Is the structure $(\mathbb{R}, *)$ "weakly minimal" in this precise sense?

Question

This is a follow-up to my previous question about definability in the structure $(\mathbb{R}. *)$ The definition of a minimal structure is that every parameter-definable subset of that structure is finite or cofinite. Now, certainly the structure $(\mathbb{R}, *)$ is not minimal. However, I conjecture that it does have the following weaker property: that every parameter-definable set is either finite, cofinite, the positive reals modulo a finite set, or the negative reals modulo a finite set. By modulo a finite set, I mean the set with some finite number of elements added and/or removed. Is this conjecture true? I would be very surprised if it was false.

Answer 1

$\mathcal{R}=(\mathbb{R};*)$, while not itself minimal, has a strongly minimal subset which really captures the whole: namely, $\mathcal{R}^+=(\mathbb{R}_{>0},*)$. $\mathcal{R}$ is basically two copies of $\mathcal{R}^+$, plus a definable singleton, glued together in a definable way, so we can often derive results about $\mathcal{R}$ from results about $\mathcal{R}^+$.

In particular, we have via an appropriate interpretation of $\mathcal{R}$ in $\mathcal{R}^+$:

If $X$ is a definable-with-parameters subset of $\mathcal{R}$, then $\{\vert c\vert: c\in X\}$ is a definable-with-parameters subset of $\mathcal{R}^+$.

Now unlike $\mathcal{R}$, the substructure $\mathcal{R}^+$ actually is minimal: it's just a $\mathbb{Q}$-vector space (think about the map $x\mapsto e^x$). This lets us rule out counterexamples to the conjecture in the OP as follows. Given a definable-with-parameters $X$ in $\mathcal{R}$, let $X_+=X\cap\mathbb{R}_{>0}$ and $X_-=X\cap\mathbb{R}_{<0}$. By the facts above, both $X_-$ and $X_+$ must be either finite or cofinite, so we're done.

More snappily, this is a general feature of any "finite cover" of a minimal structure in the appropriate sense.