Prove that set is not definable on $(\mathbb{R},<)$ with parameters. I need to prove that set $\{x:\sin x\ge\frac{1}{2}\}$ is not definable on $(\mathbb{R},<)$ with parameters.
It is easy to show that above set is not definable without parameters. But how looks procedure when parameters are involved?
 A: There is a much easier approach to this problem than via quantifier elimination - look at automorphisms of the structure. 
In any structure, every definable set has to be preserved under every automorphism. So our task amounts to showing that for every sequence of parameters $r_1, ..., r_n\in\mathbb{R}$, the set $\{x: \sin(x)\ge{1\over 2}\}$ is not preserved by some automorphism of $(\mathbb{R}; <, r_1, ..., r_n)$.
To do this, note that any map $\alpha:\mathbb{R}\rightarrow \mathbb{R}$ with the following properties is an automorphism of $(\mathbb{R}; <, r_1, ..., r_n)$:


*

*$\alpha(x)=x$ for $x\le\max\{r_1, ..., r_n\}$ (call this maximum "$m$").

*$\alpha\upharpoonright (m, \infty)$ is an order-preserving bijection from $(m, \infty)$ to itself.
Pick $u, v>m$ such that


*

*$\sin(u)\ge{1\over 2}$,

*$\sin(v)<{1\over 2}$,

*$u>v$ (this isn't essential, it just makes things a bit easier to think about).
(Why do these exist?) We'll be done if you can show that there is an automorphism $\alpha$ of $(\mathbb{R}; <, r_1, ..., r_n)$ such that $\alpha(u)=v$.
So let's think first about the case when $m=0$. Then here's such an $\alpha$:


*

*For $x\le 0$, $\alpha(x)=x$.

*For $x>0$, $\alpha(x)={vx\over u}$.
It's not hard to check that this has the desired properties.
Now do you see how to generalize this to arbitrary values of $m$?
A: First observe that $\operatorname{Th}(\mathbb{R}, <)$ has quantifier elimination. If you know that $\mathrm{DLO}_0$ - the theory of dense linear orders without endpoints, which is complete - has q.e., just note that $(\mathbb{R}, <) \models \mathrm{DLO}_0$. 
Now suppose $X \subseteq \mathbb{R}$ is definable with parameters, so there is a formula $\varphi(x, \overline{y})$ and a sequence $\overline{b} = (b_1, \ldots, b_n) \subseteq \mathbb{R}$ such that $X = \varphi(\mathbb{R}, \overline{b}) = \{ a \in \mathbb{R} : \mathbb{R} \models \varphi(a, \overline{b}) \}$. From the q.e. we get a quantifier free formula $\psi(x, \overline{y})$ such that $\mathbb{R} \models \varphi(x, \overline{y}) \leftrightarrow \psi(x, \overline{y})$. So clearly $X = \psi(\mathbb{R}, \overline{b})$.
But the family of subsets of $\mathbb{R}$ defined with parameters by quantifier free formulas is an algebra of sets generated by the subsets defined by atomic formulas:
$$\begin{align*}
\varphi(x, a) \equiv x < a & \quad \longrightarrow \quad \varphi(\mathbb{R}, a) = (-\infty, a) \\
\varphi(x, b) \equiv b < x & \quad \longrightarrow \quad \varphi(\mathbb{R}, b) = (b, \infty) \\
\varphi(x, c) \equiv x = c & \quad \longrightarrow \quad \varphi(\mathbb{R}, c) = \{ c \}
\end{align*}$$
plus trivial ones, like $x=x$ or $a<b$ where $a, b$ are parameters, which always define either $\mathbb{R}$ or $\varnothing$. It's easy to see that the above sets generate exactly finite unions of arbitrary intervals (bounded or not, open or closed from either side) and points. Hence $X = \psi(\mathbb{R}, \overline{b})$ must be a finite union of arbitrary intervals and points, which
$\{ x : \sin x \geqslant \frac{1}{2} \}$
clearly isn't. So that set is not definable.
