2Convergence of sequences
IA Analysis I
2.3 Monotone-sequences property
Recall that we characterized the least upper bound property. It turns out that
there is an alternative characterization of real number using sequences, known
as the monotone-sequences property. In this section, we will show that the two
characterizations are equivalent, and use the monotone-sequences property to
deduce some useful results.
Definition (Monotone sequence). A sequence (
a
n
) is increasing if
a
n
≤ a
n+1
for all n.
It is strictly increasing if
a
n
< a
n+1
for all
n
. (Strictly) decreasing sequences
are defined analogously.
A sequence is (strictly) monotone if it is (strictly) increasing or (strictly)
decreasing.
Definition (Monotone-sequences property). An ordered field has the monotone
sequences property if every increasing sequence that is bounded above converges.
We want to show that the monotone sequences property is equivalent to the
least upper bound property.
Lemma. Least upper bound property ⇒ monotone-sequences property.
Proof.
Let (
a
n
) be an increasing sequence and let
C
an upper bound for (
a
n
).
Then
C
is an upper bound for the set
{a
n
:
n ∈ N}
. By the least upper bound
property, it has a supremum s. We want to show that this is the limit of (a
n
).
Let
ε >
0. Since
s
=
sup{a
n
:
n ∈ N}
, there exists an
N
such that
a
N
> s−ε
.
Then since (
a
n
) is increasing,
∀n ≥ N
, we have
s − ε < a
N
≤ a
n
≤ s
. So
|a
n
− s| < ε.
We first prove a handy lemma.
Lemma. Let (
a
n
) be a sequence and suppose that
a
n
→ a
. If (
∀n
)
a
n
≤ x
, then
a ≤ x.
Proof.
If
a > x
, then set
ε
=
a − x
. Then we can find
N
such that
a
N
> x
.
Contradiction.
Before showing the other way implication, we will need the following:
Lemma. Monotone-sequences property ⇒ Archimedean property.
Proof. We prove version 2, i.e. that 1/n → 0.
Since 1
/n >
0 and is decreasing, by MSP, it converges. Let
δ
be the limit.
By the previous lemma, we must have δ ≥ 0.
If
δ >
0, then we can find
N
such that 1
/N <
2
δ
. But then for all
n ≥
4
N
,
we have 1/n ≤ 1/(4N ) < δ/2. Contradiction. Therefore δ = 0.
Lemma. Monotone-sequences property ⇒ least upper bound property.
Proof.
Let
A
be a non-empty set that’s bounded above. Pick
u
0
, v
0
such that
u
0
is not an upper bound for
A
and
v
0
is an upper bound. Now do a repeated
bisection: having chosen
u
n
and
v
n
such that
u
n
is not an upper bound and
v
n
is, if (
u
n
+
v
n
)
/
2 is an upper bound, then let
u
n+1
=
u
n
,
v
n+1
= (
u
n
+
v
n
)
/
2.
Otherwise, let u
n+1
= (u
n
+ v
n
)/2, v
n+1
= v
n
.
Then u
0
≤ u
1
≤ u
2
≤ ··· and v
0
≥ v
1
≥ v
2
≥ ···. We also have
v
n
− u
n
=
v
0
− u
0
2
n
→ 0.
By the monotone sequences property,
u
n
→ s
(since (
u
n
) is bounded above by
v
0
). Since v
n
− u
n
→ 0, v
n
→ s. We now show that s = sup A.
If
s
is not an upper bound, then there exists
a ∈ A
such that
a > s
. Since
v
n
→ s
, then there exists
m
such that
v
m
< a
, contradicting the fact that
v
m
is
an upper bound.
To show it is the least upper bound, let
t < s
. Then since
u
n
→ s
, we can
find
m
such that
u
m
> t
. So
t
is not an upper bound. Therefore
s
is the least
upper bound.
Why do we need to prove the Archimedean property first? In the proof
above, we secretly used the it. When showing that
v
n
−u
n
→
0, we required the
fact that
1
2
n
→
0. To prove this, we sandwiched it with
1
n
. But to show
1
n
→
0,
we need the Archimedean property.
Lemma. A sequence can have at most 1 limit.
Proof.
Let (
a
n
) be a sequence, and suppose
a
n
→ x
and
a
n
→ y
. Let
ε >
0
and pick
N
such that
∀n ≥ N
,
|a
n
− x| < ε/
2 and
|a
n
− y| < ε/
2. Then
|x − y| ≤ |x − a
N
|
+
|a
N
− y| < ε/
2 +
ε/
2 =
ε
. Since
ε
was arbitrary,
x
must
equal y.
Lemma (Nested intervals property). Let
F
be an ordered field with the monotone
sequences property. Let
I
1
⊇ I
2
⊇ ···
be closed bounded non-empty intervals.
Then
T
∞
n=1
I
n
6= ∅.
Proof.
Let
T
n
= [
a
n
, b
n
] for each
n
. Then
a
1
≤ a
2
≤ ···
and
b
1
≥ b
2
≥ ···
.
For each
n
,
a
n
≤ b
n
≤ b
1
. So the sequence
a
n
is bounded above. So by the
monotone sequences property, it has a limit
a
. For each
n
, we must have
a
n
≤ a
.
Otherwise, say
a
n
> a
. Then for all
m ≥ n
, we have
a
m
≥ a
n
> a
. This implies
that a > a, which is nonsense.
Also, for each fixed
n
, we have that
∀m ≥ n
,
a
m
≤ b
m
≤ b
n
. So
a ≤ b
n
.
Thus, for all n, a
n
≤ a ≤ b
n
. So a ∈ I
n
. So a ∈
T
∞
n=1
I
n
.
We can use this to prove that the reals are uncountable:
Proposition. R is uncountable.
Proof.
Suppose the contrary. Let
x
1
, x
2
, ···
be a list of all real numbers. Find
an interval that does not contain
x
1
. Within that interval, find an interval that
does not contain
x
2
. Continue ad infinitum. Then the intersection of all these
intervals is non-empty, but the elements in the intersection are not in the list.
Contradiction.
A powerful consequence of this is the Bolzano-Weierstrass theorem. This is
formulated in terms of subsequences:
Definition (Subsequence). Let (
a
n
) be a sequence. A subsequence of (
a
n
) is a
sequence of the form a
n
1
, a
n
2
, ···, where n
1
< n
2
< ···.
Example. 1, 1/4, 1/9, 1/16, ··· is a subsequence of 1, 1/2, 1/3, ···.
Theorem (Bolzano-Weierstrass theorem). Let
F
be an ordered field with the
monotone sequences property (i.e. F = R).
Then every bounded sequence has a convergent subsequence.
Proof.
Let
u
0
and
v
0
be a lower and upper bound, respectively, for a sequence
(
a
n
). By repeated bisection, we can find a sequence of intervals [
u
0
, v
0
]
⊇
[
u
1
, v
1
]
⊇
[
u
2
, v
2
]
⊇ ···
such that
v
n
− u
n
= (
v
0
− u
0
)
/
2
n
, and such that each
[u
n
, v
n
] contains infinitely many terms of (a
n
).
By the nested intervals property,
T
∞
n=1
[
u
n
, v
n
]
6
=
∅
. Let
x
belong to the
intersection. Now pick a subsequence
a
n
1
, a
n
2
, ···
such that
a
n
k
∈
[
u
k
, v
k
]. We
can do this because [
u
k
, v
k
] contains infinitely many
a
n
, and we have only picked
finitely many of them. We will show that a
n
k
→ x.
Let
ε >
0. By the Archimedean property, we can find
K
such that
v
K
−u
K
=
(
v
0
−u
0
)
/
2
K
≤ ε
. This implies that [
u
K
, v
K
]
⊆
(
x −ε, x
+
ε
), since
x ∈
[
u
K
, v
K
].
Then
∀k ≥ K
,
a
n
k
∈
[
u
k
, v
k
]
⊆
[
u
K
, v
K
]
⊆
(
x−ε, x
+
ε
). So
|a
n
k
−x| < ε
.