Totally ordered set whose every element has an immediate successor and predecessor

Question

Are there characterizations of this kind of ordering? Or at least, how is it called? Note that this is not necessarily a well-order. For instance, this kind of ordering on the set of integers: 0 < 1 < 2 < 3 < ... < -3 < -2 < -1.

Answer 1

Yes, these orderings can be completely classified. Given such an ordering $X$, define an equivalence relation $\sim$ by $x\sim y$ if $x$ is the $n$th successor of $y$ for some $n\in\mathbb{Z}$ (for negative $n$ this means $x$ is an iterated predecessor of $y$). Then the set $Y$ of equivalence classes of $\sim$ itself is totally ordered, by $[x]<[y]$ iff $x<y$ (it is easy to see this is independent of the representatives of the equivalence classes chosen). Now for each equivalence class $y\in Y$ pick a representative $f(y)\in X$. We then get a bijection $g:Y\times\mathbb{Z}\to X$ sending $(f(y),n)$ to the $n$th successor of $f(y)$. Equipping $Y\times \mathbb{Z}$ with the lexicographic order, it is easy to see that $g$ is in fact an isomorphism.

So, every such $X$ is isomorphic to $Y\times\mathbb{Z}$ with the lexicographic order for some totally ordered set $Y$. Conversely, if $Y$ is any totally ordered set, then every element of $Y\times\mathbb{Z}$ has a successor and a predecessor. Moreover, if you start with $Y\times\mathbb{Z}$ and take the set of equivalence classes as described above, you get a totally ordered set that is canonically isomorphic to $Y$. So the isomorphism classes of such $X$ are in bijection with the isomorphism classes of totally ordered sets $Y$.

Orders of this type are sometimes called "discrete total orders without endpoints". They are a basic example in model theory, and in particular the first order theory of nonempty discrete total orders without endpoints is complete (so all the nonempty examples are elementarily equivalent).