$$$$

6  The Real Numbers

Highlights of this Chapter: We see that there exists a unique complete ordered field, and use this to axiomatically define the real numbers. We then investigate properties of this number system, proving several foundational results related to the archimedean property and nested intervals:

• The real numbers do not contain any infinite numbers or infinitesimals.
• The square root of 2 is a real number.
• The rational numbers are dense in the reals.
• The real numbers are uncountable.

We have now carefully axiomatized the properties that are used in classical mathematics when dealing with the number line, defining a the structure of a complete ordered field.

Definition 6.1 A complete ordered field is an ordered field that satisfies the completeness axiom. Precisely, it is a set $\mathbb{F}$ with the following properties

• Addition: A commutative associative operation $+$, with identity $0$, where ever element has an additive inverse.
• Multiplication: A commutative associative operation $\cdot$ with identity $1\ne 0$, where every nonzero element has a multiplicative inverse.
• Distributivity: For all $a,b,c\in \mathbb{F}$ we have $a(b+c)=ab+ac$
• Order: A subset $P\subset \mathbb{F}$ called the positives containing exactly one of $x,-x$ for every nonzero $x\in \mathbb{F}$, which is closed under addition and multiplication: if $a,b\in P$ then $a+b\in P$ and $ab\in P$.
• Completeness: Every nonempty subset $A\subset \mathbb{F}$ which is bounded above has a least upper bound.

The subject of real analysis is the study of complete ordered fields and their properties, so everything that follows in this course logically follows from this set of axioms, and nothing more. The success and importance of the above definition is best exemplified by the following theorem:

Remark 6.1. This was very important work at the turn of the previous century; as neither step is a priori obvious. It’s easy to write down axiom systems that don’t describe anything because they’re inconsistent (for example, add to ordered field axioms that all polynomials have at least one zero, and there is no longer such a structure), and its also common that axioms don’t uniquely pick out a single object but rather describe an entire class (the axioms of a group define a whole subject, not a single example).

Theorem 6.1 (Uniqueness of the Reals) There exists a complete ordered field, and it is unique. We call this field the real numbers and denote it by $\mathbb{R}$.

This theorem represents the culmination of much work at the end of the 19th and beginning of the 20th century to fully understand the real number line.

While not necessarily beyond our abilities, proving existence of a structure satisfying these axioms is a job for the set theorists and logicians that we will not tackle here.

Beyond providing justification for our usual way of speaking, the uniqueness of the reals is an important result to the history of mathematics. Its statement and proof in 1903 by Huntington marked the end of the era of searching for the fundamental principles behind the real numbers, and the beginning of the modern point of view, completely specifying their structure axiomatically.

Remark 6.2. The completeness axiom is what sets analysis apart from algebra, as it does not tell us how elements behave with respect to a given operation, but rather tells us about the existence of new elements. Indeed, this assertive ability of the completeness axiom is more radical than it seems at first, and can even be captured by mathematical logic: the other axioms are all first order axioms, whereas the completeness axiom is second order.

6.0.1 Dubious Numbers

Now that we have a precise definition of the real number line, we can make precise the philosophical questions that were raised during the origin story of The Calculus.

The first of these was the concept of a nilpotent number, something so small that its square was literally equal to zero.

Definition 6.2 (Nilpotent Numbers) A number is nilpotent if $ϵ\ne 0$ but ${ϵ}^2=0$.

Such numbers were often used in justifying various calculations of the derivative (and continue to be used, as heuristic arguments in introductory calculus and science courses). But it is immediate from even just the field axioms that no such numbers exist.

Proposition 6.1 (Fields have no Nilpotent Numbers) Let $$\mathbb{F}$ be any field, and $ϵ$ some number where ${ϵ}^2=0$. By the zero-product-property (Example 3.7), this implies $ϵ=0$. Thus there are no nonzero elements that square to zero.

The other two classes of numbers proposed to help make sense of the calculus were infinite numbers (to represent the number of summands in an infinite sum, for instance) and infinitesimal numbers (as we saw with differentiation).

Definition 6.3 (Infinite Numbers) A number $x$ is finite if its bounded above and below by integers. If a number is not finite, its said to be infinite.

Equivalently, a number $x$ is infinite if either it or its negation is greater than all positive integers.

Definition 6.4 (Infinitesimal Numbers) A positive number $ϵ$ is infinitesimal if it is smaller than $1/n$ for all $n$.

So of course, the next thing we should do is figure out if these numbers exist!

6.1 Infinites and Infinitesimals

Theorem 6.2 (Infinite Numbers Do Not Exist) There are no infinite elements of $\mathbb{R}$.

Proof. Assume for the sake of contradiction that there is some infinite number: without loss of generality (perhaps after multiplying by $-1$) we may assume its positive. Thus, this number is greater than every natural number, and so the natural numbers are bounded above.

Thus, by the completeness axiom, we find that the natural numbers must have a supremum. Denote this by $X=sup\mathbb{N}$. So far, everything seems fine. But consider the number $X-1$. This is smaller than $X$, and since $X$ is the least upper bound, $X-1$ cannot be an upper bound to $\mathbb{N}$. This means there must be some element $n\in \mathbb{N}$ with $n>X-1$. But this means $X<n+1$, and as $n+1$ is a natural number whenever $n$ is, we’ve run headfirst into a contradiction: $X$ is not an upper bound at all!

It is an immediate corollary of this that infinitesimals also do not exist (but, because this is such an important result, we call it a theorem on its own.)

Theorem 6.3 (Infinitesimals Do Not Exist) There are no infinitesimal elements of $\mathbb{R}$.

Proof. Let $x$ be a positive element of $\mathbb{R}$, and consider its reciprocal $1/x$. By Theorem 6.2 $1/x$ is finite, so there’s some $n\in \mathbb{N}$ with $n>1/x$. Re-arranging the inequality shows $x>1/n$ as required, so $x$ is not infinitesimal.

This argument shows that for a field, containing infinite elements and infinitesimal elements are logically equivalent: thanks to division, you can’t have one without the other.

Exercise 6.1 (An Flawed Argument for the Nonexistence of Infinitesimals) Its possible to show that the infimum of the set of all positive real numbers is zero, just from the definition of infimum.

• Prove this
• Explain why this is not enough to conclude the nonexistence of infinitesimals.

These theorems are fundamental to the foundations of calculus. We know the rationals do not contain infinites or infinitesimals, but to go from the rationals to the reals its quite possible that such numbers were added. After all, the field of real numbers is defined axiomatically - we don’t know what the elements are we just know how to work with them! And many throughout the history of calculus assumed the reals contained infinitesimals….but it turns out they were all wrong.

6.1.1 Archimedean Property

A useful way to repackage the nonexistence of infinite numbers and infinitesimals into a usable statement known as the Archimedean property, as Archimedes took it as an axiom describing the number system in his paper The Sphere and the Cylinder. It also appears (earlier) as a definition in Euclid’s elements: Book V Definition 4:

Magnitudes are said to have a ratio to one another which can, when multiplied, exceed one another.

We rephrase this in precise modern terminology below:

Definition 6.5 (Archimedean Field) A field $\mathbb{F}$ is archimedean if for every positive $a,b\in \mathbb{F}$ there is a natural number $n$ with $$na>b$$

Remark 6.3. While Archimedes himself attributes this to Eudoxus of Cnidus, it was named after Archimedes in the 1880s.

The important applications of this property all come from the case where $b$ is really large, and $a$ is really small. In an archimedean field, no matter how small $a$ is you can always collect enough of them $na=a+a+a+\cdots +a$ to surpass $b$. A common way to remember this property is to poetically rephrase it as you can empty the ocean with a teaspoon.

Its possible to give an elementary proof (directly from the definition of rational numbers as fradtions $p/q$ for $p,q\in \mathbb{Z}$, $q\ne 0$) that $\mathbb{Q}$ is an archimedean field:

Exercise 6.2 (The Rationals are Archimedean) Prove the rationals are an archimedean field. Hint: write $a$ and $b$ as fractions, can you figure out from the inequality you want, what $n$ can be?

Such a proof is not possible for $\mathbb{R}$ as we don’t have an explicit description of its elements! All we know is its axiomatic properties. However, a proof is immediate using Theorem 6.2:

Theorem 6.4 (The Reals are Archimedean) Complete ordered fields satisfy the Archimedean property.

Proof. (In the book, Lemma 1.26, page 28) Let $a,b$ be positive real numbers. Since $b/a\in \mathbb{R}$ it is finite (by Theorem 6.2), so there is some $n\in \mathbb{N}$ with $n>\frac{b}{a}$, and thus $na>b$.

Its also a short proof to show that archimedean fields cannot contain infinite elements (and thus also cannot contain infinitesimals), providing a useful equivalence:

Theorem 6.5 The following three conditions are equivalent, for an ordered field $\mathbb{F}$:

• $\mathbb{F}$ is archimedean.
• $\mathbb{F}$ contains no infinite elements.
• $\mathbb{F}$ contains no infinitesimal elements.

Proof. We already know the existence of infinite elements and infinitesimal elements are equivalent, so all we need to show is that $\mathbb{F}$ is archimedean if and only if all elements are finite.

But the proof of Theorem 6.4 already provides an argument that a field with only finite elements is necessarily archimedean, so we seek only the converse.

If $\mathbb{F}$ is archimedean, then for any positive $b\in \mathbb{F}$ we may take $a=1$ and apply the archimedean property to get an $n\in \mathbb{N}$ with $n\cdot 1>b$. For negative $b$, applying the ame to $-b$ results in a $n\in \mathbb{N}$ where $-n<b$, and together these imply all elements of $\mathbb{F}$ are finite.

Remark 6.4. In fact one can be more precise than this: it turns out that the real numbers are the largest possible archimedean field - and every archimedean field fits somewhere between the rationals and the reals.

Originating in the foundations of analysis, the archimedean property has proven a useful guide in the general study of ordered fields. When encountering a new ordered field, one of the first questions one usually asks is is it archimedean? If so, we know immediately that it does not contain any infinites or infinitesimals, and one one can use intuition from the rationals or real numbers. Non-archimedean fields on the other hand are a totally different beast, and lead to theories rather qualitatively different from real analysis.

Exercise 6.3 Prove that the supremum of the set $S=\{\frac{n}{n+1}\mid n\in \mathbb{N}\}$ is $1$.

6.2 Irrationals

Definition 6.6 (Irrational Numbers) A number $x\in \mathbb{R}$ is irrational if it is not rational.

6.2.1 Existence of $\sqrt{2}$

Our first goal is to prove that irrational numbers exist, by exhibiting one. We will use the example of the square root of two, and rigorously prove that $\sqrt{2}$ is a real number. (Just so you don’t brush this off as trivial, its not immediately obvious: after all, $\sqrt{-2}$ is not a real number!)

Theorem 6.6 Let $\mathbb{F}$ be archimedean, and consider the set $$S=\{r\in \mathbb{F}\mid r^2<2\}$$ Then if $\sigma =supS$ exists, ${\sigma }^2=2$.

We prove this rather indirectly, showing that both ${\sigma }^2>2$ and ${\sigma }^2<2$ are impossible, so the only remaining option is ${\sigma }^2=2$.

Example 6.1 (${\sigma }^2>2$ is impossible.) To show this is impossible, we will show if you have any upper bound $b\in \mathbb{F}$ with $b^2>2$, it’s not the least upper bound, as we can make a smaller one.

Let $b$ be any upper bound with $b^2>2$. To find a smaller upper bound, one idea is to try and find a natural number $n$ where $\beta =b-1/n$ works. That is, $${(b-\frac{1}{n})}^2>2$$

Expanding this out, we see $b^2-2b/n+1/n^2>2$, or after moving terms around, $b^2-2>2b/n-1/n^2$. Now we need a little ingenuity: notice that $2b/n-1/n^2$ is less than $2b/n$ (because we’re subtracting something) so in fact, if we can find an $n$ where $2b/n<b^2-2$ we’re already good. Re-arranging this equation, we need to find $n$ with $$(b^2-2)n>2b$$ But this is possible using the Archimedean property! Since $A=b^2-2$ and $B=2b$ are both positive numbers, we can always find an $n\in \mathbb{N}$ where $nA>B$. Thus, we may choose this value of $n$, and note that $\beta =b-\frac{1}{n}$ is an upper bound for $S$ that is smaller than $b$. Thus $b$ was not the least upper bound!

Exercise 6.4 (${\sigma }^2<2$ is impossible.) Can you preform an argument similar to Example 6.1, to prove that ${\sigma }^2<2$ also leads to contradiction?

Since both the real numbers and the rationals are archimedean, the above applies to a consideration of either field

However applying the same knowledge to the reals yields the opposite conclusion, by virtue of the completeness axiom.

Theorem 6.7 ($\sqrt{2}$ is a Real Number) There exists a positive real number which squares to $2$.

Proof. Let $S=\{r\in \mathbb{R}\mid r^2<2\}$. Then, $S$ is nonempty, as $0\in S$ since $O^2=0$ and $0<2$. Next, we show that $S$ is bounded above by $10$:

Let $r\in S$ is arbitrary. Without loss of generality we may assume $r>0$ as if $r<0$ then certainly $r<10$. By the definition of $S$, we know $r^2<2$ and thus clearly $r^2<100$. But recall Proposition 4.5: for positive $a,b$ if $a^2<b^2$ then $a<b$, so from $r^2<100$ we may conclude $r<10$.

Knowing that $S$ is both nonempty and bounded above, the completeness axiom applies to furnish us with a least upper bound $\sigma =supS$. And knowing its existence, Theorem 6.6 immediately implies that ${\sigma }^2=2$, so $\sigma$ is by definition a square root of 2.

Theorem 6.8 (The Rationals are Incomplete) Within the field of rational numbers, the set $S=\{r\in \mathbb{Q}\mid r^2<2\}$ is bounded above and nonempty, but does not have a supremum.

Proof. The argument that $S$ is nonempty and bounded above is identical to that in Theorem 6.7. And, Theorem 6.6 implies that if the supremum exists it must square to $2$. But we know by Theorem 1.1 that there is no such rational number. Thus, the supremum must not exist, and so $\mathbb{Q}$ fails the completeness axiom.

There is nothing special about $2$ in the above argument, other than it is easy for us to work with. We could stop right now to prove the more general statement that all square roots exist:

Theorem 6.9 (Square Roots Exist) If $x\in \mathbb{R}$ is positive, then $\sqrt{x}$ is a real number.

Though to not be too repetitive, we will hold off and prove this a different way, to illustrate more powerful tools in CITE.

Exercise 6.5 Prove that the product of a nonzero rational and an irrational number is irrational.

Exercise 6.6 The sum of two irrational numbers need not be irrational, as the example $\sqrt{2}-\sqrt{2}=0$ shows. Prove or disprove: the sum of two positive irrational numbers is irrational.

6.2.2 Density

Definition 6.7 (Density) Let $S$ be a subset of an ordered field $\mathbb{F}$. Then $S$ is dense in $\mathbb{F}$ if between any two elements $a,b\in \mathbb{F}$ with $a<b$ there is some $s\in S$ with $$a<s<b$$

Theorem 6.10 (Density of the Rationals) The rational numbers are dense in the real numbers.

Proof. We need to start with two arbitrary real numbers $a<b$, and find a rational number $r$ between them. Let’s do some scratch work: if $r=m/n$ and we want $a<m/n<b$ then it suffices to find an integer $m$ between $na$ and $nb$. This sounds doable!

Precisely, since $b-a>0$, we can use the archimedean property to find some $n\in \mathbb{N}$ with $n(b-a)>1$. Now since $nb-na>1$, we just need to prove there’s an integer $m$ between them, and this’ll be the number we want!

To rigorously prove this $m$ exists, we can reason as follows: we know there are integers greater than $na$ (since $\mathbb{R}$ has no infinite elements), so let $m$ be the smallest such. Then by definition $m>na$, so all we need to show is $m<nb$. Since $m$ is the smallest integer greater than $na$, we know $m-1<na$, or $m<na+1$. But $na+1<nb$ so $m<nb$ as required.

Now we have a natural number $n$ and an integer $m$ with $na<m<nb$. Dividing through by $n$ gives $$a<\frac{m}{n}<b$$

As we have gotten used to being very careful in our arguments, you may think while working out the above argument to fill in a little lemma showing that every set of integers bounded below has a minimum. And, you could indeed do so by induction (try it - but fair warning, the argument is a little tricky! It’s easiest with “strong induction” - what are we inducting over?). However this fact is actually logically equivalent to the principle of induction, and in foundations of arithmetic things are often reversed: we take this as an axiom, and prove induction from it! The statement is called the well ordering principle.

Definition 6.8 (The Well Ordering Principle) Every nonempty subset of $\mathbb{N}$ has a least element.

Exercise 6.7 (Density of the Irrationals) Use Theorem 6.10 above to prove that the irrationals are also dense in the reals.

Exercise 6.8 The dyadic rationals are the subset of $\mathbb{Q}$ which have denominators that are a power of 2 when written in lowest terms.

Prove the dyadic rationals are dense in $\mathbb{R}$.

6.3 Uncountability

We can use this to prove the uncountability of the reals using Cantor’s original argument. (We will give the better known Cantor diagonalization argument later, once we’ve introduced decimals)

Theorem 6.11 ($\mathbb{R}$ is Uncountable) There is no bijection between $\mathbb{N}$ and $\mathbb{R}$

Proof. Let $f:\mathbb{N}\to [0,1]$ be any function whatsoever. We can use this function to produce a sequence of points as follows:

$$f(1)=x_1,f(2)=x_2,f(3)=x_3\dots$$

From this we can construct a set of nested intervals.
Let $I_1\subset [0,1]$ be any closed interval that doesn’t contain $x_1$. Then let $I_2\subset I_1$ be a closed interval which does not contain $x_2$ (if $x_2$ was outside $I_1$, you could just take $I_1$ again, otherwise if its inside $I_1$ just take an interval on one side or the other of it). Continuing, we can easily choose an interval $I_{n+1}\subset I_n$ which doesn’t contain $x_{n+1}$.

This gives us an infinite sequence of closed nested intervals inside a complete ordered field, so Theorem 5.2 tells us that their intersection must be nonempty. That is, there is some point $y\in [0,1]$ where $y\in I_n$ for all $n$.

What does this mean? Well, since $y\in I_1$ we know $y\ne x_1$ since $I_1$ was purpose-built to exclude $x_1$. Similarly $y\in I_2$ guarantees $y\ne x_2$, and so on…$y\in I_n$ means $y\ne x_n$. Thus, $y$ is some point in $[0,1]$ which is not in our list!

Since $y\ne f(n)$ for any $n$, we see that our original (arbitrary) function cannot have been surjective. And, since bijections are both injective and surjective, this proves there is no bijection from $\mathbb{N}$ to $[0,1]$, so $[0,1]$ is uncountable! Then, as $[0,1]\subset \mathbb{R}$ we see $\mathbb{R}$ is uncountable as well.

This has some pretty wild corollaries if you have studied countable sets before. Here’s a couple examples

Corollary 6.1 (Transcendental Numbers) There exist real numbers which are not the solution of any algebraic equation with rational coefficients.

Corollary 6.2 (Uncomputable Numbers) There exist real numbers which cannot be computed by any computer program.

These are additional motivation for why we really need a precise theory of the real numbers: with very little work we’ve already proven that there is no way to study this number system with algebra alone - or even with the most powerful computer you could imagine.

6.4 $★$ Topology

One final basic property of $\mathbb{R}$ that we will show follows from completeness is that its “connected” - it really does form a continuous line.

Definition 6.9 (Connected) Let $S$ be a subset of a topological space. Then a separation of $S$ is a pair of disjoint open sets $U,V$ whose union is $S$.

A subset is called disconnected if there is a separation, and connected if there is no way to make a separation.

Example 6.2 (A disconnected set) Let $S=\{x\in \mathbb{R}\mid x>0,x<2,\text{ and }x\ne 1\}$. Then $S$ is disconnected as we can write $$S=(0,1)\cup (1,2)$$ And note these two intervals are both open, and dont share any points in common (so they are disjoint).

It’s harder to imagine doing this for the interval $(0,2)$ however: if you try to imagine cutting it into two disjoint intervals at some point $x$, you’re going to end up with $(0,x)\cup [x,2)$ or $(0,x]\cup (x,2)$. In either case, these intervals are not both open! To make them both open you could try $(0,x)\cup (x,2)$ but now they miss the point $x$ (so their union isnt the whole space) or $(0,x+0.01)\cup (x-0.01,2)$ but now they overlap and aren’t disjoint. Intuitively there’s no way to do it - the interval $(0,2)$ is connected!

Theorem 6.12 (The Real Line is Connected)  

Proof. Assume for the sake of contradiction that $U\cup V$ is a separation of $\mathbb{R}$ (so, $U,V$ are nonempty open sets and every point of $\mathbb{R}$ is in exactly one of them).

Choose some $x\in U$ and $y\in V$ - we can do this because they’re nonempty - and without loss of generality assume that $x<y$. Considering the interval $[x,y]$ we know the left side is in $U$ and the right in $V$, so we can define the $$Z=\{z\in [x,y]\mid [x,z]\subset U\}$$

This set is nonempty (as $x\in Z$) and its bounded above (by $y$), so by completeness it has some supremum $\zeta =supZ$. Now the question is, which set is $\zeta$ in, $U$ or $V$?

If $\zeta \in V$ then we know that since $V$ is open ther’s some small interval $(\zeta -ϵ,\zeta +ϵ)$ fully contained in $V$. But this means there’s a number smaller than $\zeta$ contained in $V$, which means the interval $[0,\zeta ]$ isnt fully contained in $U$, a contradiction!

If $\zeta \in U$ then we know since $U$ is open, that there must be some tiny open interval $(\zeta -ϵ,\zeta +ϵ)$ around $\zeta$ contained $U$. This means there’s a number *larger than $\zeta$ (for example, $\zeta +ϵ/2$) where $[x,\zeta +ϵ/2]$ is contained in $U$. So, $\zeta$ can’t even be an upper bound to the set of all such numbers, a contradiction!

Both cases lead to contradiction, so there must be no such $\zeta$, and hence no such separation.

Exercise 6.9 (Open Intervals of $\mathbb{R}$ are connected) Prove that every open interval $(a,b)\subset \mathbb{R}$ is connected, mimicking the proof style above.

This fails for the rational numbers - they are not connected!

Theorem 6.13 (The Rationals are Not Connected) Consider the following two subsets of the rational numbers: $$A=\{x>0\mid x^2>2\}$$ $$B=\{x\in \mathbb{Q}\mid x\notin A\}$$

Then $A$ and $B$ form a separation of $\mathbb{Q}$.

Proof. $A$ and $B$ are open intervals in $\mathbb{Q}$ (they’re the rational points of the open intervals $(\sqrt{2},\mathrm{\infty })$ and $(-\mathrm{\infty },\sqrt{2})$). By definition every point of $\mathbb{Q}$ is in either $A$ or $B$ and they’re disjoint. Since we just showed they are open, they form a separation, so $\mathbb{Q}$ is disconnected.

In fact, $\mathbb{Q}$ is extremely disconnected - this same argument applies at every irrational number of $\mathbb{R}$.