What are the definable subsets of $(\mathbb{R}, +)$?

Question

Consider the structure $(\mathbb{R},+)$. What are the subsets of that structure that are definable without parameters? I conjecture there are only four, namely $\emptyset$, $\mathbb{R}$, $\{0\}$, and $\mathbb{R} - \{0\}$. Is this correct?

Answer 1

Turning the comment thread above into an answer:

Most of the time it's easier to analyze the orbit relation of a structure than its definable sets per se. This is the relation $a\sim b$ iff there is an automorphism sending $a$ to $b$. Since automorphisms preserve $\models$, we know that every definable set is closed with respect to $\sim$ - or, more concretely, every definable set is a union of $\sim$-classes.

For example, in the case above we have $a\sim b$ iff both $a$ and $b$ are nonzero or both $a$ and $b$ are zero, so there are two $\sim$-classes ($\{0\}$ and $\mathbb{R}\setminus\{0\}$). Consequently there are only $2^2=4$ possible candidates for definable sets and it's easy to check that each of these four is in fact definable.

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Now in general this won't always be enough to fully classify the definable subsets of a structure.. For example, the structure $(\mathbb{N};<)$ is rigid - it has no nontrivial automorphisms - but it can only have countably many (parameter-freely-)definable subsets. This is where more intricate techniques such as quantifier elimination come in. But for the specific problem above, looking at automorphisms is sufficient.