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

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.
