6.2 Building the Map from the Rationals Below

Section 6.1 gave every element $x$ of an ordered field $F$ a record, the set $\mathbb{Q}_F(x)$ of rationals below it, and showed that in an Archimedean field the record determines the element (Lemma 6.1.3). This section turns the record into a map. Given an Archimedean field $F$ and a complete field $L$ , we send $x \in F$ to the element of $L$ with the same record. The result is the main theorem of the chapter in its one-sided form: there is exactly one order-preserving field map $F \to L$ . Section 6.3 applies it twice, in both directions, to compare two complete ordered fields.

Two fields are in play throughout, so every proof here keeps $\iota_F$ and $\iota_L$ visible. Only the two lemmas, each about one field, write $q$ for $\iota_F(q)$ .

6.2.1. What the map must be

Suppose $\varphi \colon F \to L$ is an order-preserving field map (Definition 6.1.1). By Proposition 6.1.1 it fixes the rationals, $\varphi(\iota_F(q)) = \iota_L(q)$ , and it preserves and reflects order. So for $q \in \mathbb{Q}$ and $x \in F$ , $\iota_F(q) < x \iff \varphi\bigl(\iota_F(q)\bigr) < \varphi(x) \iff \iota_L(q) < \varphi(x) .$ In the notation of Section 6.1: $\mathbb{Q}_L(\varphi(x)) = \mathbb{Q}_F(x)$ . The image of $x$ has exactly the rationals below it that $x$ has.

That already shows there can be at most one such map when $L$ is Archimedean: two candidates for $\varphi(x)$ would have the same rationals below them, and rationals separate. For existence, we need an element of $L$ with a prescribed set of rationals below it. When $L$ is complete there is a natural candidate: the least element of $L$ that lies above all the copies $\iota_L(q)$ , $q \in \mathbb{Q}_F(x)$ , that is, their supremum.

The check that this candidate respects sums and products goes through the record, not through suprema. For that we need to know how the rationals below $a + b$ and below $ab$ are made from those below $a$ and $b$ .

6.2.2. Sums and products, measured by rationals

A rational $q$ lies below $a + b$ exactly when it splits into a rational below $a$ and a rational below $b$ .

Lemma 6.2.1 (Rationals below a Sum)

Let $F$ be an Archimedean ordered field, $a, b \in F$ and $q \in \mathbb{Q}$ . Then $\iota_F(q) < a + b$ if and only if there is $r \in \mathbb{Q}$ with $\iota_F(r) < a \qquad \text{and} \qquad \iota_F(q - r) < b .$ In other words, $q \in \mathbb{Q}_F(a + b) \iff \exists r \in \mathbb{Q} \ \bigl( r \in \mathbb{Q}_F(a) \wedge q - r \in \mathbb{Q}_F(b) \bigr)$ .

Idea. One direction is addition of inequalities. For the other, split $q$ as $r + (q - r)$ with $r$ a rational just below $a$ : we need $r$ in the gap between $q - b$ and $a$ , which is non-empty exactly because $q < a + b$ , and density supplies it.

Proof. We write $q$ for $\iota_F(q)$ ; since $\iota_F$ preserves sums, it preserves differences, so $\iota_F(q - r) = q - r$ in $F$ .

$(\Leftarrow)$ If $r < a$ and $q - r < b$ , then $q = r + (q - r) < a + b$ by Proposition 2.3.5 (b).

$(\Rightarrow)$ Let $q < a + b$ . Then $q - b < a$ (Proposition 2.3.5 (a)). By Theorem 4.2.5 there is $r \in \mathbb{Q}$ with $q - b < r < a$ . Then $r < a$ , and $q - r < b$ by Proposition 2.3.5 (a) again.

For products the same idea works for positive elements and positive rationals, with $q/r$ in place of $q - r$ .

Lemma 6.2.2 (Rationals below a Product)

Let $F$ be an Archimedean ordered field, $a, b \in F$ with $a > 0$ and $b > 0$ , and $q \in \mathbb{Q}$ with $q > 0$ . Then $\iota_F(q) < ab$ if and only if there is $r \in \mathbb{Q}$ with $r > 0 , \qquad \iota_F(r) < a \qquad \text{and} \qquad \iota_F(q/r) < b .$ For $q \le 0$ , $\iota_F(q) < ab$ holds always.

In words: below a product of positives, a positive rational splits as a product of a rational below each factor. Exercise B1 asks for the proof; the fold below has it.

Second readingThe proof of the product lemma (try Exercise B1 first)

Idea. As for sums, with division in place of subtraction: split $q$ as $r \cdot (q/r)$ with $r$ a rational in the gap between $q/b$ and $a$ , a gap of positive elements.

Proof. We write $q$ for $\iota_F(q)$ ; since $\iota_F$ preserves products and $\iota_F(r) \ne 0$ for $r \ne 0$ , it preserves quotients, so $\iota_F(q/r) = q/r$ in $F$ .

If $q \le 0$ , then $q \le 0 < ab$ by (O4). Let $q > 0$ .

$(\Leftarrow)$ Let $r > 0$ with $r < a$ and $q/r < b$ . Since $q/r > 0$ (Proposition 2.5.2 (a), (O4)), Proposition 2.3.5 (c) gives $q = r \cdot (q/r) < a \cdot (q/r)$ ; since $a > 0$ , it gives $a \cdot (q/r) < ab$ . So $q < ab$ .

$(\Rightarrow)$ Let $q < ab$ . Multiplying by $b^{-1} > 0$ (Proposition 2.5.2 (a), Proposition 2.3.5 (c)) gives $q/b < a$ , and $q/b > 0$ by (O4). By Theorem 4.2.5 there is $r \in \mathbb{Q}$ with $q/b < r < a$ . Then $r > q/b > 0$ , so $r > 0$ in $\mathbb{Q}$ , since $\iota_F$ reflects order. Multiplying $q/b < r$ by $b > 0$ gives $q < rb$ , and multiplying by $r^{-1} > 0$ gives $q/r < b$ .

Quick check. In $F = \mathbb{Q}$ , with $a \mathrel{:=} 3/4$ , $b \mathrel{:=} 1/2$ and $q \mathrel{:=} 1$ , find a rational $r$ as in Lemma 6.2.1. Which rationals $r$ work?

Solution. We need $r < 3/4$ and $1 - r < 1/2$ , that is, $1/2 < r < 3/4$ . For example $r = 2/3$ : then $1 - 2/3 = 1/3 < 1/2$ . Exactly the rationals of the open gap between $q - b = 1/2$ and $a = 3/4$ work, as the proof predicts.

6.2.3. The map into a complete field

Here is the theorem. Its proof is the first key proof of the chapter.

Theorem 6.2.3 (Embedding an Archimedean Field in a Complete One)

Let $F$ be an Archimedean ordered field and $L$ a complete ordered field. Then there is exactly one order-preserving field map $\varphi \colon F \to L$ . It is given by $\varphi(x) \mathrel{:=} \sup\nolimits_L \, \{ \iota_L(q) : q \in \mathbb{Q}_F(x) \} \qquad (x \in F) ,$ the supremum taken in $L$ , and it satisfies $\mathbb{Q}_L\bigl(\varphi(x)\bigr) = \mathbb{Q}_F(x) \qquad \text{for every } x \in F . \tag{$\star$}$

Try it first (ten minutes): if $\varphi$ exists and fixes $\mathbb{Q}$ , which element of $L$ must $\varphi(x)$ be?

Idea. The natural first attempt defines $\varphi$ on the copy of $\mathbb{Q}$ , by $\varphi(\iota_F(q)) \mathrel{:=} \iota_L(q)$ , and then tries to “extend”. It stalls at once: an element $x \in F$ that is not a rational has no formula, and nothing in $F$ names it. What we do know, from the previous subsection, is its record: $\varphi(x)$ must have exactly the rationals $\mathbb{Q}_F(x)$ below it. So we construct a point as a supremum: $\varphi(x)$ is the least upper bound in $L$ of the copies of those rationals, and $(\star)$ says the construction has the right record. After that, sums, products and uniqueness are checked the same way: compute the record of both sides (for sums and products by the two lemmas) and let rationals separate in $L$ . Monotonicity needs only one rational between $x$ and $y$ .

Step roadmap: ① $\varphi(x)$ exists; ② $(\star)$ ; ③ $\varphi$ fixes the rationals; ④ $\varphi$ is strictly increasing; ⑤ sums; ⑥ products and the unit; ⑦ uniqueness. Before reading on: where is completeness of $L$ used, and where the Archimedean property of $F$ ?

Proof. By Theorem 6.1.2, $L$ is Archimedean. So Theorem 4.2.5, Lemma 6.1.3, Lemma 6.2.1 and Lemma 6.2.2 apply in $L$ as well as in $F$ . For $x \in F$ write $A_x \mathrel{:=} \{ \iota_L(q) : q \in \mathbb{Q}_F(x) \} \subseteq L$ .

Step 1: $\varphi(x)$ is defined. Let $x \in F$ . Since $F$ is Archimedean, there is $n \in \mathbb{N}$ with $-x < n \cdot 1_F$ , so $\iota_F(-n) = -(n \cdot 1_F) < x$ (Proposition 2.3.5 (a)), and $-n \in \mathbb{Q}_F(x)$ : the set $A_x$ is non-empty. Applied to $x$ , the Archimedean property gives $m \in \mathbb{N}$ with $x < m \cdot 1_F = \iota_F(m)$ . For $q \in \mathbb{Q}_F(x)$ , $\iota_F(q) < x < \iota_F(m)$ , so $q < m$ because $\iota_F$ reflects order, and $\iota_L(q) < \iota_L(m)$ . So $\iota_L(m)$ is an upper bound of $A_x$ . Since $L$ is complete, $A_x$ has a supremum in $L$ , and it is unique; $\varphi(x)$ is that supremum. So $\varphi \colon F \to L$ is a function.

Step 2: $(\star)$ . Let $x \in F$ and $q \in \mathbb{Q}$ .

$(\subseteq)$ Let $q \in \mathbb{Q}_F(x)$ . By density in $F$ there is $r \in \mathbb{Q}$ with $\iota_F(q) < \iota_F(r) < x$ . Then $r \in \mathbb{Q}_F(x)$ , so $\iota_L(r) \in A_x$ and $\iota_L(r) \le \varphi(x)$ , since $\varphi(x)$ is an upper bound. Also $q < r$ (reflection in $F$ ), so $\iota_L(q) < \iota_L(r) \le \varphi(x)$ , and $q \in \mathbb{Q}_L(\varphi(x))$ .

$(\supseteq)$ Let $\iota_L(q) < \varphi(x)$ . Since $\varphi(x)$ is the least upper bound of $A_x$ , $\iota_L(q)$ is not an upper bound, so there is $r \in \mathbb{Q}_F(x)$ with $\iota_L(q) < \iota_L(r)$ (totality). Then $q < r$ (reflection in $L$ ), so $\iota_F(q) < \iota_F(r) < x$ , and $q \in \mathbb{Q}_F(x)$ .

Step 3: $\varphi$ fixes the rationals. Let $r \in \mathbb{Q}$ . By $(\star)$ and the first example after Definition 6.1.2, used in $F$ and in $L$ , $\mathbb{Q}_L\bigl(\varphi(\iota_F(r))\bigr) = \mathbb{Q}_F\bigl(\iota_F(r)\bigr) = \{ q \in \mathbb{Q} : q < r \} = \mathbb{Q}_L\bigl(\iota_L(r)\bigr) .$ By Lemma 6.1.3 in $L$ , $\varphi(\iota_F(r)) = \iota_L(r)$ . In particular $\varphi(0_F) = 0_L$ and $\varphi(1_F) = 1_L$ .

Step 4: $\varphi$ is strictly increasing. Let $x < y$ in $F$ . By density in $F$ there is $r \in \mathbb{Q}$ with $x < \iota_F(r) < y$ . Then $r \notin \mathbb{Q}_F(x)$ (by (O1)), so $r \notin \mathbb{Q}_L(\varphi(x))$ by $(\star)$ , and $\varphi(x) \le \iota_L(r)$ by totality. And $r \in \mathbb{Q}_F(y) = \mathbb{Q}_L(\varphi(y))$ , so $\iota_L(r) < \varphi(y)$ . Hence $\varphi(x) \le \iota_L(r) < \varphi(y)$ .

Step 5: sums. Let $x, y \in F$ and $q \in \mathbb{Q}$ . Then $\begin{aligned} q \in \mathbb{Q}_L\bigl(\varphi(x + y)\bigr) &\iff q \in \mathbb{Q}_F(x + y) \\ &\iff \exists r \in \mathbb{Q} \ \bigl( r \in \mathbb{Q}_F(x) \wedge q - r \in \mathbb{Q}_F(y) \bigr) \\ &\iff \exists r \in \mathbb{Q} \ \bigl( r \in \mathbb{Q}_L(\varphi(x)) \wedge q - r \in \mathbb{Q}_L(\varphi(y)) \bigr) \\ &\iff q \in \mathbb{Q}_L\bigl(\varphi(x) + \varphi(y)\bigr) , \end{aligned}$ by $(\star)$ , Lemma 6.2.1 in $F$ , $(\star)$ twice, and Lemma 6.2.1 in $L$ . By Lemma 6.1.3 in $L$ , $\varphi(x + y) = \varphi(x) + \varphi(y)$ .

Step 6: products and the unit. First let $x > 0$ and $y > 0$ . By Steps 3 and 4, $\varphi(x) > \varphi(0_F) = 0_L$ and $\varphi(y) > 0_L$ ; and $xy > 0$ , $\varphi(x)\varphi(y) > 0$ by (O4). Let $q \in \mathbb{Q}$ . If $q \le 0$ , then $q \in \mathbb{Q}_F(xy) = \mathbb{Q}_L(\varphi(xy))$ , and $q \in \mathbb{Q}_L(\varphi(x)\varphi(y))$ , by the last clause of Lemma 6.2.2 in $F$ and in $L$ . If $q > 0$ , then $\begin{aligned} q \in \mathbb{Q}_L\bigl(\varphi(xy)\bigr) &\iff q \in \mathbb{Q}_F(xy) \\ &\iff \exists r \in \mathbb{Q}_{>0} \ \bigl( r \in \mathbb{Q}_F(x) \wedge q/r \in \mathbb{Q}_F(y) \bigr) \\ &\iff \exists r \in \mathbb{Q}_{>0} \ \bigl( r \in \mathbb{Q}_L(\varphi(x)) \wedge q/r \in \mathbb{Q}_L(\varphi(y)) \bigr) \\ &\iff q \in \mathbb{Q}_L\bigl(\varphi(x)\varphi(y)\bigr) , \end{aligned}$ by $(\star)$ , Lemma 6.2.2 in $F$ , $(\star)$ twice, and Lemma 6.2.2 in $L$ (with $\varphi(x), \varphi(y) > 0$ ). By Lemma 6.1.3 in $L$ , $\varphi(xy) = \varphi(x)\varphi(y)$ for $x, y > 0$ .

Now the signs. By Steps 5 and 3, $\varphi(x) + \varphi(-x) = \varphi(0_F) = 0_L$ , so $\varphi(-x) = -\varphi(x)$ (Proposition 2.3.2 (a)). If $x = 0$ or $y = 0$ , both sides of $\varphi(xy) = \varphi(x)\varphi(y)$ are $0_L$ (Step 3 and Proposition 2.3.2 (b)). If $x < 0 < y$ , then $-x > 0$ (Proposition 2.3.5 (a)) and $xy = -\bigl((-x)y\bigr)$ (Proposition 2.3.2 (c)), so $\varphi(xy) = -\varphi\bigl((-x)y\bigr) = -\bigl(\varphi(-x)\varphi(y)\bigr) = -\bigl(-\varphi(x)\bigr)\varphi(y) = \varphi(x)\varphi(y) .$ The case $x > 0 > y$ follows by (F6), and if $x, y < 0$ , then $xy = (-x)(-y)$ and $\varphi(xy) = \bigl(-\varphi(x)\bigr)\bigl(-\varphi(y)\bigr) = \varphi(x)\varphi(y)$ . Finally $\varphi(1_F) = 1_L$ by Step 3. With Steps 4 and 5, $\varphi$ is an order-preserving field map.

Step 7: uniqueness. Let $\chi \colon F \to L$ be any order-preserving field map, and let $x \in F$ , $q \in \mathbb{Q}$ . By Proposition 6.1.1, $\chi(\iota_F(q)) = \iota_L(q)$ . If $\iota_F(q) < x$ , then $\iota_L(q) = \chi(\iota_F(q)) < \chi(x)$ . If $\iota_L(q) < \chi(x)$ , that is, $\chi(\iota_F(q)) < \chi(x)$ , then $\iota_F(q) < x$ , because $\chi$ reflects order (the remark after Definition 6.1.1). So $\mathbb{Q}_L\bigl(\chi(x)\bigr) = \mathbb{Q}_F(x) = \mathbb{Q}_L\bigl(\varphi(x)\bigr) ,$ the second equality by $(\star)$ . By Lemma 6.1.3 in $L$ , $\chi(x) = \varphi(x)$ . Since $x$ was arbitrary, $\chi = \varphi$ . This proves the theorem.

The answer to the prompt. Completeness of $L$ is used in Step 1, where the supremum must exist, and in Step 2 $(\supseteq)$ , where its leastness turns “below $\varphi(x)$ ” into “below some $\iota_L(r)$ ”. The Archimedean property of $F$ is used in Step 1 for the bounds $-n$ and $m$ , in Steps 2 and 4 through density in $F$ , and in Steps 5 and 6 through the two lemmas applied in $F$ . Completeness of $L$ is used a third time, indirectly: it makes $L$ Archimedean (Theorem 6.1.2), which Steps 3, 5, 6 and 7 need for rationals to separate in $L$ and for the lemmas in $L$ .

Remark. Step 7 is the whole of “uniqueness”. Its idea, comparing the rationals below two candidate values, needs only that $L$ is Archimedean, not that it is complete: Exercise C1 runs it for any two order-preserving field maps $F \to L$ , with no appeal to $(\star)$ . Section 6.3 leans on the uniqueness clause throughout: to show that the maps between two complete fields are inverse to each other, that the isomorphism they give is the only one, and that a complete ordered field has no order-preserving field automorphism other than the identity.

Remark. Neither hypothesis can be dropped from the existence half. Without completeness of $L$ : take $F = \mathbb{R}$ , which is Archimedean (Theorem 5.4.2), and $L = \mathbb{Q}$ . An order-preserving field map $\varphi \colon \mathbb{R} \to \mathbb{Q}$ would fix $2$ (Proposition 6.1.1), so for $\sqrt 2 \in \mathbb{R}$ (the element of Example (The Class of the Newton Sequence), as in Section 6.1), $\varphi(\sqrt 2)^2 = \varphi((\sqrt 2)^2) = \varphi(2) = 2$ in $\mathbb{Q}$ , against Theorem 2.6.1. Without the Archimedean property of $F$ : take $F = \mathbb{Q}(t)$ of Example (A Non-Archimedean Ordered Field) and any complete $L$ . An order-preserving field map $\varphi$ would fix each $n \cdot 1$ and be strictly increasing, so $n \cdot 1_L = \varphi(n \cdot 1) < \varphi(t)$ for every $n \in \mathbb{N}$ , against the Archimedean property of $L$ (Theorem 6.1.2).

Warning. The supremum in the definition of $\varphi$ is taken in $L$ , of the copies $\iota_L(q)$ . It cannot be taken in $\mathbb{Q}$ , or in $F$ . In $\mathbb{Q}$ : the set $\mathbb{Q}_{\mathbb{R}}(\sqrt 2)$ of Example (The Rationals below the Square Root of Two) (with $\sqrt 2$ as in Section 6.1) has no supremum in $\mathbb{Q}$ . It contains the set $A = \{ p \in \mathbb{Q} : p^2 < 2 \}$ of Theorem 2.6.2, and its other elements are negative, so below $0 \in A$ ; hence the two sets have the same upper bounds in $\mathbb{Q}$ , and $A$ has no least one. In $F$ : the supremum of $\iota_F(\mathbb{Q}_F(x))$ always exists and is $x$ itself, since $x$ is an upper bound and, by density in $F$ (Theorem 4.2.5), every $u < x$ lies below some $\iota_F(q) < x$ . So it is an element of the wrong field.

Quick check. Take $L \mathrel{:=} F$ , a complete ordered field. Which map $F \to F$ does Theorem 6.2.3 produce, and why, without computing a single supremum?

Solution. The identity. The theorem says there is exactly one order-preserving field map $F \to F$ ( $F$ is Archimedean by Theorem 6.1.2), and $\operatorname{id}_F$ is one (the first example after Definition 6.1.1). So $\varphi = \operatorname{id}_F$ . This is the argument Section 6.3 builds on.

6.2.4. Exercises

A. Check your understanding

Exercise (A1)

These questions are about the proof of Theorem 6.2.3.

(a) Which steps use that $L$ is complete, and which that $F$ is Archimedean? Which steps use that $L$ is Archimedean, and where does that come from?
(b) Run the definition of $\varphi$ with $F = \mathbb{Q}$ and $L = \mathbb{R}$ , the constructed reals. What is $\varphi$ ?
(c) State the idea of the proof in two sentences.
(d) Why may Step 1 not define $\varphi(x)$ as a supremum taken in $F$ ?

Solution.

(a) $L$ complete: Step 1 (the supremum exists) and Step 2 $(\supseteq)$ (its leastness). $F$ Archimedean: Step 1 (the bounds $-n$ , $m$ ), Steps 2 and 4 (density in $F$ ), and Steps 5 and 6 (Lemma 6.2.1 and Lemma 6.2.2 applied in $F$ ). $L$ Archimedean: Steps 3, 5, 6 and 7, through Lemma 6.1.3 in $L$ and the two lemmas in $L$ . It comes from the completeness of $L$ , by Theorem 6.1.2.
(b) $\varphi = \iota_{\mathbb{R}} = \kappa$ . By Step 3, $\varphi(\iota_{\mathbb{Q}}(r)) = \iota_{\mathbb{R}}(r)$ for every $r \in \mathbb{Q}$ , and $\iota_{\mathbb{Q}}$ is the identity (Chapter 4, after Theorem 4.2.2). So $\varphi(r) = \iota_{\mathbb{R}}(r)$ , which is $\kappa(r)$ by Theorem 5.4.1 (d). Concretely, $\varphi(r)$ is the supremum in $\mathbb{R}$ of $\{ \kappa(q) : q < r \}$ , and that supremum is $\kappa(r)$ .
(c) An order-preserving map that fixes the rationals must send $x$ to an element with the same rationals below it, and in a complete $L$ the supremum of their copies is such an element. Every field identity for $\varphi$ is then checked by comparing the rationals below the two sides, which the sum and product lemmas compute, and rationals separate in $L$ .
(d) In $F$ the supremum of $\iota_F(\mathbb{Q}_F(x))$ always exists, since $F$ is Archimedean, and it is $x$ itself (Exercise B2 of Section 6.1): an element of $F$ , not of $L$ . So it does not give a map into $L$ . The supremum must be of the copies $\iota_L(q)$ , taken in $L$ .

B. Practice

Exercise (B1)

Prove Lemma 6.2.2: let $F$ be an Archimedean ordered field, $a, b \in F$ with $a, b > 0$ , and $q \in \mathbb{Q}$ with $q > 0$ ; show that $\iota_F(q) < ab$ if and only if some rational $r > 0$ has $\iota_F(r) < a$ and $\iota_F(q/r) < b$ . Show also that $\iota_F(q) < ab$ for every rational $q \le 0$ .

Hint: imitate the proof of Lemma 6.2.1, with division in place of subtraction.

Solution. Write $q$ for $\iota_F(q)$ . Since $\iota_F$ preserves products, $\iota_F(r)\,\iota_F(r^{-1}) = \iota_F(1) = 1$ for $r \ne 0$ , so $\iota_F(r^{-1}) = \iota_F(r)^{-1}$ (Lemma 4.1.1 (a)) and $\iota_F(q/r) = q/r$ in $F$ .

If $q \le 0$ , then $q \le 0 < ab$ by (O4).

Let $q > 0$ . $(\Leftarrow)$ Let $r > 0$ , $r < a$ , $q/r < b$ . Then $q/r > 0$ (Proposition 2.5.2 (a), (O4)), so Proposition 2.3.5 (c) gives $q = r(q/r) < a(q/r)$ , and, as $a > 0$ , $a(q/r) < ab$ . Hence $q < ab$ .

$(\Rightarrow)$ Let $q < ab$ . Since $b^{-1} > 0$ , Proposition 2.3.5 (c) gives $q/b < a$ , and $q/b > 0$ . By Theorem 4.2.5 there is $r \in \mathbb{Q}$ with $q/b < \iota_F(r) < a$ . Then $\iota_F(r) > 0 = \iota_F(0)$ , so $r > 0$ since $\iota_F$ reflects order. From $q/b < r$ and $b > 0$ , $q < rb$ ; multiplying by $r^{-1} > 0$ , $q/r < b$ . This $r$ is as required.

Exercise (B2: Cube roots)

Let $F$ be a complete ordered field and $a \in F$ with $a > 0$ . Let $S \mathrel{:=} \{ y \in F : y \ge 0 \text{ and } y^3 < a \}$ . The following steps prove that some $s \in F$ has $s^3 = a$ .

(1) $S$ is non-empty ( $0 \in S$ ) and bounded above by $1 + a$ : if $y > 1 + a$ , then $y > 1$ , so $y^3 > y > a$ . Let $s \mathrel{:=} \sup S$ . Then $s > 0$ , since $m \mathrel{:=} \min\{1, a/2\}$ has $0 < m \le 1$ , so $m^3 \le m \le a/2 < a$ , and $m \in S$ .
(2) For $y, z \in F$ , $z^3 - y^3 = (z - y)(z^2 + zy + y^2)$ and $z^2 + zy + y^2 = (z + y/2)^2 + (3/4)y^2$ , which is $> 0$ unless $y = z = 0$ (if $y \ne 0$ , then $(3/4)y^2 > 0$ by Proposition 2.3.5 (e), (c) and the square is $\ge 0$ , so the sum is positive by (b); if $y = 0$ and $z \ne 0$ , the expression is $z^2 > 0$ by (e)). Hence $y < z \iff y^3 < z^3$ , for all $y, z \in F$ .
(3) The case $s^3 < a$ is impossible. Let $h \mathrel{:=} \min\{1, (a - s^3)/(2(3s^2 + 3s + 1))\} > 0$ . Since $0 < h \le 1$ and $s \ge 0$ , $(s + h)^3 = s^3 + 3s^2h + 3sh^2 + h^3 \le s^3 + h(3s^2 + 3s + 1) \le s^3 + (a - s^3)/2 < a$ , so $s + h \in S$ , against $s$ being an upper bound.
(4) The case $s^3 > a$ is impossible. Name the move, then complete it.
(5) Hence $s^3 = a$ .

Then explain how to get a cube root of $a = 0$ and of $a < 0$ , and why every $a \in F$ has exactly one cube root.

Solution. The move is construct a point as a supremum: $s$ is built as $\sup S$ , and each of the two wrong cases of trichotomy is refuted by exhibiting a point that contradicts one clause of the supremum. In step (3) it was an element of $S$ above $s$ , against (S1). In step (4) it is an upper bound below $s$ , against (S2).

Suppose $s^3 > a$ . Let $h \mathrel{:=} (s^3 - a)/(3s^2)$ ; it is positive, since $s > 0$ by (1). Also $h < s$ : this says $s^3 - a < 3s^3$ , which holds because $a > 0$ and $2s^3 > 0$ . So $s - h > 0$ , and $(s - h)^3 - a = (s^3 - a) - 3s^2h + 3sh^2 - h^3 = h^2(3s - h) > 0 ,$ since $3s^2 h = s^3 - a$ , $h^2 > 0$ and $3s - h > 2s > 0$ . Let $y > s - h$ . By (2), $y^3 > (s - h)^3 > a$ , so $y \notin S$ . Hence every element of $S$ is $\le s - h$ , and $s - h$ is an upper bound of $S$ smaller than $s$ , against $s$ being the least upper bound.
Steps (3) and (4) leave only $s^3 = a$ , by trichotomy (O1).

For $a = 0$ , $0^3 = 0$ . For $a < 0$ , $-a > 0$ has a cube root $c$ by the above, and $(-c)^3 = -c^3 = a$ (Proposition 2.3.2 (c)). Uniqueness: if $b^3 = c^3$ , then neither $b < c$ nor $c < b$ can hold, since by (2) either would make the cubes differ; so $b = c$ . So every element of a complete ordered field has exactly one cube root.

C. Going deeper

Exercise (C1: Exit check: at most one map into an Archimedean field)

Let $F$ and $L$ be Archimedean ordered fields, and let $\chi_1, \chi_2 \colon F \to L$ be order-preserving field maps. Prove that $\chi_1 = \chi_2$ .

Solution (Hint 1). Compare the rationals below $\chi_1(x)$ and below $\chi_2(x)$ .

Solution (Hint 2). By Proposition 6.1.1, $\chi_i(\iota_F(q)) = \iota_L(q)$ for $i = 1, 2$ and every $q \in \mathbb{Q}$ .

Solution. Let $x \in F$ and $i \in \{1, 2\}$ . By Proposition 6.1.1 (which needs no order hypothesis), $\chi_i(\iota_F(q)) = \iota_L(q)$ for every $q \in \mathbb{Q}$ . Since $\chi_i$ is strictly increasing and reflects order (the remark after Definition 6.1.1), $\iota_F(q) < x \iff \chi_i\bigl(\iota_F(q)\bigr) < \chi_i(x) \iff \iota_L(q) < \chi_i(x) .$ So $\mathbb{Q}_L(\chi_1(x)) = \mathbb{Q}_F(x) = \mathbb{Q}_L(\chi_2(x))$ . Since $L$ is Archimedean, Lemma 6.1.3 gives $\chi_1(x) = \chi_2(x)$ . As $x$ was arbitrary, $\chi_1 = \chi_2$ . (This is Step 7 of Theorem 6.2.3 with $\chi_2$ in place of $\varphi$ ; completeness of $L$ is not needed, and neither is the Archimedean property of $F$ .)

Why $L$ must be Archimedean (remark, not proved here). The last step separates $\chi_1(x)$ from $\chi_2(x)$ by a rational strictly between them, which is density of $\iota_L(\mathbb{Q})$ in $L$ , and density needs Archimedes (Theorem 4.2.5). In the non-Archimedean field $\mathbb{Q}(t)$ of Example (A Non-Archimedean Ordered Field), the elements $1/t$ and $2/t$ are different and have the same rationals below them, namely all $q \le 0$ : a positive rational $q$ has $q > 1/n$ in $\mathbb{Q}$ for some $n \in \mathbb{Z}_{>0}$ (Theorem 2.5.3), so $\iota_{\mathbb{Q}(t)}(q) > \iota_{\mathbb{Q}(t)}(1/n) = 1/(n \cdot 1) > 2/t$ , because $t > 2n \cdot 1$ (Example (A Non-Archimedean Ordered Field)). So the separating step fails in such a field. For the stronger statement proved above, with $F$ arbitrary, the hypothesis on $L$ cannot be dropped: on $\mathbb{Q}(t)$ , the substitution $t \mapsto t + 1$ keeps degrees and leading coefficients, so it is an order-preserving field map $\mathbb{Q}(t) \to \mathbb{Q}(t)$ different from the identity (school polynomial algebra, as flagged in Example (A Non-Archimedean Ordered Field)). An example with $F$ Archimedean needs more algebra than this volume has, and we give none.

Marking scheme. (10 points) - 2: each $\chi_i$ fixes $\mathbb{Q}$ , with the reason (Proposition 6.1.1, or directly: induction for $\mathbb{N}$ , then negatives and quotients). - 3: links the rationals below $x$ to those below $\chi_i(x)$ , using that $\chi_i$ fixes $\mathbb{Q}$ and reflects order. Either gives the full equivalence $\iota_F(q) < x \iff \iota_L(q) < \chi_i(x)$ , or, for the squeeze route of the next item, applies reflection by $\chi_1$ and $\chi_2$ to a rational strictly between $\chi_1(x)$ and $\chi_2(x)$ . - 3: concludes $\chi_1(x) = \chi_2(x)$ by rationals separating, or by a rational strictly between the two, invoking the Archimedean property of $L$ . - 2: states the conclusion for every $x \in F$ . Fatal errors (0 for the step): using density in $L$ without a hypothesis that gives it; assuming $\chi_i$ surjective.

Exercise (C2: Dense image, and when the map is onto)

Let $F$ be an Archimedean ordered field, $L$ a complete ordered field, and $\varphi \colon F \to L$ the map of Theorem 6.2.3.

(a) Prove that $\varphi(F)$ is dense in $L$ : for all $u < v$ in $L$ there is $x \in F$ with $u < \varphi(x) < v$ .
(b) Prove that if $F$ is complete, then $\varphi$ is surjective.
(c) Prove that if $\varphi$ is surjective, then $F$ is complete.

Solution (Hint 1). For (b), given $y \in L$ , look at the set $S \mathrel{:=} \mathbb{Q}_L(y)$ and its copy $\iota_F(S)$ in $F$ .

Solution (Hint 2). For (b): $\iota_F(S)$ is non-empty and bounded above in $F$ ; let $x \mathrel{:=} \sup_F \iota_F(S)$ , and show $\mathbb{Q}_F(x) = S$ .

Solution. By Theorem 6.1.2, $L$ is Archimedean.

(a) Let $u < v$ in $L$ . By Theorem 4.2.5 in $L$ there is $q \in \mathbb{Q}$ with $u < \iota_L(q) < v$ . By Step 3 of the proof of Theorem 6.2.3, $\iota_L(q) = \varphi(\iota_F(q))$ , so $x \mathrel{:=} \iota_F(q)$ works.
(b) Let $F$ be complete and $y \in L$ . Let $S \mathrel{:=} \mathbb{Q}_L(y)$ . Since $L$ is Archimedean, there are $n, m \in \mathbb{N}$ with $-y < \iota_L(n)$ and $y < \iota_L(m)$ . So $-n \in S$ , and every $q \in S$ has $\iota_L(q) < y < \iota_L(m)$ , hence $q < m$ and $\iota_F(q) < \iota_F(m)$ . So $\iota_F(S)$ is non-empty and bounded above in $F$ , and $x \mathrel{:=} \sup_F \iota_F(S)$ exists.

We show $\mathbb{Q}_F(x) = S$ . $(\supseteq)$ Let $q \in S$ . By density in $L$ there is $r \in \mathbb{Q}$ with $\iota_L(q) < \iota_L(r) < y$ ; then $r \in S$ and $q < r$ , so $\iota_F(q) < \iota_F(r) \le x$ . $(\subseteq)$ Let $\iota_F(q) < x$ . Then $\iota_F(q)$ is not an upper bound of $\iota_F(S)$ , so $\iota_F(q) < \iota_F(r)$ for some $r \in S$ ; hence $q < r$ and $\iota_L(q) < \iota_L(r) < y$ , so $q \in S$ .

By $(\star)$ , $\mathbb{Q}_L(\varphi(x)) = \mathbb{Q}_F(x) = S = \mathbb{Q}_L(y)$ , so $\varphi(x) = y$ by Lemma 6.1.3 in $L$ .
(c) Let $\varphi$ be surjective. It is injective (strictly increasing), so bijective, with an inverse $\varphi^{-1} \colon L \to F$ (Sets, Relations, and Cardinality, Theorem 1.5.4). Let $A \subseteq F$ be non-empty with an upper bound $b$ . Then $\varphi(A)$ is non-empty and bounded above by $\varphi(b)$ , since $a \le b$ gives $\varphi(a) \le \varphi(b)$ . As $L$ is complete, $s \mathrel{:=} \sup \varphi(A)$ exists; let $c \mathrel{:=} \varphi^{-1}(s)$ . Upper bound: for $a \in A$ , $\varphi(a) \le s = \varphi(c)$ , so $a \le c$ , because $\varphi$ reflects order (and is injective, for the case of equality). Least: if $u$ is an upper bound of $A$ , then $\varphi(u)$ is an upper bound of $\varphi(A)$ , so $\varphi(c) = s \le \varphi(u)$ , and $c \le u$ as before. So $c = \sup A$ (Sets, Relations, and Cardinality, Definition 2.6.5), and $F$ is complete.

So $\varphi$ is an order-preserving field isomorphism exactly when $F$ is complete. Section 6.3 reaches the “if” direction by a different route, which computes no supremum at all.

Marking scheme. (10 points) - 2: (a) $L$ is Archimedean (Theorem 6.1.2), so some $\iota_L(q)$ lies strictly between $u$ and $v$ , and $\iota_L(q) = \varphi(\iota_F(q))$ (Step 3). - 2: (b) $\iota_F(S)$ is non-empty and bounded above in $F$ (Archimedes in $L$ , transferred through $\iota_F$ ), so $x = \sup_F \iota_F(S)$ exists by completeness of $F$ . - 2: (b) $\mathbb{Q}_F(x) = S$ , both inclusions; $(\supseteq)$ uses density in $L$ . - 1: (b) $\mathbb{Q}_L(\varphi(x)) = \mathbb{Q}_F(x) = \mathbb{Q}_L(y)$ by $(\star)$ , hence $\varphi(x) = y$ by Lemma 6.1.3 in $L$ . - 3: (c) for non-empty $A \subseteq F$ bounded above by $b$ , $\varphi(A)$ is bounded above by $\varphi(b)$ , and $\varphi^{-1}(\sup \varphi(A))$ is an upper bound of $A$ and the least one, both using that $\varphi$ reflects order and is bijective. Fatal errors (0 for the step): using density of $\iota_L(\mathbb{Q})$ in $L$ without the Archimedean property of $L$ ; in (b), taking a supremum in $F$ before checking that the set is non-empty and bounded above; in (c), assuming $\varphi^{-1}$ preserves order without justification.