NESTED INTERVALS
KENT MERRYFIELD
Ross introduces the concepts of limsup and liminf in the middle of section 10 for one specialpurpose: to prove that all Cauchy sequences converge. Now we will want to see these concepts,and we will see them soon, but at ﬁrst glance I am sure that most of you are quite confused bythe idea. Let me try to prove that all Cauchy sequences converge by a completely diﬀerent means,a means that itself has a great deal of power and leads eventually to the mathematical concept of “compactness”.1.
Notation and previous results:
I won’t say everything in this handout, especially if it is also in Ross. Please read the deﬁnitionof a subsequence and the discussion of subsequences in section 11. Read the statement and proof of Theorem 11.2: if a sequence converges, every subsequence of that sequence converges to thesame limit. We will also be referring constantly the Monotone Sequence Theorem: all boundedmonotone sequences converge.Much of what follows concerns intervals, speciﬁcally closed, bounded intervals. To say that
I
isa closed, bounded interval in
R
is to say that there are two numbers
a,b
with
a
≤
b
such that
I
can be written as the interval [
a,b
]
,
which is in turn deﬁned as
{
x
:
a
≤
x
≤
b
}
.
Please note thatthe single point
a
can be considered to be the closed interval [
a,a
]
.
If
I
= [
a,b
]
,
then we deﬁne thelength of
I
in the obvious way: length(
I
) = length([
a,b
]) =
b
−
a.
As a minor result that we will make some use of later, let me prove the following almostobviouslemma:
Lemma 1.
The Interval Diameter Lemma:
If
x,y
∈
I
for some closed bounded interval
I,
then

x
−
y
 ≤
length(
I
)
.
Proof.
Let
I
= [
a,b
]
,
so that length(
I
) =
b
−
a.
Then
a
≤
x
and
y
≤
b
so that
y
−
x
≤
y
−
a
≤
b
−
a.
Similarly,
a
≤
y
and
x
≤
b
so that
x
−
y
≤
x
−
a
≤
b
−
a.
Thus
−
(
b
−
a
)
≤
x
−
y
≤
b
−
a,
which isthe same as saying

x
−
y
 ≤
b
−
a,
which is what we wanted to prove.
2.
Nested intervals:
Let us consider a sequence of intervals
I
1
,I
2
,I
2
,...,
where
I
n
= [
a
n
,b
n
]
.
We will call this asequence of
nested intevvals
if each interval is contained in the one before it?
i.e.
, if
I
1
⊇
I
2
⊇
I
3
⊇
...
It is clear that in order for the intervals to be nested, we must have
a
1
≤
a
2
≤
a
3
≤
...
and
b
1
≥
b
2
≥
b
3
≥
....
We have a result concerning such intervals – a result that is on the one hand sointuitively obvious that you are not sure at ﬁrst that it even needs a proof, but on the other handso strong that it is fully equivalent to the completeness axiom for the real numbers.
Theorem 1.
The Nested Interval Theorem:
Let
I
1
⊇
I
2
⊇
I
3
⊇
...
be a nested sequence of closed bounded intervals in
R
.
Then the intersection of all of these intervals is nonempty. Phrased in another way: there exists some
x
∈
R
such that
x
∈
I
n
for every
n
∈
N
.
Proof.
Consider the sequence (
a
n
) of lefthand endpoints of the intervals
I
n
.
As is discussed above,
a
1
≤
a
2
≤
a
3
≤
...
so that (
a
n
) is a nondecreasing sequence. Since
a
n
≤
b
n
≤
b
1
for all
n,
(
a
n
)is bounded above. Hence, (
a
n
) is a bounded monotone sequence, and by the Monotone SequenceTheorem, it converges. Let
x
= lim
a
n
.
If we examine the proof of the Monotone Sequence Theorem,
1
we see that
x
is actually equal to sup
{
a
n
:
n
∈
N
}
,
so that
x
≥
a
n
for all
n.
Furthermore,
a
n
≤
b
m
for all
m
and for all
n.
(Proof: let
k
= max(
m,n
)
.
Then
a
n
≤
a
k
≤
b
k
≤
b
m
.
) Hence we can applythe result one of our homework exercises to show that
x
= lim
a
n
≤
b
m
for all
m.
We now havethat for each
n, x
≥
a
n
and
x
≤
b
n
.
Thus for each
n, x
∈
I
n
and we are done.
Corollary 1.
The Nested Interval Corollary:
Let
I
1
⊇
I
2
⊇
I
3
⊇
...
be a nested sequence of closed bounded intervals in
R
and suppose that
limlength(
I
n
) = 0
.
Then the intersection of all these intervals consists exactly of a single point. More explicitly, there exists a unique
x
∈
R
such that
x
∈
I
n
for every
n
∈
N
, and for every
>
0
there exists an
N
such that for all
n > N,
the interval
I
n
is contained in the open interval
(
x
−
,x
+
)
.
Proof.
By the Nested Interval Theorem, there is some
x
such that
x
∈
I
n
for every
n
∈
N
.
Now welet
be any positive number and we consider the open interval (
x
−
,x
+
)
.
This interval can berewritten as
{
y
:

y
−
x

<
}
.
Choose
N
such that for all
n > N,
length(
I
n
)
< .
Now suppose
y
is any element of
I
n
for some
n > N.
We also know that
y
∈
I
n
.
Apply the result that we calledthe Interval Diameter Lemma – it tells us that

y
−
x
 ≤
length(
I
n
)
<
, so that
y
∈
(
x
−
,x
+
)
.
Since this is true for all
y
∈
I
n
,
we have that
I
n
⊆
(
x
−
,x
+
)
.
Finally, suppose that, besides
x,
there is also
y
such that
y
∈
I
n
for every
n
∈
N
.
Then by the proof we have just given,

y
−
x

< ,
for all
>
0
.
By a problem we have done – this is essentially Homework 1 Problem 8 – this makes
y
=
x,
so that there is only one number with that property.
We will have many occasions to use this theorem and this corollary throughout this course. Mostof the time, we will be making use of it through the following construction:
The Successive Bisection Construction.
Start with some
I
=
I
1
= [
a,b
]
.
“Cut this intervalin half” – that is consider the left half of it, which is the interval
a, a
+
b
2
,
and the right half of it, which is the interval
a
+
b
2
,b
.
By some means, which depends on the purpose to which youare putting this construction, choose one or the other of these two halves, and call it the interval
I
2
.
Now cut
I
2
in half and repeat the process. In general, if we have already chosen
I
n
= [
a
n
,b
n
]
,
then we consider its two halves: the left half
a
n
, a
n
+
b
n
2
and the right half
a
n
+
b
n
2
,b
n
.
Wechoose one of these halves and call it
I
n
+1
,
and continue the construction. We obtain a sequence of intervals (
I
n
) with the property that there exists a unique
x
∈
R
such that
x
∈
I
n
for every
n
∈
R
,
and for every
>
0 there exists an
N
such that for all
n > N,
the interval
I
n
is contained in theopen interval (
x
−
,x
+
)
.
Discussion.
It is clear from the construction that the sequence (
I
n
) is a sequence of nested closedintervals and we can also see that length(
I
n
) = 2
−
n
+1
·
length(
I
1
)
,
which goes to 0 as
n
→ ∞
.
Thusthe hypotheses of the Nested Interval Corollary are satisﬁed, and the conclusions of that corollarythen follow.3.
Theorems about sequences:
Now we are ready to apply this machinery. What major theorems can we obtain from it?Certainly two of the biggest theorems in analysis, as follows:
Theorem 2.
The BolzanoWeierstrass Theorem:
Every bounded sequence has a convergent subsequence.Proof.
Of course, this theorem is in Ross. It is Theorem 11.5, and there is a proof on that pagethat depends on the existence of a monotone subsequence. Let me give a very diﬀerent proof here.
2
To say that a sequence (
s
n
) is bounded is to say that it is bounded below and bounded above;that is, there is some
a
∈
R
and some
b
∈
R
such that for all
n
∈
N
, a
≤
s
n
≤
b.
In other words,
s
n
∈
[
a,b
]
,
which is a closed, bounded interval. Let
I
1
= [
a,b
]
.
choose any element of the sequenceand call it
s
n
1
.
Since (
s
n
) is a sequence, there are inﬁnitely many elements in it, including inﬁnitelymany elements that come later in the sequence than
s
n
1
.
Now split
I
1
in half, as in the SuccessiveBisection Construction. One or the other of the halves of the interval must still contain inﬁnitelymany terms of the sequence. Choose that half as
I
2
.
(Of course, it may well happen that both of the halves of the interval contain inﬁnitely many terms. We have no problem with that – we justﬂip a coin and take one half or the other, it doesn’t matter which.) Choose an
n
2
> n
1
such that
s
n
2
∈
I
2
.
(With inﬁnitely many terms to choose from, we can always do that.) Continue. Oncewe’ve chosen the interval
I
i
and the term
s
n
i
∈
I
i
,
we divide
I
i
into its two halves. Since there areinﬁnitely many terms of the sequence in
I
i
, there are inﬁnitely many terms of the sequence in atleast one of the halves. Let that half be
I
i
+1
,
and choose some
n
i
+1
> n
i
so that
s
n
i
+1
∈
I
i
+1
.
(which we can always do because we have inﬁnitely many terms of the sequence to choose from.)Now we claim that the subsequence (
s
n
i
) converges. By the Nested Interval Corollary, there is anumber
x
which is in all of the intervals
I
i
.
Furthermore, for each
>
0
,
we can ﬁnd an
N
such thatif
i > N,
then
I
i
⊆
(
x
−
,x
+
)
.
During the course of our proof of that corollary, we rewrote thislast condition as the following: for all
y
∈
I
i
,

y
−
x

< .
But
s
n
i
∈
I
i
,
so for all
i > N,

s
n
i
−
x

< .
This is exactly the same as saying that lim
i
→∞
s
n
i
=
x,
so this subsequence converges.
Finally, we want to show the other version of completeness – the version of completeness that wewill want to carry with us when we start talking about sequences of things that aren’t necessarilyreal numbers:
Theorem 3.
The Cauchy Completeness Theorem:
All Cauchy sequences converge.Proof.
This is proved in Chapter 10 in Ross, and there is a proof there that involves limsup’s andliminf’s. We must prove that if (
s
n
) is a Cauchy sequence, then it converges. We do this in threesteps:
Step 1:
Since (
s
n
) is a Cauchy sequence, it is a bounded sequence. This is Lemma 10.10 in Ross,and its proof is there.
Step 2:
Since (
s
n
) is a bounded sequence, it has a convergent subsequence (
s
n
i
)
.
That is, it has asubsequence (
s
n
i
) such that lim
i
→∞
s
n
i
=
x.
This, of course, is the BolzanoWeierstrass Theorem andwe have just proved it.
Step 3:
We now must show that lim
n
→∞
s
n
=
x.
We do this as follows: let
>
0
.
Choose
J
such thatfor all
i > J,

s
n
i
−
x

<
2
.
(We can do this since the subsequence converges.) Next, choose
M
such that if
n
and
m
are both greater than
M,
then

s
n
−
s
m

<
2
.
(We can do this because it isa Cauchy sequence.) Now let
N
= max(
M,n
J
)
.
For
n > N
we choose as well some
n
i
with
i > J,
so that
n
i
> n
J
.
Then:

s
n
−
x

=

s
n
−
s
n
i
+
s
n
i
−
x
 ≤ 
s
n
−
s
n
i

+

s
n
i
−
x

<
2 +
2 =
Thisproves the theorem.
4.
Versions of completeness:
What are the real numbers? They are a “complete ordered ﬁeld”. But what ways are there tosay “complete”? I oﬀer the following set of ideas.We take the real numbers to be an ordered ﬁeld (they satisfy the axioms A1 through O5 insection 3 of Ross), and then we must make one more assumption – one more axiom. We need toassume something that will make this set
complete.
I claim that we could make any one of the
3
following seven statements that axiom, and that no matter which one we started with, we couldprove the other six as theorems.(1)
The Least Upper Bound Property:
Every nonempty, bounded above set of real numbers has a least upper bound (a supremum).(This is assumed in Ross as an axiom.)(2)
The Dedekind Separation Property:
If there are two nonempty sets
A
and
B
withthe following properties:
A
∩
B
=
∅
, A
∪
B
=
R
,
and for all
a
∈
A
and
b
∈
B, a < b,
theneither
A
has a maximum element or
B
has a minimum element.(This isn’t in Ross anywhere.)(3)
The Monotone Sequence Theorem
(4)
The Cauchy Completeness Theorem
(5)
The BolzanoWeierstrass Theorem
(6)
The Nested Interval Theorem
(7)
The Decimal Expansion Property:
Every decimal expansion stands for a real numberand every real number has at least one decimal expansion.(We will discuss this in connection with section 16. The second half of this – that everyreal number has a decimal expansion – is not really a completeness property, but we needit anyway to assure us the the real numbers have the Archimedean property – alternatively,we could restate (7) as “Every decimal expansion stands for a real number and the realnumbers are Archimedean.”)Ross contains the following parts of the proofs that all of these properties are equivalent:(1) =
⇒
(3), in 10.1; (3) =
⇒
(4), in 10.11, (with the help of the concepts of limsup and liminf);(3) =
⇒
(5), in 11.5, (using 11.3 along the way), and (3) =
⇒
(7), split between the discussion in10.3 and section 16. In this handout, we have provided proofs of the following steps: (3) =
⇒
(6),(6) =
⇒
(5), and (5) =
⇒
(4). I can fairly readily think of proofs that (4) =
⇒
(5), (4) =
⇒
(3),(7) =
⇒
(3), (1) =
⇒
(2), (2) =
⇒
(1), and (absolutely vital to complete the connection)(6) =
⇒
(2). I may eventually assign several of these as homework exercises.There are basically three constructions of the real numbers (assurances that through the humanintellectual processes of mathematics, there is such a mathematical object that satisﬁes the axiomsof a complete ordered ﬁeld): One is simply to declare that the real numbers are inﬁnite decimalexpansions. This approach is fairly grubby, since you must deal with the fact that some numbershave more than one such expansion, and since you must describe in excruciating detail haw youwould add and multiply inﬁnite decimal expansions – but it can be done. Once we’ve done thisconstruction, we have automatically that axiom (7) is true, from which we can prove the other sixproperties. The next construction is Dedekind’s – the famous “Dedekind cuts” that Ross brieﬂydiscusses in section 6. (If anyone is curious about what 19thcentury mathematics looks like writtenout, I have a reprinted translation of Dedekind’s srcinal article available.) Once we’ve done thisconstruction, we have (quite readily) either axiom (1) or axiom (2), from which (as in this book)we can prove all the others. Finally, there is Weierstrass’s construction of the real numbers asequivalence classes of Cauchy sequences of rational numbers – an idea that generalizes to all metricspaces. Under this construction, it is axiom (4) that we start with on our way to proving all theothers.
4