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"Hypercontinuity" is a cardinality of a continuous set's power set (set of all subsets). When talking about fields, I mean the cardinality of field's set. At first glance there is nothing preventing them from existing, however I have never seen an example of one, a non-constructive proof that they exist, or a proof that they don't exist. Either of these would be appreciated.

That is basically all there is to the question itself, and the only thing I can add is my attempts to answer it myself.

  1. An attempt to construct one. Assume any continuous set $A$. We can construct a structure $\langle 2^A, \Delta, \cap \rangle$, where $2^A$ is a power set of $A$, which is hypercontinuous by the Cantor's theorem, $\Delta$ is symmetric difference of sets, which acts as addition of elements, and $\cap$ is intersection, which acts as multiplication. In this structure $\varnothing$ is an additive identity, and $A$ is a multiplicative identity. In such structure all of the field axioms are satisfied, except for the axiom of multiplicative inverses existance, making it a commutative ring instead of a field. In addition by assuming $A$ being a set of any cardinality, we can prove that commutative ring of any cardinality, representable as cardinality of some other set's power set, exists. Nothing about fields, however.
  2. Frobenius theorem. It states that under certain conditions any division ring is isomorphic either to $\Bbb R$, or to $\Bbb C$, or to $\Bbb H$. However one of the theorem conditions is that division ring has to be a finite-dimensional vector space over $\Bbb R$. And any such vector space is continuous, so this theorem is also not relevant to the question.
  3. Non-standard analysis. Unfortunately I am completely unfamiliar with this area of mathematics, however while researching on the question I stumbled upon a statement that gave me a vague idea that non-standard analysis might have an answer. I don't know if any of the structures it studies are fields, or if any are hypercontinuous, but I still thought it might be useful to mention it.
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    $\begingroup$ What is a "hypercontinuous field" ? $\endgroup$
    – Peter
    Commented Dec 10, 2021 at 12:23
  • $\begingroup$ Maybe this is a well-known to others, but what is the definition of hypercontinuity? $\endgroup$
    – Vsotvep
    Commented Dec 10, 2021 at 12:23
  • $\begingroup$ @Vsotvep, I edited the post to elaborate on that. $\endgroup$
    – nbvehbectw
    Commented Dec 10, 2021 at 12:33
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    $\begingroup$ Does the following definition fit: you're searching for a field with a cardinality larger than the cardinality of $\Bbb R$ that is also a linear continuum (i.e., it has the least-upper-bound property and is dense in the sense that between any two distinct elements of the field there is a third element) $\endgroup$
    – Vsotvep
    Commented Dec 10, 2021 at 12:51
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    $\begingroup$ Just to clear some things up, "continuity" is a property of functions, not of spaces, and it also is not related to cardinality. Cardinality is the "size" of a set, where sets have the same size if there exists a bijection. Continuity refers to a topology, where a function is continuous if every preimage of an open set is open. You can make topologies of any cardinality, and there will be continuous functions for those topological spaces. Hence "larger than continuous" does not make sense. There is the concept of a "(linear) continuum", which is different from "continuous". $\endgroup$
    – Vsotvep
    Commented Dec 10, 2021 at 13:25

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The "upwards" direction of the Lowenheim-Skolem theorem (this direction is really just the compactness theorem in disguise) gives a positive answer to a wide variety of questions of this form: roughly speaking, any first-order theory has arbitrarily large models (assuming it has at least one infinite model in the first place). This gives, for example, the existence of real closed fields of cardinality $>\vert\mathbb{R}\vert$ as an immediate corollary. Bringing in the downwards Lowenheim-Skolem theorem (which is a genuinely different result) lets us further control the cardinality of the structures we build, so e.g. we can prove "There is a real closed field of cardinality $\vert\mathbb{R}\vert^{++++++}$ exactly."

Things get a little more complicated if we add non-first-order desiderata, and indeed certain properties are not so malleable - for examlpe, there is no Archimedean ordered field of cardinality $>\vert\mathbb{R}\vert$, and indeed every Archimedean ordered field is isomorphic to a unique subfield of $\mathbb{R}$ - but for what you've specifically asked, the Lowenheim-Skolem theorem(s) is a convenient "nuke."

As an aside, your use of the terms "continuous" and "hypercontinuous" is incorrect - continuity is a property of functions, not a name for a cardinality, and there is sadly no snappy term for "set with cardinality $>\vert\mathbb{R}\vert$" or "set with cardinality $2^{\vert\mathbb{R}\vert}$" or similar.

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Judging by the comments we exchanged, you're searching for a field with a cardinality larger than $\Bbb R$.

Such fields do indeed exist.


Most notably, there are the surreal numbers, discovered by Conway, which forms a field that contains both the reals, the hyperreals and all ordinal numbers (which implies that it is a proper class).

But perhaps you want a field that is also a set, and not a proper class. For this, you can consider the surreal numbers with a "birthday" before any epsilon ordinal $\epsilon$, so the set of surreal numbers that are constructed recursively in a step before $\epsilon$. Such epsilon numbers can be arbitrarily large, hence we can find fields of arbitrarily large cardinality.


You may even request several nice properties for the field to obey, such as being Cauchy-complete, or being real closed. See for example Chapter 2 of Lorenzo Galeotti's PhD thesis.

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    $\begingroup$ The cardinality of $\Bbb{Q}(X)$ is the same as $|X|$ for any infinite set of variables $X$. $\endgroup$
    – reuns
    Commented Dec 10, 2021 at 14:10
  • $\begingroup$ @reuns, sounds like you could make a separate answer out of that :) I presented the surreals mostly because I'm more familiar with them and they behave a lot like $\Bbb R$, hence they make more intuitive sense to me personally than extensions such as $\Bbb Q(X)$ do. $\endgroup$
    – Vsotvep
    Commented Dec 10, 2021 at 15:19

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