Topologically, how does a "super continuum" differ from the reals? 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?
 A: 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.
