# Construction of a Complete Ordered Field

I'm studying the textbook of General Theory of Functions and Integration by Angus Taylor. I got to the section of "Construction of a Complete Ordered Field" and I got stuck. Wonder if I could get help.

From the textbook:

In this section we shall describe briefly how one may construct a system of object forming a complete ordered field. The construction starts by assuming the ordered field of rational numbers as a known system.

We now consider sections $(L,R)$ in $F$ such that $L$ has no greatest number. Let $\mathcal{P}$ be the collection of all such sections in $F$, and let $\mathcal{F}$ be the collection of all the left parts $L$ coming from section belonging to $\mathcal{P}$. Then each $L$ is a certain set of rational numbers. If $(L, R)$ is in $\mathcal{P}$ observe that $R = F - L$.

When $L$ is in $\mathcal{F}$ such that $R$ has a smallest member $r$ (where $R = F - L)$, this rational number $r$ completely determines and is determined by $L$. Hence the set of all $L$'s of this particular kind is one-to-one correspondence with the set of all elements of the rational field $\mathcal{F}$. These elements $L$ will be called the rational element of $\mathcal{F}$ corresponding to the rational number $r$ in $F$, then $L = \{x: x \in F , x < r\}$, while $R = F - L = \{ y: y \in F, r \leq y\}$.

An element $L$ of $\mathcal{F}$ such that $R(F - L)$ has no smallest member will be called an irrational element of $\mathcal{F}$

Questions:

1) "Then each $L$ is a certain set of rational numbers." -- is this a definition or derived from somewhere? If it is defined, why is it defined so?

2) Why is it that: "When $L$ is in $\mathcal{F}$ such that $R$ has a smallest member $r$ (where $R = F - L$), this rational number $r$ completely determines and is determined by $L$. Hence the set of all $L$'s of this particular kind is one-to-one correspondence with the set of all elements of the rational field $\mathcal{F}$".

I'm completely lost here. Especially, how it is that the set of all $L$'s of this kind is one-to-one correspondence with the set of all elements of the rational field $\mathcal{F}$??

3) "An element $L$ of $\mathcal{F}$ such that $R(F - L)$ has no smallest member will be called an irrational element of $\mathcal{F}$" -- can someone explain this? I do not follow.

Thank you!

The usual presentation is to take all pairs $(L,R)$ of subsets of $\mathbb Q$ such that (1)... $L\cup R=\mathbb Q$ and (2)... $L\cap R=\emptyset$ and (3)... $L\ne \emptyset \ne R$ and (4)... $\forall x\in L\;\forall y\in R\;(x<y)$ and (5)... $L$ has no largest member. Such pairs $(L,R)$ are called Dedekind cuts on $\mathbb Q.$ Let $D$ be the set of all Dedekind cuts on $\mathbb Q.$
For each $x\in \mathbb Q$ there is a unique $d_x=(L_x,R_x)\in D$ such that $\min R_x=x,$ namely: $L_x=\{y\in \mathbb Q:y<x\}$ and $R_x=\{y\in \mathbb Q: y\geq x\}.$ For $x,y \in \mathbb Q$ define $d_x+d_y=d_{(x+y)}$ and $d_x\times d_y=d_{(xy)}.$ And define $d_x<d_y\iff x<y.$ This makes $\{d_x: x\in \mathbb Q\}$ an ordered field, isomorphic to the ordered field $\mathbb Q.$
There are members of $D$ that are not equal to $d_x$ for any $x\in \mathbb Q.$ For example $(L,R)$ where $R=\{x\in \mathbb Q:0<x\land 2<x^2\}$ and $L=\mathbb Q$ \ $R.$ Such members of $D$ are just those $(L,R)\in D$ for which $\min R$ does not exist.
The rest of the construction is to define $+$ and $\times$ and $<$ on all of $D$ to make it an ordered field, and to prove that it is complete.
A further stage is to prove that if $E$ is any complete ordered field then $E$ is isomorphic to $D.$ So up to isomorphism there is just one complete ordered field. So we speak of "the" real-number system, not "a" real-number system.