I know using $D_1=\{(x,y)|x<y\}$, $D_2=\{(x,y)|x=y\}$, $D_3=\{(x,y)|y<x\}$, in $(\mathbb{R};<)$, $2^3$ binary relations are definable.
But how do I know that any other binary relations are undefinable?
professor gave me hint; using automorphism, if $D$ is definable in $\mathbb{R}\times\mathbb{R}$, then for $i=1,2,3,$ $D\cap D_i \neq \emptyset$ iff $D_i \subseteq D$.
I don't get any ideas how to prove this hint(but I know using hint, any other binary relations are not definable)
Hint: assume $D$ is defined by some formula $\phi(x, y)$, that $(a, b) \in D \cap D_1$, (so $a < b$) and that $(c, d) \in D_1$ (so that $c < d$). Find an order-preserving map $f : \Bbb{R} \to \Bbb{R}$, such that $(c, d) = (f(a), f(b))$ (you can do this with an $f$ of the form $f(x) = \alpha x + \beta$ where $\alpha > 0$). Since $f$ is order-preserving, $\phi(f(x), f(y))$ holds iff $\phi(x, y)$ holds (i.e., $f$ is a homomorphism, and hence an automorphism of $(\Bbb{R}, <)$). In particular, $\phi(f(a), f(b)) = \phi(c, d)$ must hold, $(c, d) \in D$. This gives us $D_1 \subseteq D$. The proofs for $D_2$ and $D_3$ are similar.