RA The set of real numbers

Section 1.2 The set of real numbers

Note: 2 lectures, the extended real numbers are optional

Subsection 1.2.1 The set of real numbers

We finally get to the real number system. To simplify matters, instead of constructing the real number set from the rational numbers, we simply state their existence as a theorem without proof. Notice that $\Q$ is an ordered field.

Theorem 1.2.1.

There exists a unique

Uniqueness is up to isomorphism, but we wish to avoid excessive use of algebra. For us, it is simply enough to assume that a set of real numbers exists. See Rudin [R2] for the construction and more details.

ordered field $\R$ with the least-upper-bound property such that $\Q \subset \R\text{.}$

Note also that $\N \subset \Q\text{.}$ We saw that $1 > 0\text{.}$ By induction (exercise), we can prove that $n > 0$ for all $n \in \N\text{.}$ Similarly, we verify simple statements about rational numbers. For example, we proved that if $n > 0\text{,}$ then $\nicefrac{1}{n} > 0\text{.}$ Then $m < k$ implies $\nicefrac{m}{n} < \nicefrac{k}{n}\text{.}$

Analysis consists of proving inequalities, and the following proposition, or one of its many variations, is how an analyst proves a nonstrict inequality.

Proposition 1.2.2.

If $x \in \R$ is such that $x \leq \epsilon$ for all $\epsilon \in \R$ where $\epsilon > 0\text{,}$ then $x \leq 0\text{.}$

Proof.

If $x > 0\text{,}$ then $0 < \nicefrac{x}{2} < x$ (why?). Take $\epsilon = \nicefrac{x}{2}$ to get a contradiction. Thus $x \leq 0\text{.}$

For nonnegative $x\text{,}$ equality results: If $x \geq 0$ is such that $x \leq \epsilon$ for all $\epsilon > 0\text{,}$ then $x = 0\text{.}$ A common version uses the absolute value (see Section 1.3): If $\sabs{x} \leq \epsilon$ for all $\epsilon > 0\text{,}$ then $x = 0\text{.}$ To prove $x \geq 0\text{,}$ an analyst might prove that $x \geq -\epsilon$ for all $\epsilon > 0\text{.}$ From now on, when we say $x \geq 0$ or $\epsilon > 0\text{,}$ we automatically mean that $x \in \R$ and $\epsilon \in \R\text{.}$

The idea behind the proposition above is that any time we have two real numbers $a < b\text{,}$ there is another real number $c$ such that $a < c < b\text{.}$ Infinitely many such $c$ exist. One of them is, for example, $c = \frac{a+b}{2}$ (why?). We will use this fact in the next example.

The most useful property of $\R$ for analysts is not just that it is an ordered field, but that it has the least-upper-bound property. Essentially, we want $\Q\text{,}$ but we also want to take suprema (and infima) willy-nilly. So what we do is take $\Q$ and throw in enough numbers to obtain $\R\text{.}$

We mentioned already that $\R$ contains elements that are not in $\Q$ because of the least-upper-bound property. Let us prove it. We saw there is no rational square root of two. The set $\{ x \in \Q : x^2 < 2 \}$ implies the existence of the real number $\sqrt{2}\text{,}$ although this fact requires a bit of work. See also Exercise 1.2.14.

Example 1.2.3.

Claim: There exists a unique positive $r \in \R$ such that $r^2 = 2\text{.}$ We denote $r$ by $\sqrt{2}\text{.}$

Proof.

Take the set $A \coloneqq \{ x \in \R : x^2 < 2 \}\text{.}$ We first show that $A$ is bounded above and nonempty. The inequality $x \geq 2$ implies $x^2 \geq 4$ (see Exercise 1.1.3). So if $x^2 < 2\text{,}$ then $x < 2\text{.}$ So $A$ is bounded above. As $1 \in A\text{,}$ the set $A$ is nonempty. The supremum, therefore, exists.

Let $r \coloneqq \sup\, A\text{.}$ We will show that $r^2 = 2$ by showing that $r^2 \geq 2$ and $r^2 \leq 2\text{.}$ This is the way analysts show equality, by showing two inequalities. We already know that $r \geq 1 > 0\text{.}$

In the following, it may seem we are pulling certain expressions out of a hat. When writing a proof such as this, we would, of course, come up with the expressions only after playing around with what we wish to prove. The order in which we write the proof is not necessarily the order in which we come up with the proof.

Let us first show that $r^2 \geq 2\text{.}$ Take a positive number $s$ such that $s^2 < 2\text{.}$ We wish to find an $h > 0$ such that ${(s+h)}^2 < 2\text{.}$ As $2-s^2 > 0\text{,}$ we have $\frac{2-s^2}{2s+1} > 0\text{.}$ Choose an $h \in \R$ such that $0 < h < \frac{2-s^2}{2s+1}\text{.}$ Furthermore, assume $h < 1\text{.}$ Estimate,

\begin{equation*} \begin{aligned} {(s+h)}^2 - s^2 & = h(2s + h) \\ & < h(2s+1) & & \quad \bigl(\text{since } h < 1\bigr) \\ & < 2-s^2 & & \quad \bigl(\text{since } h < \tfrac{2-s^2}{2s+1} \bigr) . \end{aligned} \end{equation*}

Therefore, ${(s+h)}^2 < 2\text{.}$ Hence $s+h \in A\text{,}$ but as $h > 0\text{,}$ we have $s+h > s\text{.}$ So $s < r = \sup\, A\text{.}$ As $s$ was an arbitrary positive number such that $s^2 < 2\text{,}$ it follows that $r^2 \geq 2\text{.}$

Now take a positive number $s$ such that $s^2 > 2\text{.}$ We wish to find an $h > 0$ such that ${(s-h)}^2 > 2$ and $s-h$ is still positive. As $s^2-2 > 0\text{,}$ we have $\frac{s^2-2}{2s} > 0\text{.}$ Let $h \coloneqq \frac{s^2-2}{2s}\text{,}$ and check $s-h=s-\frac{s^2-2}{2s} = \frac{s}{2}+\frac{1}{s} > 0\text{.}$ Estimate,

\begin{equation*} \begin{aligned} s^2 - {(s-h)}^2 & = 2sh - h^2 \\ & < 2sh & & \quad \bigl( \text{since } h^2 > 0 \text{ as } h \not= 0 \bigr) \\ & = s^2-2 & & \quad \bigl( \text{since } h = \tfrac{s^2-2}{2s} \bigr) . \end{aligned} \end{equation*}

By subtracting $s^2$ from both sides and multiplying by $-1\text{,}$ we find ${(s-h)}^2 > 2\text{.}$ Therefore, $s-h \notin A\text{.}$

Moreover, if $x \geq s-h\text{,}$ then $x^2 \geq {(s-h)}^2 > 2$ (as $x > 0$ and $s-h > 0$) and so $x \notin A\text{.}$ Thus, $s-h$ is an upper bound for $A\text{.}$ However, $s-h < s\text{,}$ or in other words, $s > r = \sup\, A\text{.}$ Hence, $r^2 \leq 2\text{.}$

Together, $r^2 \geq 2$ and $r^2 \leq 2$ imply $r^2 = 2\text{.}$ The existence part is finished. We still need to handle uniqueness. Suppose $s \in \R$ such that $s^2 = 2$ and $s > 0\text{.}$ Thus, $s^2 = r^2\text{.}$ However, if $0 < s < r\text{,}$ then $s^2 < r^2\text{.}$ Similarly, $0 < r < s$ implies $r^2 < s^2\text{.}$ Hence, $s = r\text{.}$

The number $\sqrt{2} \notin \Q\text{.}$ The set $\R \setminus \Q$ is called the set of irrational numbers. We just proved that $\R \setminus \Q$ is nonempty. Not only is it nonempty, as we will see, it is very large indeed.

Using the same technique as above, we can show that a positive real number $x^{1/n}$ exists for all $n\in \N$ and all $x > 0\text{.}$ That is, for each $x > 0\text{,}$ there exists a unique positive real number $r$ such that $r^n = x\text{.}$ The proof is left as an exercise.

Subsection 1.2.2 Archimedean property

As we have seen, there are plenty of real numbers in any interval. But there are also infinitely many rational numbers in any interval. The following is one of the fundamental facts about the real numbers. The two parts of the next theorem are actually equivalent, even though it may not seem like that at first sight.

Theorem 1.2.4.

(Archimedean property)
²
Named after the Ancient Greek mathematician Archimedes of Syracuse (c. 287 BC – c. 212 BC). This property is Axiom V from Archimedes’ “On the Sphere and Cylinder” 225 BC.
If $x, y \in \R$ and $x > 0\text{,}$ then there exists an $n \in \N$ such that

\begin{equation*} nx > y . \end{equation*}
($\Q$ is dense in $\R$) If $x, y \in \R$ and $x < y\text{,}$ then there exists an $r \in \Q$ such that $x < r < y\text{.}$

Proof.

Let us prove i. Divide through by $x\text{.}$ Then i says that for every real number $t\coloneqq \nicefrac{y}{x}\text{,}$ we can find $n \in \N$ such that $n > t\text{.}$ In other words, i says that $\N \subset \R$ is not bounded above. Suppose for contradiction that $\N$ is bounded above. Let $b \coloneqq \sup \N\text{.}$ The number $b-1$ cannot possibly be an upper bound for $\N$ as it is strictly less than $b$ (the least upper bound). Thus there exists an $m \in \N$ such that $m > b-1\text{.}$ Add one to obtain $m+1 > b\text{,}$ contradicting $b$ being an upper bound.

Figure 1.2. Idea of the proof of the density of $\Q\text{:}$ Find $n$ such that $y-x > \nicefrac{1}{n}\text{,}$ then take the least $m$ such that $\nicefrac{m}{n} > x\text{.}$

Let us tackle ii. See Figure 1.2 for a picture of the idea behind the proof. First assume $x \geq 0\text{.}$ Note that $y-x > 0\text{.}$ By i, there exists an $n \in \N$ such that

\begin{equation*} n(y-x) > 1 \qquad \text{or} \qquad y-x > \nicefrac{1}{n}. \end{equation*}

Again by i the set $A \coloneqq \{ k \in \N : k > nx \}$ is nonempty. By the well ordering property of $\N\text{,}$ $A$ has a least element $m\text{,}$ and as $m \in A\text{,}$ then $m > nx\text{.}$ Divide through by $n$ to get $x < \nicefrac{m}{n}\text{.}$ As $m$ is the least element of $A\text{,}$ $m-1 \notin A\text{.}$ If $m > 1\text{,}$ then $m-1 \in \N\text{,}$ but $m-1 \notin A$ and so $m-1 \leq nx\text{.}$ If $m=1\text{,}$ then $m-1 = 0\text{,}$ and $m-1 \leq nx$ still holds as $x \geq 0\text{.}$ In other words,

\begin{equation*} m-1 \leq nx \qquad \text{or} \qquad m \leq nx+1 . \end{equation*}

On the other hand, from $n(y-x) > 1$ we obtain $ny > 1+nx\text{.}$ Hence $ny > 1+nx \geq m\text{,}$ and therefore $y > \nicefrac{m}{n}\text{.}$ Putting everything together, we obtain $x < \nicefrac{m}{n} < y\text{.}$ So take $r = \nicefrac{m}{n}\text{.}$

Now assume $x < 0\text{.}$ If $y > 0\text{,}$ then just take $r=0\text{.}$ If $y \leq 0\text{,}$ then $0 \leq -y < -x\text{,}$ and we find a rational $q$ such that $-y < q < -x\text{.}$ Then take $r = -q\text{.}$

Let us state and prove a simple but useful corollary of the Archimedean property.

Corollary 1.2.5.

$\inf \{ \nicefrac{1}{n} : n \in \N \} = 0\text{.}$ See Figure 1.3.

Proof.

Let $A \coloneqq \{ \nicefrac{1}{n} : n \in \N \}\text{.}$ Obviously, $A$ is not empty. Furthermore, $\nicefrac{1}{n} > 0$ for all $n \in \N\text{,}$ so $0$ is a lower bound and $b \coloneqq \inf\, A$ exists. As $0$ is a lower bound, then $b \geq 0\text{.}$ Take an arbitrary $a > 0\text{.}$ By the Archimedean property, there exists an $n$ such that $na > 1\text{,}$ that is, $a > \nicefrac{1}{n} \in A\text{.}$ Therefore, $a$ cannot be a lower bound for $A\text{.}$ Hence $b=0\text{.}$

Figure 1.3. The set $\{ \nicefrac{1}{n} : n \in \N \}$ and its infimum $0\text{.}$

Subsection 1.2.3 Using supremum and infimum

Suprema and infima are compatible with algebraic operations. For a set $A \subset \R$ and $x \in \R$ define

\begin{equation*} \begin{aligned} x + A & \coloneqq \{ x+y \in \R : y \in A \} , \\ xA & \coloneqq \{ xy \in \R : y \in A \} . \end{aligned} \end{equation*}

For example, if $A = \{ 1,2,3 \}\text{,}$ then $5+A = \{ 6,7,8 \}$ and $3A = \{ 3,6,9 \}\text{.}$

Proposition 1.2.6.

Let $A \subset \R$ be nonempty.

If $x \in \R$ and $A$ is bounded above, then $\sup (x+A) = x + \sup\, A\text{.}$
If $x \in \R$ and $A$ is bounded below, then $\inf (x+A) = x + \inf\, A\text{.}$
If $x > 0$ and $A$ is bounded above, then $\sup (xA) = x ( \sup\, A )\text{.}$
If $x > 0$ and $A$ is bounded below, then $\inf (xA) = x ( \inf\, A )\text{.}$
If $x < 0$ and $A$ is bounded below, then $\sup (xA) = x ( \inf\, A )\text{.}$
If $x < 0$ and $A$ is bounded above, then $\inf (xA) = x ( \sup\, A )\text{.}$

Do note that multiplying a set by a negative number switches supremum for an infimum and vice versa. Also, as the proposition implies that supremum (resp. infimum) of $x+A$ or $xA$ exists, it also implies that $x+A$ or $xA$ is nonempty and bounded above (resp. below).

Proof.

Let us only prove the first statement. The rest are left as exercises.

Suppose $b$ is an upper bound for $A\text{.}$ That is, $y \leq b$ for all $y \in A\text{.}$ Then $x+y \leq x+b$ for all $y \in A\text{,}$ and so $x+b$ is an upper bound for $x+A\text{.}$ In particular, if $b = \sup\, A\text{,}$ then

\begin{equation*} \sup (x+A) \leq x+b = x+ \sup\, A . \end{equation*}

The opposite inequality is similar: If $c$ is an upper bound for $x+A\text{,}$ then $x+y \leq c$ for all $y \in A$ and so $y \leq c-x$ for all $y \in A\text{.}$ So $c-x$ is an upper bound for $A\text{.}$ If $c = \sup (x+A)\text{,}$ then

\begin{equation*} \sup\, A \leq c-x = \sup (x+A) -x . \end{equation*}

The result follows.

Sometimes we need to apply supremum or infimum twice. Here is an example.

Proposition 1.2.7.

Let $A, B \subset \R$ be nonempty sets such that $x \leq y$ whenever $x \in A$ and $y \in B\text{.}$ Then $A$ is bounded above, $B$ is bounded below, and $\sup\, A \leq \inf\, B\text{.}$

Proof.

Any $x \in A$ is a lower bound for $B\text{.}$ Therefore, $x \leq \inf\, B$ for all $x \in A\text{,}$ so $\inf\, B$ is an upper bound for $A\text{.}$ Hence, $\sup\, A \leq \inf\, B\text{.}$

We must be careful about strict inequalities and taking suprema and infima. Note that $x < y$ whenever $x \in A$ and $y \in B$ still only implies $\sup\, A \leq \inf\, B\text{,}$ and not a strict inequality. For example, take $A \coloneqq \{ 0 \}$ and $B \coloneqq \{ \nicefrac{1}{n} : n \in \N \}\text{.}$ Then $0 < \nicefrac{1}{n}$ for all $n \in \N\text{.}$ However, $\sup\, A = 0$ and $\inf\, B = 0\text{.}$ This important subtle point comes up often.

The proof of the following often used fact is left to the reader. A similar result holds for infima.

Proposition 1.2.8.

If $S \subset \R$ is nonempty and bounded above, then for every $\epsilon > 0$ there exists an $x \in S$ such that $(\sup\, S) - \epsilon < x \leq \sup\, S\text{.}$

To make using suprema and infima even easier, we may want to write $\sup\, A$ and $\inf\, A$ without worrying about $A$ being bounded and nonempty. We make the following natural definitions.

Definition 1.2.9.

Let $A \subset \R$ be a set.

If $A$ is empty, then $\sup\, A \coloneqq -\infty\text{.}$
If $A$ is not bounded above, then $\sup\, A \coloneqq \infty\text{.}$
If $A$ is empty, then $\inf\, A \coloneqq \infty\text{.}$
If $A$ is not bounded below, then $\inf\, A \coloneqq -\infty\text{.}$

For convenience, $\infty$ and $-\infty$ are sometimes treated as if they were numbers, except we do not allow arbitrary arithmetic with them. We make $\R^* \coloneqq \R \cup \{ -\infty , \infty\}$ into an ordered set by letting

\begin{equation*} -\infty < \infty \quad \text{and} \quad -\infty < x \quad \text{and} \quad x < \infty \quad \text{for all $x \in \R$}. \end{equation*}

The set $\R^*$ is called the set of extended real numbers. It is possible to define some arithmetic on $\R^*\text{.}$ Most operations are extended in an obvious way, but we must leave $\infty-\infty\text{,}$ $0 \cdot (\pm\infty)\text{,}$ and $\frac{\pm\infty}{\pm\infty}$ undefined. We refrain from using this arithmetic, it leads to easy mistakes as $\R^*$ is not a field. Now we can take suprema and infima without fear of emptiness or unboundedness. In this book, we mostly avoid using $\R^*$ outside of exercises and leave such generalizations to the interested reader.

Subsection 1.2.4 Maxima and minima

By Exercise 1.1.2, a finite set of numbers always has a supremum and an infimum and they are both contained in the set itself. In this case, we usually do not use the words supremum or infimum. When a set $A$ of real numbers is bounded above and $\sup\, A \in A\text{,}$ we can use the word maximum and the notation $\max\, A$ to denote the supremum. Similarly for infimum: When $A$ is bounded below and $\inf\, A \in A\text{,}$ we can use the word minimum and the notation $\min\, A\text{.}$ For example,

\begin{equation*} \begin{aligned} & \max \{ 1,2.4,\pi,100 \} = 100 , \\ & \min \{ 1,2.4,\pi,100 \} = 1 . \end{aligned} \end{equation*}

While writing $\sup$ and $\inf$ may be technically correct in this situation, $\max$ and $\min$ are generally used to emphasize that the supremum or infimum is in the set itself, especially when the set is finite.

Exercises 1.2.5 Exercises

1.2.1.

Prove that if $t > 0$ ($t \in \R$), then there exists an $n \in \N$ such that $\dfrac{1}{n^2} < t\text{.}$

1.2.2.

Prove that if $t \geq 0$ ($t \in \R$), then there exists an $n \in \N$ such that $n-1 \leq t < n\text{.}$

1.2.3.

Finish the proof of Proposition 1.2.6.

1.2.4.

Let $x, y \in \R\text{.}$ Suppose $x^2 + y^2 = 0\text{.}$ Prove that $x = 0$ and $y = 0\text{.}$

1.2.5.

Show that $\sqrt{3}$ is irrational.

1.2.6.

Let $n \in \N\text{.}$ Show that $\sqrt{n}$ is either an integer or it is irrational.

1.2.7.

Prove the arithmetic-geometric mean inequality. For two positive real numbers $x,y\text{,}$

\begin{equation*} \sqrt{xy} \leq \frac{x+y}{2} . \end{equation*}

Furthermore, equality occurs if and only if $x=y\text{.}$

1.2.8.

Show that for every pair of real numbers $x$ and $y$ such that $x < y\text{,}$ there exists an irrational number $s$ such that $x < s < y\text{.}$ Hint: Apply the density of $\Q$ to $\dfrac{x}{\sqrt{2}}$ and $\dfrac{y}{\sqrt{2}}\text{.}$

1.2.9.

Let $A$ and $B$ be two nonempty bounded sets of real numbers. Let $C \coloneqq \{ a+b : a \in A, b \in B \}\text{.}$ Show that $C$ is a bounded set and that

\begin{equation*} \sup\,C = \sup\,A + \sup\,B \qquad \text{and} \qquad \inf\,C = \inf\,A + \inf\,B . \end{equation*}

1.2.10.

Let $A$ and $B$ be two nonempty bounded sets of nonnegative real numbers. Define the set $C \coloneqq \{ ab : a \in A, b \in B \}\text{.}$ Show that $C$ is a bounded set and that

\begin{equation*} \sup\,C = (\sup\,A )( \sup\,B) \qquad \text{and} \qquad \inf\,C = (\inf\,A )( \inf\,B). \end{equation*}

1.2.11.

(Hard) Given $x > 0$ and $n \in \N\text{,}$ show that there exists a unique positive real number $r$ such that $x = r^n\text{.}$ Usually, $r$ is denoted by $x^{1/n}\text{.}$

1.2.12.

(Easy) Prove Proposition 1.2.8.

1.2.13.

Prove the so-called Bernoulli’s inequality

Named after the Swiss mathematician Jacob Bernoulli (1655–1705).

: If $1+x > 0\text{,}$ then for all $n \in \N\text{,}$ we have $(1+x)^n \geq 1+nx\text{.}$

1.2.14.

Prove $\sup \{ x \in \Q : x^2 < 2 \} = \sup \{ x \in \R : x^2 < 2 \}\text{.}$

1.2.15.

Prove that given $y \in \R\text{,}$ we have $\sup \{ x \in \Q : x < y \} = y\text{.}$
Let $A \subset \Q$ be a set that is bounded above such that whenever $x \in A$ and $t \in \Q$ with $t < x\text{,}$ then $t \in A\text{.}$ Further suppose $\sup\, A \not\in A\text{.}$ Show that there exists a $y \in \R$ such that $A = \{ x \in \Q : x < y \}\text{.}$ A set such as $A$ is called a Dedekind cut.
Show that there is a bijection between $\R$ and Dedekind cuts.

Note: Dedekind used sets as in part b) in his construction of the real numbers.

1.2.16.

Prove that if $A \subset \Z$ is a nonempty subset bounded below, then there exists a least element in $A\text{.}$ Now describe why this statement would simplify the proof of Theorem 1.2.4 part ii so that you do not have to assume $x \geq 0\text{.}$

1.2.17.

Let us suppose we know $x^{1/n}$ exists for every $x > 0$ and every $n \in \N$ (see Exercise 1.2.11 above). For integers $p$ and $q > 0$ where $\nicefrac{p}{q}$ is in lowest terms, define $x^{p/q} \coloneqq {(x^{1/q})}^p\text{.}$

Show that the power is well-defined even if the fraction is not in lowest terms: If $\nicefrac{p}{q} = \nicefrac{m}{k}$ where $m$ and $k > 0$ are integers, then ${(x^{1/q})}^p = {(x^{1/k})}^m\text{.}$
Let $x$ and $y$ be two positive numbers and $r$ a rational number. Assuming $r > 0\text{,}$ show $x < y$ if and only if $x^r < y^r\text{.}$ Then suppose $r < 0$ and show: $x < y$ if and only if $x^r > y^r\text{.}$
Suppose $x > 1$ and $r,s$ are rational where $r < s\text{.}$ Show $x^r < x^s\text{.}$ If $0 < x < 1$ and $r < s\text{,}$ show that $x^r > x^s\text{.}$ Hint: Write $r$ and $s$ with the same denominator.
(Challenging)
⁴
In Section 5.4 we will define the exponential and the logarithm and define $x^z \coloneqq \exp(z \ln x)\text{.}$ We will then have sufficient machinery to make proofs of these assertions far easier. At this point, however, we do not yet have these tools.
For an irrational $z \in \R \setminus \Q$ and $x > 1$ define $x^z \coloneqq \sup \{ x^r : r \leq z, r \in \Q \}\text{,}$ for $x=1$ define $1^z = 1\text{,}$ and for $0 < x < 1$ define $x^z \coloneqq \inf \{ x^r : r \leq z, r \in \Q \}\text{.}$ Prove the two assertions of part b) for all real $z\text{.}$

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