So then
n > N
implies

s
n

s

<
and
s
n
→
s
as required.
Another proof that Cauchy sequences converge.
Suppose (
s
n
) is a Cauchy sequence. Then (
s
n
) is
bounded and so lim inf
s
n
and lim sup
s
n
are real numbers. We’ll show that lim inf
s
n
= lim sup
s
n
.
Let
>
0. There exists an
N
so that
n, m > N
=
⇒ 
s
n

s
m

<
2
.
Fix
m > N
. Then
n > N
=
⇒
s
m

2
< s
n
< s
m
+
2
and so
s
m

2
≤
inf
{
s
n
:
n > N
} ≤
lim inf
s
n
≤
lim sup
s
n
≤
sup
{
s
n
:
n > N
} ≤
s
m
+
2
.
Thus 0
≤
lim sup
s
n

lim inf
s
n
≤
. Since
was an arbitrary positive number this implies that
lim inf
s
n
= lim sup
s
n
. The previous theorem tells us that (
s
n
) converges to lim inf
s
n
= lim sup
s
n
.
2
Subsequential limits
Definition.
Suppose (
s
n
) is a sequence of real numbers.
A
subsequential limit
is an element of
R
∪ {∞
,
+
∞}
, which is a limit of a subsequence of (
s
n
).
Theorem.
Let
(
s
n
)
be a sequence.
1. Suppose
t
∈
R
. Then
t
is a subsequential limit of
(
s
n
)
if and only if the set
{
n
∈
N
:

s
n

t

<
}
is infinite for all
>
0
.
2.
+
∞
is a subsequential limit if and only if the sequence is unbounded above.
3.
∞
is a subsequential limit if and only if the sequence is unbounded below.
2
Proof.
1. Suppose (
s
n
k
) is a subsequence converging to
t
. Then given an
>
0, we can find an
N
so
that
k > N
implies

s
n
k

t

<
. So
{
n
k
:
k > N
}
is an infinite subset of
{
n
∈
N
:

s
n

t

<
}
.
Suppose conversely, that the set
{
n
∈
N
:

s
n

t

<
}
is infinite for all
>
0 and suppose we
have
n
1
< n
2
< . . . < n
k

1
with

s
n
r

t

<
1
r
for each
r < k
. Since the set
n
∈
N
:

s
n

t

<
1
k
is infinite, we can find
n
k
> n
k

1
with

s
n
k

t

<
1
k
.
Continuing in this way we obtain a
subsequence (
s
n
k
) with lim
k
→∞
s
n
k
=
t
.
2. Suppose (
s
n
) is unbounded above.
Suppose we have
n
1
< n
2
< . . . < n
k

1
with
s
n
r
> r
for each
r < k
.
Since
{
s
n
:
n
∈
N
}
is not bounded above we can find
n
k
∈
N
so that
s
n
k
>
max[
{
s
n
:
n
≤
n
k

1
} ∪ {
k
}
].
We
necessarily have
n
k
> n
k

1
, and
s
n
k
> k
. Continuing this way we obtain a subsequence (
s
n
k
)
with lim
k
→∞
s
n
k
= +
∞
.
Suppose +
∞
is a sequential limit. Then there is a subsequence (
s
n
k
) diverging to
∞
. Given
M >
0 we can find
k
so that
s
n
k
> M
and so (
s
n
) is unbounded above.
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 Fall '08
 hitrik
 lim, Metric space, Limit of a sequence, subsequence, lim sup sn