The Real Numbers | An Introduction to Real Analysis

Introduction

Recall that a rational number is a number \(x\) that can be written in the form \(x = \frac{p}{q}\) where \(p, q\) are integers with \(q\neq 0\). 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 \(\tfrac{1}{16}\) of an inch. However, to do mathematical analysis the rational numbers have some very serious shortcomings; here is a an example.

If \(x^2 = 2\) then \(x\) is not a rational number.

Suppose by contradiction that there exists \(x\in\mathbb{Q}\) such that \(x^2 = 2\). We may write \(x=\frac{p}{q}\) for some integers \(p,q\), and we can assume that \(p\) and \(q\) have no common factor other than \(1\) (that is, \(p\) and \(q\) are relatively prime). Now, since \(x^2 = 2\) then \(p^2 = 2q^2\) and thus \(p^2\) is an even number. This implies that \(p\) is also even. Since \(p\) is even, we may write \(p=2k\) for some \(k\in\mathbb{N}\) and therefore \((2k)^2 = 2q^2\), from which it follows that \(2k^2 = q^2\). Hence, \(q^2\) is even and thus \(q\) is also even. Thus, both \(p\) and \(q\) are even, which contradicts the fact that \(p\) and \(q\) are relatively prime.

The previous theorem highlights that the set of rational numbers are in some sense incomplete, or that there are gaps in \(\rat\), 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 \(\pi\) and \(e\). 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 \(\real\) 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 \(\{A, B\}\) of \(\rat\) such that both \(A\) and \(B\) are non-empty and

if \(b\in A\) and \(a \lt b\) then \(a\in A\), and
for any \(a\in A\) there exists \(b\in A\) such that \(a \lt b\).

The set of real numbers \(\real\) is then defined to be the set of all Dedekind cuts. For example, let \(A = \{x\in\rat\;|\; x^2 \lt 2 \text{ or } x \lt 0\}\) and thus \(B = \rat\backslash\hspace{-0.3em} A\). Then one can show that \(\{A, B\}\) is a Dedekind cut of \(\rat\) and the idea is that \(x=\{A, B\}\) represents the real number \(x\) such that \(x^2 = 2\), that is, the irrational number \(\sqrt{2}\). Having defined \(\real\) as the set of Dedekind cuts we then proceed to define all the usual operations of arithmetic and arrive at the familiar model of \(\real\) \cite{DG:01}. Additionally, if \(x = \{A, B\}\) and \(y=\{C, D\}\) then we write that \(x\leq y\) if \(A\subset C\) and write \(x \lt y\) if \(A\subset C\) and \(A\neq C\). Refer to \cite{DG:01} for further details. In this book, we instead adopt the familiar viewpoint that the real numbers \(\real\) 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 \(\real\) is that there are no holes in the real number system. This view of \(\real\) allows us to quickly start learning the properties of \(\real\) instead of focusing on the details of constructing a model for \(\real\). Naturally, the rational numbers \(\rat\) are a subset of \(\real\) and we say that a real number \(x\in\real\) is irrational if it is not rational. As we saw in Thereom 2.1.1, the positive number \(x\in\real\) such that \(x^2=2\) is irrational.

Exercises

Let \(x\in\rat\) be fixed. Prove the following statements without using proof by contradiction.

Prove that if \(y\in\real\backslash\hspace{-0.3em}\rat\) then \(x+y\in\real\backslash\hspace{-0.3em}\rat\).
Suppose in addition that \(x \gt 0\). Prove that if \(y\in\real\backslash\hspace{-0.3em}\rat\) then \(xy\in\real\backslash\hspace{-0.3em}\rat\).

Prove that if \(0 \lt x \lt 1\) then \(x^n \lt x\) for all natural numbers \(n\geq 2\). Do not assume that \(x\) is rational.

Algebraic and Order Properties

We will soon see the main difference between \(\rat\) and \(\real\) from an analysis point of view but in this section we will discuss one important thing that \(\rat\) and \(\real\) have in common, namely, both are ordered fields. We begin then with the definition of a field.

A field is a set \(\mathbb{F}\) with two binary operations \(+:\mathbb{F}\times\mathbb{F}\rightarrow\mathbb{F}\) and \(\times:\mathbb{F}\times\mathbb{F}\rightarrow\mathbb{F}\), the former called addition and the latter multiplication, satisfying the following properties:

\(a+b=b+a\) for all \(a,b\in\mathbb{F}\)
\((a+b)+c = a+(b+c)\) for all \(a,b,c\in\mathbb{F}\)
\(a\times (b+c) = a\times b + a\times c\)
There exists an element \(1\in\mathbb{F}\) such that \(a\times 1 = 1\times a = a\) for all \(a\in\mathbb{F}\)
There exists an element \(0\in\mathbb{F}\) such that \(a + 0 = 0 + a = a\) for all \(a\in\mathbb{F}\)
For each \(a\in\mathbb{F}\), there exists an element \(-a\in\mathbb{F}\) such that \(a + (-a) = (-a) + a =0\).
For each \(a\in\mathbb{F}\), there exists an element \(a^{-1}\in\mathbb{F}\) such that \(a\times a^{-1} = a^{-1}\times a =1\).

It is not hard to see that \(\N\) and \(\mathbb{Z}\) are not fields. In each case, what property of a field fails to hold?

Both \(\rat\) and \(\real\) are fields.

Besides being fields, both \(\rat\) and \(\real\) are totally ordered sets. By totally ordered we mean that for any \(a, b\in\real\) either \(a=b\), \(a \lt b\), or \(b \lt a\). This property of \(\real\) is referred to as the Law of Trichotomy. For \(a, b\in \real\), the relation \(a\leq b\) means that either \(a \lt b\) or \(a=b\). Similarly, \(a\geq b\) means that \(b\leq a\). From our number line viewpoint of \(\real\), if \(a \lt b\) then \(a\) is on the left of \(b\). We now present some very important rules of inequalities that we will use frequently in this book.

Let \(a,b,c\in\real\).

If \(a \lt b\) and \(b \lt c\) then \(a \lt c\). (transitivity)
If \(a \lt b\) then \(a+c \lt b+c\).
If \(a \lt b\) and \(c \gt 0\) then \(ac \lt bc\).
If \(a \lt b\) and \(c \lt 0\) then \(ac \gt bc\).
If \(ab \gt 0\) then either \(a, b \gt 0\) or \(a,b \lt 0\).
If \(a\neq 0\) then \(a^2 \gt 0\).

The two inequalities \(a\leq b\) and \(b\leq c\) as sometimes combined as \[ a\leq b\leq c. \]

Suppose that \(a \gt 0\) and \(b \gt 0\), or written more compactly as \(a,b \gt 0\). Prove that if \(a \lt b\) then \(\frac{1}{b} \lt \frac{1}{a}\).

Suppose that \(a\leq x\) and \(b\leq y\). Prove that \(a+b\leq x+y\). Deduce that if \(a\leq x\leq \xi\) and \(b\leq y\leq \zeta\) then \[ a+b\leq x+y\leq \zeta + \xi. \]

We will encounter situations where we will need to prove that if two numbers \(a, b\in\real\) satisfy a certain property then \(a=b\). Proving that \(a=b\) is equivalent to proving that \(x=a-b = 0\). In such situations, the following theorem will be very useful.

Let \(x\in\real\) be non-negative, that is, \(x\geq 0\). If for every \(\varepsilon \gt 0\) it holds that \(x \lt \varepsilon\) then \(x=0\).

We prove the contrapositive, that is, we prove that if \(x \gt 0\) then there exists \(\eps \gt 0\) such that \(\eps \lt x\). Assume then that \(x \gt 0\) and let \(\eps = \frac{x}{2}\). Then \(\eps \gt 0\) and clearly \(\eps \lt x\).

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

Let \(\veps = 0.0001\). Find a natural number \(n\in\mathbb{N}\) such that \[ \frac{1}{n+1} \lt \veps \]

Let \(\veps = 0.0001\). Find analytically a natural number \(n\in\mathbb{N}\) such that \[ \frac{n+2}{n^2+3} \lt \veps \]

Let \(\veps = 0.001\). Find analytically a natural number \(n\in\mathbb{N}\) such that \[ \frac{5n-4}{2n^3-n} \lt \veps \]

Exercises

Let \(\varepsilon=0.0001\) and find analytically a natural number \(n\in\mathbb{N}\) such that \[ \frac{3n-2}{n^3 + 2n} \lt \varepsilon. \]

Let \(\varepsilon=0.0001\) and find analytically a natural number \(n\in\mathbb{N}\) such that \[ \frac{3n+2}{4n^3 - n} \lt \varepsilon. \]

Let \(\varepsilon=0.0001\) and find analytically a natural number \(n\in\mathbb{N}\) such that \[ \frac{\cos^2(3n)+1}{\arctan(n)+n} \lt \varepsilon. \]

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 \(a\in\real\), we define the absolute value of \(a\) as the number \[ |a|:=\begin{cases} a, & \text{ if } a\geq 0\\[2ex] -a, & \text{ if } a \lt 0.\end{cases} \] Clearly, \(|a| = 0\) if and only if \(a=0\), and \(0\leq |a|\) for all \(a\in\real\). Below we record some important properties of the absolute value function.

Let \(a,b\in\real\) and \(c\geq 0\).

\(|ab| = |a|\cdot |b|\)
\(|a|^2 = a^2\)
\(|a|\leq c\) if and only if \(-c\leq a\leq c\)
\(-|a|\leq a \leq |a|\)

Statements (i) and (ii) are trivial. To prove (iii), first suppose that \(|a|\leq c\). If \(a \gt 0\) then \(a\leq c\). Hence, \(-a\geq -c\) and since \(a \gt 0\) then \(a \gt -a\geq -c\). Hence, \(-c \leq a \leq c\). If \(a \lt 0\) then \(-a\leq c\), and thus \(a\geq -c\). Since \(a \lt 0\) then \(a \lt -a\leq c\). Thus, \(-c\leq a\leq c\). Now suppose that \(-c\leq a \leq c\). If \(0 \lt a\leq c\) then \(|a|=a\leq c\). If \(a \lt 0\) then from multiplying the inequality by \((-1)\) we have \(c\geq -a \geq -c\) and thus \(|a|=-a \leq c\). To prove part (iv), notice that \(|a|\leq |a|\) and thus applying (iii) we get \(-|a|\leq a\leq |a|\).

If \(b\neq 0\) prove that \(\left|\frac{a}{b}\right| = \frac{|a|}{|b|}\).

From Theorem 2.3.1 part (i), we have that \(|a^2| = |a\cdot a| = |a|\cdot |a| = |a|^2\). Therefore, \(|a^2| = |a|^2\). Similarly, one can show that \(|a^3| = |a|^3\). By induction, for each \(n\in\N\) it holds that \(|a^n| = |a|^n\).

Below is the most important inequality in this book.

For any \(x,y\in\real\) it holds that \[ |x+y|\leq |x|+|y|. \]

We have that \begin{gather*} -|x|\leq x\leq |x|\\ -|y|\leq y \leq |y| \end{gather*} from which it follows that \[ -(|x|+|y|)\leq x+y\leq |x|+|y| \] and thus \[ |x+y|\leq |x| + |y|. \]

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

For any \(x_1, x_2, \ldots, x_n \in\real\) it holds that \[ |x_1+x_2+\cdots+x_n| \leq |x_1| + |x_2| + \cdots + |x_n|. \]

A compact way to write the triangle inequality using summation notation is \[ \left|\sum_{i=1}^n x_i\right| \leq \sum_{i=1}^n |x_i|. \] Here is another consequence of the Triangle inequality.

For \(x,y\in\real\) it holds that

\(|x-y|\leq |x| + |y|\)
\(||x|-|y|| \leq |x-y|\)

For part (i), we have \begin{align*} |x-y| &= |x + (-y)|\\ & \leq |x| + |-y| \\ &= |x|+|y|. \end{align*} For part (ii), consider \[ |x| = |x-y+y| \leq |x-y| + |y| \] and therefore \(|x|-|y| \leq |x-y|\). Switching the role of \(x\) and \(y\) we obtain \(|y|-|x| \leq |y-x| = |x-y|\), and therefore multiplying this last inequality by \(-1\) yields \(-|x-y| \leq |x|-|y|\). Therefore, \[ -|x-y| \leq |x|-|y| \leq |x-y| \] which is the stated inequality.

For \(a,b\in\real\) prove that \(|a+b|\geq |a| - |b|\).

Let \(f(x) = 2x^2 - 3x + 7\) for \(x\in [-2,2]\). Find a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(-2\leq x\leq 2\).

Clearly, if \(-2\leq x\leq 2\) then \(|x|\leq 2\). Apply the triangle inequality and the properties of the absolute value: \begin{align*} |f(x)| &= |2x^2 - 3x+7| \\ &\leq |2x^2| + |3x| + |7|\\ &= 2|x|^2 + 3|x| + 7\\ &\leq 2 (2)^2 + 3(2) + 7\\ &= 21. \end{align*} Therefore, if \(M=21\) then \(|f(x)|\leq M\) for all \(x\in [-2,2]\).

Let \(f(x) = \frac{2x^2+3x+1}{1-2x}\). Find a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(2\leq x\leq 3\).

It is clear that if \(2\leq x\leq 3\) then \(|x|\leq 3\). Using the properties of the absolute value and the triangle inequality repeatedly on the numerator: \begin{align*} |f(x)| &= \left| \frac{2x^2+3x+1}{1-2x}\right|\\[2ex] &= \frac{|2x^2+3x+1|}{|1-2x|}\\[2ex] &\leq \frac{|2x^2|+|3x|+|1|}{|1-2x|}\\[2ex] &= \frac{2|x|^2+3|x|+1|}{|1-2x|}\\[2ex] &\leq \frac{2\cdot 3^2 + 3\cdot 3 + 1}{|1-2x|}\\[2ex] &= \frac{28}{|2x-1|}. \end{align*} Now, for \(2\leq x\leq 3\) we have that \(-5\leq 1-2x\leq -3\) and therefore \(3\leq |1-2x|\leq 5\) and then \(\frac{1}{|2x-1|}\leq \frac{1}{3}\). Therefore, \[ |f(x)|\leq \frac{28}{|1-2x|} \leq \frac{28}{3}. \] Hence, we can take \(M=\frac{28}{3}\).

Let \(f(x) = \frac{\sin(2x)-3}{x^2+1}\). Find a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(-3\leq x\leq 2\).

In analysis, the absolute value is used to measure distance between points in \(\real\). For any \(a\in\real\), the absolute value \(|a|\) is the distance from \(a\) to \(0\). This interpretation of the absolute value can be used to measure the difference (in magnitude) between two points. That is, given \(x,y\in\real\), the distance between \(x\) and \(y\) is \(|x-y|\). From the properties of the absolute value, this distance is also \(|y-x|\).

figures/absolute-value-distance.svg — The number \(|x-y|\) is the distance between \(x\) and \(y\).

We will often be concerned with how close a given number \(x\) is to a fixed number \(a\in\real\). To do this, we introduce the notion of neighborhoods based at \(a\).

Let \(a\in\real\) and let \(\varepsilon \gt 0\). The \(\varepsilon\)-neighborhood of \(a\) of radius is the set \[ B_\varepsilon(a) = \{x\in\real\;|\; |x-a| \lt \varepsilon\} = (a-\varepsilon, a + \varepsilon) \]

Notice that if \(\varepsilon_1 \lt \varepsilon_2\) then \(B_{\varepsilon_1}(a) \subset B_{\varepsilon_2}(a)\).

Let \(f(n) = \frac{3n+1}{2n+3}\) and let \(\varepsilon = 0.0001\). From calculus, we know that \(\displaystyle\lim_{n\rightarrow\infty} f(n) = \tfrac{3}{2}\). Find a natural number \(N\) such that \(|f(n) - \frac{3}{2}| \lt \varepsilon\) for every \(n\geq N\).

The inequality \(|f(n)-\frac{3}{2}| \lt \varepsilon\) means that \(f(n)\) is within \(\varepsilon\) of \(\frac{3}{2}\). That is, \[ \frac{3}{2}-\varepsilon \lt f(n) \lt \frac{3}{2}+\varepsilon. \] This inequality does not hold for all \(n\), but it will eventually hold for some \(N\in\N\) and for all \(n\geq N\). For example, \(f(1) = \frac{4}{5} = 0.8\) and \(|f(1)-\frac{3}{2}| = 0.7 \gt \varepsilon\), and similarly for \(f(2) = \frac{7}{7} = 1\) and \(|f(2)-\frac{3}{2}| = 0.5\). In fact, \(|f(20) - \frac{3}{2}| = 0.081 \gt \varepsilon\). However, because we know that \(\lim_{n\rightarrow\infty} f(n) = \frac{3}{2}\), eventually \(|f(n)-\frac{3}{2}| \lt \varepsilon\) for large enough \(n\). To find out how large, let's analyze the magnitude \(|f(n)-\frac{3}{2}|\): \begin{align*} |f(n)-\tfrac{3}{2}| &= \left|\frac{3n+1}{2n+3}-\frac{3}{2}\right|\\[2ex] &= \left|\frac{6n+2 - 6n - 9}{2(2n+3)} \right|\\ &= \frac{7}{2(2n+3)} \end{align*} Hence, \(|f(n)-\frac{3}{2}| \lt \varepsilon\) if and only if \[ \frac{7}{2(2n+3)} \lt \varepsilon \] which after re-arranging can be written as \[ n \gt \frac{7}{4\varepsilon} -\frac{3}{2}. \] With \(\varepsilon=0.0001\) we obtain \[ n \gt 17,498.5. \] Hence, if \(N = 17,499\) then if \(n\geq N\) then \(|f(n)-\frac{3}{2}| \lt \varepsilon\).

Let \(\varepsilon_1 \gt 0\) and \(\eps_2 \gt 0\), and let \(a\in\real\). Show that \(B_{\eps_1}(a)\cap B_{\eps_2}(a)\) and \(B_{\eps_1}(a)\cup B_{\eps_2}(a)\) are \(\eps\)-neighborhoods of \(a\) for some appropriate value of \(\eps\).

Exercises

Prove that if \(a \lt x \lt b\) and \(a \lt y \lt b\) then \(|x-y| \lt b-a\). Draw a number line with points \(a,b,x,y\) satisfying the inequalities and graphically interpret the inequality \(|x-y| \lt b-a\).

Let \(a_0,a_1,a_2,\ldots,a_n\) be positive real numbers and consider the polynomial \[ f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n. \] Prove that \[ |f(x)|\leq f(|x|) \] for all \(x\in\real\). {Hint: For example, if say \(f(x) = 2 + 3x^2 + 7x^3\) then you are asked to prove that \[ \underbrace{|2+3x^2+7x^3|}_{|f(x)|} \leq \underbrace{2 + 3|x|^2 + 7|x|^3}_{f(|x|)}. \] However, prove the claim for a general polynomial \(f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n\) with \(a_i \gt 0\).}

Let \(f(x) = 3x^2 - 7x + 11\) for \(x\in [-4,2]\). Find analytically a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(x\in[-4,2]\). Do not use calculus to find \(M\).

Let \(\displaystyle f(x) = \frac{x - 1}{x^2 + 7}\) for \(x\in [0,10]\). Find analytically a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(x\in [0,10]\). Do not use calculus to find \(M\).

Let \(\displaystyle f(x) = \frac{3\cos(\pi x)}{x^2-2x+3}\) for \(x\in [0,2]\). Find analytically a number \(M \gt 0\) such that \(|f(x)|\leq M\) for all \(x\in [0,2]\). Do not use calculus to find \(M\). (Hint: Complete the square.)

Let \(a,b\in\real\) be distinct points. Show that there exists neighborhoods \(B_\varepsilon(a)\) and \(B_\delta(b)\) such that \(B_\epsilon(a)\cap B_\delta(b)\neq\emptyset\).

The Completeness Axiom

In this section, we introduce the Completeness Axiom of \(\real\). 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 \(S\subset\real\) be a non-empty set.

We say that \(S\) is bounded above if there exists \(u\in\real\) such that \(x\leq u\) for all \(x\in S\). We then say that \(u\) is an upper bound of \(S\).
We say that \(S\) is bounded below if there exists \(w\in\real\) such that \(w\leq x\) for all \(x\in S\). We then say that \(w\) is a lower bound of \(S\).
We say that \(S\) is bounded if it is both bounded above and bounded below.
We say that \(S\) is unbounded if it is not bounded.

For each case, determine if \(S\) 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.

\(S=[0,1]\)
\(S=(-\infty, 3)\)
\(\N=\{1,2,3,\ldots,\}\)
\(S = \{\frac{1}{x^2+1}\;|\; -\infty \lt x \lt \infty\}\)
\(S=\{x\in\real\;|\; x^2 \lt 2\}\)

Let \(A\) and \(B\) be sets and suppose that \(A\subset B\).

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

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

Let \(S\subset\real\) be non-empty.

Let \(S\) be bounded above. An upper bound \(u\) of \(S\) is said to be a least upper bound of \(S\) if \(u\leq u'\) for any upper bound \(u'\) of \(S\). In this case we also say that \(u\) is a supremum of \(S\) and write \(u=\sup(S)\).
Let \(S\) be bounded below. A lower bound \(w\) of \(S\) is said to be a greatest lower bound of \(S\) if \(w' \leq w\) for any lower bound \(w'\) of \(S\). In this case we also say that \(w\) is an infimum of \(S\) and write \(w=\inf(S)\).

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, \(\sup(S)\) is a number that is an upper bound of \(S\) and is the smallest possible upper bound of \(S\). Therefore, \(x\leq \sup(S)\) for all \(x\in S\), and any number less than \(\sup(S)\) is not an upper bound of \(S\) (because \(\sup(S)\) is the least upper bound!). Similarly, when it exists, \(\inf(S)\) is a number that is a lower bound of \(S\) and is the largest possible lower bound of \(S\). Therefore, \(\inf(S)\leq x\) for all \(x\in S\) and any number greater than \(\inf(S)\) is not a lower bound of \(S\) (because \(\inf(S)\) is the greatest lower bound of \(S\)!).

In some analysis texts, \(\sup(S)\) is written as \(\text{lub}(S)\) and \(\inf(S)\) is written as \(\text{glb}(S)\). In other words, \(\sup(S)=\text{lub}(S)\) and \(\inf(S)=\text{glb}(S)\).

Does every non-empty bounded set \(S\) 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 \(\real\) that is bounded above has a least upper bound (a supremum) in \(\real\). Similarly, every non-empty subset of \(\real\) that is bounded below has a greatest lower bound (an infimum) in \(\real\).

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 \(\sup(S)\) and \(\inf(S)\), if they exist.

\(S=\{-5, -9, 2, -1, 11, 0, 4\}\)
\(S=[0,\infty)\)
\(S=(-\infty,3)\)

The Completeness Axiom is sometimes called the supremum property of \(\real\) or the least upper bound property of \(\real\). The Completeness property makes \(\real\) into a complete ordered field. The following example shows that \(\rat\) does not have the completeness property.

The set of rational numbers is an ordered field but it is not complete. Consider the set \(S=\{x\in\mathbb{Q}\;|\; x^2 \lt 2\}\). By definition, \(S\subset\rat\). Clearly, \(S\) is bounded above, for example \(u=10\) is an upper bound of \(S\), but the least upper bound of \(S\) is \(u=\sqrt{2}\) which is not a rational number. Therefore \(S\subset\mathbb{Q}\) does not have a supremum in \(\mathbb{Q}\) and therefore \(\mathbb{Q}\) does not have the Completeness property. From the point of view of analysis, this is the main distinction between \(\mathbb{Q}\) and \(\real\) (note that both are ordered fields).

In some cases, it is obvious what \(\sup\) and \(\inf\) are, however, to do analysis rigorously, we need systematic ways to determine \(\sup(S)\) and \(\inf(S)\). 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 \(S\subset\real\) be non-empty and suppose that \(u\in\real\) is an upper bound of \(S\). Then \(u\) is the least upper bound of \(S\) if and only if for any \(\varepsilon \gt 0\) there exists \(x\in S\) such that \[ u-\varepsilon \lt x \leq u. \]

Suppose that \(u\) is the supremum of \(S\), that is, \(u\) is the least upper bound of \(S\). Since \(u-\varepsilon \lt u\) then \(u-\varepsilon\) is not an upper bound of \(S\). Thus, there exists \(x\in S\) such that \(u-\varepsilon \lt x\). Now suppose that for any \(\varepsilon \gt 0\) there exists \(x\in S\) such that \(u-\varepsilon \lt x \leq u\). Now let \(v\in\real\) be such that \(v \lt u\). Then there exists \(\varepsilon \gt 0\) such that \(v = u - \varepsilon\), and thus by assumption, there exists \(x\in S\) such that \(v \lt x\). Hence, \(v\) is not an upper bound of \(S\) and this shows that \(u\) is the least upper bound of \(S\), that is, \(u=\sup(S)\).

If \(A\subset B\subset\real\), and \(B\) is bounded above, prove that \(A\) is bounded above and that \(\sup(A)\leq \sup(B)\).

Since \(B\) is bounded above, \(\sup(B)\) exists by the Completeness property of \(\real\). Let \(x\in A\). Then \(x\in B\) and therefore \(x\leq \sup(B)\). This proves that \(A\) is bounded above by \(\sup(B)\) and therefore \(\sup(A)\) exists. Since \(\sup(A)\) is the least upper bound of \(A\) we must have \(\sup(A)\leq \sup(B)\). For example, if say \(B=[1,3]\) and \(A=[1,2]\) then \(\sup(A) \lt \sup(B)\), while if \(A=[2,3]\) then \(\sup(A)=\sup(B)\).

Let \(A\subset\real\) be non-empty and bounded above. Let \(c\in\real\) and define the set \[ cA = \{y\in\real \; |\; \exists\; x\in A\;\text{ s.t. } y = cx\}. \] Prove that if \(c \gt 0\) then \(cA\) is bounded above and \(\sup(cA) = c\sup(A)\). Show by example that if \(c \lt 0\) then \(cA\) need not be bounded above even if \(A\) is bounded above.

Let \(y\in cA\) be arbitrary. Then there exists \(x\in A\) such that \(y=cx\). By the Completeness property, \(\sup(A)\) exists and \(x\leq \sup(A)\). If \(c \gt 0\) then \(cx \leq c\sup(A)\) which is equivalent to \(y \leq c\sup(A)\). Since \(y\) is arbitrary, this shows that \(c\sup(A)\) is an upper bound of the set \(cA\) and thus \(cA\) is bounded above and thus \(\sup(cA)\) exists. Because \(\sup(cA)\) is the least upper bound of \(cA\) then \begin{equation} \sup(cA)\leq c\sup(A). \end{equation} Now, by definition, \(y\leq \sup(cA)\) for all \(y\in cA\). Thus, \(cx \leq \sup(cA)\) for all \(x\in A\) and therefore \(x \leq \frac{1}{c}\sup(cA)\) for all \(x\in A\). Therefore, \(\frac{1}{c}\sup(cA)\) is an upper bound of \(A\) and consequently \(\sup(A) \leq \frac{1}{c}\sup(cA)\) because \(\sup(A)\) is the least upper bound of \(A\). We have therefore proved that \begin{equation}\label{eqn:cA-2} c\sup(A) \leq \sup(cA). \end{equation} Combining \eqref{eqn:cA-1} and \eqref{eqn:cA-2} we conclude that \[ \sup(cA) = c\sup(A). \]

Suppose that \(A\) and \(B\) are non-empty and bounded above. Prove that \(A\cup B\) is bounded above and that \[ \sup(A\cup B) = \sup\{\sup(A),\sup(B)\}. \]

Let \(u=\sup\{\sup(A),\sup(B)\}\). Then clearly \(\sup(A)\leq u\) and \(\sup(B)\leq u\). We first show that \(A\cup B\) is bounded above by showing that \(u\) is an upper bound of \(A\cup B\). Let \(x\in A\cup B\). Then either \(z\in A\) or \(x\in B\) (or both). If \(z\in A\) then \(z\leq \sup(A)\leq u\) and if \(z\in B\) then \(z\leq \sup(B)\leq u\). Hence, \(A\cup B\) is bounded above and \(u\) is an upper bound of \(A\cup B\). Consequently, \(\sup(A\cup B)\) exists by the Completeness axiom and moreover \(\sup(A\cup B)\leq u\), that is, \begin{equation} \sup(A\cup B) \leq \sup\{\sup(A),\sup(B)\}. \end{equation} Now, by definition of the supremum, \(z\leq \sup(A\cup B)\) for all \(z\in A\cup B\). Now since \(A\subset A\cup B\) this implies that \(x\leq \sup(A\cup B)\) for all \(x\in A\) and \(y\leq \sup(A\cup B)\) for all \(y\in B\). In other words, \(\sup(A\cup B)\) is an upper bound of both \(A\) and \(B\) and thus \(\sup(A) \leq \sup(A\cup B)\) and \(\sup(B) \leq \sup(A\cup B)\). Then clearly \begin{equation}\label{eqn:AcupB-2} \sup\{\sup(A),\sup(B)\}\} \leq \sup(A\cup B) \end{equation} and combining \eqref{eqn:AcupB-1} and \eqref{eqn:AcupB-2} we have proved that \(\sup(A\cup B) = \sup\{\sup(A),\sup(B)\}\).

Suppose that \(A\) and \(B\) are non-empty and bounded below, and suppose that \(A\cap B\) is non-empty. Prove that \(A\cap B\) is bounded below and that \[ \sup\{\inf(A),\inf(B)\}\leq \inf(A\cap B). \]

If \(x\in A\cap B\) then \(x\in A\) and therefore \(\inf(A)\leq x\), and also \(x\in B\) and thus \(\inf(B)\leq x\). Therefore, both \(\inf(A)\) and \(\inf(B)\) are lower bounds of \(A\cap B\), and by definition of \(\inf(A\cap B)\) we have that \(\inf(A)\leq \inf(A\cap B)\) and \(\inf(B)\leq \inf(A\cap B)\), and consequently \(\sup\{\inf(A),\inf(B)\}\leq \inf(A\cap B)\).

For any set \(A\) define the set \[ -A=\{y\in\real\;|\;\exists\; x\in A\text{ s.t. } y = -x\}. \] Prove that if \(A\subset\real\) is non-empty and bounded then \(\sup(-A) = -\inf(A)\).

It holds that \(\inf(A)\leq x\) for all \(x\in A\) and therefore, \(-\inf(A)\geq -x\) for all \(x\in A\), which is equivalent to \(-\inf(A) \geq y\) for all \(y\in -A\). Therefore, \(-\inf(A)\) is an upper bound of the set \(-A\) and therefore \(\sup(-A) \leq -\inf(A)\). Now, \(y \leq \sup(-A)\) for all \(y\in -A\), or equivalently \(-x \leq \sup(-A)\) for all \(x\in A\). Hence, \(x \geq -\sup(-A)\) for all \(x\in A\). This proves that \(-\sup(-A)\) is a lower bound of \(A\) and therefore \(-\sup(-A)\leq \inf(A)\), or \(\sup(-A) \geq -\inf(A)\). This proves that \(\sup(-A) = -\inf(A)\).

Let \(A\subset\real\) be non-empty and bounded below. Let \(c\in\real\) and define the set \[ c + A = \{y\in\real\;|\;\exists\; x\in A\text{ s.t. } y = c + x\}. \] Prove that \(c+A\) is bounded below and that \(\inf(c+A) = c + \inf(A)\).

For all \(x\in A\) it holds that \(\inf(A)\leq x\) and therefore \(c+\inf(A)\leq c+ x\). This proves that \(c+\inf(A)\) is a lower bound of the set \(c+A\), and therefore \(c+\inf(A) \leq \inf(c+A)\). Now, \(\inf(c+A) \leq y\) for all \(y\in c+A\) and thus \(\inf(c+A) \leq c + x\) for all \(x\in A\), which is the same as \(\inf(c+A) - c \leq x\) for all \(x\in A\). Hence, \(\inf(c+A) - c \leq \inf(A)\) or equivalently \(\inf(c+A) \leq c + \inf(A)\). This proves the claim.

Let \(A\) and \(B\) be non-empty subsets of \(\real_{ \gt 0}=\{x\in\real\,:\, x \gt 0\}\), and suppose that \(A\) and \(B\) are bounded below. Define the set \(AB=\{xy\,:\, x\in A, \, y\in B\}\).

Prove that \(AB\) is bounded below.
Prove that \(\inf(AB)=\inf(A)\cdot\inf(B)\). Hint: Consider two cases, when say \(\inf(A)\inf(B)=0\) and when \(\inf(A)\inf(B)\neq 0\).
How do things change if we do not assume \(A\) and \(B\) are subsets of \(\real_{ \gt 0}\).

Give an example of a non-empty set \(A\subset\real_{ \gt 0}\) such that \(\inf(A)=0\).

For any two non-empty sets \(P\) and \(Q\) of \(\real\) let us write that \(P\leq Q\) if, for each \(x\in P\), there exists \(y\in Q\) such that \(x\leq y\).

Prove that if \(P\leq Q\) then \(\sup(P)\leq \sup(Q)\).
Show via an example that if \(P\leq Q\) then it is not necessarily true that \(\inf(P)\leq\inf(Q)\).

Let \(A\) and \(B\) be non-empty bounded sets of positive real numbers. Define the set \[ \frac{A}{B} = \left\{ z\in\real \; | \; \exists\; x\in A, \exists\; y\in B \text{ s.t. } z=\tfrac{x}{y} \right\}. \] Assume that \(\inf(B) \gt 0\). Prove that \[ \sup\left(\tfrac{A}{B}\right) = \frac{\sup(A)}{\inf(B)}. \]

Since \(A\) is bounded above, \(\sup(A)\) exists and \(x\leq\sup(A)\) for all \(x\in A\). Since \(B\) is bounded below, \(\inf(B)\) exists and \(\inf(B)\leq y\) for all \(y\in B\). Let \(z=x/y\) be an arbitrary point in \(\frac{A}{B}\). Then since \(\inf(B)\leq y\) and \(y \gt 0\) and \(\inf(B) \gt 0\) we obtain that \[ \frac{1}{y} \leq \frac{1}{\inf(B)}. \] Since \(x\leq\sup(A)\) and \(x \gt 0\) (and then clearly \(\sup(A) \gt 0\)) we obtain \[ \frac{x}{y} \leq \frac{x}{\inf(B)} \leq \frac{\sup(A)}{\inf(B)}. \] This proves that \(z \leq \frac{\sup(A)}{\inf(B)}\) and since \(z\in\frac{A}{B}\) was arbitrary we have proved that \(\frac{\sup(A)}{\inf(B)}\) is an upper bound of the set \(\frac{A}{B}\). This proves that \(\sup(\frac{A}{B})\) exists. Moreover, by the definition of the supremum, we also have that \begin{equation} \sup\left(\tfrac{A}{B}\right) \leq \frac{\sup(A)}{\inf(B)}. \end{equation} Now, \(z \leq \sup(\tfrac{A}{B})\) for all \(z\in \tfrac{A}{B}\) and thus \(\tfrac{x}{y} \leq \sup(\tfrac{A}{B})\) for all \(x\in A\) and all \(y\in B\). If \(y\) is held fixed then \(x\leq y \cdot \sup(\tfrac{A}{B})\) for all \(x\in A\) and thus \(\sup(A) \leq y \cdot \sup(\tfrac{A}{B})\). Therefore, \(\sup(A) / \sup(\tfrac{A}{B}) \leq y\), which holds for all \(y\in B\). Therefore, \(\sup(A) / \sup(\tfrac{A}{B}) \leq \inf(B)\) and consequently \begin{equation}\label{eqn:AoverB-2} \frac{\sup(A)}{\inf(B)} \leq \sup\left(\tfrac{A}{B}\right). \end{equation} Combining \eqref{eqn:AoverB-1} and \eqref{eqn:AoverB-2} completes the proof.

Exercises

Let \(S\subset\real\) be a non-empty set and suppose that \(u\) is an upper bound of \(S\). Prove that if \(u\in S\) then necessarily \(u=\sup S\).

Let \(A\) and \(B\) be non-empty subsets of \(\real\), and suppose that \(A\subset B\). Prove that if \(B\) is bounded below then \(\inf B\leq \inf A\).

If \(P\) and \(Q\) are non-empty subsets of \(\real\) such that \(\sup P=\sup Q\) and \(\inf P = \inf Q\) does it follow that \(P=Q\)? Support your answer with either a proof or a counterexample.

Let \(A\subset\real\) be a bounded set, let \(x\in\real\) be fixed, and define the set \[ x+A = \{y\in\real\;|\;\exists\; a\in A\text{ s.t. } y = x+a\}. \] Prove that \(\sup (x+A) = x+\sup A\).

Let \(A,B\subset\real\) be bounded above. Let \[ A+B=\{z\in\real\;|\;\exists\; x\in A,\;\exists\; y\in B\text{ s.t. } z = x + y\}. \] Prove that \(A+B\) is bounded above and that \(\sup(A+B)=\sup A+\sup B\).

Let \(\real_{ \gt 0}\) denote the set of all positive real numbers and let \(A,B\subset\real_{ \gt 0}\) be bounded. Assume that \(\inf(B) \gt 0\). Define the set \[ \frac{A}{B} = \left\{z\in\real\;|\;\exists\; x\in A,\;\exists\; y\in B\text{ s.t. } z=\tfrac{x}{y}\right\}. \] Prove that \[ \sup\left(\tfrac{A}{B}\right) = \frac{\sup(A)}{\inf(B)}. \]

Applications of the Supremum

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

Let \(D\subset\real\) be a non-empty set and let \(f:D\rightarrow\real\) be a function.

\(f\) is bounded below if the range \(f(D) =\{ f(x)\;|\; x\in D\}\) is bounded below.
\(f\) is bounded above if the range \(f(D) =\{ f(x)\;|\; x\in D\}\) is bounded above.
\(f\) is bounded if the range \(f(D) =\{ f(x)\;|\; x\in D\}\) is bounded.

Hence, boundedness of a function \(f\) is boundedness of its range.

Suppose \(f,g:D\rightarrow\real\) are bounded functions and \(f(x)\leq g(x)\) for all \(x\in D\). Show that \(\sup(f(D))\leq \sup(g(D))\).

Since \(g\) is bounded, \(\sup(g(D))\) exists and is by definition an upper bound for the set \(g(D)\), that is, \(g(x)\leq \sup(g(D))\) for all \(x\in D\). Now, by assumption, for all \(x\in D\) it holds that \(f(x)\leq g(x)\) and therefore \(f(x)\leq \sup(g(D))\). This shows that \(\sup(g(D))\) is an upper bound of the set \(f(D)\), and therefore by definition of the supremum, \(\sup(f(D))\leq \sup(g(D))\).

Let \(f,g:D\rightarrow\real\) be bounded functions and suppose that \(f(x)\leq g(y)\) for all \(x,y\in D\). Show that \(\sup(f(D))\leq \inf(g(D))\).

Fix \(x^*\in D\). Then, \(f(x^*)\leq g(y)\) for all \(y\in D\). Therefore, \(f(x^*)\) is a lower bound of \(g(D)\), and thus \(f(x^*)\leq \inf(g(D))\) by definition of the infimum. Since \(x^*\in D\) was arbitrary, we have proved that \(f(x)\leq \inf(g(D))\) for all \(x\in D\). Hence, \(\inf(g(D))\) is an upper bound of \(f(D)\) and thus by definition of the supremum we have that \(\sup(f(D))\leq \inf(g(D))\).

The following sequence of results will be used to prove an important property of the rational numbers \(\mathbb{Q}\) as seen from within \(\real\).

If \(x\in\real\) then there exists \(n\in \mathbb{N}\) such that \(x\leq n\).

Suppose not. Hence, \(n\leq x\) for all \(n\in\mathbb{N}\), and thus \(x\) is an upper bound for \(\mathbb{N}\), and therefore \(\mathbb{N}\) is bounded. Let \(u=\sup(\mathbb{N})\). By definition of \(u\), \(u-1\) is not an upper bound of \(\mathbb{N}\) and therefore there exists \(m\in\mathbb{N}\) such that \(u-1 \lt m\). But then \(u \lt m+1\) and this contradicts the definition of \(u\).

If \(S=\{\frac{1}{n}\;|\; n\in\mathbb{N}\}\) then \(\inf(S) = 0\).

Since \(0 \lt \frac{1}{n}\) for all \(n\) then \(0\) is a lower bound of \(S\). Now suppose that \(0 \lt y\). By the Archimedean Property, there exists \(n\in\mathbb{N}\) such that \(\frac{1}{y} \lt n\) and thus \(\frac{1}{n} \lt y\). Hence, \(y\) is not a lower bound of \(S\). Therefore \(0\) is the greatest lower bound of \(S\), that is, \(\inf(S)=0\).

For any \(y \gt 0\) there exists \(n\in\mathbb{N}\) such that \(\frac{1}{n} \lt y\).

Given \(y \gt 0\) there exists \(n\in\mathbb{N}\) such that \(n-1\leq y \lt n\).

Let \(E=\{k\in\real\;|\; y \lt k\}\). By the Archimedean property, \(E\) is non-empty. By the Well-Ordering Principle of \(\mathbb{N}\), \(E\) has a least element, say that it is \(m\). Hence, \(m-1\notin E\) and thus \(m-1 \leq y \lt m\).

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

If \(x,y\in\real\) and \(x \lt y\) then there exists \(r\in\mathbb{Q}\) such that \(x \lt r \lt y\).

We first prove the claim for the case that \(0 \lt x \lt y\). Suppose that \(y - x \gt 1\) and thus \(x + 1 \lt y\). There exists \(m\in\mathbb{N}\) such that \(m -1 \leq x \lt m\) and thus \(m \leq x + 1\). Therefore, \[ x \lt m \leq x + 1 \lt y \] and thus \(x \lt m \lt y\) and we may take \(r=m\). In general, if \(y - x \gt 0\) then there exists \(n\in \mathbb{N}\) such that \(\frac{1}{n} \lt y - x\) and thus \(1 + nx \lt ny\). Since \(ny - nx \gt 1\) there exists \(m\in\mathbb{N}\) such that \(1+nx \lt m \lt ny\) and thus \[ nx \lt 1 + nx \lt m \lt ny \] and dividing by \(n\) yields \(x \lt \frac{m}{n} \lt y\) and thus we may take \(r=\frac{m}{n}\). This proves the claim when both \(x\) and \(y\) are positive. If \(x \lt 0 \lt y\) then take \(r = 0\) and if \(x \lt y \lt 0\) then apply the previous arguments to \(0 \lt -y \lt -x\).

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 \(\zeta\in\real\backslash\hspace{-0.3em}\rat\) be an irrational number and let \(\varepsilon \gt 0\) be arbitrary. Prove that there exists a rational number \(x\in\rat\) such that \(\zeta - \eps \lt x \lt \zeta + \eps\), that is, \(x\) is in the \(\eps\)-neighborhood of \(\zeta\).

If \(x,y\in\real\) and \(x \lt y\) then there exists \(\xi\in\real\backslash\hspace{-0.3em}\mathbb{Q}\) such that \(x \lt \xi \lt y\).

We have that \(\sqrt{2}x \lt \sqrt{2}y\). By the Density Theorem, \(\sqrt{2}x \lt r \lt \sqrt{2}y\) for some \(r\in\mathbb{Q}\). Then \(\xi=r/\sqrt{2}\).

Exercises

Let \(D\subset\real\) be non-empty and let \(f,g:D\rightarrow\real\) be functions. Let \(f+g\) denote the function defined by \[ (f+g)(x) = f(x) + g(x) \] for any \(x\in D\). If \(f(D)\) and \(g(D)\) are bounded above prove that \((f+g)(D)\) is also bounded above and that \[ \sup (f+g)(D) \leq \sup f(D) + \sup g(D) \]

If \(y \gt 0\) prove that there exists \(n\in\N\) such that \(\frac{1}{2^n} \lt y\). (Note: If you begin with \(\frac{1}{2^n} \lt y\) and solve for \(n\) then you are assuming that such an \(n\) exists. You are not asked to find an \(n\), you are asked to prove that such an \(n\) exists.)

Nested Interval Theorem

If \(a,b\in\real\) and \(a \lt b\) define \[ (a,b):=\{x\in\real\;|\; a \lt x \lt b\}. \] The set \((a,b)\) is called an open interval from \(a\) to \(b\). Define also \[ [a,b]:=\{x\in\real\;|\; a\leq x\leq b\} \] which we call the closed interval from \(a\) to \(b\). The following are called half-open (or half-closed) intervals: \begin{align*} [a,b)&:=\{x\in\real\;|\; a\leq x \lt b\}\\[2ex] (a,b] &:=\{x\in\real\;|\; a \lt x \leq b\}. \end{align*} If \(a=b\) then \((a,a)=\emptyset\) and \([a,a]=\{a\}\). Infinite intervals are \begin{align*} (a,\infty) &= \{x\in\real\;|\; x \gt a\}\\ (-\infty,b) &= \{x\in\real\;|\; x \lt b\}\\ [a,\infty) &= \{x\in\real\;|\; x\geq a\}\\ (-\infty,b] &= \{x\in\real\;|\; x\leq b\}. \end{align*} Below is a characterization of intervals, we will omit the proof.

Let \(S\subset\real\) contain at least two points. Suppose that if \(x,y\in S\) and \(x \lt y\) then \([x,y]\subset S\). Then \(S\) is an interval.

A sequence \(I_1, I_2, I_3, I_4, \ldots\) of intervals is nested if \[ I_1 \supseteq I_2\supseteq I_3 \supseteq I_4 \supseteq \cdots \] As an example, consider \(I_n = [0,\frac{1}{n}]\) where \(n\in\N\). Then \(I_1, I_2, I_3, \ldots\) is nested: \[ [0,1] \supseteq [0, \tfrac{1}{2}] \supseteq [0,\tfrac{1}{3}] \supseteq [0,\tfrac{1}{4}] \cdots \] Notice that since \(0\in I_n\) for each \(n\in\N\) then \[ 0 \in \bigcap_{n=1}^\infty I_n. \] Is there another point in \(\bigcap_{n=1}^\infty I_n\)? Suppose that \(x\neq 0\) and \(x\in\bigcap_{n=1}^\infty I_n\). Then necessarily \(x \gt 0\). Then there exists \(m\in\N\) such that \(\frac{1}{m} \lt x\). Thus, \(x\notin [0,\frac{1}{m}]\) and therefore \(x\notin\bigcap_{n=1}^\infty I_n\). Therefore, \[ \bigcap_{n=1}^\infty I_n = \{0\}. \] In general, we have the following.

Let \(I_1, I_2, I_3, \ldots\) be a sequence of nested closed bounded intervals. Then there exists \(\xi \in\real\) such that \(\xi\in I_n\) for all \(n\in\N\), that is, \(\xi\in \bigcap_{n=1}^\infty I_n\). In particular, if \(I_n=[a_n,b_n]\) for \(n\in \N\), and \(a= \sup\{a_1,a_2,a_3,\ldots\}\) and \(b=\inf\{b_1,b_2,b_3,\ldots\}\) then \[ \{a, b\} \subset \bigcap_{n=1}^\infty I_n. \]

Since \(I_n\) is a closed interval, we can write \(I_n = [a_n, b_n]\) for some \(a_n, b_n \in \real\) and \(a_n \leq b_n\) for all \(n\in\N\). The nested property can be written as \[ [a_1, b_1]\supseteq [a_2, b_2] \supseteq [a_3, b_3] \supseteq [a_4,b_4] \supseteq \cdots \] Since \([a_n, b_n]\subseteq [a_1, b_1]\) for all \(n\in\N\) then \(a_n \leq b_1\) for all \(n\in\N\). Therefore, the set \(S=\{a_n\;|\; n\in\N\}\) is bounded above. Let \(\xi = \sup(S)\) and thus \(a_n \leq \xi\) for all \(n\in \N\). We will show that \(\xi \leq b_n\) for all \(n\in\N\) also. Let \(n\in\N\) be arbitrary. If \(k\leq n\) then \([a_n,b_n]\subseteq [a_k,b_k]\) and therefore \(a_k\leq a_n\leq b_n\). On the other hand, if \(n \lt k\) then \([a_k,b_k]\subset [a_n,b_n]\) and therefore \(a_n\leq a_k\leq b_n\). In any case, \(a_k \leq b_n\) for all \(k\in\N\). Hence, \(b_n\) is an upper bound of \(S\), and thus \(\xi \leq b_n\). Since \(n\in\N\) was arbitrary, we have that \(\xi \leq b_n\) for all \(n\in\N\). Therefore, \(a_n\leq \xi\leq b_n\) for all \(n\in\N\), that is \(\xi \in \bigcap_{n=1}^\infty [a_n,b_n]\). The proof that \(\inf\{b_1,b_2,b_3,\ldots\} \in \bigcap_{n=1}^\infty I_n\) is similar.

The following theorem gives a condition when \(\bigcap_{n=1}^\infty I_n\) contains a single point.

Let \(I_n=[a_n,b_n]\) be a sequence of nested closed bounded intervals. If \[ \inf\{b_n - a_n \;|\; n\in\N\} = 0 \] then \(\bigcap_{n=1}^n I_n\) is a singleton set.

Let \(I_n = \left[1-\frac{1}{n}, 1+\frac{1}{n}\right]\) for \(n\in \N\).

Prove that \(I_1,I_2,I_3,\ldots\) is a sequence of nested intervals.
Find \(\bigcap_{n=1}^\infty I_n\).

Using the Nested Interval property of \(\real\), we give a proof that \(\real\) is uncountable

The real numbers \(\real\) are uncountable.

We will prove that the interval \([0,1]\) is uncountable. Suppose by contradiction that \(I=[0,1]\) is countable, and let \(I=\{x_1,x_2,x_3,\ldots,\}\) be an enumeration of \(I\) (formally this means we have a bijection \(f:\N\rightarrow [0,1]\) and \(f(n) = x_n\)). Since \(x_1 \in I=[0,1]\), there exists a closed and bounded interval \(I_1\subset [0,1]\) such that \(x_1\notin I_1\). Next, consider \(x_2\). There exists a closed and bounded interval \(I_2\subset I_1\) such that \(x_2 \notin I_2\). Next, consider \(x_3\). There exists a closed and bounded interval \(I_3\subset I_2\) such that \(x_3\notin I_3\). By induction, there exists a sequence \(I_1, I_2, I_3,\ldots\) of closed and bounded intervals such that \(x_n \notin I_n\) for all \(n\in\N\). Moreover, by construction the sequence \(I_n\) is nested and therefore \(\bigcap_{n=1}^\infty I_n\) is non-empty, say it contains \(\xi\). Clearly, since \(\xi\in I_n\subset[0,1]\) for all \(n\in\N\) then \(\xi\in [0,1]\). Now, since \(x_n\notin I_n\) for each \(n\in\N\) then \(x_n \notin \bigcap_{n=1}^\infty I_n\). Therefore, \(\xi \neq x_n\) for all \(n\in\N\) and thus \(\xi \notin I =\{x_1,x_2,\ldots\} = [0,1]\), which is a contradiction since \(\xi\in [0,1]\). Therefore, \([0,1]\) is uncountable and this implies that \(\real\) is also uncountable.

We now give an alternative proof that \(\real\) is uncountable. To that end, consider the following subset \(S\subset\real\): \[ S = \{0.a_1a_2a_3a_4\cdots \in \real\;|\; a_k = 0\text{ or } a_k = 1,\; k\in \N\}. \] In other words, \(S\) consists of numbers \(x\in [0,1)\) whose decimal expansion consists of only 0's and 1's. For example, some elements of \(S\) are \begin{align*} x &= 0.000000 \ldots\\ x &= 0.101010 \ldots\\ x &= 0.100000 \ldots\\ x &= 0.010100 \ldots \end{align*} If we can construct a bijection \(f:S\rightarrow\mathcal{P}(\N)\), then since \(\mathcal{P}(\N)\) is uncountable then by Example 1.4.7 this would show that \(S\) is uncountable. Since \(S\subset\real\) then this would show that \(\real\) is uncountable (by Theorem 1.4.8). To construct \(f\), given \(x=0.a_1a_2a_3\ldots\) in \(S\) define \(f(x)\in\mathcal{P}(\N)\) as \[ f(0.a_1a_2a_3\cdots) = \{k\in\N\;|\; a_k = 1\}. \] In other words, \(f(x)\) consists of the decimal places in the decimal expansion of \(x\) that have a value of \(1\). For example, \begin{align*} f(0.000000\ldots) &= \emptyset \\ f(0.101010\ldots) &= \{1,3,5,7,\ldots\} \\ f(0.100100\ldots) &= \{1,4\} \\ f(0.011000\ldots) &= \{2,3\} \end{align*} It is left as an exercise to show that \(f\) is a bijection (see Exercise 1.4.4).

Exercises

Let \(I_n=\left[0,\frac{1}{n}\right]\) for \(n\in\mathbb{N}\). Prove that \(\bigcap_{n=1}^\infty I_n = \{0\}\).