The Real Numbers | An Introduction to Real Analysis

Introduction

Recall that a rational number is a number

that can be written in the form

where

are integers with

. The rational number system is all you need to accomplish most everyday tasks. For instance, to measure distances when building a house it suffices to use a tape measure with an accuracy of about

of an inch. However, to do mathematical analysis the rational numbers have some very serious shortcomings; here is a an example.

If

then

is not a rational number.

Suppose by contradiction that there exists

such that

. We may write

for some integers

, and we can assume that

and

have no common factor other than

(that is,

and

are relatively prime). Now, since

then

and thus

is an even number. This implies that

is also even. Since

is even, we may write

for some

and therefore

, from which it follows that

. Hence,

is even and thus

is also even. Thus, both

and

are even, which contradicts the fact that

and

are relatively prime.

The previous theorem highlights that the set of rational numbers are in some sense incomplete, or that there are gaps in

, and that a larger number system is needed to enlarge the set of math problems that can be analyzed and solved. Although mathematicians in the 1700s were using the real number system and resorting to limiting processes to analyze problems in physics, it was not until the late 1800s that mathematicians gave a rigorous construction of the real number system. Motivated by Theorem 2.1.1, we might be tempted to define the real numbers as the set of solutions of all polynomial equations with integer coefficients. As it turns out, however, this definition of the reals would actually leave out almost all the real numbers, including some of our favorites like

and

. In fact, the set of all numbers that are solutions to polynomial equations with rational coefficients is countable! There are two standard ways to construct the set of real numbers. One standard method to construct

uses the notion of Cauchy sequences of rational numbers and is attributed to Georg Cantor \cite{DG:01}. We will cover Cauchy sequences in Section and therefore postpone describing some of the details of the construction until then. The second standard method to construct the reals relies on the notion of a Dedekind cut and is attributed to Richard Dedekind (1831-1916). A Dedekind cut is a partition

of

such that both

and

are non-empty and

if and then , and
for any there exists such that .

The set of real numbers

is then defined to be the set of all Dedekind cuts. For example, let

and thus

. Then one can show that

is a Dedekind cut of

and the idea is that

represents the real number

such that

, that is, the irrational number

. Having defined

as the set of Dedekind cuts we then proceed to define all the usual operations of arithmetic and arrive at the familiar model of

\cite{DG:01}. Additionally, if

and

then we write that

if

and write

if

and

. Refer to \cite{DG:01} for further details. In this book, we instead adopt the familiar viewpoint that the real numbers

are in a one-to-one correspondence with the points on an infinite line:

figures/real-line.svg — The real numbers are in a one-to-one correspondence with points on an infinite line

The essential feature that we want to capture with this view of

is that there are no holes in the real number system. This view of

allows us to quickly start learning the properties of

instead of focusing on the details of constructing a model for

. Naturally, the rational numbers

are a subset of

and we say that a real number

is irrational if it is not rational. As we saw in Thereom 2.1.1, the positive number

such that

is irrational.

Exercises

Let

be fixed. Prove the following statements without using proof by contradiction.

Prove that if then .
Suppose in addition that . Prove that if then .

Prove that if

then

for all natural numbers

. Do not assume that

is rational.

Algebraic and Order Properties

We will soon see the main difference between

and

from an analysis point of view but in this section we will discuss one important thing that

and

have in common, namely, both are ordered fields. We begin then with the definition of a field.

A field is a set

with two binary operations

and

, the former called addition and the latter multiplication, satisfying the following properties:

for all
for all
There exists an element such that for all
There exists an element such that for all
For each , there exists an element such that .
For each , there exists an element such that .

It is not hard to see that

and

are not fields. In each case, what property of a field fails to hold?

Both

and

are fields.

Besides being fields, both

and

are totally ordered sets. By totally ordered we mean that for any

either

,

, or

. This property of

is referred to as the Law of Trichotomy. For

, the relation

means that either

or

. Similarly,

means that

. From our number line viewpoint of

, if

then

is on the left of

. We now present some very important rules of inequalities that we will use frequently in this book.

Let

.

If and then . (transitivity)
If then .
If and then .
If and then .
If then either or .
If then .

The two inequalities

and

as sometimes combined as

Suppose that

and

, or written more compactly as

. Prove that if

then

.

Suppose that

and

. Prove that

. Deduce that if

and

then

We will encounter situations where we will need to prove that if two numbers

satisfy a certain property then

. Proving that

is equivalent to proving that

. In such situations, the following theorem will be very useful.

Let

be non-negative, that is,

. If for every

it holds that

then

.

We prove the contrapositive, that is, we prove that if

then there exists

such that

. Assume then that

and let

. Then

and clearly

.

The next few examples will give us practice with working with inequalities.

Let

. Find a natural number

such that

Let

. Find analytically a natural number

such that

Let

. Find analytically a natural number

such that

Exercises

Let

and find analytically a natural number

such that

Let

and find analytically a natural number

such that

Let

and find analytically a natural number

such that

The Absolute Value

To solve problems in calculus you need to master differentiation and integration. To solve problems in analysis you need to master inequalities. The content of this section, mostly on inequalities, is fundamental to everything else that follows in this book. Given any

, we define the absolute value of

as the number

Clearly,

if and only if

, and

for all

. Below we record some important properties of the absolute value function.

Let

and

.

if and only if

Statements (i) and (ii) are trivial. To prove (iii), first suppose that

. If

then

. Hence,

and since

then

. Hence,

. If

then

, and thus

. Since

then

. Thus,

. Now suppose that

. If

then

. If

then from multiplying the inequality by

we have

and thus

. To prove part (iv), notice that

and thus applying (iii) we get

.

If

prove that

.

From Theorem 2.3.1 part (i), we have that

. Therefore,

. Similarly, one can show that

. By induction, for each

it holds that

.

Below is the most important inequality in this book.

For any

it holds that

We have that

from which it follows that

and thus

By induction, one can prove the following corollary to the Triangle inequality.

For any

it holds that

A compact way to write the triangle inequality using summation notation is

Here is another consequence of the Triangle inequality.

For

it holds that

For part (i), we have

For part (ii), consider

and therefore

. Switching the role of

and

we obtain

, and therefore multiplying this last inequality by

yields

. Therefore,

which is the stated inequality.

For

prove that

.

Let

for

. Find a number

such that

for all

.

Clearly, if

then

. Apply the triangle inequality and the properties of the absolute value:

Therefore, if

then

for all

.

Let

. Find a number

such that

for all

.

It is clear that if

then

. Using the properties of the absolute value and the triangle inequality repeatedly on the numerator:

Now, for

we have that

and therefore

and then

. Therefore,

Hence, we can take

.

Let

. Find a number

such that

for all

.

In analysis, the absolute value is used to measure distance between points in

. For any

, the absolute value

is the distance from

to

. This interpretation of the absolute value can be used to measure the difference (in magnitude) between two points. That is, given

, the distance between

and

is

. From the properties of the absolute value, this distance is also

.

figures/absolute-value-distance.svg — The number is the distance between and .

We will often be concerned with how close a given number

is to a fixed number

. To do this, we introduce the notion of neighborhoods based at

.

Let

and let

. The

-neighborhood of of radius is the set

Notice that if

then

.

Let

and let

. From calculus, we know that

. Find a natural number

such that

for every

.

The inequality

means that

is within

of

. That is,

This inequality does not hold for all

, but it will eventually hold for some

and for all

. For example,

and

, and similarly for

and

. In fact,

. However, because we know that

, eventually

for large enough

. To find out how large, let's analyze the magnitude

:

Hence,

if and only if

which after re-arranging can be written as

With

we obtain

Hence, if

then if

then

.

Let

and

, and let

. Show that

and

are

-neighborhoods of

for some appropriate value of

.

Exercises

Prove that if

and

then

. Draw a number line with points

satisfying the inequalities and graphically interpret the inequality

.

Let

be positive real numbers and consider the polynomial

Prove that

for all

. {Hint: For example, if say

then you are asked to prove that

However, prove the claim for a general polynomial

with

.}

Let

for

. Find analytically a number

such that

for all

. Do not use calculus to find

.

Let

for

. Find analytically a number

such that

for all

. Do not use calculus to find

.

Let

for

. Find analytically a number

such that

for all

. Do not use calculus to find

. (Hint: Complete the square.)

Let

be distinct points. Show that there exists neighborhoods

and

such that

.

The Completeness Axiom

In this section, we introduce the Completeness Axiom of

. Recall that an axiom is a statement or proposition that is accepted as true without justification. In mathematics, axioms are the first principles that are accepted as truths and are used to build mathematical theories; in this case real analysis. Roughly speaking, the Completeness Axiom is a way to say that the real numbers have no gaps or no holes, contrary to the case of the rational numbers. As you will see below, the Completeness Axiom is centered around the notions of bounded sets and least upper bounds; let us begin then with some definitions.

Let

be a non-empty set.

We say that is bounded above if there exists such that for all . We then say that is an upper bound of .
We say that is bounded below if there exists such that for all . We then say that is a lower bound of .
We say that is bounded if it is both bounded above and bounded below.
We say that is unbounded if it is not bounded.

For each case, determine if

is bounded above, bounded below, bounded, or unbounded. If the set is bounded below, determine the set of lower bounds, and similarly if it is bounded above.

Let

and

be sets and suppose that

.

Prove that if is bounded above (below) then is bounded above (below).
Give an example of sets and such that is bounded below but is not bounded below.

We now come to an important notion that will be at the root of what we do from now.

Let

be non-empty.

Let be bounded above. An upper bound of is said to be a least upper bound of if for any upper bound of . In this case we also say that is a supremum of and write .
Let be bounded below. A lower bound of is said to be a greatest lower bound of if for any lower bound of . In this case we also say that is an infimum of and write .

It is straightforward to show that a set that is bounded above (bounded below) can have at most one supremum (infimum). At the risk of being repetitive, when it exists,

is a number that is an upper bound of

and is the smallest possible upper bound of

. Therefore,

for all

, and any number less than

is not an upper bound of

(because

is the least upper bound!). Similarly, when it exists,

is a number that is a lower bound of

and is the largest possible lower bound of

. Therefore,

for all

and any number greater than

is not a lower bound of

(because

is the greatest lower bound of

!).

In some analysis texts,

is written as

and

is written as

. In other words,

and

.

Does every non-empty bounded set

have a supremum/infimum? You might say Yes, of course!! and add that It is a self-evident principle and needs no justification!. Is not that what an axiom is?

Every non-empty subset of

that is bounded above has a least upper bound (a supremum) in

. Similarly, every non-empty subset of

that is bounded below has a greatest lower bound (an infimum) in

.

As you will see in the pages that follow, The Completeness Axiom is the key notion upon which the theory of real analysis depends on.

Determine

and

, if they exist.

The Completeness Axiom is sometimes called the supremum property of

or the least upper bound property of

. The Completeness property makes

into a complete ordered field. The following example shows that

does not have the completeness property.

The set of rational numbers is an ordered field but it is not complete. Consider the set

. By definition,

. Clearly,

is bounded above, for example

is an upper bound of

, but the least upper bound of

is

which is not a rational number. Therefore

does not have a supremum in

and therefore

does not have the Completeness property. From the point of view of analysis, this is the main distinction between

and

(note that both are ordered fields).

In some cases, it is obvious what

and

are, however, to do analysis rigorously, we need systematic ways to determine

and

. To start with, we first need to be a bit more rigorous about what it means to be the least upper bound, or at least have a more concrete description that we can work with, i.e., using inequalities. The following lemma does that and, as you will observe, it is simply a direct consequence of the definition of the supremum.

Let

be non-empty and suppose that

is an upper bound of

. Then

is the least upper bound of

if and only if for any

there exists

such that

Suppose that

is the supremum of

, that is,

is the least upper bound of

. Since

then

is not an upper bound of

. Thus, there exists

such that

. Now suppose that for any

there exists

such that

. Now let

be such that

. Then there exists

such that

, and thus by assumption, there exists

such that

. Hence,

is not an upper bound of

and this shows that

is the least upper bound of

, that is,

.

If

, and

is bounded above, prove that

is bounded above and that

.

Since

is bounded above,

exists by the Completeness property of

. Let

. Then

and therefore

. This proves that

is bounded above by

and therefore

exists. Since

is the least upper bound of

we must have

. For example, if say

and

then

, while if

then

.

Let

be non-empty and bounded above. Let

and define the set

Prove that if

then

is bounded above and

. Show by example that if

then

need not be bounded above even if

is bounded above.

Let

be arbitrary. Then there exists

such that

. By the Completeness property,

exists and

. If

then

which is equivalent to

. Since

is arbitrary, this shows that

is an upper bound of the set

and thus

is bounded above and thus

exists. Because

is the least upper bound of

then

Now, by definition,

for all

. Thus,

for all

and therefore

for all

. Therefore,

is an upper bound of

and consequently

because

is the least upper bound of

. We have therefore proved that

Combining

and

we conclude that

Suppose that

and

are non-empty and bounded above. Prove that

is bounded above and that

Let

. Then clearly

and

. We first show that

is bounded above by showing that

is an upper bound of

. Let

. Then either

or

(or both). If

then

and if

then

. Hence,

is bounded above and

is an upper bound of

. Consequently,

exists by the Completeness axiom and moreover

, that is,

Now, by definition of the supremum,

for all

. Now since

this implies that

for all

and

for all

. In other words,

is an upper bound of both

and

and thus

and

. Then clearly

and combining

and

we have proved that

.

Suppose that

and

are non-empty and bounded below, and suppose that

is non-empty. Prove that

is bounded below and that

If

then

and therefore

, and also

and thus

. Therefore, both

and

are lower bounds of

, and by definition of

we have that

and

, and consequently

.

For any set

define the set

Prove that if

is non-empty and bounded then

.

It holds that

for all

and therefore,

for all

, which is equivalent to

for all

. Therefore,

is an upper bound of the set

and therefore

. Now,

for all

, or equivalently

for all

. Hence,

for all

. This proves that

is a lower bound of

and therefore

, or

. This proves that

.

Let

be non-empty and bounded below. Let

and define the set

Prove that

is bounded below and that

.

For all

it holds that

and therefore

. This proves that

is a lower bound of the set

, and therefore

. Now,

for all

and thus

for all

, which is the same as

for all

. Hence,

or equivalently

. This proves the claim.

Let

and

be non-empty subsets of

, and suppose that

and

are bounded below. Define the set

.

Prove that is bounded below.
Prove that . Hint: Consider two cases, when say and when .
How do things change if we do not assume and are subsets of .

Give an example of a non-empty set

such that

.

For any two non-empty sets

and

of

let us write that

if, for each

, there exists

such that

.

Prove that if then .
Show via an example that if then it is not necessarily true that .

Let

and

be non-empty bounded sets of positive real numbers. Define the set

Assume that

. Prove that

Since

is bounded above,

exists and

for all

. Since

is bounded below,

exists and

for all

. Let

be an arbitrary point in

. Then since

and

we obtain that

Since

and

(and then clearly

) we obtain

This proves that

and since

was arbitrary we have proved that

is an upper bound of the set

. This proves that

exists. Moreover, by the definition of the supremum, we also have that

Now,

for all

and thus

for all

and all

. If

is held fixed then

for all

and thus

. Therefore,

, which holds for all

. Therefore,

and consequently

Combining

and

completes the proof.

Exercises

Let

be a non-empty set and suppose that

is an upper bound of

. Prove that if

then necessarily

.

Let

and

be non-empty subsets of

, and suppose that

. Prove that if

is bounded below then

.

If

and

are non-empty subsets of

such that

and

does it follow that

? Support your answer with either a proof or a counterexample.

Let

be a bounded set, let

be fixed, and define the set

Prove that

.

Let

be bounded above. Let

Prove that

is bounded above and that

.

Let

denote the set of all positive real numbers and let

be bounded. Assume that

. Define the set

Prove that

Applications of the Supremum

Having defined the notion of boundedness for a set, we can define a notion of boundedness for a function.

Let

be a non-empty set and let

be a function.

is bounded below if the range is bounded below.
is bounded above if the range is bounded above.
is bounded if the range is bounded.

Hence, boundedness of a function

is boundedness of its range.

Suppose

are bounded functions and

for all

. Show that

.

Since

is bounded,

exists and is by definition an upper bound for the set

, that is,

for all

. Now, by assumption, for all

it holds that

and therefore

. This shows that

is an upper bound of the set

, and therefore by definition of the supremum,

.

Let

be bounded functions and suppose that

for all

. Show that

.

Fix

. Then,

for all

. Therefore,

is a lower bound of

, and thus

by definition of the infimum. Since

was arbitrary, we have proved that

for all

. Hence,

is an upper bound of

and thus by definition of the supremum we have that

.

The following sequence of results will be used to prove an important property of the rational numbers

as seen from within

.

If

then there exists

such that

.

Suppose not. Hence,

for all

, and thus

is an upper bound for

, and therefore

is bounded. Let

. By definition of

,

is not an upper bound of

and therefore there exists

such that

. But then

and this contradicts the definition of

.

If

then

.

Since

for all

then

is a lower bound of

. Now suppose that

. By the Archimedean Property, there exists

such that

and thus

. Hence,

is not a lower bound of

. Therefore

is the greatest lower bound of

, that is,

.

For any

there exists

such that

.

Given

there exists

such that

.

Let

. By the Archimedean property,

is non-empty. By the Well-Ordering Principle of

,

has a least element, say that it is

. Hence,

and thus

.

We now come to an important result that we will use frequently.

If

and

then there exists

such that

.

We first prove the claim for the case that

. Suppose that

and thus

. There exists

such that

and thus

. Therefore,

and thus

and we may take

. In general, if

then there exists

such that

and thus

. Since

there exists

such that

and thus

and dividing by

yields

and thus we may take

. This proves the claim when both

and

are positive. If

then take

and if

then apply the previous arguments to

.

Hence, between any two distinct real numbers there is a rational number. This implies that any irrational number can be approximated by a rational number to within any degree of accuracy.

Let

be an irrational number and let

be arbitrary. Prove that there exists a rational number

such that

, that is,

is in the

-neighborhood of

.

If

and

then there exists

such that

.

We have that

. By the Density Theorem,

for some

. Then

.

Exercises

Let

be non-empty and let

be functions. Let

denote the function defined by

for any

. If

and

are bounded above prove that

is also bounded above and that

If

prove that there exists

such that

. (Note: If you begin with

and solve for

then you are assuming that such an

exists. You are not asked to find an

, you are asked to prove that such an

exists.)

Nested Interval Theorem

If

and

define

The set

is called an open interval from

to

. Define also

which we call the closed interval from

to

. The following are called half-open (or half-closed) intervals:

If

then

and

. Infinite intervals are

Below is a characterization of intervals, we will omit the proof.

Let

contain at least two points. Suppose that if

and

then

. Then

is an interval.

A sequence

of intervals is nested if

As an example, consider

where

. Then

is nested:

Notice that since

for each

then

Is there another point in

? Suppose that

and

. Then necessarily

. Then there exists

such that

. Thus,

and therefore

. Therefore,

In general, we have the following.

Let

be a sequence of nested closed bounded intervals. Then there exists

such that

for all

, that is,

. In particular, if

for

, and

and

then

Since

is a closed interval, we can write

for some

and

for all

. The nested property can be written as

Since

for all

then

for all

. Therefore, the set

is bounded above. Let

and thus

for all

. We will show that

for all

also. Let

be arbitrary. If

then

and therefore

. On the other hand, if

then

and therefore

. In any case,

for all

. Hence,

is an upper bound of

, and thus

. Since

was arbitrary, we have that

for all

. Therefore,

for all

, that is

. The proof that

is similar.

The following theorem gives a condition when

contains a single point.

Let

be a sequence of nested closed bounded intervals. If

then

is a singleton set.

Let

for

.

Prove that is a sequence of nested intervals.
Find .

Using the Nested Interval property of

, we give a proof that

is uncountable

The real numbers

are uncountable.

We will prove that the interval

is uncountable. Suppose by contradiction that

is countable, and let

be an enumeration of

(formally this means we have a bijection

and

). Since

, there exists a closed and bounded interval

such that

. Next, consider

. There exists a closed and bounded interval

such that

. Next, consider

. There exists a closed and bounded interval

such that

. By induction, there exists a sequence

of closed and bounded intervals such that

for all

. Moreover, by construction the sequence

is nested and therefore

is non-empty, say it contains

. Clearly, since

for all

then

. Now, since

for each

then

. Therefore,

for all

and thus

, which is a contradiction since

. Therefore,

is uncountable and this implies that

is also uncountable.

We now give an alternative proof that

is uncountable. To that end, consider the following subset

:

In other words,

consists of numbers

whose decimal expansion consists of only 0's and 1's. For example, some elements of

are

If we can construct a bijection

, then since

is uncountable then by Example 1.4.7 this would show that

is uncountable. Since

then this would show that

is uncountable (by Theorem 1.4.8). To construct

, given

in

define

as

In other words,

consists of the decimal places in the decimal expansion of

that have a value of

. For example,

It is left as an exercise to show that

is a bijection (see Exercise 1.4.4).

Exercises

Let

for

. Prove that

.