A a totally ordered set without maximum and minimum such that any bounded subset is finite is isomorphic to $Z$.

Question

Theorem

Let be $X$ a set totally ordered by the relation $\preceq$ respect which it has not maximum and minimum. So if any bounded subset of $X$ is finite then $X$ is isomorphic to the set $\Bbb Z$ of the integer numbers.

I point out that I defined the set of the integer numbers through the differences: here if you like you can see my formalism. So I ask to prove the theorem. Could someone help me, please?

Answer 1

Choose $x_0 \in X$. For any $x \geqslant x_0$, define $f(x)= Card \{y \in X \, | \, x_0 < y \leqslant x \} \in \mathbb{Z}_{\geqslant 0}$. Show that $f$ is an increasing bijection from $\{ x \in X \, | \, x \geqslant x_0 \}$ to $\mathbb{Z}_{\geqslant 0}$. And do the same thing on $\{ x \in X \, | \, x<x_0 \}$.