If for two sets $S, R$ there exists a bijection between them and they are totally ordered, then there exists an order isomorphism between them?

Question

I am trying to prove that some two sets with their respective total orders are similar (there exists an order isomorphism between them) but I haven't made a lot of progress and I have only proved that there exist a bijection between them. Is there any theorem that could guarantee that there exists said order isomorphism?

Answer 1

No. For instance, there is a bijection between $\mathbb{N}$ and $\mathbb{Z}$, but they are not order-isomorphic with their usual orders (for instance, $\mathbb{N}$ has a least element and $\mathbb{Z}$ does not).

It is true if your sets are finite: if $S$ is a totally ordered set with $n$ elements, then it is order-isomorphic to $\{1,\dots,n\}$ with the usual order. You can prove this by induction on $n$. As a sketch, since $S$ is finite, it has a greatest element $s$, and then by the induction hypothesis $S\setminus\{s\}$ is order-isomorphic to $\{1,\dots,n-1\}$, and you can extend the isomorphism by mapping $s$ to $n$.