Proof of Ordered Fields in Which Every Nonempty Set Bounded Above Has a Supremum Are Unique up to Unique Isomorphism

Conventions. Throughout, $c$, $p$ and $q$ each stand for one of the ordered fields $r$ and $s$ of the statement; $p$ and $q$ may be the same field. We use the hypothesis of the statement in the form (S): for $c\in\{r,s\}$, every subset of $c$ other than the empty set that is bounded above has a supremum. By (S), the hypotheses 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 hold with $c$ in place of $r$, so its clauses may be applied in $c$. Each of them is an ordered field, hence a set with a total order, and $\mathbb{Q}$ is an ordered field by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §numbers. In each of these fields $<$ is the strict relation of its order, and negatives, differences, reciprocals and quotients are as in Negatives, Differences, Reciprocals and Quotients §negative and Negatives, Differences, Reciprocals and Quotients §reciprocal. Following the reading convention of Homomorphisms and Isomorphisms of Ordered Fields: operations and orders applied to $x$ and $y$ (elements of $p$) are those of $p$; those applied to $a$, $b$, $z$ (elements of $q$, or in Step 0 of $c$), to values of $\kappa_{q}$ (in Step 0 of $\kappa_{c}$) and to values of maps into $q$ are those of $q$ (in Step 0 of $c$); and those applied to $u$, $v$, $w$ (rational numbers) are those of $\mathbb{Q}$. The rules of Rules of Arithmetic and Order in an Ordered Field are used in each of these ordered fields, and sums and products are rearranged by the laws of a commutative ring without further comment, and $t\cdot t^{-1}=1_{c}$ for $t\in c$ with $t\neq0_{c}$ by Negatives, Differences, Reciprocals and Quotients §reciprocal.

(T) Total orders. In each of these fields: if $a\le b$ and $b<z$, or $a<b$ and $b\le z$, then $a<z$, 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; $a\le b$ fails if and only if $b<a$, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §total-negation; and exactly one of $a<b$, $a=b$, $b<a$ holds, by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §trichotomy.

(K) The canonical embeddings. By The Canonical Embedding of the Rational Numbers into an Ordered Field §embedding, $\kappa_{c}$ is the map of The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique, so $\kappa_{c}(1)=1_{c}$, $\kappa_{c}(u+v)=\kappa_{c}(u)+\kappa_{c}(v)$ and $\kappa_{c}(u\cdot v)=\kappa_{c}(u)\cdot\kappa_{c}(v)$ for all $u,v\in\mathbb{Q}$. By The Rational Numbers Embed in Exactly One Way into Every Ordered Field §order, $u<v$ if and only if $\kappa_{c}(u)<\kappa_{c}(v)$; with $\kappa_{c}(0)=0_{c}$ from The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero, $0<u$ if and only if $0_{c}<\kappa_{c}(u)$. By The Rational Numbers Embed in Exactly One Way into Every Ordered Field §negative, $\kappa_{c}(-u)=-\kappa_{c}(u)$ and $\kappa_{c}(u-v)=\kappa_{c}(u)-\kappa_{c}(v)$; by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §reciprocal, if $u\neq0$ then $\kappa_{c}(u)\neq0_{c}$ and $\kappa_{c}(v/u)=\kappa_{c}(v)/\kappa_{c}(u)$.

Step 0: four facts in $c$. Let $a,b\in c$.

(0a) $a<b$ if and only if $-b<-a$. Indeed $a\le b$ if and only if $-b\le-a$ by Rules of Arithmetic and Order in an Ordered Field §order-negative, and $a=b$ if and only if $-b=-a$ because $-(-a)=a$ and $-(-b)=b$ by Rules of Arithmetic and Order in an Ordered Field §signs; combine with Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §strict-characterization.

(0b) For $w\in\mathbb{Q}$: $\kappa_{c}(w)<a+b$ if and only if there are $u,v\in\mathbb{Q}$ with $u+v=w$, $\kappa_{c}(u)<a$ and $\kappa_{c}(v)<b$. Suppose $\kappa_{c}(w)<a+b$. Adding $-b$ (Rules of Arithmetic and Order in an Ordered Field §order-sum) gives $\kappa_{c}(w)-b<a$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §density in $c$ there is $u\in\mathbb{Q}$ with $\kappa_{c}(w)-b<\kappa_{c}(u)<a$. Put $v=w-u$, so $u+v=w$ and, by (K), $\kappa_{c}(v)=\kappa_{c}(w)-\kappa_{c}(u)$. Adding $b-\kappa_{c}(u)$ to both sides of $\kappa_{c}(w)-b<\kappa_{c}(u)$ (Rules of Arithmetic and Order in an Ordered Field §order-sum) gives $\kappa_{c}(w)-\kappa_{c}(u)<b$, that is $\kappa_{c}(v)<b$. Conversely, given such $u,v$, Rules of Arithmetic and Order in an Ordered Field §order-sum gives $\kappa_{c}(u)+\kappa_{c}(v)<a+\kappa_{c}(v)$ and $a+\kappa_{c}(v)<a+b$, so $\kappa_{c}(w)=\kappa_{c}(u)+\kappa_{c}(v)<a+b$ by (K) and (T).

(0c) Suppose $0_{c}<a$ and $0_{c}<b$, and let $w\in\mathbb{Q}$ with $0<w$. Then $\kappa_{c}(w)<a\cdot b$ if and only if there are $u,v\in\mathbb{Q}$ with $0<u$, $0<v$, $u\cdot v=w$, $\kappa_{c}(u)<a$ and $\kappa_{c}(v)<b$. Suppose $\kappa_{c}(w)<a\cdot b$. As $0_{c}<b$, $b\neq0_{c}$ and $0_{c}<b^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal, so multiplying by $b^{-1}$ (Rules of Arithmetic and Order in an Ordered Field §order-product) gives $\kappa_{c}(w)\cdot b^{-1}<a$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §density in $c$ there is $u\in\mathbb{Q}$ with $\kappa_{c}(w)\cdot b^{-1}<\kappa_{c}(u)<a$. Since $0_{c}<\kappa_{c}(w)$ by (K), multiplying by $b^{-1}$ and using $0_{c}\cdot b^{-1}=0_{c}$ (Rules of Arithmetic and Order in an Ordered Field §order-product, Rules of Arithmetic and Order in an Ordered Field §zero) gives $0_{c}<\kappa_{c}(w)\cdot b^{-1}$, so $0_{c}<\kappa_{c}(u)$ by (T) and $0<u$ by (K); in particular $u\neq0$ and $0_{c}<\kappa_{c}(u)^{-1}$ by Rules of Arithmetic and Order in an Ordered Field §positive-reciprocal. Put $v=w/u$, so $u\cdot v=w$ and $\kappa_{c}(v)=\kappa_{c}(w)\cdot\kappa_{c}(u)^{-1}$ by (K). Multiplying $\kappa_{c}(w)\cdot b^{-1}<\kappa_{c}(u)$ by $b$ and then by $\kappa_{c}(u)^{-1}$ (Rules of Arithmetic and Order in an Ordered Field §order-product) gives $\kappa_{c}(w)<\kappa_{c}(u)\cdot b$ and then $\kappa_{c}(v)<b$. Multiplying $0_{c}<\kappa_{c}(w)$ by $\kappa_{c}(u)^{-1}$ gives $0_{c}<\kappa_{c}(v)$, so $0<v$ by (K). Conversely, given such $u,v$, $0_{c}<\kappa_{c}(v)$ by (K), and Rules of Arithmetic and Order in an Ordered Field §order-product gives $\kappa_{c}(u)\cdot\kappa_{c}(v)<a\cdot\kappa_{c}(v)$ (multiplying by $\kappa_{c}(v)$) and $\kappa_{c}(v)\cdot a<b\cdot a$ (multiplying by $a$), so $\kappa_{c}(w)=\kappa_{c}(u)\cdot\kappa_{c}(v)<a\cdot b$ by (K) and (T).

(0d) If $a\neq b$, there is $w\in\mathbb{Q}$ such that exactly one of $\kappa_{c}(w)<a$ and $\kappa_{c}(w)<b$ holds; if moreover $0_{c}\le a$ and $0_{c}\le b$, there is such a $w$ with $0<w$. By (T), $a<b$ or $b<a$; the claim is symmetric in $a$ and $b$, so let $a<b$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §density in $c$ there is $w\in\mathbb{Q}$ with $a<\kappa_{c}(w)<b$. Then $\kappa_{c}(w)<b$ holds and $\kappa_{c}(w)<a$ fails by (T). If $0_{c}\le a$, then $0_{c}<\kappa_{c}(w)$ by (T), so $0<w$ by (K).

Step 1: every homomorphism fixes the rationals. Let $h:p\to q$ be a homomorphism of ordered fields. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition, $h\circ\kappa_{p}:\mathbb{Q}\to q$ is a map with $(h\circ\kappa_{p})(u)=h(\kappa_{p}(u))$ for every $u\in\mathbb{Q}$. By (K) for $p$ and Homomorphisms and Isomorphisms of Ordered Fields §homomorphism, $h(\kappa_{p}(1))=h(1_{p})=1_{q}$, $h(\kappa_{p}(u+v))=h(\kappa_{p}(u)+\kappa_{p}(v))=h(\kappa_{p}(u))+h(\kappa_{p}(v))$ and likewise $h(\kappa_{p}(u\cdot v))=h(\kappa_{p}(u))\cdot h(\kappa_{p}(v))$ for all $u,v\in\mathbb{Q}$. By (K) for $q$, $\kappa_{q}:\mathbb{Q}\to q$ has the same three properties. By the uniqueness in The Rational Numbers Embed in Exactly One Way into Every Ordered Field §unique, applied to the ordered field $q$, $h\circ\kappa_{p}=\kappa_{q}$; hence $h(\kappa_{p}(u))=\kappa_{q}(u)$ for every $u\in\mathbb{Q}$.

Step 2: there is at most one homomorphism $p\to q$. Let $h,h':p\to q$ be homomorphisms of ordered fields and suppose $h(x)\neq h'(x)$ for some $x\in p$. By (T) and the symmetry of the roles of $h$ and $h'$, let $h(x)<h'(x)$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §density in $q$ there is $u\in\mathbb{Q}$ with $h(x)<\kappa_{q}(u)<h'(x)$. The order of $p$ is total, so $\kappa_{p}(u)\le x$ or $x\le\kappa_{p}(u)$. In the first case Homomorphisms and Isomorphisms of Ordered Fields §homomorphism and Step 1 give $\kappa_{q}(u)=h(\kappa_{p}(u))\le h(x)$, contradicting $h(x)<\kappa_{q}(u)$ by (T). In the second case they give $h'(x)\le h'(\kappa_{p}(u))=\kappa_{q}(u)$, contradicting $\kappa_{q}(u)<h'(x)$ by (T). Hence $h(x)=h'(x)$ for every $x\in p$; both maps have domain $p$ by Functions, Values of a Function, and Functions from One Class to Another §map, so $h=h'$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §equality.

Step 3: construction of a map $h:p\to q$. For $x\in p$ let

$$L(x)=\{z\in q:\exists u\,(u\in\mathbb{Q}\wedge z=\kappa_{q}(u)\wedge\kappa_{p}(u)<x)\},$$

formed by restricted class abstraction with the parameters $q$, $\mathbb{Q}$, $\kappa_{p}$, $\kappa_{q}$, $x$ and the strict relation of $p$; its formula quantifies over sets only, the values $\kappa_{p}(u)$ and $\kappa_{q}(u)$ being defined set symbols used at $u\in\mathbb{Q}$. It is a subclass of the set $q$, hence a subset of $q$ 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.

(3a) $L(x)$ is nonempty and bounded above. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §archimedean in $p$, applied to $-x$, there is $v\in\mathbb{Q}$ (a natural number read as a rational number by The Integers and the Rational Numbers, with the Natural Numbers and the Integers Identified with Subsets of the Rationals §identification) with $-x<\kappa_{p}(v)$; by (0a) and $-(-x)=x$ (Rules of Arithmetic and Order in an Ordered Field §signs), $-\kappa_{p}(v)<x$, and $-\kappa_{p}(v)=\kappa_{p}(-v)$ by (K), so $\kappa_{q}(-v)\in L(x)$. Next, by Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §archimedean in $p$ there is $v'\in\mathbb{Q}$ with $x<\kappa_{p}(v')$. We show more generally: if $u'\in\mathbb{Q}$ and $x\le\kappa_{p}(u')$, then $\kappa_{q}(u')$ is an upper bound of $L(x)$. Let $z\in L(x)$, so $z=\kappa_{q}(u)$ for some $u\in\mathbb{Q}$ with $\kappa_{p}(u)<x$. Then $\kappa_{p}(u)<\kappa_{p}(u')$ by (T), so $u<u'$ and $\kappa_{q}(u)<\kappa_{q}(u')$ by (K), and $z\le\kappa_{q}(u')$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict. With $u'=v'$ this shows that $L(x)$ is bounded above.

By (3a) and (S) for $q$, $L(x)$ has a supremum $\sup L(x)$, an element of $\operatorname{Ub}(L(x))\subseteq q$, for every $x\in p$. So $\sup L(x)$ is a defined set symbol used properly at every $x\in p$, and by Maps and Relations Given by Formulas §map there is exactly one map $h:p\to q$ with $h(x)=\sup L(x)$ for every $x\in p$. Fix this $h$.

(3b) Let $x\in p$ and $u\in\mathbb{Q}$. (F1) If $\kappa_{p}(u)<x$, then $\kappa_{q}(u)\le h(x)$: indeed $\kappa_{q}(u)\in L(x)$, and $h(x)\in\operatorname{Ub}(L(x))$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum and Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §bounds. (F2) If $x\le\kappa_{p}(u)$, then $h(x)\le\kappa_{q}(u)$: indeed $\kappa_{q}(u)\in\operatorname{Ub}(L(x))$ by (3a), and $h(x)$ is the least element of $\operatorname{Ub}(L(x))$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum.

(3c) $h(\kappa_{p}(v))=\kappa_{q}(v)$ for every $v\in\mathbb{Q}$. By (F2) with $x=\kappa_{p}(v)$ and $u=v$, $h(\kappa_{p}(v))\le\kappa_{q}(v)$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §rational-supremum in $q$, $\kappa_{q}(v)$ is the supremum of the subset $D=\{z\in q:\exists u\,(u\in\mathbb{Q}\wedge z=\kappa_{q}(u)\wedge z<\kappa_{q}(v))\}$ of $q$. If $z\in D$, then $z=\kappa_{q}(u)$ with $\kappa_{q}(u)<\kappa_{q}(v)$, so $u<v$ and $\kappa_{p}(u)<\kappa_{p}(v)$ by (K), and $z\le h(\kappa_{p}(v))$ by (F1). Thus $h(\kappa_{p}(v))\in\operatorname{Ub}(D)$, so $\kappa_{q}(v)\le h(\kappa_{p}(v))$ by Bounds, Least and Greatest Elements, Suprema and Infima for a Partial Order §supremum. Antisymmetry of the order of $q$ (Partial and Total Orders on a Set and the Associated Strict Relation §partial) gives the equality.

(3d) For all $x,y\in p$: $x<y$ if and only if $h(x)<h(y)$, and $x\le y$ if and only if $h(x)\le h(y)$. Let $x<y$. By Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema §density in $p$, first choose $u\in\mathbb{Q}$ with $x<\kappa_{p}(u)<y$, then $v\in\mathbb{Q}$ with $\kappa_{p}(u)<\kappa_{p}(v)<y$. By (F2) (as $x\le\kappa_{p}(u)$), $h(x)\le\kappa_{q}(u)$; by (K), $u<v$ and $\kappa_{q}(u)<\kappa_{q}(v)$; by (F1), $\kappa_{q}(v)\le h(y)$. Hence $h(x)<h(y)$ by (T). Conversely, if $x<y$ fails, then $y<x$ or $y=x$ by (T), so $h(y)<h(x)$ or $h(y)=h(x)$, and $h(x)<h(y)$ fails by (T). For the second equivalence: if $x\le y$, then $x<y$ or $x=y$ by Uniqueness of Least and Greatest Elements, Properties of the Strict Order, and Trichotomy for Total Orders §weak-strict, so $h(x)\le h(y)$; if $x\le y$ fails, then $y<x$ by (T), so $h(y)<h(x)$ and $h(x)\le h(y)$ fails by (T).

(3e) For all $x\in p$ and $u\in\mathbb{Q}$: $\kappa_{p}(u)<x$ if and only if $\kappa_{q}(u)<h(x)$. By (3d), $\kappa_{p}(u)<x$ if and only if $h(\kappa_{p}(u))<h(x)$, and $h(\kappa_{p}(u))=\kappa_{q}(u)$ by (3c).

Step 4: $h$ is a homomorphism of ordered fields. By (K) and (3c), $h(1_{p})=h(\kappa_{p}(1))=\kappa_{q}(1)=1_{q}$, and $h(0_{p})=h(\kappa_{p}(0))=\kappa_{q}(0)=0_{q}$ by The Rational Numbers Embed in Exactly One Way into Every Ordered Field §zero. Monotonicity is (3d). Let $x,y\in p$.

(4a) Sums. For every $w\in\mathbb{Q}$, the following are equivalent: $\kappa_{q}(w)<h(x+y)$; $\kappa_{p}(w)<x+y$ (by (3e)); there are $u,v\in\mathbb{Q}$ with $u+v=w$, $\kappa_{p}(u)<x$ and $\kappa_{p}(v)<y$ (by (0b) with $c=p$); there are $u,v\in\mathbb{Q}$ with $u+v=w$, $\kappa_{q}(u)<h(x)$ and $\kappa_{q}(v)<h(y)$ (by (3e) for each of $u$ and $v$); $\kappa_{q}(w)<h(x)+h(y)$ (by (0b) with $c=q$). Hence no $w\in\mathbb{Q}$ satisfies exactly one of $\kappa_{q}(w)<h(x+y)$ and $\kappa_{q}(w)<h(x)+h(y)$, and $h(x+y)=h(x)+h(y)$ by (0d) with $c=q$. In particular $h(x)+h(-x)=h(x+(-x))=h(0_{p})=0_{q}$, so $h(-x)=-h(x)$ by the uniqueness in Negatives, Differences, Reciprocals and Quotients §negative.

(4b) Products of positive elements. Let $0_{p}<x$ and $0_{p}<y$. Multiplying $0_{p}<x$ by $y$ (Rules of Arithmetic and Order in an Ordered Field §order-product, Rules of Arithmetic and Order in an Ordered Field §zero) gives $0_{p}<x\cdot y$. By (3d) and $h(0_{p})=0_{q}$, $0_{q}<h(x)$, $0_{q}<h(y)$ and $0_{q}<h(x\cdot y)$, and likewise $0_{q}<h(x)\cdot h(y)$. For every $w\in\mathbb{Q}$ with $0<w$, the following are equivalent: $\kappa_{q}(w)<h(x\cdot y)$; $\kappa_{p}(w)<x\cdot y$ (by (3e)); there are $u,v\in\mathbb{Q}$ with $0<u$, $0<v$, $u\cdot v=w$, $\kappa_{p}(u)<x$ and $\kappa_{p}(v)<y$ (by (0c) with $c=p$); the same with $\kappa_{q}(u)<h(x)$ and $\kappa_{q}(v)<h(y)$ (by (3e)); $\kappa_{q}(w)<h(x)\cdot h(y)$ (by (0c) with $c=q$, using $0_{q}<h(x)$ and $0_{q}<h(y)$). Since $0_{q}\le h(x\cdot y)$ and $0_{q}\le h(x)\cdot h(y)$, (0d) with $c=q$ gives $h(x\cdot y)=h(x)\cdot h(y)$.

(4c) Products in general. If $x=0_{p}$ or $y=0_{p}$, then $x\cdot y=0_{p}$ and $h(x)\cdot h(y)=0_{q}$ by Rules of Arithmetic and Order in an Ordered Field §zero, as $h(0_{p})=0_{q}$; so $h(x\cdot y)=h(x)\cdot h(y)$. Otherwise each of $x$, $y$ is positive or negative by (T), and an element $t$ of $p$ with $t<0_{p}$ satisfies $0_{p}<-t$ by (0a), as $-0_{p}=0_{p}$ by the uniqueness in Negatives, Differences, Reciprocals and Quotients §negative, since $0_{p}+0_{p}=0_{p}$. If $0_{p}<x$ and $0_{p}<y$, then $h(x\cdot y)=h(x)\cdot h(y)$ is (4b). The remaining cases are treated as follows. If $x<0_{p}<y$, then $x\cdot y=-((-x)\cdot y)$ by Rules of Arithmetic and Order in an Ordered Field §signs twice, so by (4a), (4b) and Rules of Arithmetic and Order in an Ordered Field §signs in $q$, $h(x\cdot y)=-h((-x)\cdot y)=-(h(-x)\cdot h(y))=-((-h(x))\cdot h(y))=-(-(h(x)\cdot h(y)))=h(x)\cdot h(y)$. The case $y<0_{p}<x$ follows by commutativity. If $x<0_{p}$ and $y<0_{p}$, then $x\cdot y=(-x)\cdot(-y)$ by Rules of Arithmetic and Order in an Ordered Field §signs, so $h(x\cdot y)=h(-x)\cdot h(-y)=(-h(x))\cdot(-h(y))=h(x)\cdot h(y)$. Together with (4a), (3d) and $h(1_{p})=1_{q}$, $h$ is a homomorphism of ordered fields.

Step 5: conclusion. Apply Steps 3 and 4 with $(p,q)=(r,s)$ to obtain a homomorphism of ordered fields $f:r\to s$, and with $(p,q)=(s,r)$ to obtain a homomorphism of ordered fields $g:s\to r$.

Clause Ordered Fields in Which Every Nonempty Set Bounded Above Has a Supremum Are Unique up to Unique Isomorphism §unique: $f$ exists, and by Step 2 with $(p,q)=(r,s)$ every homomorphism of ordered fields $r\to s$ equals $f$.

Clause Ordered Fields in Which Every Nonempty Set Bounded Above Has a Supremum Are Unique up to Unique Isomorphism §rationals: by Step 1 with $(p,q)=(r,s)$, $f(\kappa_{r}(u))=\kappa_{s}(u)$ for every $u\in\mathbb{Q}$.

Clause Ordered Fields in Which Every Nonempty Set Bounded Above Has a Supremum Are Unique up to Unique Isomorphism §isomorphism: by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §composition, $g\circ f:r\to r$ is a map with $(g\circ f)(x)=g(f(x))$. It is a homomorphism of ordered fields: $g(f(1_{r}))=g(1_{s})=1_{r}$, $g(f(x+y))=g(f(x)+f(y))=g(f(x))+g(f(y))$, likewise for products, and $x\le y$ implies $f(x)\le f(y)$, which implies $g(f(x))\le g(f(y))$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §identity, $\mathrm{id}_{r}:r\to r$ is a map with $\mathrm{id}_{r}(x)=x$, which is a homomorphism of ordered fields, since $\mathrm{id}_{r}(1_{r})=1_{r}$, $\mathrm{id}_{r}(x+y)=x+y=\mathrm{id}_{r}(x)+\mathrm{id}_{r}(y)$, likewise for products, and $x\le y$ implies $\mathrm{id}_{r}(x)\le\mathrm{id}_{r}(y)$. By Step 2 with $p=q=r$, $g\circ f=\mathrm{id}_{r}$; in the same way, with the roles of $r$ and $s$ exchanged and Step 2 applied with $p=q=s$, $f\circ g=\mathrm{id}_{s}$. By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction §inverse-criterion, $f$ is bijective. Finally, for $x,y\in r$, $f(x)\le f(y)$ implies $x\le y$ by (3d) with $(p,q)=(r,s)$. Hence $f$ is an isomorphism of ordered fields.