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^{ 1 } ordered field \(\R\) with the leastupperbound property such that \(\Q \subset \R\text{.}\)
Note that also \(\N \subset \Q\text{.}\) 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 \(\nicefrac{m}{n} < \nicefrac{k}{n}\text{.}\)
Let us prove one of the most basic but useful results about the real numbers. The following proposition is essentially how an analyst proves an 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?). Taking \(\epsilon = \nicefrac{x}{2}\) obtains a contradiction. Thus \(x \leq 0\text{.}\)
Another useful version of this idea is the following equivalent statement for nonnegative numbers: If \(x \geq 0\) is such that \(x \leq \epsilon\) for all \(\epsilon > 0\text{,}\) then \(x = 0\text{.}\) And to prove that \(x \geq 0\) in the first place, an analyst might prove that all \(x \geq \epsilon\) 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 \(\epsilon \in \R\text{.}\)
A related simple fact is that any time we have two real numbers \(a < b\text{,}\) then there is another real number \(c\) such that \(a < c < b\text{.}\) Take, for example, \(c = \frac{a+b}{2}\) (why?). In fact, there are infinitely many real numbers between \(a\) and \(b\text{.}\) 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 leastupperbound property. Essentially, we want \(\Q\text{,}\) but we also want to take suprema (and infima) willynilly. 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 leastupperbound 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 \(A := \{ x \in \R : x^2 < 2 \}\text{.}\) We first show that \(A\) is bounded above and nonempty. The equation \(x \geq 2\) implies \(x^2 \geq 4\) (see Exercise 1.1.3), so if \(x^2 < 2\text{,}\) then \(x < 2\text{,}\) and \(A\) is bounded above. As \(1 \in A\text{,}\) the set \(A\) is nonempty. We can therefore find the supremum.
Let \(r := \sup\, A\text{.}\) 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 \(2s^2 > 0\text{,}\) we have \(\frac{2s^2}{2s+1} > 0\text{.}\) Choose an \(h \in \R\) such that \(0 < h < \frac{2s^2}{2s+1}\text{.}\) Furthermore, assume \(h < 1\text{.}\) Estimate,
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 \({(sh)}^2 > 2\) and \(sh\) is still positive. As \(s^22 > 0\text{,}\) we have \(\frac{s^22}{2s} > 0\text{.}\) Let \(h := \frac{s^22}{2s}\text{,}\) and check \(sh=s\frac{s^22}{2s} = \frac{s}{2}+\frac{1}{s} > 0\text{.}\) Estimate,
By subtracting \(s^2\) from both sides and multiplying by \(1\text{,}\) we find \({(sh)}^2 > 2\text{.}\) Therefore, \(sh \notin A\text{.}\)
Moreover, if \(x \geq sh\text{,}\) then \(x^2 \geq {(sh)}^2 > 2\) (as \(x > 0\) and \(sh > 0\)) and so \(x \notin A\text{.}\) Thus, \(sh\) is an upper bound for \(A\text{.}\) However, \(sh < s\text{,}\) 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 saw that \(\R \setminus \Q\) is nonempty. Not only is it nonempty, we will see later that 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 } 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 \(t:= \nicefrac{y}{x}\text{,}\) 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 \(b := \sup \N\text{.}\) The number \(b1\) 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 \(m > b1\text{.}\) Add one to obtain \(m+1 > b\text{,}\) contradicting \(b\) being an upper bound.
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 \(yx > 0\text{.}\) By i, there exists an \(n \in \N\) such that
Again by i the set \(A := \{ k \in \N : k > nx \}\) 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 \(A\text{,}\) \(m1 \notin A\text{.}\) If \(m > 1\text{,}\) then \(m1 \in \N\text{,}\) but \(m1 \notin A\) and so \(m1 \leq nx\text{.}\) If \(m=1\text{,}\) then \(m1 = 0\text{,}\) and \(m1 \leq nx\) still holds as \(x \geq 0\text{.}\) In other words,
On the other hand, from \(n(yx) > 1\) 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.
\(\inf \{ \nicefrac{1}{n} : n \in \N \} = 0\text{.}\)
Proof.
Let \(A := \{ \nicefrac{1}{n} : n \in \N \}\text{.}\) Obviously \(A\) is not empty. Furthermore, \(\nicefrac{1}{n} > 0\) for all \(n \in \N\text{,}\) and so 0 is a lower bound, and \(b := \inf\, A\) 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 \(na > 1\text{,}\) or in other words \(a > \nicefrac{1}{n} \in A\text{.}\) Therefore, \(a\) cannot be a lower bound for \(A\text{.}\) Hence \(b=0\text{.}\)
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
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
The opposite inequality is similar. If \(b\) is an upper bound for \(x+A\text{,}\) then \(x+y \leq b\) for all \(y \in A\) and so \(y \leq bx\) for all \(y \in A\text{.}\) So \(bx\) is an upper bound for \(A\text{.}\) If \(b = \sup (x+A)\text{,}\) then
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. This is an important subtle point that comes up often. For example, take \(A := \{ 0 \}\) and take \(B := \{ \nicefrac{1}{n} : n \in \N \}\text{.}\) Then \(0 < \nicefrac{1}{n}\) for all \(n \in \N\text{.}\) However, \(\sup\, A = 0\) and \(\inf\, B = 0\text{.}\)
The proof of the following often used elementary fact is left to the reader. A similar statement 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 \(\sup\, A := \infty\text{.}\)
If \(A\) is not bounded above, then \(\sup\, A := \infty\text{.}\)
If \(A\) is empty, then \(\inf\, A := \infty\text{.}\)
If \(A\) is not bounded below, then \(\inf\, A := \infty\text{.}\)
For convenience, \(\infty\) and \(\infty\) are sometimes treated as if they were numbers, except we do not allow arbitrary arithmetic with them. We make \(\R^* := \R \cup \{ \infty , \infty\}\) into an ordered set by letting
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 or an infimum that is 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, such that \(\sup\, A \in A\text{,}\) then we can use the word maximum and the notation \(\max\, A\) to denote the supremum. Similarly for infimum: When a set \(A\) is bounded below and \(\inf\, A \in A\text{,}\) then we can use the word minimum and the notation \(\min\, A\text{.}\) For example,
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.
Subsection 1.2.5 Exercises
Exercise 1.2.1.
Prove that if \(t > 0\) (\(t \in \R\)), then there exists an \(n \in \N\) such that \(\dfrac{1}{n^2} < t\text{.}\)
Exercise 1.2.2.
Prove that if \(t \geq 0\) (\(t \in \R\)), then there exists an \(n \in \N\) such that \(n1 \leq t < n\text{.}\)
Exercise 1.2.3.
Finish the proof of Proposition 1.2.6.
Exercise 1.2.4.
Let \(x, y \in \R\text{.}\) Suppose \(x^2 + y^2 = 0\text{.}\) Prove that \(x = 0\) and \(y = 0\text{.}\)
Exercise 1.2.5.
Show that \(\sqrt{3}\) is irrational.
Exercise 1.2.6.
Let \(n \in \N\text{.}\) Show that either \(\sqrt{n}\) is either an integer or it is irrational.
Exercise 1.2.7.
Prove the arithmeticgeometric mean inequality. That is, for two positive real numbers \(x,y\text{,}\) we have
Furthermore, equality occurs if and only if \(x=y\text{.}\)
Exercise 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 \(\dfrac{y}{\sqrt{2}}\text{.}\)
Exercise 1.2.9.
Let \(A\) and \(B\) be two nonempty bounded sets of real numbers. Let \(C := \{ a+b : a \in A, b \in B \}\text{.}\) Show that \(C\) is a bounded set and that
Exercise 1.2.10.
Let \(A\) and \(B\) be two nonempty bounded sets of nonnegative real numbers. Define the set \(C := \{ ab : a \in A, b \in B \}\text{.}\) Show that \(C\) is a bounded set and that
Exercise 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 \(x^{1/n}\text{.}\)
Exercise 1.2.12.
(Easy) Prove Proposition 1.2.8.
Exercise 1.2.13.
Prove the socalled Bernoulli's inequality^{ 4 }: If \(1+x > 0\text{,}\) then for all \(n \in \N\text{,}\) we have \((1+x)^n \geq 1+nx\text{.}\)
Exercise 1.2.14.
Prove \(\sup \{ x \in \Q : x^2 < 2 \} = \sup \{ x \in \R : x^2 < 2 \}\text{.}\)
Exercise 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.
Exercise 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 \(x \geq 0\text{.}\)
Exercise 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 \(x^{p/q} := {(x^{1/q})}^p\text{.}\)
Show that the power is welldefined even if the fraction is not in lowest terms: If \(\nicefrac{p}{q} = \nicefrac{m}{k}\) where \(m\) and \(k > 0\) are integers, then \({(x^{1/q})}^p = {(x^{1/m})}^k\text{.}\)
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)^{ 6 } For an irrational \(z \in \R \setminus \Q\) and \(x > 1\) define \(x^z := \sup \{ x^r : r \leq z, r \in \Q \}\text{,}\) for \(x=1\) define \(1^z = 1\text{,}\) and for \(0 < x < 1\) define \(x^z := \inf \{ x^r : r \leq z, r \in \Q \}\text{.}\) Prove the two assertions of part b) for all real \(z\text{.}\)
https://en.wikipedia.org/wiki/Archimedes
https://en.wikipedia.org/wiki/Jacob_Bernoulli