Is $\mathbb{R}$ as a complete field a/the “smallest” complete field containing the field $\mathbb{Q}$

abstract-algebraanalysiscomplete-spacesfield-theory

I know that the metric space of real numbers equipped with the Euclidean metric $(\mathbb{R}, d_{\mathrm{Eu}})$ is the Euclidean-, not a p-adic, completion of the metric space of rational numbers equipped with the Euclidean metric $(\mathbb{Q}, d_{\mathrm{Eu}})$. Then…

Question Out of Curiosity: Is $\mathbb{R}$ as a (Cauchy-) complete field (in terms of the Euclidean metric) a/the "smallest" (Cauchy-) complete field containing (an isomorphic copy of) the field $\mathbb{Q}$ (in terms of the Euclidean metric)?

By "smallest" I mean two things out of curiosity:

(1) If $\mathbb{F}$ is some (Cauchy-) complete field (in terms of the Euclidean metric), must $\mathbb{F}$ contain (an isomorphic copy of) the field $\mathbb{R}$ as a subfield?

(2) If $\mathbb{F}$ is some (Cauchy-) complete field (in terms of the Euclidean metric) containing (an isomorphic copy of) the field $\mathbb{Q}$ (in terms of the Euclidean metric), must $\mathbb{F}$ have cardinality strictly greater than $\mathbb{Q}$ (i.e., must $\mathbb{F}$ be uncountable)?

Actually, leading by my own curiosity, I have found this paper "Analysis in the Computable Number Field" by Oliver Aberth published in 1968: https://sci-hub.se/10.1145/321450.321460.

After giving it a read, I think Oliver Aberth constructed a countable (Cauchy-) complete field (in terms of the Euclidean metric) containing (an isomorphic copy of) the field $\mathbb{Q}$ as a subfield, hence the field $\mathbb{R}$ is not a/the "smallest". But an online friend who is much more mathematically matured than me said (nearly a month ago) the paper is misleading so I am misunderstanding the facts. Until now I still don't know why, so I am asking here and wishing someone can help about my curiosity and/or clarify what Oliver Aberth actually did in the paper.

Best Answer

Not only are the real numbers the smallest complete ordered field containing $\mathbb{Q}$, they are in fact the only complete ordered field. The proof is surprisingly straightforward: you show that any two such have to have the same prime subfield, and then you lift the isomorphism between their prime subfields to one between the fields themselves. You can find the full details of the proof here.

As mentioned in a comment, the issue with "Analysis in the Computable Number Field" is that they're using a different notion of completeness than "mainstream" non-constructive analysis. If I were to describe a sequence that doesn't converge to an element of Aberth's field, Aberth would object that in his axioms the sequence does not in fact converge. And he would be right, but he's using a different set of axioms than the majority of mathematicians.