extension of group operation from $\mathbb{Q}$ to $\mathbb{R}$ 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!
 A: 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 $\sqrt{2} - 1$.
A: 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.
