extension of group operation from $\mathbb{Q}$ to $\mathbb{R}$

Question

I'm having a hard time with this (seems easy, but could be misleading) problem:

Let $A \subseteq \mathbb{Q}$ be a convex subset, and let $+$ group operation on $A$. Let $\overline{A} := \{x \in \mathbb{R} : \exists a,b \in A \; a<x<b\}$. Can we extend this group operation to $\overline{A}$ ?

I added also model-theory tag, because I believe it maybe can be solved throw there. Thanks!

Answer 1

The answer is no if $+$ is required to be continuous with respect to the order topology.

Let $A = \mathbb{Q}$, so $\overline{A} = \mathbb{R}$. Let $f\colon\mathbb{Q}\to\mathbb{Q}$ be a homeomorphism that acts as the identity on $\mathbb{Q}\cap (-\infty,\sqrt{2})$, but acts as an order-reversing homeomorphism of $\mathbb{Q}\cap(\sqrt{2},\infty)$, and define a binary operation $\oplus$ on $\mathbb{Q}$ by $$ x \oplus y \;=\; f^{-1}(f(x) + f(y)), $$ where $+$ denotes the usual addition on $\mathbb{Q}$.

Then $\oplus$ is clearly a continuous group operation on $\mathbb{Q}$. However, this operation cannot be extended to all of $\mathbb{R}$. In particular, the set $$ \{x \oplus 1 \mid x\in I\} $$ is unbounded for any interval $I$ containing $\sqrt{2} - 1$.

Answer 2

Without any assumptions on continuity, the answer is yes, for purely combinatorial reasons: notice that $\overline A$ has cardinality continuum, as does $A^{\bf N}$, and the latter has a natural group structure with $A$ embedded as the set of functions which are zero everywhere except the first coordinate, and $A$ is countable.

Now, if you fix a bijection between $\overline A\setminus A$ and $A^{\bf N}\setminus A$ (they have the same cardinality), you can just transport the group operation through the bijection and you're done.