This question was originally asked at [MSE](https://math.stackexchange.com/questions/4420627/transfinitely-iterating-the-puiseux-levi-civita-or-hahn-series-constructions) but seems too advanced, so I'm reposting it here.


In short, the idea is that many constructions for non-Archimedean fields can naturally be iterated, in some way, to arrive at a transfinite sequence of increasingly large non-Archimedean fields, each of which embeds into the last. For instance, if we start with the reals, we can take a transfinite sequence of iterated ultrapowers to arrive at a proper class sized field, which is the maximal real-closed field. The surreal numbers have something similar with their "birthday structure," or what Philip Ehrlich calls an "$s$-hierarchical" structure, so that we can keep adding new levels of infinitesimals for each new ordinal and again arrive at the maximal real-closed field.


The question, in a sense, is if we can get the same results iterating simpler power series-type constructions such as the Levi-Civita, Hahn, or Puiseux series. They also can easily be iterated: given any real-closed field, we can generate the Puiseux series in that field, which is a new real-closed field for which we can generate another Puiseux series (with an even smaller infinitesimal), etc. We can basically do the same thing with iterating the Levi-Civita construction, which is just the Cauchy completion of the Puiseux series. We get something similar with Hahn series, which also let us iterate the *exponents* at each level, so that they also can take values in the previous field, rather than only taking values in the rationals.


On some level these fields are all just minor variations of one another, differing only in what kinds of rational exponents and infinite series are permitted, so the question is: **to what extent can we iterate these kinds of things and ultimately arrive at the same, maximal, proper class sized real-closed field as the surreals and hyperreals?**


Or, put another way: **what is the simplest power series-style construction like this which *can* be iterated transfinitely many times to get that same field?**


I say this because there are a few quirks - such as for Hahn series, we can either iterate the field of coefficients, or the group of exponents, or both, at each step, and I'm not sure just how important these details are. For instance, the surreals have about the simplest iteration procedure possible, requiring no ultrapowers or anything else, yet they succeed in this task. So, I'm curious if iterating something very simple like the Levi-Civita or Puiseux fields lead to a similar result.




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LATER EDIT: one important note is that the Levi-Civita construction, if iterated, only seems to add increasingly smaller infinitesimals and larger infinities, but doesn't quite add *larger* infinitesimals and *smaller* infinities in the right way. For instance, the Levi-Civita field has $\omega^1, \omega^{1/2}, \omega^{1/3}, \omega^{1/4}, ...$ at the first stage of iteration, but doesn't contain anything smaller than all of those but larger than the natural numbers. These numbers are never created at any successive stages either, because we aren't adding any new exponents to get something like $\omega^{1/\omega}$. So the short answer is that we will at least need something like Hahn series, so that we can modify the exponents at each stage. Alling apparently built something isomorphic to the surreals via a modified Hahn series construction in these papers: <https://doi.org/10.1090%2FS0002-9947-1962-0146089-X> <https://doi.org/10.1090%2Fs0002-9947-1985-0766225-7>



Transfinitely iterating the Levi-Civita, Hahn or Puiseux constructions
Let us work in NBG set theory with global choice. There is, up to non unique isomorphism, a unique real-closed field that is $\kappa$-saturated for all infinite cardinals $\kappa$. Let's denote it by $\mathbf{K}$. For real-closed fields, being $\kappa$-saturated is the same as having no cut of size $<\kappa$, by which I mean an ordered pair $(L,R)$ of subsets $L,R$ of size $<\kappa$ such that $L<R$ and that there is no element between $L$ and $R$.


Since both Hahn series, Levi-Civita series and Puiseux series constructions you mention give a real-closed field as a result, and since being real-closed is preserved by increasing unions, your question reduces to the following one:


Which of those three processes ends up filling all set-sized cuts?




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Let us start with the Levi-Civita one. Writing $\alpha$ for the ordinal step of the iteration process, one can see that the $\alpha$-th field $\mathbb{F}\_{\alpha}$ in the construction is contained in the Hahn series field $\mathbb{L}\_{<\alpha}$ of series with real coefficients and monomials in the group $\mathfrak{L}\_{<\alpha}$. This is the group of formal products $\prod \limits\_{\gamma<\alpha} {x\_{\gamma}}^{r\_{\gamma}}$ where $(r\_{\gamma})\_{\gamma<\alpha}$ is a family of real numbers with finite support, i.e. which is zero outside of a finite subset of $\alpha$. (Or depending on your conventions for the construction, you could replace $\alpha$ with $\alpha+1$). The group $\mathfrak{L}\_{<\alpha}$ is anti-lexicographically ordered, defining a non trivial product $\prod \limits\_{\gamma<\alpha} {x\_{\gamma}}^{r\_{\gamma}}$ to be larger than $1$ if the last non-zero exponent $r\_{\gamma\_0}$ is strictly negative. To see that this is the case, show that the Levi-Civita field with coefficients in $\mathbb{L}\_{<\alpha}$ embeds into $\mathbb{L}\_{<\alpha+1}$, and that $\bigcup \limits\_{\beta<\alpha} \mathbb{L}\_{<\beta}$ is naturally contained in $ \mathbb{L}\_{<\alpha}$ for all non-zero limit $\alpha$.


It follows that the union $\mathbb{F}\_{\infty}$ of all $\mathbb{F}\_{\alpha}$'s is contained in a field $\mathbb{L}$, which is the same as $\mathbb{L}\_{<\alpha}$ except the $\gamma$'s can now be arbitrary ordinals. This field is real-closed, but not at all $\kappa$-saturated for all infinite cardinals $\kappa$. For instance, there is a countable cut $(\mathbb{N},\{...,{x\_0}^{-\frac{1}{4}},{x\_0}^{-\frac{1}{2}},{x\_0}^{-1}\})$ in $\mathbb{L}$.


So $\mathbb{F}\_{\infty}$ is not isomorphic to $\mathbf{K}$.




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The process with Puiseux series yields a smaller field than $\mathbb{F}\_{\infty}$ which contains the previous cut, hence it is also not isomorphic to $\mathbf{K}$.




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Now let's turn to the Hahn series construction. If you start with the value group $\mathbb{R}$ (i.e. group of monomials $x^{\mathbb{R}}$) and iterate by extending coefficients, then you'll still end up in $\mathbb{L}$ and contain the same cut as before. So I assume we are now taking the underlying ordered group of the stage $\alpha$ field $\mathbb{H}\_{\alpha}$ as the value group for the next ordered field $\mathbb{H}\_{\alpha+1}$. At limit stages, one can either take the union of the previous monomial groups as the new monomial group, or just take unions, without chancing the end result. In any case, this construction, starting with $\mathbb{H}\_0=\mathbb{R}$, can be done within the field $\mathbf{No}$ of surreal numbers, where $\mathbb{H}\_{\alpha+1}$ will simply be the class $\mathbb{R}[[\omega^{\mathbb{H}\_{\alpha}}]]$ of surreal numbers whose Conway normal form has exponents in $\mathbb{H}\_{\alpha}$. One can see that the union $\mathbb{H}$ of all such fields still contains countable cuts. For instance $(\{\omega,{\omega}^{\omega},{\omega}^{{\omega}^{\omega}},...\},\varnothing)$. In fact in $\mathbf{No}$, there are many monomials of the form $\mathfrak{m}=\omega^{a\_1\pm\omega^{a\_2\pm\omega^{...}}}$. If all $a\_i$'s are in the field $\mathbb{H}\_{\infty}$, then the simplest such monomials generate a set-sized cut over $\mathbb{H}\_{\infty}$, since no iteration of the Hahn series construction gives such transfinite "$\omega$ expansions". See Denis Lemire's PhD thesis for more information.


