A "complete" ordered field of ordinals

Question

First, a note regarding proper classes: One can construct a tuple of proper classes, for example with the Morse definition. Same goes for proper-class-sized algebraic structures, as in the field of surreal numbers. One can construct the class of subclasses, or power class, of a proper class. Finally, one can construct proper-class-sized relations and therefore proper-class-sized functional relations, which I'll just refer to as functions.

Let $\text{Ord} = \mathbb{N}_\text{Ord}$ be the class of ordinals. Let $+$ and $\times$ be the Hessenberg sum and product, respectively. Let $<$ be the usual order (set membership). Then $(\mathbb{N}_\text{Ord}, <, +, \times)$ is an ordered rig. Let $\mathbb{Z}_\text{Ord} = (\mathbb{N}_\text{Ord} \times \mathbb{N}_\text{Ord}) / \sim_\mathbb{Z}$ where \begin{align} (a_1, a_2) \sim_\mathbb{Z} (b_1, b_2) &\leftrightarrow a_1 + b_2 = a_2 + b_1 \\ [(a_1, a_2)] < [(b_1, b_2)] &\leftrightarrow a_1 + b_2 < a_2 + b_1 \\ [(a_1, a_2)] + [(b_1, b_2)] &= [(a_1 + b_1, a_2 + b_2)] \\ [(a_1, a_2)] [(b_1, b_2)] &= [(a_1 b_1 + a_2 b_2, a_1 b_2 + a_2 b_1)] \end{align}

Then $(\mathbb{Z}_\text{Ord}, <, +, \times)$ is an ordered ring. Let $\mathbb{Q}_\text{Ord} = (\mathbb{Z}_\text{Ord} \times \mathbb{Z}_\text{Ord} {\setminus} \{0\}) / \sim_\mathbb{Q}$ where \begin{align} (a_1, a_2) \sim_\mathbb{Q} (b_1, b_2) &\leftrightarrow a_1 b_2 = a_2 b_1 \\ [(a_1, a_2)] < [(b_1, b_2)] &\leftrightarrow \begin{cases} a_1 b_2 < a_2 b_1 & a_2 b_2 > 0 \\ a_1 b_2 < a_2 b_1 & a_2 b_2 < 0 \\ \end{cases} \\ [(a_1, a_2)] + [(b_1, b_2)] &= [(a_1 b_2 + a_2 b_1, a_2 b_2)] \\ [(a_1, a_2)] [(b_1, b_2)] &= [(a_1 b_1, a_2, b_2)] \end{align}

Then $(\mathbb{Q}_\text{Ord}, <, +, \times)$ is an ordered field. How can we extend this construction to a "complete" ordered field $\mathbb{R}_\text{Ord}$? In the direction of Dedekind cuts, we might have $\mathbb{R}_\text{Ord} \subseteq \mathcal{P}(\mathbb{Q}_\text{Ord})$ be the class of nonempty, proper, downward closed subclasses $a$ without a greatest element: \begin{align} &a \neq \{\} \\ &a \neq \mathbb{Q}_\text{Ord} \\ &\forall x: \forall y: (x < y \in a) \rightarrow x \in a \\ &\forall x: (x \in a) \rightarrow \exists y: (x < y \in a) \end{align}

with \begin{align} a < b &\leftrightarrow a \subset b \\ a + b &= \{x + y : x \in a, y \in b\} \end{align}

and a more complex formula for the product like that for Dedekind cuts on $\mathbb{Q}$. In the direction of Cauchy sequences, we might have $\mathbb{R}_\text{Ord} = X / \sim$ where $X \subseteq \mathbb{N}_\text{Ord} \rightarrow \mathbb{Q}_\text{Ord}$ is the class of convergent transfinite sequences $a$: \begin{align} \forall \varepsilon \in \mathbb{Q}_\text{Ord}: \varepsilon > 0 \rightarrow \exists n \in \mathbb{N}_\text{Ord}: \forall i, j \in \mathbb{N}_\text{Ord}: i, j > n \rightarrow |a(i) - a(j)| < \varepsilon \end{align}

with \begin{align} a \sim b &\leftrightarrow \forall \varepsilon \in \mathbb{Q}_\text{Ord}: \varepsilon > 0 \rightarrow \exists n \in \mathbb{N}_\text{Ord}: \forall i \in \mathbb{N}_\text{Ord}: i > n \rightarrow |a(i) - b(i)| < \varepsilon \\ [a] < [b] &\leftrightarrow \exists n \in \mathbb{N}_\text{Ord}: \forall i \in \mathbb{N}_\text{Ord}: i > n \rightarrow a(i) < b(i) \\ [a]+[b] &= [n \mapsto a(n) + b(n)] \\ [a][b] &= [n \mapsto a(n) b(n)] \end{align}

Are these two directions valid (e.g. are they class-theoretically sound, ordered fields, and "complete" in a reasonable sense)? Are they equivalent or isomorphic? What is their relationship with the field of surreal numbers?

Answer 1

Taking Dedekind cuts in a non-Archimedean ordered abelian group will never yield an abelian group. Indeed, letting $a$ be the Dedekind cut consisting of all elements that are less than some integer, then $a+a=a$ by your definition, and so $a$ cannot have an additiive inverse. More generally, any Dedekind-complete ordered abelian group must be Archimedean by a similar argument (if $x$ is a positive element and there is an element that is greater than every integer multiple of $x$, let $a$ be the least such element and conclude that $a+a$ must be $a$).

Your Cauchy sequences construction does give an ordered field; more generally, given a directed set $I$ (here I am ignoring size issues; more on that below) and an ordered field $K$, the equivalence classes of Cauchy sequences $I\to K$ form an ordered field (the proofs are basically identical to the case $I=\mathbb{N}$ and $K=\mathbb{Q}$ that constructs the reals). I don't see any reason to think the field you get from this is related to the surreals in any natural way (besides them both containing $\mathbb{Q}_{\mathrm{Ord}}$). In fact, I would guess (but don't know how to prove) that every $\mathbb{N}_{\mathrm{Ord}}$-indexed Cauchy sequence in $\mathbb{Q}_{\mathrm{Ord}}$ already converges, so this "completion" is just isomorphic to $\mathbb{Q}_{\mathrm{Ord}}$.

Note that for both of your constructions, elements of your "completion" are themselves proper classes (and cannot be encoded with sets in any obvious way, though if my guess about Cauchy sequences is correct then they can be encoded with sets since they are equivalent to just single elements of $\mathbb{Q}_{\mathrm{Ord}}$). This means that in ZFC you cannot even talk about the collection of all elements of the completion, and the ordered field properties must be stated as metatheorems rather than theorems in the language of ZFC. This also means you can't actually form equivalence classes of Cauchy sequences; instead you have to just deal with individual Cauchy sequences under their equivalence relation.

In general, though, these size issues are a red herring. Very few natural things you might want to do with these sorts of fields actually depend on their constructions using all the ordinals; instead, you could just fix some uncountable cardinal $\kappa$ and only use the ordinals below $\kappa$, say. Or if you really want something that looks like the entire ordinals, you could require $\kappa$ to be inaccessible, so you really are using all the ordinals in the Grothendieck universe $V_\kappa$.

Answer 2

$\DeclareMathOperator{cof}{cof}$Some elements:

1. For any ordered field $F$, if $\cof(F)$ denotes the cofinality of $F$, then the following are equivalent:

-That every Cauchy-sequence $\cof(F) \longrightarrow F$ converge.

-That $F$ have no "dense Dedekind cut", i.e. Dedekind cut $(L,R)$ where $\{r-l: (l,r) \in L \times R\}$ is coinitial in $F^{>0}$.

-That $F$ have no proper dense (ordered field) extension.

I say that $F$ is Cauchy-complete if those conditions hold.

2. Given said ordered field $F$, the set of dense Dedekind cuts with Dedekind's operations forms a Cauchy-complete dense extension $\widetilde{F}$ of $F$, which is final among dense extensions of $F$ and initial among cofinal Cauchy-complete extensions of $F$. It is canonically isomorphic (over $F$) to the set of equivalences of Cauchy-sequences $\cof(F) \rightarrow F$ in $F$ with the natural operations. However I believe the axiom of choice is required to properly define that ordered field whereas $\widetilde{F}$ exists in ZF.
3. $\mathbf{No}$ itself, seen as an ordered field with cofinality $\mathbf{Ord}$, is not Cauchy-complete, hence not obtained as a Cauchy-completion of a subfield.
4. The field $\mathbb{Q}_{\mathbf{Ord}}$ is indeed already Cauchy-complete. The proof is more involved than I though so I only include a proof scheme below.
5. Not every Dedekind cut in an ordered field is dense, and one has to fill all cuts of set-sized cofinality $\cof(L,R):= (\cof(L,<),\cof(R,>))$ if one wants to obtain a class-sized saturated ordered field such as $\mathbf{No}$.
6. By a theorem of Philip Ehrlich (or maybe Norman Alling?), $\mathbf{No}$ is up to isomorphism the unique real-closed field without cuts of set-sized cofinality. So to construct $\mathbf{No}$ inductively, start with $\mathbb{Q}$, and take real-closures while filling every cut in the previously defined fields inductively. Of course, a more natural and elegant way to do so is Conway's definition.

---

So, here's a proof of 4.

A truncation of a Hahn series $s \in k[[\mathfrak{M}]]$ below a monomial $\mathfrak{m} \in \mathfrak{M}$ is the series $s_{\succ \mathfrak{m}}:=\sum \limits_{\mathfrak{n} \succ \mathfrak{m}} s_{\mathfrak{n}} \mathfrak{n}$. A subfield $F$ of $k[[\mathfrak{M}]]$ is said truncation-closed if every truncation of an element of $F$ lies in $F$.

First, I claim that $\mathbb{Q}_{\mathbf{Ord}}$ is a truncation-closed subfield of the field $\mathbb{Q}[[x^{\mathbb{Z}_{\mathbf{Ord}}}]]$ of Hahn series with rationnal coefficients and monomial group $x^{\mathbb{Z}_{\mathbf{Ord}}}$ (which is a multiplicative copy of $\mathbb{Z}_{\mathbf{Ord}}$).

To see this, first notice that being truncation-closed is preserved under increasing unions of subfields. We can construe $\mathbb{Q}_{\mathbf{Ord}}$ as the union of subfields $S_{\alpha}:=\mathbb{Q}({\omega}^{\omega^{\beta}},\beta<\alpha)$ generated by $\{{\omega}^{\omega^{\beta}}:\beta<\alpha\}$ for $\alpha\in \mathbf{Ord}$, so it is enough to prove that each of those fields is truncation-closed. We proeed by induction. This is clear if for $\alpha=0$ and the limit cases follow from the above remark. For successor $\alpha=\beta+1$, we have $S_{\alpha}=S_{\beta}({\omega}^{\omega^{\beta}})$ and any truncation $[F({\omega}^{\omega^{\beta}})]$ of an element $F({\omega}^{\omega^{\beta}})$ of $S_{\alpha}$ (where $F=\sum \limits_{k=-d}^{+\infty} p_kZ^{-k}$ lies in $S_{\beta}(Z)$ and is written as a Laurent series) has the form $[F({\omega}^{\omega^{\beta}})]=\sum \limits_{k=-d}^{n-1} p_k{\omega}^{-k\omega^{\beta}}+[p_n]{\omega}^{-n\omega^{\beta}}$ where $n\in \mathbb{N}$ and $[p_n]$ is a truncation of $p_n$, hence $[p_n] \in S_{\beta}$ by the inductive hypothesis. We see that $[F({\omega}^{\omega^{\beta}})]$ lies in $S_{\alpha}$, which proves the claim.

Now, given a Cauchy-sequence $(u_{\gamma})_{\gamma \in \mathbf{Ord}}$, for $\lambda \in \mathbf{Ord}$, there is a least ordinal $\gamma_{\lambda}$ such that the distance between any two terms of the sequence with indexes above $\gamma_{\lambda}$ is below $\omega^{-\lambda-1}$. This means that the truncations of those terms above $\omega^{-\lambda}$ are the same. We denote this common value $s_{\lambda}=[u_{\gamma}]_{\succ \omega^{-\lambda}}$ for all $\gamma \geq \gamma_{\lambda}$. By the previous argument, the sequence $(s_{\lambda})_{\lambda \in \mathbf{Ord}}$ ranges in $\mathbb{Q}_{\mathbf{Ord}}$. Moreover, for $\lambda \in \mathbf{Ord}$, the series $s_{\lambda}$ is a truncation of $s_{\lambda+1}$ so the order type of the support of the series $s_{\lambda}$ increases with $\lambda$. This order type is bounded by $\omega_1$ (in fact, by $\omega^{\omega}$) because of the formula $\frac{1}{1+\varepsilon}=\sum \limits_{k=0}^{+\infty} (-1)^k\varepsilon^k$ in Hahn series fields for infinitesimal $\varepsilon$. This implies that the sequence of order types is eventually constant, and thus that $(s_{\lambda})_{\lambda \in \mathbf{Ord}}$ is eventually constant. It is straight-forward to see that the eventual value is the limit of $(u_{\gamma})_{\gamma \in \mathbf{Ord}}$.