Proof of Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts

Conventions. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §ordered-field, $\le$ is a total order on $\mathbb{Q}$ with strict relation $<$, $\mathbb{Q}$ is an ordered field, and $u+(-u)=0$ for $u\in\mathbb{Q}$. Hence the results of Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders apply to $\le$ on $\mathbb{Q}$, and the rules of Rules of Arithmetic and Order in an Ordered Field hold in $\mathbb{Q}$, and so do the identities of a commutative ring (by Fields §field). Rearrangements of sums and products below use only these identities, $u+(-u)=0$, Rules of Arithmetic and Order in an Ordered Field §signs, and $(u/w)\cdot w=u$ for $w\neq0$, which holds by Negatives, Differences, Reciprocals and Quotients §reciprocal; as in Negatives, Differences, Reciprocals and Quotients §negative, $-u-v$ is $(-u)+(-v)$. By Rules of Arithmetic and Order in an Ordered Field §midpoint, $0<1+1$, so $1+1\neq0$ and $w/(1+1)$ is defined for every $w\in\mathbb{Q}$, with $w/(1+1)+w/(1+1)=w$.

The four conditions defining $C$ are called (C1) $x\neq\emptyset$; (C2) $x\neq\mathbb{Q}$; (C3) if $u\in x$, $v\in\mathbb{Q}$ and $v<u$, then $v\in x$; (C4) every $u\in x$ has some $v\in x$ with $u<v$.

Preliminaries. Let $x\in C$. (P1) $x\subseteq\mathbb{Q}$, since $x\in\mathcal{P}(\mathbb{Q})$ (The Union Set and the Power Set of a Set §power). (P2) There is $a\in\mathbb{Q}$ with $a\notin x$: otherwise, by (P1), $x$ and $\mathbb{Q}$ have the same elements, so $x=\mathbb{Q}$ by Axiom of Extensionality for Classes, contrary to (C2). (P3) If $u\in x$, $a\in\mathbb{Q}$ and $a\notin x$, then $u<a$: by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, otherwise $a=u$ or $a<u$, and each gives $a\in x$, the second by (C3). (P4) $\mathbb{Q}$ is a set (The Rational Numbers §rationals), so every subclass $y$ of $\mathbb{Q}$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper §subclass and lies in $\mathcal{P}(\mathbb{Q})$ by The Union Set and the Power Set of a Set §power; so for such $y$, $y\in C$ follows once (C1)-(C4) hold for $y$. The classes $u^{*}$, $x\oplus y$, $\ominus x$ and $P=\{w\in\mathbb{Q}:\exists u\,\exists v\,(u\in\mathbb{Q}\wedge v\in\mathbb{Q}\wedge u\in x\wedge v\in y\wedge0\le u\wedge0\le v\wedge w=u\cdot v)\}$ are formed by abstraction over elements of $\mathbb{Q}$, so they are subclasses of $\mathbb{Q}$, and then so is $x\odot y=0^{*}\cup P$ by The Boolean Operations on Classes, Disjointness, and the Universal Class §operations.

Clause set. $C$ is formed by abstraction over elements of $\mathcal{P}(\mathbb{Q})$, so $C\subseteq\mathcal{P}(\mathbb{Q})$. The power set $\mathcal{P}(\mathbb{Q})$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper §power and The Union Set and the Power Set of a Set §power, since $\mathbb{Q}$ is a set. Hence $C$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper §subclass.

Clause order. By clause set and Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §set, $C\times C$ is a set; $\le_{C}\subseteq C\times C$, so $\le_{C}$ is a set by Subclasses of Sets Are Sets, the Union and Power Set of a Set Exist Uniquely, Binary Unions of Sets Are Sets, and the Universal Class Is Proper §subclass. Every element of $C\times C$ is an ordered pair by The Cartesian Product of Two Classes §product, so $\le_{C}$ is a relation and a relation on $C$. For $x,y\in C$, $(x,y)\in C\times C$ by Membership in a Cartesian Product, and the Cartesian Product of Two Sets Is a Set §membership, and $(x,y)\in{\le_{C}}$ if and only if $x\subseteq y$, because a pair $(x',y')$ equal to $(x,y)$ has $x'=x$ and $y'=y$ by The Characteristic Property of Ordered Pairs and Nested Tuples of Sets §characteristic. Now let $x,y,z\in C$. Reflexivity: $x\subseteq x$ by Subclasses and Subsets §subclass. Antisymmetry: if $x\subseteq y$ and $y\subseteq x$, then $x$ and $y$ have the same elements, so $x=y$ by Axiom of Extensionality for Classes. Transitivity: if $x\subseteq y$ and $y\subseteq z$, every element of $x$ is in $y$, hence in $z$, so $x\subseteq z$. Thus $\le_{C}$ is a partial order on the set $C$. Totality: suppose $x\subseteq y$ fails. Choose $u\in x$ with $u\notin y$; $u\in\mathbb{Q}$ by (P1). For every $v\in y$, (P3) for $y$ gives $v<u$, so $v\in x$ by (C3) for $x$. Hence $y\subseteq x$. So $x\le_{C}y$ or $y\le_{C}x$, and $\le_{C}$ is a total order on $C$.

Clause rational. Let $u\in\mathbb{Q}$; by (P4) it suffices to check (C1)-(C4) for $u^{*}$. (C1): $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, so adding $u-1$ gives $u-1<u$ by Rules of Arithmetic and Order in an Ordered Field §order-sum; thus $u-1\in u^{*}$. (C2): $u\in\mathbb{Q}$ and $u\notin u^{*}$, since $u<u$ fails by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive. (C3): if $v\in u^{*}$, $w\in\mathbb{Q}$ and $w<v$, then $w<u$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive, so $w\in u^{*}$. (C4): if $v\in u^{*}$, then $v<u$, so $v<(v+u)/(1+1)<u$ by Rules of Arithmetic and Order in an Ordered Field §midpoint, and $(v+u)/(1+1)\in u^{*}$.

Clause sum. Let $x,y\in C$; by (P4) we check (C1)-(C4) for $x\oplus y$. (C1): choose $u\in x$ and $v\in y$ by (C1) for $x$ and $y$; they are rational by (P1), and $u+v\in x\oplus y$. (C2): choose $a\in\mathbb{Q}\setminus x$ and $b\in\mathbb{Q}\setminus y$ by (P2). If $u\in x$ and $v\in y$, then $u<a$ and $v<b$ by (P3), so $u+v<a+v<a+b$ by Rules of Arithmetic and Order in an Ordered Field §order-sum (and commutativity), whence $u+v<a+b$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive. So $a+b=u+v$ is impossible by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive, and $a+b\in\mathbb{Q}\setminus(x\oplus y)$. (C3): let $w=u+v$ with $u\in x$, $v\in y$, and let $t\in\mathbb{Q}$ with $t<w$. Adding $-v$ gives $t-v<u$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, so $t-v\in x$ by (C3) for $x$, and $t=(t-v)+v\in x\oplus y$. (C4): for $w=u+v$ as before, choose $u'\in x$ with $u<u'$ by (C4) for $x$; then $w<u'+v$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, and $u'+v\in x\oplus y$.

Clause negative. Let $x\in C$; by (P4) we check (C1)-(C4) for $\ominus x$. (C1): choose $p\in\mathbb{Q}\setminus x$ by (P2) and put $u=-p-1$. With $v=1$, which satisfies $0<1$ by Rules of Arithmetic and Order in an Ordered Field §squares, we have $-u-v=(p+1)-1=p\notin x$, so $u\in\ominus x$. (C2): choose $q\in x$ by (C1); $q\in\mathbb{Q}$ by (P1). We show $-q\notin\ominus x$. Let $v\in\mathbb{Q}$ with $0<v$. Then $-(-q)-v=q-v$, and adding $q-v$ to $0<v$ gives $q-v<q$ by Rules of Arithmetic and Order in an Ordered Field §order-sum, so $q-v\in x$ by (C3). Thus no $v$ witnesses $-q\in\ominus x$, and $\ominus x\neq\mathbb{Q}$. (C3): let $u\in\ominus x$ and choose $v\in\mathbb{Q}$ with $0<v$ and $-u-v\notin x$; let $w\in\mathbb{Q}$ with $w<u$. Adding $-u-w-v$ to $w<u$ gives $-u-v<-w-v$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. If $-w-v\in x$, then $-u-v\in x$ by (C3) for $x$, a contradiction; so $-w-v\notin x$ and $v$ witnesses $w\in\ominus x$. (C4): let $u\in\ominus x$, with $v$ chosen as in (C3), and put $h=v/(1+1)$. By Rules of Arithmetic and Order in an Ordered Field §midpoint applied to $0<v$, $0<h$; hence $u<u+h$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Moreover $-(u+h)-h=-u-(h+h)=-u-v\notin x$, so $h$ witnesses $u+h\in\ominus x$.

Clause absolute. Let $x\in C$. Since $0^{*}\in C$ by clause rational, clause order gives $0^{*}\subseteq x$ or $x\subseteq0^{*}$. In the second case let $u\in0^{*}$, so $u\in\mathbb{Q}$ and $u<0$. Adding $-u$ gives $0<-u$ by Rules of Arithmetic and Order in an Ordered Field §order-sum. Put $v=(-u)/(1+1)$; then $0<v$ by Rules of Arithmetic and Order in an Ordered Field §midpoint, and $-u-v=(v+v)-v=v$. Since $0<v$, $v<0$ fails by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, so $v\notin0^{*}$, hence $v\notin x$. Thus $-u-v\notin x$ and $u\in\ominus x$. So $0^{*}\subseteq\ominus x$.

Clause product. Let $x,y\in C$ with $0^{*}\subseteq x$ and $0^{*}\subseteq y$, and let $P$ be as in (P4), so $x\odot y=0^{*}\cup P$. Then $0^{*}\subseteq x\odot y$ by The Boolean Operations on Classes, Disjointness, and the Universal Class §operations. We use two facts. (F1) If $0\le s$ and $0\le t$ in $\mathbb{Q}$, then $0\le s\cdot t$, as $\mathbb{Q}$ is an ordered ring (Ordered Fields §ordered-field). (F2) If $0\le v$ and $u<a$, then $u\cdot v\le a\cdot v$: if $v=0$ both sides are $0$ by Rules of Arithmetic and Order in an Ordered Field §zero, and if $0<v$ then $u\cdot v<a\cdot v$ by Rules of Arithmetic and Order in an Ordered Field §order-product; one of the two holds by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict. By (P4) we check (C1)-(C4) for $x\odot y$.

(C1): $0^{*}\neq\emptyset$ by clause rational, and $0^{*}\subseteq x\odot y$.

(C2): choose $a\in\mathbb{Q}\setminus x$ and $b\in\mathbb{Q}\setminus y$ by (P2). As $0^{*}\subseteq x$, $a\notin0^{*}$, so $a<0$ fails and $0\le a$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation; likewise $0\le b$. Then $0\le a\cdot b$ by (F1), so $a\cdot b<0$ fails by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation and $a\cdot b\notin0^{*}$. Let $w\in P$, with $w=u\cdot v$, $u\in x$, $v\in y$, $0\le u$, $0\le v$. By (P3), $u<a$ and $v<b$; from $0\le u<a$ we get $0<a$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive. By (F2), $u\cdot v\le a\cdot v$, and $a\cdot v<a\cdot b$ by Rules of Arithmetic and Order in an Ordered Field §order-product and commutativity; so $w<a\cdot b$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive. Hence $a\cdot b\neq w$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive, so $a\cdot b\notin P$, and $a\cdot b\in\mathbb{Q}\setminus(x\odot y)$.

(C3): let $w\in x\odot y$ and $t\in\mathbb{Q}$ with $t<w$. If $t<0$, then $t\in0^{*}\subseteq x\odot y$. Otherwise $0\le t$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, so $0<w$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive; then $w\notin0^{*}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy, so $w\in P$, say $w=u\cdot v$ with $u\in x$, $v\in y$, $0\le u$, $0\le v$. If $u=0$ then $w=0$ by Rules of Arithmetic and Order in an Ordered Field §zero, contrary to $0<w$; so $0<u$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict), $u^{-1}$ exists, and $0<u^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. Put $s=t/u$. From $t<u\cdot v$, Rules of Arithmetic and Order in an Ordered Field §order-product with $u^{-1}$ gives $s<(u\cdot v)\cdot u^{-1}=v$, so $s\in y$ by (C3) for $y$; and $0\le s$ by (F1), since $0\le t$ and $0\le u^{-1}$ (from $0<u^{-1}$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict). Since $t=u\cdot s$ with $u\in x$, $s\in y$, $0\le u$, $0\le s$, we get $t\in P\subseteq x\odot y$.

(C4): let $w\in x\odot y$. If $w\in0^{*}$, then $w<0$, so $w<(w+0)/(1+1)<0$ by Rules of Arithmetic and Order in an Ordered Field §midpoint, and $(w+0)/(1+1)\in0^{*}\subseteq x\odot y$. Otherwise $w\in P$, say $w=u\cdot v$ with $u\in x$, $v\in y$, $0\le u$, $0\le v$. Choose $u'\in x$ with $u<u'$ and then $v'\in y$ with $v<v'$, by (C4) for $x$ and $y$. As before $0<u'$ and $0<v'$, so $0\le u'$ and $0\le v'$, and $u'\cdot v'\in P$. By (F2), $u\cdot v\le u'\cdot v$, and $u'\cdot v<u'\cdot v'$ by Rules of Arithmetic and Order in an Ordered Field §order-product and commutativity; so $w<u'\cdot v'$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive, and $u'\cdot v'\in P\subseteq x\odot y$.

Hence $x\odot y\in C$ and $0^{*}\subseteq x\odot y$.