A definable set in $(\mathbb{R}, +,-,*,0,1,<)$ that is not definable without parameters.

Question

Consider the structure $(\mathbb{R},+,-,*,0,1,<)$. Can someone exhibit a subset of $\mathbb{R}$ that is definable with parameters in that structure, but not definable without parameters?

Answer 1

Let $\mathcal{R}$ be the field of real numbers. I claim that the parameter-freely-definable elements of $\mathcal{R}$ are exactly the algebraic reals; consequently, sets like $\{\pi\}$ are definable with parameters but are not definable without parameters.

The parameter-free-definability of the algebraic reals follows from elementary algebra. Suppose $f$ is a nonconstant polynomial with rational coefficients, and $a$ is a solution to $f(x)=0$. Since nonconstant polynomials have finite zero sets, there are rationals $p,q$ such that $a$ is the unique zero of $f$ in $(p,q)$. But each rational is obviously definable in $\mathcal{R}$, so "the unique zero of $f$ in $(p,q)$" is in fact a parameter-free definition of $a$ in $\mathcal{R}$.

The hard direction is the converse, and this is where model theory properly enters the picture. We know that the theory of real closed fields admits quantifier elimination. It's easy to check that every real definable in $\mathcal{R}$ by a quantifier-and-parameter-free formula is in fact algebraic, so we're done. Alternatively, quantifier elimination implies model completeness, so since the algebraic reals are a real closed field we get by elementarity that every parameter-freely-definable element of $\mathcal{R}$ is algebraic.