# Ordered Fields without a Well-Ordered Cofinal Subset

I understand how with the axiom of choice, every ordered field (in fact every ordered set) can quite easily be given a well-ordered cofinal subset.

I also understand that without the axiom of choice, it is possible for many examples of ordered sets that do not have well-ordered cofinal subsets to exist (for example, any infinite Dedekind-finite set). However this is for ordered sets in general, and I was wondering what could be said for ordered fields - which require a lot more structure to be built into them.

I am not much of an expert when it comes to proving statements to be independent of ZF (which would be the case if an example exists without choice, as no example exists with choice) so I am not sure where to even begin to tackle this. If anyone could point me in the right direction, recommend some literature to read, or even quote a useful theorem or two - it would be much appreciated.

• Note that $\omega$ embeds into any ordered field by repeatedly adding $1$, so we can't kill this with a "quasi-finite" set. – Noah Schweber Apr 13 '17 at 19:31
• If you want to learn how to prove things are independent of the axiom of choice, read Jech's "Axiom of Choice", all relevant papers by Gordon P. Monro, my paper about embedding orders into cardinals, and of course learn a lot about forcing, about basic model theory, a lot of different mathematics, and generally, have an open mind for independence. – Asaf Karagila Apr 13 '17 at 19:44
• @AsafKaragila Thank you for the references. Mathematical independence is a topic I've been meaning to read up on, so I will keep a note of these papers - including your own - to add to my reading list. – Sam Forster Apr 16 '17 at 15:39

The statement is independent from ZF as well. To see this, let us consider a few related constructions:

Let $E$ be an infinite linear order, let $\mathbb{Z}^{(E)}$ denote the ordered group of maps $E \rightarrow \mathbb{Z}$ with finite support ordered lexicographically according to $E$, that is $f < g$ if for their maximum index of disagreement $e\in E$, we have $f(e) < g(e)$.

$\DeclareMathOperator{\dd}{d}$

One can define the degree $\dd^{\circ} f$ of a non zero element $f \in \mathbb{Z}^{(E)}$ (or in a similar way $\mathbb{N}^{(E)}$) as the maximum of its support.

Let $\mathbb{Q}(E)$ denote the field of fractions of $\mathbb{Q}[(X_e)_{e \in E}]$ ordered so that $\mathbb{Q}[(X_e)_{e < e_0}] < X_{e_0}$ for each $e_0 \in E$.

$\mathbb{Q}(E)$ can be seen as the ordered field of fractions of the ordered ring $\mathbb{Z}[E]$, itself defined as $\mathbb{Z}^{(\mathbb{N}^{(E)})}$ and endowed with the product $ab := f \mapsto \sum \limits_{g+h = f} a(g)b(h)$. Note that the sum is finite because $a,b$ have finite support, and $ab$ has finite support included in the pointwise sum of their supports. The isomorphism sends $X_e$ to $\chi_{\{\chi_{\{e\}}\}}$ ($\chi$ denote characteristic functions).

Now for $1 < x = \frac{a}{b} \in \mathbb{Q}(E)$ where some $X_e$ divides $a$ and $b$, let $e(x)$ denote $\dd^{\circ} (\dd^{\circ} a - \dd^{\circ} b)$. $e$ is well defined, non decreasing and has a cofinal range, therefore $e$ maps well-ordered cofinal sets of positive infinite elements of $\mathbb{Q}(E)$ onto well-ordered cofinal subsets of $E$.

Each well-ordered cofinal subset of $\mathbb{Q}(E)$ gives rise to a well-ordered cofinal subset of $\{x \in \mathbb{Q}(E) \ | \ 1 < x \}$, and thus to a well-ordered coffin subset of $E$.

Conversely, $E$ embeds cofinaly in $\mathbb{Q}(E)$ via $e \mapsto X_e$, so $\mathbb{Q}(E)$ has a well-ordered cofinal subset iff $E$ has one, and the independency from ZFC can be recovered from linear orders to ordered fields.

• After posting my question but before seeing this answer, I came up with a different but similar poof leading to the same if and only if conclusion. Your answer is still much appreciated as it uses some nice notion and just generally thank you for your help. – Sam Forster Apr 16 '17 at 15:38
• Do you assume something about $E$? – Asaf Karagila Apr 16 '17 at 18:18
• @AsafKaragila : I don't think I need to. The fact that Hahn series fields are stable under product is usually proven assuming dependant choice, but I don't think you need choice to prove that $\sum \limits_{n \in \mathbb{N}} (-\varepsilon)^n$ exists in $\mathbb{Q}((x^{\mathbb{Z}^{(E)}}))$ when $\varepsilon \in \mathbb{Q}(E)$ is infinitesimal. I have to check though, and it might be possible to find a simpler map $e(.)$: – nombre Apr 16 '17 at 19:51
• @SamForster : As Asaf Karagila implied, there still are things to justify, so if you have a simpler answer you might as well post it. I have a solution involving an integer part on $\mathbb{Q}(E)$, but it will be a bit long. – nombre Apr 16 '17 at 20:00
• @nombre: Assuming choice, every linear order has a cofinal well-ordered subset. So I'm not sure what you mean by that. – Asaf Karagila Apr 16 '17 at 21:19

Given that $E$ is a linearly ordered set, such that $E$ has no maximal element.

Then we want to consider the set of formal products of elements of $E$ (i.e. elements of the form $e_1 * e_2 * ... * e_n$ such that $e_1, e_2, ..., e_n \in E$). To do this (using my personal approach), I consider the set: $$E^{(1)} = \bigcup_{n \in \mathbb{N}} E^n$$

(where $E^n = \{$functions from $n$ to $E \}$).

In order to account for repeated permutations, for each $f \in E^{(1)}$ define the function $\#_f \colon E \to \mathbb{N}$ to be given by $\#_f(e) = \mathrm{Card}(\{n \in \mathbb{N} : f(n) = e\})$. Then consider the equivalence relation $\sim$ on $E^{(1)}$ given by $f \sim g$ if and only if $\#_f = \#_g$, so that we can take the quotient $E^{(2)} = E^{(1)} / \sim$.

Then we define the linear order on $E^{(2)}$ to be given by the following equivalence:

1. $[f]_\sim \leq [g]_\sim$
2. For all $e \in E$, either $\#_f(e) \leq \#_g(e)$ or there exists $\epsilon \in E$ with $e < \epsilon$ and $\#_f(\epsilon) < \#_g(\epsilon)$.

This gives us a well-defined linear order on the set of formal products of $E$ (where all elements of $E$ are treated as positive, and the empty product represents $1$). We quickly add in negative elements by taking a copy of $E^{(2)}$ with reverse ordering (which we can denote by $\mathrm{Neg}(E^{(2)})$), and define $E^{(3)} = \mathrm{Neg}(E^{(2)}) \sqcup E^{(2)}$ with linear order given by having all the elements of $\mathrm{Neg}(E^{(2)})$ precede all the elements of $E^{(2)}$.

Given this ordering, there is a canonical way to extend our linear order to the set of formal sums of $E^{(3)}$ (i.e. elements of the form $e_{1,1}*...*e_{1,n_1} + ... + e_{m,1}*...*e_{m,n_m}$), such that if $x, y \in E^{(3)}$ with $x < y$ and $n \in \mathbb{N} \backslash \{0\}$ then $n*x < y$. The details are left out due to being tedious, but can be added if requested.

Which leaves us with an ordered ring $\mathrm{Ring}(E)$, that in turn gives us an ordered field $\mathrm{Field}(E)$ (the field of fractions of $\mathrm{Ring}(E)$).

1. Note that $E$ is cofinal in $\mathrm{Field}(E)$. So if $E$ has a well-ordered cofinal subset, then $\mathrm{Field}(E)$ has a well-ordered cofinal subset.

2. If $\mathrm{Field}(E)$ has a well-ordered cofinal subset (say $S$). Without loss of generality, we can assume that every element of $S$ is positive. Then there exists a unique ordinal $\alpha$ and a unique order isomorphism $h \colon \alpha \to S$.

• Then for all $s \in S$ there exists unique $p_s, q_s \in \mathrm{Ring}(E)$ with $s = \frac{p_s}{q_s}$, such that $\frac{p_s}{q_s}$ is in reduced form.
• Then $P = \{ p_s \in \mathrm{Ring}(E) : s \in S \}$ is a cofinal subset of $\mathrm{Field}(E)$ as $p_s \geq s$ for all $s \in S$, due to $1$ being the smallest positive element in $\mathrm{Ring}(E)$. Further we have that $h^{*} \colon \alpha \to P$ given by $h^{*}(a) = p_{h(a)}$ is a well-defined surjective function.
• Then for all $p \in P$ there exists a unique $t_p \in E^{(2)}$ such that $t_p$ is the leading term in $p$, as every element of $P$ is a unique sum of finitely many elements in $E^{(2)}$.
• Then $T = \{ t_p \in E^{(2)} : p \in P \}$ is a cofinal subset of $\mathrm{Field}(E)$, as $E$ has no maximal element. Further we have that $h^{**} \colon \alpha \to T$ given by $h^{**}(a) = t_{h^{*}(a)}$ is a well-defined surjective function.
• Then for all $t \in T$ there exists a unique $e_t \in E$ such that $e_t$ is the maximal value in the product, as every element of $T$ is a unique product of finitely many elements of $E$.
• Then $X = \{ e_t \in E : t \in T \}$ is a cofinal subset of $\mathrm{Field}(E)$, as $E$ has no maximal element. Further we have that $h^{***} \colon \alpha \to X$ given by $h^{***}(a) = e_{h^{**}(a)}$ is a well-defined surjective function.
• Then $X$ is a cofinal subset of $E$, as $E$ is cofinal in $\mathrm{Field}(E)$.
• Then we can consider the subset $Y = \{ h^{***}(a) \in X : \forall b \in a, h^{***}(b) < h^{***}(a) \}$, which is cofinal in $E$ and well-ordered by $h^{***}|_Y$ with the order of $h^{***}|_Y$ agreeing with the order of $E$.
• Then $Y$ is a well-ordered cofinal subset of $E$.

Then $E$ has a well-ordered cofinal subset, if and only if, $\mathrm{Field}(E)$ has a well-ordered cofinal subset.

So if every ordered field has a well-ordered cofinal subset, then every linearly ordered set without a maximal element has a well-ordered cofinal subset (as $E$ is arbitrary).

So if every ordered field has a well-ordered cofinal subset, then every linearly ordered set has a well-ordered cofinal subset (as any linearly ordered set with a maximal element clearly has a well-ordered cofinal subset).

Hence every ordered field has a well-ordered cofinal subset, if and only if, every linearly ordered set has a well-ordered cofinal seubset (as the reverse claim is trivial).

• It seems our ordered fields are the same and our argument is almost exactly the same. However the subset $P$ might not be well-ordered and one modification should be made there. – nombre Apr 17 '17 at 8:28
• @nombre $P$ would always have been well-ordered as it is indexed by the well-ordered set $S$, but you're right that this well-ordering might not agree with the order of $\mathrm{Ring}(E)$. And the exact same argument applies to $T$ and the set I have now called $X$, so I have tweaked my answer to reflect this. – Sam Forster Apr 17 '17 at 13:42