So, in fact,
X
= sup
S.
♥
Corollary 1.2.
For a bounded sequence of real numbers
{
s
n
}
,
the limit inferior is
equal to the infimum of all subsequential limits of
{
s
n
}
.
Proof:
The slick proof is the following.
(1) The supremum of a set
S
of real numbers is equal to

inf
{
x
:
x
∈
S
}
.
(2) Consequently, show that
lim
n
→∞
inf
k
≥
n
s
k
=

lim
n
→∞
sup
k
≥
n

s
k
.
(3) By the theorem, this is equal to

sup
{
subsequential limits of
{
s
n
}}
,
which by step 1 is equal to
inf
{
subsequential limits of
{
s
n
}}
.
Here is a more direct proof. Let
>
0
.
Let
x
= lim inf
s
n
.
By definition of limit,
there exists
N
∈
N
such that when
n
≥
N
,
then

x

x
n

<
2
.
Note that
x
n
≤
s
k
for all
k
≥
n
by definition of infimum. So, by Theorem 17.4,
x
n
≤
L
for any subsequential limit
L.
Consequently,
x
≤
x
n
+
≤
L
+
for any subsequential limit
L.
This is true for
any
>
0
,
which shows that
x
≤
L.
Now, the subsequential limit
L
was arbitrary, which shows that
x
is a lower bound
for
S.
Consequently, by definition of infimum,
(1.3)
x
≤
inf
S.
On the other hand, by definition of supremum,
x
n
+
2
is
not
a lower bound for
{
s
k
}
k
≥
n
,
so there exists
s
n
k
with
n
k
≥
n
such that
x
n
≤
s
n
k
≤
x
n
+
2
.
By the triangle inequality,

s
n
k

x
 ≤ 
s
n
k

x
n

+

x
n

x
 ≤
for
n
k
≥
n
≥
N.
Let
k
= 2

k
.
Then, for each
k,
there exists
s
n
k
such that

s
n
k

x
 ≤
k
,
n
k
< n
k
+1
.
Thus, we have constructed a subsequence which converges to
x.
Hence,
x
∈
S
so
inf
S
≤
x.
Together with (1.3), this shows that inf
S
=
x.
MATH 117 LECTURE NOTES FEBRUARY 24, 2009
3
♥
Remark:
Within this proof, we have shown Corollary 19.12, and in particular,
given a bounded sequence
{
s
n
}
, there exists a subsequence whose limit is
lim inf
s
n
and a subsequence whose limit is
lim sup
s
n
.
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 Fall '08
 Akhmedov,A
 Real Numbers, Limits, Supremum, Limit of a sequence, Limit superior and limit inferior, Xn, subsequence, DR. JULIE ROWLETT