Proof of Addition and Order of Dedekind Cuts: an Ordered Abelian Group in Which Every Nonempty Set Bounded Above Has a Supremum

Throughout, for $x,y\in C$ the sum of The Real Numbers §operations is $x+y=x\oplus y$, and $0_{\mathbb{R}}=0^{*}$ by The Real Numbers §constants; we write $\oplus$ for the sum of cuts, so that $+$ always denotes the sum of rational numbers. Sums, negatives and the cuts $u^{*}$ lie in $C$ by Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §sum, Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §negative and Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §rational. Equalities of cuts are proved by showing that the two classes have the same elements and applying Axiom of Extensionality for Classes; inclusions are as in Subclasses and Subsets §subclass. For $x,y\in C$ we have $x\le_{C}y$ if and only if $x\subseteq y$, by the definition of $\le_{C}$ in Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts.

Rules used in $\mathbb{Q}$. 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 $<$, and $\mathbb{Q}$ is an ordered field, hence a field and a commutative ring (Ordered Fields §ordered-field, Fields §field); so the identities of Commutative Rings §ring (associativity and commutativity of $+$ and $\cdot$, $w+0=w$, $w\cdot1=w$, distributivity) hold in $\mathbb{Q}$, together with $w+(-w)=0$. We call these the ring identities. Sign rules come from Rules of Arithmetic and Order in an Ordered Field §signs; irreflexivity, transitivity and the weak/strict comparison of $<$ from Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-irreflexive, Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-transitive and Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict. We record three consequences of Rules of Arithmetic and Order in an Ordered Field §order-sum, which gives $a<b$ if and only if $a+c<b+c$. Let $a,b,e\in\mathbb{Q}$.

(R1) $a<b$ if and only if $a-b<0$, and if and only if $0<b-a$: add $-b$, respectively $-a$, and use the ring identities.

(R2) If $0<e$, then $a-e<a$: add $a-e$ to $0<e$ and use the ring identities. In particular $-e<0$.

(R3) If $0<e$ and $d=e/(1+1)$, then $0<d$ and $d+d=e$. Indeed $0<1+1$ by Rules of Arithmetic and Order in an Ordered Field §midpoint, so $1+1\neq0$ (Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization) and $(1+1)^{-1}$ exists (Negatives, Differences, Reciprocals and Quotients §reciprocal); Rules of Arithmetic and Order in an Ordered Field §midpoint applied to $0<e$ gives $0<(0+e)/(1+1)=d$; and $d+d=d\cdot1+d\cdot1=d\cdot(1+1)=e\cdot((1+1)^{-1}\cdot(1+1))=e\cdot1=e$ by the ring identities.

Two facts about a cut. Let $x\in C$. Then $x\subseteq\mathbb{Q}$, $x$ has an element, and some $q\in\mathbb{Q}$ satisfies $q\notin x$ (otherwise $x$ and $\mathbb{Q}$ would have the same elements and $x=\mathbb{Q}$ by Axiom of Extensionality for Classes).

(F1) If $p\in x$, $q\in\mathbb{Q}$ and $q\notin x$, then $p<q$. Indeed, otherwise $q\le p$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, so $q<p$ or $q=p$; the first gives $q\in x$ by downward closure, and the second gives $q\in x$ directly, a contradiction either way.

(F2) If $v\in\mathbb{Q}$ and $0<v$, then there is $a\in x$ with $a+v\notin x$. Proof: first choose $p_{0}\in x$, then $q_{0}\in\mathbb{Q}$ with $q_{0}\notin x$. Since $0<v$, $v\neq0$ and $v^{-1}$ exists. By The Rational Numbers Form an Archimedean Ordered Field Containing the Integers §archimedean, applied to $(q_{0}-p_{0})/v$, choose $l\in\mathbb{N}$ with $(q_{0}-p_{0})/v<l$, natural numbers being read as rational numbers as in The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification. By Rules of Arithmetic and Order in an Ordered Field §order-product, multiplying by $v$, and the ring identities, $q_{0}-p_{0}<l\cdot v$; adding $p_{0}$ (Rules of Arithmetic and Order in an Ordered Field §order-sum) gives $q_{0}<p_{0}+l\cdot v$. Hence $p_{0}+l\cdot v\notin x$, since otherwise $q_{0}\in x$ by downward closure. Let

$$S=\{n\in\mathbb{N}:p_{0}+n\cdot v\notin x\},$$

formed by restricted class abstraction from a formula quantifying over sets only, with parameters $p_{0}$, $v$ and $x$; it is a subclass of the set $\mathbb{N}$, hence 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 $l\in S$. By Arithmetic and Order of the Natural Numbers §well-order, choose $m\in S$ with $m\le n$ for every $n\in S$. If $m=1$, put $a=p_{0}$: then $a\in x$, and $a+v=p_{0}+1\cdot v\notin x$, using that $1\in\mathbb{N}$ is read as $1_{\mathbb{Q}}$ (The Natural Numbers and the Integers inside the Rational Numbers §naturals). If $m\neq1$, choose by Arithmetic and Order of the Natural Numbers §predecessor some $d\in\mathbb{N}$ with $m=d+1$. Then $d<m$ by Arithmetic and Order of the Natural Numbers §successor, so $d\notin S$: otherwise $m\le d$, so $m<d$ or $m=d$ by Arithmetic and Order of the Natural Numbers §partial-order; with $d<m$, the first gives $m<m$ by the transitivity of $<$ in Arithmetic and Order of the Natural Numbers §partial-order, and the second gives $d<d$, each contradicting Arithmetic and Order of the Natural Numbers §trichotomy. Put $a=p_{0}+d\cdot v$; then $a\in x$, and since $d+1$ read in $\mathbb{Q}$ is the sum of $d$ and $1$ read in $\mathbb{Q}$ (The Natural Numbers and the Integers inside the Rational Numbers §naturals), the ring identities give $a+v=p_{0}+(d+1)\cdot v=p_{0}+m\cdot v\notin x$.

Proof of lem:dedekind-cut-addition-order-nbg-2026a#group. Let $x,y,z\in C$.

Associativity. Let $w\in(x\oplus y)\oplus z$. Then $w=c+r$ with $c\in x\oplus y$ and $r\in z$, and $c=p+q$ with $p\in x$ and $q\in y$; so $w=(p+q)+r=p+(q+r)$ by the ring identities, where $q+r\in y\oplus z$, hence $w\in x\oplus(y\oplus z)$. Symmetrically, if $w=p+c'$ with $p\in x$ and $c'=q+r$ for some $q\in y$ and $r\in z$, then $w=(p+q)+r\in(x\oplus y)\oplus z$. So $(x\oplus y)\oplus z=x\oplus(y\oplus z)$.

Commutativity. If $w=p+q$ with $p\in x$ and $q\in y$, then $w=q+p\in y\oplus x$; exchanging the roles of $x$ and $y$ gives the reverse inclusion, so $x\oplus y=y\oplus x$.

Identity. Let $w\in x\oplus0^{*}$, say $w=p+q$ with $p\in x$ and $q<0$. By Rules of Arithmetic and Order in an Ordered Field §order-sum and the ring identities, $q+p<0+p$, that is $w<p$, so $w\in x$ by downward closure. Conversely let $u\in x$, and choose $u'\in x$ with $u<u'$ (as $x$ has no greatest element). Put $q=u-u'$; then $q<0$ by (R1), so $q\in0^{*}$, and $u'+q=u'+(u+(-u'))=u+(u'+(-u'))=u$ by the ring identities; so $u\in x\oplus0^{*}$. Hence $x\oplus0^{*}=x$.

Inverse. Let $w\in x\oplus\ominus x$, say $w=p+q$ with $p\in x$ and $q\in\ominus x$, and choose $v\in\mathbb{Q}$ with $0<v$ and $-q-v\notin x$. By (F1), $p<-q-v$; adding $q$ (Rules of Arithmetic and Order in an Ordered Field §order-sum) and using the ring identities, $p+q<-v$. As $-v<0$ by (R2), $w<0$ by transitivity, so $w\in0^{*}$. Conversely let $t\in0^{*}$, so $t<0$, and $0<-t$ by (R1). Let $v=(-t)/(1+1)$; by (R3), $0<v$ and $v+v=-t$. By (F2), choose $a\in x$ with $a+v\notin x$, and put $q=-((a+v)+v)$. Then $-q-v=((a+v)+v)+(-v)=a+v$ by Rules of Arithmetic and Order in an Ordered Field §signs and the ring identities, and this is not in $x$; since $0<v$, $q\in\ominus x$. Moreover, by Rules of Arithmetic and Order in an Ordered Field §signs and the ring identities,

$$a+q=a+\bigl((-a)+(-(v+v))\bigr)=-(v+v)=-(-t)=t,$$

so $t\in x\oplus\ominus x$. Hence $x\oplus\ominus x=0^{*}=0_{\mathbb{R}}$.

Proof of lem:dedekind-cut-addition-order-nbg-2026a#order. Let $x\le_{C}y$, so $x\subseteq y$. If $w\in x\oplus z$, then $w=p+r$ with $p\in x$ and $r\in z$; as $p\in y$, $w\in y\oplus z$. So $x\oplus z\subseteq y\oplus z$, and since both lie in $C$, $x\oplus z\le_{C}y\oplus z$.

Proof of lem:dedekind-cut-addition-order-nbg-2026a#complete. The set $C$ (Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §set) is totally ordered by $\le_{C}$ (Construction of the Dedekind Cuts: They Form a Set Totally Ordered by Inclusion, Closed under Sums, Negatives and Products of Nonnegative Cuts §order), so Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order applies to $s$. Choose an upper bound $b\in C$ of $s$ (Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounded, Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds), so $w\subseteq b$ for every $w\in s$; then choose $w_{0}\in s$, which exists since $s$ is not the empty set. The union $U=\bigcup s$ is a set by The Union Set and the Power Set of a Set §union (with 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 §union), and $u\in U$ if and only if $u\in w$ for some $w\in s$. We check that $U\in C$.

$U\subseteq b$: if $u\in w\in s$, then $u\in b$ as $w\subseteq b$. Hence $U\subseteq\mathbb{Q}$, so $U\in\mathcal{P}(\mathbb{Q})$ by The Union Set and the Power Set of a Set §power. $U\neq\emptyset$: $w_{0}\in C$ has an element, which lies in $U$. $U\neq\mathbb{Q}$: some $q\in\mathbb{Q}$ satisfies $q\notin b$, and then $q\notin U$. Downward closure: if $v\in\mathbb{Q}$, $u\in U$ and $v<u$, choose $w\in s$ with $u\in w$; as $w\in C$, $v\in w$, so $v\in U$. No greatest element: if $u\in U$, choose $w\in s$ with $u\in w$, then $v\in w$ with $u<v$; so $v\in U$. Hence $U\in C$.

For every $w\in s$, $w\subseteq U$, so $w\le_{C}U$; thus $U\in\operatorname{Ub}(s)$. If $b'\in\operatorname{Ub}(s)$, then $w\subseteq b'$ for every $w\in s$, so every element of $U$ lies in $b'$, that is $U\le_{C}b'$. So $U$ is a least element of $\operatorname{Ub}(s)$, which is the supremum of $s$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum.

Proof of lem:dedekind-cut-addition-order-nbg-2026a#rational-sum. Let $u,v\in\mathbb{Q}$. If $w\in u^{*}\oplus v^{*}$, say $w=p+q$ with $p<u$ and $q<v$, then by Rules of Arithmetic and Order in an Ordered Field §order-sum and the ring identities $p+q<u+q$ and $u+q<u+v$, so $w<u+v$ by transitivity and $w\in(u+v)^{*}$. Conversely let $w\in(u+v)^{*}$, so $w<u+v$, and $0<(u+v)-w$ by (R1). Let $d=((u+v)-w)/(1+1)$; by (R3), $0<d$ and $d+d=(u+v)-w$. By (R2), $u-d<u$ and $v-d<v$, so $u-d\in u^{*}$ and $v-d\in v^{*}$, and by Rules of Arithmetic and Order in an Ordered Field §signs and the ring identities

$$(u-d)+(v-d)=(u+v)-(d+d)=(u+v)-((u+v)-w)=w.$$

So $w\in u^{*}\oplus v^{*}$. Hence $(u+v)^{*}=u^{*}\oplus v^{*}$, which is $u^{*}+v^{*}$.

Proof of lem:dedekind-cut-addition-order-nbg-2026a#rational-order. Let $u,v\in\mathbb{Q}$. If $u\le v$ and $w\in u^{*}$, then $w<u$, and $u<v$ or $u=v$; so $w<v$, by transitivity in the first case and directly in the second. Hence $u^{*}\subseteq v^{*}$, that is $u^{*}\le_{C}v^{*}$. If $u\le v$ fails, then $v<u$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation, so $v\in u^{*}$, while $v\notin v^{*}$ by irreflexivity; hence $u^{*}\not\subseteq v^{*}$ and $u^{*}\le_{C}v^{*}$ fails. So $u\le v$ if and only if $u^{*}\le_{C}v^{*}$.