Axiomatic introduction to the Real Numbers

I am sure that you probably know, intuitively, about the Real Numbers. However, this resource will disregard intuition of the Real Numbers, and will give a formal, Axiomatic approach to the Real Numbers.

The Rationals

We start off from the rationals, as later we use them to define the Real Numbers. We can define the rationals as

$\mathbb {Q} =\{{\frac {p}{q}}:p,q\in \mathbb {Z} ,q\neq 0\}$. We can add, subtract, multiply and divide (except by 0) in the usual way. We call $\mathbb {Q}$ a field.

On Order

I add to this in which the rational numbers are also an ordered field. That is, a field with a strict order $<$ such that $a<b\rightarrow a+c<b+c$, and also $0<a$ and $0<b$ implies $0<ab$.

Limitations of the rational numbers

Why do we need the Real numbers at all? This section goes over the limitations of the rationals, which help explain why we need the real numbers.

On the Least-upper-bound property

Definitions

We go over the property of Least-upper-bound first. We start by defining the "prerequisite definitions": Say if we have a subset $S$ of an ordered field. It is bounded above if there exists a number $b$ in said ordered field s.t. $\forall x\in S,b\geq x$. $b$ is known as an upper bound of $S$.

A supremum $\beta$ of $S$ satisfies 2 properties: It is an upper bound of $S$, and all other upper bounds of S are greater than or equal to $\beta$. In other words, a supremum of a set is the least upper bound of a set. As an example, take the set $\{x\in \mathbb {N} :x^{-1},x\neq 0\}$, in which we denote by $F$. It is bounded above, as there exists a number (say ${\frac {3}{2}}$) for which it is greater for all elements in $F$, and ${\frac {3}{2}}$ is an upper bound. The supremum of $F$ is 1.

Also, we look back at $S$. It is bounded below if there exists a number $c$ in said ordered field s.t. $\forall x\in S,c\leq x$, and $c$ is a lower bound. The infimum, or greatest lower bound of $S$ is a lower bound of $S$ that is greater that all other lower bounds of $S$.

The property of least-upper-bound then states that any subset $X\subseteq Y$ (whatever set $Y$ may be) that is bounded above also has a supremum. The rational numbers do NOT fulfill this property.

Theorem. $\mathbb {Q}$ does NOT have the property of Least-upper-bound.

Proof. We give the example of the subset $\{x\in \mathbb {Q} :x^{2}\leq 2\}$. This subset has an upper bound, with the example of $x>1.5$. However, it does NOT have a supremum, as ${\sqrt {2}}\not \in \mathbb {Q}$, even being bounded above. Therefore, $\mathbb {Q}$ does NOT have the property of Least-upper-bound. $\Box$

Completeness

The least-upper-bound property is one version of the completeness property. Intuitively, completeness (of the real numbers) states that there are no "gaps" or "holes" in the real line. Intuitively also, the "holes" in the rational line are equivalent to irrationals. [1] There are other ways of defining this property.

Cauchy Completeness

This definition of completeness states that every Cauchy Sequence of numbers in a set converge to another number in a set. (Mainly the real numbers are used in this example.) The Rationals are not cauchy complete.

Theorem. The rational numbers are not Cauchy complete.

Proof. Take the Cauchy sequence

$1,1.4,1.41,1.414...$

This sequence is comprised of rationals, but converges to ${\sqrt {2}}$, but we know that ${\sqrt {2}}\not \in \mathbb {Q}$. Therefore, not every Cauchy Sequence of rational numbers converge to another rational number. $\Box$