Let $(\mathbb{X},<)$ be a linear order. Assume that there exists $g:\mathbb{X} \times \mathbb{R} \rightarrow \mathbb{R}$ such that for each fixed $x<x'$ the function $y \mapsto g(x,y)-g(x',y)$ is a strictly increasing bijection of $\mathbb{R}$ onto itself.
I am trying to prove (or disprove) that then $(\mathbb{X},<)$ is necessarily order-isomorphic to a subset of $\mathbb{R}$ with the usual order. That is, there exists a strictly increasing function $f:\mathbb{X} \rightarrow \mathbb{R}$.
Thank you.
Define $F\colon\Bbb X\to \Bbb R$ as $$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof. Let $h(y)=g(x,y)-g(x',y)$. We are given that $h\colon\Bbb R\to\Bbb R$ is strictly increasing. Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$ as desired. $\square$