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

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.