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This is a question I've wondered about for a while. Consider the following setup. Suppose you take a non-Archimedean ordered field, a strict field extension of the reals, and then you form its Dedekind completion. We are interested in these structures in terms of the topology that appears; not algebra - algebraically, they are generally rather badly behaved due to the presence of "sticking points" (nonzero elements $a$ such that $a + b = a$ for some other, nonzero $b$). The purpose of the algebra is just that it provides a convenient base from which to build them on.

We call such a construction a super continuum:

Definition: A super-continuum or super-line is a Dedekind completion of a non-Archimedean ordered field extension of the reals.

This therefore gives us a whole class of "line-like" spaces that are similar to the reals in that they are connected and thus "continuous" in some way, but may be quite different, and what I'm interested in is just how, from a topological point of view. Moreover, such a thing seems of interest because it could potentially permit the generalization of some hitherto "reals-only" topological concepts like path-connectedness and manifolds to more general kinds of spaces.

In particular, we can at the very least construct such a space that cannot be homeomorphic to the reals by creating a suitably large ordered field extension of $\mathbb{R}$ that is itself so big that it is bigger than $\mathbb{R}$, then performing the Dedekind completion. As its cardinality will then be larger, homeomorphism will not be possible. What, then, may be the properties of such spaces? Moreover, do their attendant extensions of the ideas of path-continuity, say, have any use?

The_Sympathizer
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    You are assuming that the completion has larger cardinality, but have not shown it to be true. Your non-Archimedian field still has continuum many elements. The completion talks of sequences of these elements indexed by the naturals, so there are only continuum many of them. There are then only continuum many limits of these sequences, so the closure of your set is still of size continuum. – Ross Millikan Aug 01 '20 at 03:26
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    Note that your set of pos/neg polynomials is not a field. It does not have inverses, as $1+\omega$ has no inverse. The way you show it you are limited to a finite number of terms in each direction. If you want to expand to an infinite number of terms, you need to define convergence of these infinite series. – Ross Millikan Aug 01 '20 at 03:30
  • So non-Archimedean ordered field is simply ordered field with each element smaller than some natural number? Would you be nice to explain why is such a field important? – Charlie Chang Aug 01 '20 at 04:14
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    @CharlieChang: No, that is not what a non-Archimedean ordered field is; you'll find the definition and some examples here. – Brian M. Scott Aug 01 '20 at 04:31
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    @CharlieChang: no, an Archimedean field guarantees that you can multiply any element by an integer and get something greater than $1$. In the reals $\frac 12$ is less than any natural (or $-\frac 12$ if you think $0$ is a natural-I flip on this). In this model you can't multiply $\epsilon$ by any natural and have it greater than $1$. The nice thing is you have infinitesimals for calculus. The bad thing is you don't have completeness-there are Cauchy sequences that do not converge. One can translate theorems between the models. Most mathematicians like the standard reals better. – Ross Millikan Aug 01 '20 at 04:32
  • So I guess it's sort of including $0_+$, similar to that $\infty$ is included in the complex plane. A consequent eof such a definition is that we can have e.g. 2+$0_+$, 2+2$\times 0_+$ etc, which are not equal to each other. So it's a property for informal calculus – Charlie Chang Aug 01 '20 at 04:38
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    @CharlieChang: If we add $\infty$ to the complex plane, we give up the field structure. OP is trying to maintain a field by adding all the elements that are required by adding $\omega$. – Ross Millikan Aug 01 '20 at 04:41
  • @Ross Millikan: Hence why I mentioned about starting from a field with cardinal already larger than the continuum as an alternative if that one didn't pan out - I just gave that as a candidate possibility because I didn't know the cardinal for the completion. – The_Sympathizer Aug 01 '20 at 04:49
  • It sounded to me like you assumed that the completion of your space (which may not be a field) had a larger cardinal, which is not correct. I think the completion of the space you defined has the same topology as the reals, but am not sure. If you want to start from a field of larger cardinality, you should say so and I will check out because I don't know about those. – Ross Millikan Aug 01 '20 at 04:52
  • @Ross Millikan : Yes, that's what I was after. It was conditional on the (then unknown) cardinal of that space. – The_Sympathizer Aug 01 '20 at 05:24
  • Another way to make such hyperreal fields is to take $X$ a Tychonoff space, $C(X)$ its ring of continuous real-valued functions, and mod out modulo a maximal ideal. These are ordered fields that contain the reals via the classes of constant functions. See Gillman & Jerrison's book for more info on such fields. – Henno Brandsma Aug 01 '20 at 06:06

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Below, "separable" is always used in the topological sense - that is, meaning "has a countable dense subset" - as opposed to its field-theoretic sense.

Your question is a bit open-ended, but I think the following is relevant. For $F$ an ordered field, let $F^+$ be its Dedekind completion (thought of as either a linear order or a space with the order topology). Let me address the specific question of when $F^+$ is path connected.

A key point is that ordered fields are "self-similar:" in an ordered field $F$, every nontrivial open interval is homeomorphic to the whole. Since $F$ is dense in its Dedekind completion $F^+$, this means that if $F$ has cardinality $>2^{\aleph_0}$ then every nontrivial interval in $F^+$ also has cardinality $>2^{\aleph_0}$. Not only does this force non-homeomorphism with $\mathbb{R}$, it also means that $F^+$ is not path connected if $F$ has cardinality $>2^{\aleph_0}$.

What if $F$ has cardinality continuum? Well, we still might not have $F^+\cong\mathbb{R}$. Specifically, if $F$ has uncountable cofinality (which can happen even if $\vert F\vert=2^{\aleph_0}$) then $\omega_1$ embeds into $F$, hence into $F^+$ a fortiori, so the latter can't be homeomorphic to (or even embed into) $\mathbb{R}$. In fact, per the self-similarity point above, such an $F$ results in a non-path-connected $F^+$ again (note that self-similarity is crucial here since there are path-connected linear orders with uncountable cofinality such as the long line). Ordered fields of uncountable cofinality and size continuum can be produced by taking an arbitrary ordered field of uncountable cofinality (e.g. an appropriate ultrapower of $\mathbb{R}$) and then taking the subfield generated by the reals and some fixed increasing $\omega_1$-sequence.

This leaves only the case of ordered fields of size continuum into which $\omega_1$ doesn't embed. Suppose $F$ is such a field and $F^+$ is path connected. Fix $a<b$ in $F$, let $i:[0,1]\rightarrow F^+$ be a path connecting $a$ and $b$, and let $X=im(i)\cap F$. The preimage $i^{-1}(X)$ is separable since $[0,1]$ is hereditarily separable, and separability being preserved by continuous surjections we have that $X$ itself is separable. The self-similarity point above means that $F$ is separable, which immediately implies that $F^+\cong\mathbb{R}$.

Consequently, we have a complete answer to the question of when $F^+$ is path-connected:

For an ordered field $F\supseteq\mathbb{R}$, the Dedekind completion $F^+$ of $F$ is path-connected iff $F^+\cong\mathbb{R}$, and this happens iff $F$ is separable.

Finally, it's a good exercise to show that there do exist separable non-Archimedean extensions of $\mathbb{R}$ in the first place.

Noah Schweber
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    "This leaves only the case of ordered fields of size continuum with a countable dense subset." That seems a bit misleading. You can have an ordered field of size continuum with no countable dense subset but also into which $\omega_1$ does not embed. For instance, take $\mathbb{R}$ and adjoin a bunch of transcendental infinite elements that are ordered like an Aronszajn line. – Eric Wofsey Jun 03 '21 at 05:49
  • @EricWofsey Oh nasty, I overlooked that! It's too late at night for me to make real improvements to this answer, but I'll note the error and come back to it tomorrow. – Noah Schweber Jun 03 '21 at 05:52
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    The final conclusion should still be correct, though. If there is a nontrivial path in $F^+$, then the image of that path must be order-isomorphic to $[0,1]$, and $F$ must have countable cofinality. You can then use this to cover $F^+$ by countably many scaled copies of the path and conclude that $F$ is separable. – Eric Wofsey Jun 03 '21 at 05:53