A linearly ordered set without endpoints such that every closed interval is finite is isomorphic to set of integers.

Question

Suppose $A$ is a linearly ordered set without maximum or minimum and every closed interval is a finite set. I want to show $A$ is isomorphic to the set of integers with the usual order.

I know that if $A$ is countable then I can use induction to construct partial isomorphisms and hence an isomorphism.

Any help is appreciated.

Answer 1

It follows from your hypothesis that every element of $A$ has a predecessor and a successor. So picking any $a\in A$, there is an embedding $f\colon \mathbb{Z}\hookrightarrow A$ sending $0$ to $a$, and whose image is a convex set. Suppose for contradiction that $f$ is not surjective. Then if $b$ is not in the image of $f$, either $b$ is greater than every element in the image of $f$ or less than every element. So one of the closed intervals $[b,a]$ or $[a,b]$ is infinite.

Answer 2

Consider $x_0\in A$ and the function \begin{align}f:A&\to \Bbb Z\\ f(x)&=\lvert[x_0,x)\rvert-\lvert (x,x_0]\rvert\end{align}

This function is injective (which is sufficient for the specific passage you need). It's also surjective and an isomorphism of orders, however.