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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!

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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.

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  • $\begingroup$ There is a clear presentation of this in the book "Topology" by Dieudonne, which, in spite of the title, is an introductory text on analysis. (Analysis is the "sprawling octopus" that calculus grew into.) $\endgroup$ – DanielWainfleet Sep 15 '17 at 6:53

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