying (11.1) with
N
=
n
k
gives
n
k
+1
> n
k
, such that
s
n
k
+1
≥
s
n
k
.
By Mathematical Induction, (
s
n
k
) is a nondecreasing subsequence.
Corollary 11.9.
Let
(
s
n
)
be a sequence, then there exists
•
a monotone subsequence with limit
lim sup
s
n
, and
•
a monotone subsequence with limit
lim inf
s
n
.
Proof.
We only prove the first statement for lim sup
s
n
.
Let
v
N
= sup
{
s
n
:
s > N
}
,
N
∈
N
, then
v
= lim
v
N
= lim sup
s
n
.
If
v
=
∞
, then lim
s
n
=
∞
, so any monotone sequence will converge to
lim sup
s
n
, and we finished.
If
v
6
=
∞
, select an arbitrary monotone increasing sequence (
t
N
), such
that lim
t
N
=
v
:
•
if
v
is finite, let
t
N
=
v

1
N
;
•
if
v
= +
∞
, let
t
N
=
N
.
Now we discuss the two cases as in the proof of the above theorem.
Case 1: by assumption,
s
n
k
= sup
{
s
m
:
m
≥
n
k
}
=
v
n
k

1
, so
{
s
n
k
}
is decreasing
and
lim
k
→∞
s
n
k
= lim
v
N
= lim sup
s
n
.
Case 2: Given
N > n
1
as above, by (11.1), there exists
m
1
> N
, such that
s
m
1
≥
s
N
. Since
t
N
< v
≤
v
N
= sup
{
s
m
:
m > N
}
,
38
XIN ZHOU
there exists
m
2
> N
, such that
s
m
2
> t
N
.
either
s
m
1
≥
s
m
2
,
=
⇒
s
m
1
≥
s
N
and
s
m
1
> t
N
;
or
s
m
2
≥
s
m
1
,
=
⇒
s
m
2
≥
s
N
and
s
m
2
> t
N
.
Then we have
(11.2) Given
N
≥
n
1
, there exists
m > N
, such that
s
m
≥
s
N
and
s
m
> t
N
.
We are going to use “inductive construction” to find the monotone subse
quence.
(1) Applying (11.2) with
N
=
n
1
gives
n
2
> n
1
, such that
s
n
2
≥
s
n
1
and
s
n
2
> t
n
1
;
(2) Suppose
n
1
,
· · ·
, n
k
have been selected such that
n
1
< n
2
<
· · ·
< n
k
,
s
n
1
≤
s
n
2
≤ · · · ≤
s
n
k
,
and
s
n
j
+1
> t
n
j
,
for all 1
≤
j
≤
k

1
.
Applying (11.2) with
N
=
n
k
gives
n
k
+1
> n
k
, such that
s
n
k
+1
≥
s
n
k
and
s
n
k
+1
> t
n
k
. Moreover,
s
n
k
1
≤
sup
{
s
m
:
m > n
k
}
=
v
n
k
, so
t
n
k
< s
n
k
+1
≤
v
n
k
.
By Mathematical Induction, (
s
n
k
) is a nondecreasing sequence and lim
s
n
k
=
lim sup
s
n
.
Definition 11.10.
Let (
s
n
) be a sequence. A
subsequential limit
is any real
number, or +
∞
, or
∞
, that is the limit of some subsequence.
Example 11.11.
If
lim
s
n
=
s
, then the set of subsequential limit is
{
s
}
.
Example 11.12.
Let
(
r
n
)
be the list of rational numbers, then the set of
subsequential limits are
R
∪ {
+
∞
,
∞}
.
Theorem 11.13.
Let
(
s
n
)
be a sequence, and
S
be the set of subsequential
limits of
(
s
n
)
. Then
(i)
S
6
=
∅
;
(ii)
sup
S
= lim sup
s
n
,
inf
S
= lim inf
s
n
;
(iii)
lim
s
n
exists if and only if
S
consists of only one element, i.e.
S
=
{
s
}
.
MATH 117 FALL 2017 UCSB
39
Proof.
(i) follows from the above corollary since lim sup
s
n
,
lim inf
s
n
∈
S
, and
(iii) follows from (ii).
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 Fall '08
 Akhmedov,A
 Math, Natural Numbers