In $(\mathbb{R},<)$ $D_{1}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,x<y\} , D_{2}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,x=y\} , D_{3}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,y<x\} D_{4}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,x\neq y\}, D_{5}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,x\leq y\} , D_{6}:=\{(x,y)\in\mathbb{R}\times\mathbb{R}|\,y\leq x\} , D_{7}=\mathbb{R}\times\mathbb{R}$ , and $D_{8}:=\emptyset $
I want to Show none of else is definable in $(\mathbb R,<)$. Using Hint: . Using if h is an automorphism, then it preserves setwise,i.e. h[D]=D.
And hint also says, first check that if D is definable in $\mathbb{R}\times\mathbb{R}$ , then for i=1,2,3 . $D\cap D_{i}\neq\emptyset$ iff $D_{i}\subseteq D$
And left hand side is done. The problem is r.h.s.
For doing $\Rightarrow $, I need an automorphism $h$ which preserves < and for any $(x,y) \in D_i$, and for (a,b) in $D_i\cap D$, $ (h(a),h(b))=(x,y)$ so that $D_i\subseteq D$. But I couldn't find such automorphism.
Hint: the affine function $h(x) = \alpha x + \beta$, where $\alpha, \beta \in \Bbb{R}$ and $\alpha > 0$ is strictly order-preserving, i.e., an automorphism of $(\Bbb{R}, <)$. Given $(x, y) \in D_i$, $(a, b) \in D_i \cap D$, use what that tells you about $x, y, a$ and $b$ to find $\alpha$ and $\beta$ such that $(h(a), h(b)) = (x, y)$.