Completeness in Levi-Civita field

Question

I've been wondering for quite a time about Levi-Civita field (you can read it simply in https://en.wikipedia.org/wiki/Levi-Civita_field). I remember that I've read somewhere that Levi-Civita field is the smallest complete Non-Archimedean field. I am not too sure what it means by "complete" there. Does it mean that every number is in that field? I was thinking, for example, $\exp^{\pi}$ won't be there as the power is not rational. Is that right though? Is every number included in LC field? Cheers!

Answer 1

Let $LC$ denote the Levi-Civita ordered field.

Here complete means Cauchy-complete, that is, any sequence $u: \mathbb{N}\rightarrow LC$ which is Cauchy in the sense that

$\forall 0<\varepsilon \in LC, \exists n \in \mathbb{N}, \forall n\leq p,q\in \mathbb{N}, |u_p-u_q|<\varepsilon$

converges in $LC$ (for the order topology).

This is the same as saying $LC$ has no proper dense ordered field extension. In particular the completeness of $LC$ has no bearing on questions regarding the essence of a number.

By the way $LC$ is not the smallest Cauchy-complete non archimedean ordered field, which would be the field of Laurent series. Perhaps you read that it is smallest as a Cauchy complete and real closed non archimedean extension of $\mathbb{R}$. **edit: as Chilote pointed out, this is not true either; take for instance the field $\mathbb{Q}((X_n)_{n \in \mathbb{N}})$ ordered imposing $X_n >\mathbb{Q}((X_k)_{0\leq k<n})$: the completion of its real closure does not contain a copy of $LC$)

I suggest you read the Wikipedia entry for real closed fields.

One idea of proof is as follows: (most of those arguments are not trivial to justify properly!)

Let $F$ be a non archimedean ordered field containing $\mathbb{R}$ and such that there is $0<\varepsilon \in F$ with $(\varepsilon^n)_{n\in \mathbb{N}} \to 0$. Equivalently, the rank of value group of $F$ under natural valuation has a maximum. Fix such an infinitesimal $\varepsilon$.

a) The subfield $\mathbb{R}(\varepsilon)$ of $F$ is isomorphic to the smallest non archimedean ordered extension of $\mathbb{R}$, here denoted $\mathbb{R}(x)$.

b) Since $\mathbb{R}(\varepsilon)$ is cofinal in $F$ which is Cauchy-complete, and by functorial properties of the Cauchy-completion (see here), $F$ contains a unique copy of the Cauchy completion of $\mathbb{R}(x)$ over this field, which can be seen as the field $\mathbb{R}((x^{\mathbb{Z}}))$ of Laurent series.

c) Now since $F$ is real closed and by similar properties of real closure, $F$ contains a unique copy of the real closure $R$ of $\mathbb{R}((x^{\mathbb{Z}}))$ over $\mathbb{R}((x^{\mathbb{Z}}))$, which can be seen as the field of Puiseux series $R = \bigcup \limits_{n \in \mathbb{N}^{>0}}\mathbb{R}((x^{\frac{1}{n}.\mathbb{Z}}))$.

d) Again by Cauchy-completion of $F$, the Cauchy completion of this last field, which is the Levi-Civita field (indeed it is Cauchy-complete and contains Puiseux series as a dense subfield), embeds in it.

e) All the embeddings except that of $\mathbb{R}(x)$ as $\mathbb{R}(\varepsilon)$ are unique over the respective incomplete fields, so the Levi-Civita field is initial among non-archimedean Cauchy complete real closed extensions of $\mathbb{R}$ with prescribed infinitesimal $\varepsilon$ satisfying the condition.

As for the containment of all numbers, I beleive the point of Asaf Karagila is very solid: that depends on what you call a "number".

Regardless, it is hard to make numbers coming from different domains of mathematics work together in a single field while retaining all their proper caracteristics and related notions. I'd rather say the Levi-Civita field "sells" numbers. It sells a pretty classical foam of real numbers, some flavors of infinitesimals, not unrelated infinite numbers, nice looking roots of all those, while the field $\mathbb{C}$ of complex numbers sells you real numbers but urges you to look around them, also sells you solutions to problems you might not have concieved before and markets them as plain numbers; I could go on...