The Real Numbers

Here, we define what the Real Number system is, and its properties. We begin by introducing some common notations for sets of numbers.

$$\begin{array}{rl}& \mathbb{N}=\{1,2,3,\dots \}\\ & \mathbb{Z}=\{\dots ,-2,-1,0,1,2,\dots \}\\ & \mathbb{Q}=\{p/q:p,q\in \mathbb{Z},q\ne 0\}\end{array}$$

The Real Number System §

We define the Real Numbers, $\mathbb{R}$, as the number system satisfying the following 3 groups of axioms: field axioms, positivity axioms, and completeness axioms.

Field Axioms §

Part of the definition of the Real Numbers $\mathbb{R}$ is that it is a field.

A field is a set of numbers with two binary operations, called addition ($+$) and multiplication ($\times$), that satisfy rules known as field axioms. These axioms are split into addition, multiplication, and distributive axioms, and are described below.

Addition and multiplication don’t necessarily have to stand for the addition and multiplication we’re used to, though in the purposes of real numbers they do.

Addition Axioms §

The addition axioms include properties of commutativity, associativity, identity, and invertibility.

$$\begin{array}{rlr}\forall x,y\in \mathbb{F}& x+y=y+x& \text{Commutativity}\\ \forall x,y,z\in \mathbb{F}& (x+y)+z=x+(y+z)& \text{Associativity}\\ \exists 0\in \mathbb{F}:\forall x\in \mathbb{F}& x+0=x& \text{Identity}\\ \forall x\in \mathbb{F},\exists (-x)\in \mathbb{F}& x+(-x)=0& \text{Invertibility}\end{array}$$

Multiplication Axioms §

Just like the addition axioms, the multiplication axioms include properties of commutativity, associativity, identity, and invertibility.

$$\begin{array}{rlr}\forall x,y\in \mathbb{F}& x\cdot y=y\cdot x& \text{Commutativity}\\ \forall x,y,z\in \mathbb{F}& (x\cdot y)\cdot z=x\cdot (y\cdot z)& \text{Associativity}\\ \exists 1\in \mathbb{F}:1\ne 0\wedge \forall x\in \mathbb{F}& x\cdot 1=x& \text{Identity}\\ \forall x\in \mathbb{F},x\ne 0,\exists x^{-1}\in \mathbb{F}& x\cdot x^{-1}=1& \text{Invertibility}\end{array}$$

Distributivity and Nontriviality §

Additionally, a field relates the addition and multiplication operation through a property known as distributivity, defined below.

$$\begin{array}{rlr}\forall x,y,z\in \mathbb{F}& x\cdot (y+z)=x\cdot y+x\cdot z& \text{Distributivity}\end{array}$$

Finally, a field has the property of nontriviality, which states that $0\ne 1$.

$$\begin{array}{rlr}& 0\ne 1& \text{Nontriviality}\end{array}$$

From these base axioms, we can derive all of the rules for standard algebraic manipulations which we can perform in the Real Numbers.

Positivity Axioms §

Let $P$ be the subset of $\mathbb{R}$ containing positive numbers. Then, as $\mathbb{R}$ satisfies the positivity axioms, the following hold:

$$\begin{array}{rlr}\forall a,b\in P& ab\in P,a+b\in P& \text{Closure}\\ \forall a\in \mathbb{R}& a\in P\vee -a\in P\vee a=0& \text{Trichotomy}\end{array}$$

These axioms can be used to define the inequality operations, which we often use in the real numbers.

Completeness Axioms §

Bounds §

We say that a subset of the real numbers, $S\in \mathbb{R}$, is bounded above if

$$\exists \beta \in \mathbb{R}:\forall x\in S,x\le \beta$$

Where $\beta$ is known as an upper bound. In other words, there exists some $\beta$ that is greater than or equal to all elements in the set $S$.

Similarly, we say that $S$ is bounded below if

$$\exists \beta \in \mathbb{R}:\forall x\in S,x\ge \beta$$

Where $\beta$ is known as a lower bound. In other words, there exists some $\beta$ that is less than or equal to all elements in the set $S$.

Supremum and Infimum §

Suppose we have a subset of the real numbers, $S\in \mathbb{R}$.

Then, a value $x\in S$ is a least upper bound (supremum) of $S$, denoted $\text{sup}(S)$, when:

1. $x$ is an upper bound of $S$.
2. If $y\in X$ is also an upper bound of $S$, then $x\le y$.

If the supremum is within $S$ itself, then it is known as a maximum.

Similarly, a value $s\in S$ is a greatest lower bound (infumum) of $S$, denoted $\text{inf}(S)$, when:

1. $x$ is a lower bound of $S$.
2. If $y\in X$ is also a lower bound of $S$, then $y\le x$.

If the infimum is within $S$ itself, then it is known as a minimum.

We can think of supremum and infimum as the “closest” value in $X$ that serves as an upper (or lower) bound for $S$! However, they are not necessarily maximums or minimums, as supremums and infimums do not have to be in $S$!

Theorem: Alternative Definition of Supremum and Infimum

Let $S\subseteq \mathbb{R}$, and bounded above (by $M$). Then,

$$M=\text{sup}(S)⟺\forall ϵ>0,\exists x_{ϵ}\in S:x_{ϵ}>M-ϵ$$

In other words, $M$ is a supremum if and only if for all positive epsilon, there exists some point in the set that is greater than the supremum minus epsilon.

Similarly, let $S$ be bounded below (by $m$). Then,

$$m=\text{inf}(S)⟺\forall ϵ>0,\exists x_{ϵ}\in S:x_{ϵ}<m+ϵ$$

These statements are basically saying that there is no such smaller upper bound, or no such smaller lower bound., as we can find an $x\in S$ which fails to satisfy the bound definition.

Example: Supremums

1. Suppose $S=(0,1)$. Then, $\text{sup}(S)=1$.
2. Suppose $S=\{-1/x:x>0\}$. Then, $\text{sup}(S)=0$.
3. Suppose $S=\mathbb{N}$. Then, $\text{sup}(S)$ does not exist, as the naturals aren’t bounded above.

Example: Infimum Proof

Suppose $S=(0,1)$. Prove that $\text{inf}(S)=0$. We need to show that:

1. 0 is a lower bound of $S$.

As $0\le 0<x<1$ for all $x\in S$, 0 is a lower bound for $S$.

2. For any other arbitrary lower bound $y$, $y\le 0$.

Let $y$ be any lower bound of $S$, and suppose by way of contradiction that $y>0$. We do a case analysis to show that no value $y>0$ can possibly be a lower bound.

• If $0<y<1$, we can pick an element in $S$ that is smaller than $y$, by choosing $y/2$. This yields a contradiction!
• If $y\ge 1$, we can simply pick any element in $S$, as it will be smaller than $y$. Choosing 0.5, we get a contraditiction!

Thus, if $y$ is a lower bound of $S$, it must be $\le 0$.

Completeness Axiom §

Suppose we have any subset of the real numbers $S\in \mathbb{R}$ that is bounded above. Then, $S$ is guaranteed to have a supremum, which is a property known as the least upper bound property (completeness axiom) of $\mathbb{R}$.

We can use this property to prove the existence of infimums as well.

Theorem: Existence of Infimum

Suppose $X$ is a set with the least upper bound property. Then, every subset of $X$ with a lower bound has a greatest lower bound.

By virtue of this theorem, all subsets $S\in \mathbb{R}$ must have infimums

Proof (Sketch)

Suppose we have a subset $S$ whose existence of a supremum is guaranteed. Then, if we form a new set $T$ by multiplying the subset $S$ by -1 (reflecting it), the lower bounds of $S$ becomes the upper bounds of $T$. We thus show that there exists a supremum for $T$, and this supremum is the inverse of the infimum of $S$.

Example: Uniqueness of Supremum

Prove $\text{sup}(S)$ is unique. We do this by assuming $\text{sup}(S)=a,\text{sup}(s)=b,a\ne b$, and getting a contradiction.

Properties of the Real Numbers §

Below, we describe some interesting and useful properties of the Real Numbers.

Archimedian Property of $\mathbb{R}$ §

The Archimedian Property states the following:

Theorem: Archimedian Property

1. For all $c\in \mathbb{R}$ such that $c>0$, there $\exists n\in \mathbb{N}$ such that $n>c$.
2. For all $ϵ>0$, there $\exists n\in \mathbb{N}$ such that $1/n<ϵ$.

Note that we can obtain one from the other by setting $ϵ=1/c$.

Proof (1)

We will prove lemma (1) using contradiction. Suppose by way of contradiction that

$$\neg (\forall c>0,\exists n\in \mathbb{N}:n>c)$$

In other words, there exists a $c>0$ such that $\forall n\in \mathbb{N}$, $n\le c$. By this assumption, $\mathbb{N}$ must be bounded above by definition, which by the completeness axiom, guarantees the existence of a supremum.

Let this supremum be $b=\text{sup}(\mathbb{N})$. We obtain a contradiction by finding a natural number $n\in \mathbb{N}$ that is strictly larger than $b$, violating our definition of a supremum.

We show this in 2 ways.

1. By Lemma 1.8 (see later), we know that there must exist a unique integer $k$ such that

   $$b<b+1\le k<b+2$$

   Then $k$ is an element of $\mathbb{N}$, and is strictly larger than $b$, givingus a contradiction.

2. Since $b=\text{sup}(\mathbb{N})$, we know that there exists an $n\in \mathbb{N}$ such that

   $$n>b-\frac{1}{2}$$

   This follows by our epsilon definition of a supremum. Note that we could have chosen any value in place of $1/2$, as if there does not exist an $n$, then $b$ is not the least upper bound.

   We rearrange this to find a natural number larger than b (add 1).

   $$n+1>b+\frac{1}{2}>b$$

Density in $\mathbb{R}$ §

A set $S$ is dense in $\mathbb{R}$ if for any interval $(a,b)$, you can find some element of $S$ inside that interval. More formally, for any $a,b\in \mathbb{R}$ such that $a<b$,

$$\exists s\in S:a<s<b$$

See the example below.

Example: Density of $\mathbb{Z}$ in $\mathbb{R}$

Is $\mathbb{Z}$ dense in $\mathbb{R}$?

No, as we can choose an interval in $\mathbb{R}$ that does not contain any integers as a counterexample. For example, the interval $(0.5,0.6)$ does not contain any integers.

Density of $\mathbb{Q}$ in $\mathbb{R}$ §

We use the following lemmas below to show that $\mathbb{Q}$ is dense in $\mathbb{R}$. Such lemmas can be proven, but these proofs are ommitted for convenience.

Lemmas

(1.6) For all $n\in \mathbb{Z}$, there is no integer in the interval $(n,n+1)$.

(1.7) Let $S$ be a nonempty set in $\mathbb{Z}$, and bounded above. Then, $S$ has a maximum.

(1.8) For all real numbers $c\in \mathbb{R}$, there exists a unique integer $k\in \mathbb{Z}$ such that

$$k\in [c,c+1)$$

Theorem: Density of $\mathbb{Q}$ in $\mathbb{R}$

$\mathbb{Q}$ is dense in $\mathbb{R}$. This means that, for any $x,y\in \mathbb{R}$, $x<y$, then there must exist a $q\in \mathbb{Q}$ such that

$$x<q<y$$

Proof (Sketch)

Let $(a,b)$ be any interval in $\mathbb{R}$. By definition, we need to show that there exists some $q\in \mathbb{Q}$ that is in $(a,b)$.

First, by the Archimedian property (showed later), we choose a natural number $n\in \mathbb{N}$ such that

$$\frac{1}{n}<b-a$$

We use this to find the interval $[b-1/n,b)$, which is within $(a,b)$.

By Lemma (3), applied to $[nb-1,nb)$, we know that there exists an $m\in \mathbb{Z}$ that is within $[nb-1,nb)$, so

$$nb-1\le m<nb⟹a<b-\frac{1}{n}\le \frac{m}{n}<b$$

Density of the Irrationals in $\mathbb{R}$ §

Theorem: Density of the Irrationals in $\mathbb{R}$

The irrational numbers are also dense in $\mathbb{R}$.

Proof

Suppose we have an interval $(a,b)$. We want to find an irrational number $z$ such that $a<z<b$.

To solve this, we find a different interval and use it to find our $(a,b)$ interval. Namely, we will use the interval

$$\left(\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}\right)$$

And want to use the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$ to find a $q\in \mathbb{Q}$ within that interval. We can then multiply by $\sqrt{2}$ to find $z=q\sqrt{2}$