Is the structure $(\mathbb{R}, +)$ minimal?

Question

Is the structure $(\mathbb{R},+)$ minimal, that is, every definable-with-parameter subset is finite or cofinite?

Answer 1

The atomic formulas are of the form $$ \tag{1} a_n x_n + \ldots + a_1 x_1 = b_k y_k + \ldots + b_1 y_1, $$ where $a_1, \ldots, a_n, b_1, \ldots, b_k \in \mathbb{Z}_{\geq 0}$ and $x_1, \ldots, x_n, y_1, \ldots, y_k$ are the free variables. Of course, $n x$ is an abbreviation for $$ \underbrace{x + \ldots + x}_{n \text{ times}}. $$ Using this description it should not be hard to prove that this structure has quantifier elimination.

Every formula is thus a Boolean combination of equations of the form (1). In one variable all such equations have a finite or cofinite set of solutions, hence any definable subset (in one variable) is finite or cofinite.

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This is in essence just the theory of $\mathbb{Q}$-vector spaces axiomatised in the language of groups. More precisely, the theory of torsion-free divisible abelian groups (in the language $\{0, +\}$). This theory is strongly minimal by the same proof as above. Your question is just about one specific model, namely $\mathbb{R}$. Although you do not include $0$ as a constant in your language, but this is quantifier-free definable anyway since $x = 0$ iff $x + x = x$.