Proof of Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema

Each result cited is universally quantified over the data in its own statement.

Notation. Below we write $r$ for $F$, $0_{r}$ and $1_{r}$ for the zero and the unit of $F$, and $\kappa$ for the canonical embedding $\kappa_{F}:\mathbb{Q}\to r$. Operations and orders applied to rational numbers are those of $\mathbb{Q}$, and those applied to elements of $r$, in particular to values of $\kappa$, are those of $r$, as The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §overloading allows. By Sets and Maps: Ordinary Notation §orders, $<$ is the strict relation of $\le$, and bounds, suprema and infima of subsets of $r$ are as in Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order, formed in $r$. For $n\in\mathbb{N}_{0}$ we write $n_{r}$ for the image of $n$ in $r$, which is the element of $r$ that $n$ denotes by Commutative Rings, Fields and Ordered Fields: Standard Notation §numerals; thus clause archimedean of the statement says that $x<n_{r}$ for some $n\in\mathbb{N}$. A natural number standing where a rational number is required, such as an operand of a difference or a quotient in $\mathbb{Q}$ or an argument of $\kappa$, denotes its image in $\mathbb{Q}$, by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification. Clauses of the statement are cited below by their names.

Preliminaries. By hypothesis, $r$ is an ordered field; so $r$ is a set, $+$ and $\cdot$ are binary operations on it, and $\le$ is a total order on $r$, by Fields §field and Ordered Fields §ordered-field. Hence Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders apply with $a=r$, and Rules of Arithmetic and Order in an Ordered Field applies to $r$, with $0_{r}$, $1_{r}$ in place of $0$, $1$. Negatives, differences, reciprocals and quotients in $r$ are as in Negatives, Differences, Reciprocals and Quotients §negative and Negatives, Differences, Reciprocals and Quotients §reciprocal. Rearrangements in $r$ that use only associativity, commutativity, distributivity, $z+0_{r}=z$, $z\cdot1_{r}=z$ (all from Commutative Rings §ring), $z+(-z)=0_{r}$ (Negatives, Differences, Reciprocals and Quotients §negative) and $z\cdot z^{-1}=1_{r}$ for $z\neq0_{r}$ (Negatives, Differences, Reciprocals and Quotients §reciprocal) are called ring rearrangements below.

Facts about $\kappa$. By The Canonical Embedding of the Rational Numbers into an Ordered Field §embedding, $\kappa$ is the map of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique, so for all $u,v\in\mathbb{Q}$: $u<v$ if and only if $\kappa(u)<\kappa(v)$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §order; $\kappa(0)=0_{r}$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero; $\kappa(u-v)=\kappa(u)-\kappa(v)$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §negative; and, if $v\neq0$, $\kappa(u/v)=\kappa(u)/\kappa(v)$ with $\kappa(v)\neq0_{r}$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §reciprocal. Here $\mathbb{Q}$ is an ordered field by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §rationals, so its order is a total order.

Natural numbers in $r$. Let $n\in\mathbb{N}$. Then $n+1\in\mathbb{N}$, as $1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets and sums of natural numbers are natural numbers by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §operations. As $r$ is a commutative ring by Fields §field, the image of $1$ in $r$ is the unit $1_{r}$ of $r$, by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants, and

$$(n+1)_{r}=n_{r}+1_{r},\qquad0_{r}<n_{r},\qquad n_{r}=\kappa(n),$$

the first by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §sum and The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products §constants, the second by Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §naturals, and the third, with $n$ read as a rational number, by Natural Numbers Read in an Ordered Field Agree with the Canonical Embedding of the Rationals §numerals. These three facts, together with the fact that the image $1_{r}$ of $1$ in $r$ is the unit of $r$, are the facts (N).

Clause infimum. Let $s$ be a nonempty subset of $r$ that is bounded below, and fix a lower bound $b\in r$ of $s$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded) and an element $v_{0}\in s$. Let

$$t=\{z\in r:\exists v\,(v\in s\wedge z=-v)\},$$

formed by class abstraction from a formula quantifying over sets only, in which $-v$ is used only for $v\in s\subseteq r$, where it is defined; $t$ is a subset of $r$ by Sets and Maps: Ordinary Notation §set-builder, as are the sets formed in the same way below. It is nonempty, since $-v_{0}\in t$. It is bounded above by $-b$: if $z\in t$, say $z=-v$ with $v\in s$, then $b\le v$, so $-v\le-b$ by Rules of Arithmetic and Order in an Ordered Field §order-negative. By the hypothesis on $r$, $t$ has a supremum $m$, that is, a least element of $\operatorname{Ub}(t)$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum). We show that $-m$ is a greatest element of $\operatorname{Lb}(s)$, i.e. an infimum of $s$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum.

First, $-m\in\operatorname{Lb}(s)$: for $v\in s$ we have $-v\in t$, so $-v\le m$, hence $-m\le-(-v)=v$ by Rules of Arithmetic and Order in an Ordered Field §order-negative and Rules of Arithmetic and Order in an Ordered Field §signs. Second, let $c\in\operatorname{Lb}(s)$. Then $-c\in\operatorname{Ub}(t)$: for $z\in t$, $z=-v$ with $v\in s$, so $c\le v$ and $z=-v\le-c$ by Rules of Arithmetic and Order in an Ordered Field §order-negative. As $m$ is a least element of $\operatorname{Ub}(t)$, $m\le-c$, hence $c=-(-c)\le-m$ by Rules of Arithmetic and Order in an Ordered Field §order-negative and Rules of Arithmetic and Order in an Ordered Field §signs. So $-m$ is a greatest element of $\operatorname{Lb}(s)$, and $s$ has an infimum.

Clause archimedean. Suppose, for a contradiction, that $x<n_{r}$ fails for every $n\in\mathbb{N}$. Then $n_{r}\le x$ for every $n\in\mathbb{N}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Let

$$K=\{z\in r:\exists n\,(n\in\mathbb{N}\wedge z=n_{r})\},$$

formed from a formula quantifying over sets only, with $n_{r}$ defined for every $n\in\mathbb{N}$ by The Image of a Natural Number in a Commutative Ring §image; $K$ is a subset of $r$. It is nonempty, as $1_{r}\in K$ ($1\in\mathbb{N}$ by The Natural Numbers and the Natural Numbers with Zero: Arithmetic, Order, Induction and Recursion §sets), and $x$ is an upper bound of it. By the hypothesis on $r$ it has a supremum $m$, that is, a least element of $\operatorname{Ub}(K)$.

From $0_{r}<1_{r}$ (Rules of Arithmetic and Order in an Ordered Field §squares), adding $m-1_{r}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum and using ring rearrangements, $m-1_{r}<m$. Hence $m\le m-1_{r}$ fails by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, so $m-1_{r}\notin\operatorname{Ub}(K)$, since otherwise $m\le m-1_{r}$ as $m$ is least in $\operatorname{Ub}(K)$. So there is $z\in K$ for which $z\le m-1_{r}$ fails; fix one and fix $n\in\mathbb{N}$ with $z=n_{r}$. Then $m-1_{r}<n_{r}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, and adding $1_{r}$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, $m<n_{r}+1_{r}=(n+1)_{r}$ by (N). But $n+1\in\mathbb{N}$ by (N), so $(n+1)_{r}\in K$ and $(n+1)_{r}\le m$, which by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation contradicts $m<(n+1)_{r}$. Hence some $n\in\mathbb{N}$ has $x<n_{r}$, which is clause archimedean. Since $x\in r$ was arbitrary, this holds for every element of $r$; it is used below for elements other than $x$.

Clause density. Let $x<y$. The choices are made in the following order.

Step 1: the gap. Let $d=y-x$. By Rules of Arithmetic and Order in an Ordered Field §order-sum, adding $-x$ to $x<y$ gives $0_{r}<d$. So $d\neq0_{r}$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization), $d^{-1}$ is defined, and $0_{r}<d^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal.

Step 2: the scale. By clause archimedean applied to $d^{-1}$, fix $m\in\mathbb{N}$ with $d^{-1}<m_{r}$, and put $M=m_{r}$; $0_{r}<M$ by (N). Multiplying by $d$, with $0_{r}<d$, by Rules of Arithmetic and Order in an Ordered Field §order-product, $1_{r}=d^{-1}\cdot d<M\cdot d$. By ring rearrangements and Rules of Arithmetic and Order in an Ordered Field §signs, $M\cdot d=M\cdot y-M\cdot x$, and adding $M\cdot x$ by Rules of Arithmetic and Order in an Ordered Field §order-sum,

$$M\cdot x+1_{r}<M\cdot y.$$

Step 3: the shift. By clause archimedean applied to $-(M\cdot x)$, fix $p\in\mathbb{N}$ with $-(M\cdot x)<p_{r}$, and put $w=M\cdot x+p_{r}$. Adding $M\cdot x$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, $0_{r}<w$.

Step 4: the least natural number above $w$. Let $A=\{k\in\mathbb{N}:w<k_{r}\}$, formed from a formula quantifying over sets only; it is a subset of $\mathbb{N}$, and nonempty by clause archimedean applied to $w$. By Arithmetic and Order of the Natural Numbers §well-order, fix $k\in A$ with $k\le v$ for every $v\in A$. Then $w<k_{r}$, and we claim $k_{r}\le w+1_{r}$. If $k=1$, then $k_{r}=1_{r}=0_{r}+1_{r}\le w+1_{r}$ by (N), $0_{r}\le w$ (from $0_{r}<w$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict) and Rules of Arithmetic and Order in an Ordered Field §order-sum (with $1_{r}\le1_{r}$, by reflexivity of the partial order $\le$, for the second summand). If $k\neq1$, fix $e\in\mathbb{N}$ with $k=e+1$ by Arithmetic and Order of the Natural Numbers §predecessor. Then $e<k$ by Arithmetic and Order of the Natural Numbers §successor, so $k\le e$ fails by Arithmetic and Order of the Natural Numbers §trichotomy and Arithmetic and Order of the Natural Numbers §partial-order; hence $e\notin A$, i.e. $w<e_{r}$ fails, so $e_{r}\le w$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, and $k_{r}=e_{r}+1_{r}\le w+1_{r}$ by (N) and Rules of Arithmetic and Order in an Ordered Field §order-sum. In both cases

$$w<k_{r}\le w+1_{r}.$$

Step 5: the rational number. By (N) and the facts about $\kappa$, $\kappa(0)=0_{r}<M=m_{r}=\kappa(m)$, so $0<m$ in $\mathbb{Q}$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §order, and $m\neq0$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization. Let $u=(k-p)/m\in\mathbb{Q}$, with $k$, $p$ and $m$ read as rational numbers (The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification, The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §agreement) and the difference and quotient formed in $\mathbb{Q}$. By the facts about $\kappa$ and (N), $\kappa(u)=(\kappa(k)-\kappa(p))/\kappa(m)=(k_{r}-p_{r})/M$; as $M\neq0_{r}$ (from $0_{r}<M$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization), ring rearrangements give $\kappa(u)\cdot M=k_{r}-p_{r}$. Adding $-p_{r}$ to the display of Step 4 by Rules of Arithmetic and Order in an Ordered Field §order-sum, and using $w-p_{r}=M\cdot x$ (ring rearrangements), we get

$$M\cdot x<\kappa(u)\cdot M\le M\cdot x+1_{r}.$$

With Step 2 and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares §mixed, $\kappa(u)\cdot M<M\cdot y$. Since $0_{r}<M$, Rules of Arithmetic and Order in an Ordered Field §order-product and commutativity turn $x\cdot M<\kappa(u)\cdot M$ and $\kappa(u)\cdot M<y\cdot M$ into $x<\kappa(u)$ and $\kappa(u)<y$. So $u\in\mathbb{Q}$ satisfies $x<\kappa(u)<y$, which is clause density.

Clause rational-supremum. Let $S=\{u\in\mathbb{Q}:\kappa(u)<x\}$, a subset of $\mathbb{Q}$ by Sets and Maps: Ordinary Notation §set-builder, and let $L=\kappa(S)$, the set of the statement. Here $\kappa(S)$ denotes the image of $S$ and not a value of $\kappa$, because an element of $\mathbb{Q}$ is used as a number only, never as a set (The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §numbers-only), so $S$ is not an element of $\mathbb{Q}$ in this reading; by Sets and Maps: Ordinary Notation §images it is the image $\kappa[S]$ of The Image and the Preimage of a Class under a Class §image, a set. Its elements are exactly the $\kappa(u)$ with $u\in\mathbb{Q}$ and $\kappa(u)<x$: by The Image and the Preimage of a Class under a Class §image, $z\in L$ if and only if $(u,z)\in\kappa$ for some $u\in S$, and for $u\in S\subseteq\mathbb{Q}=\operatorname{dom}\kappa$ (Functions, Values of a Function, and Functions from One Class to Another §map), $(u,z)\in\kappa$ holds if and only if $z=\kappa(u)$, by Functions, Values of a Function, and Functions from One Class to Another §value. In particular $L$ is a subset of $r$, as $\kappa$ maps into $r$. We show that $x$ is a least element of $\operatorname{Ub}(L)$, i.e. a supremum of $L$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum; as $L$ has at most one supremum by the same clause, $x$ is then the supremum of $L$, which is clause rational-supremum.

$x\in\operatorname{Ub}(L)$: every $z\in L$ has $z<x$, hence $z\le x$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict.

$x$ is least in $\operatorname{Ub}(L)$: let $b\in\operatorname{Ub}(L)$ and suppose $x\le b$ fails. Then $b<x$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, and by clause density fix $u\in\mathbb{Q}$ with $b<\kappa(u)<x$. Then $\kappa(u)\in L$, so $\kappa(u)\le b$, contradicting $b<\kappa(u)$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation. Hence $x\le b$, so $x$ is a least element of $\operatorname{Ub}(L)$, and hence the supremum of $L$, as explained above.