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

For this theorem I wouldn’t worry about using any specific formalism for the integers: all that matters is their order properties, which I would simply take as known. Here’s a suggestion. Fix some $x_0\in X$ arbitrarily; your isomorphism $\varphi$ is going to take $x_0$ to $0$. Now for each $x\in X\setminus\{x_0\}$, either $x_0<x$, or $x<x_0$. If $x_0<x$, let $\varphi(x)=\big|[x_0,x]\big|-1$, and if $x<x_0$, let $\varphi(x)=1-\big|[x,x_0]\big|$. Then $\varphi$ is clearly a well-defined function from $X$ to $\Bbb Z$, so all of the real work will be proving that it’s an order-isomorphism.

I would begin by proving that every $x\in X$ has an immediate successor and an immediate predecessor. If $x\in X$, there is a $y\in X$ such that $x<y$. Let $n=\big|[x,y]\big|$. Show that $y$ is an immediate successor of $x$ if $n=2$. If $n>2$, show that $\big|[x,z]\big|<n$ for each $z\in(x,y)$, pick $z\in(x,y)$ such that $\big|[x,z]\big|$ is minimal, and show that $\big|[x,z]\big|=2$ and hence that $z$ is an immediate successor of $x$. Finish up by showing that the immediate successor is unique.

Once you’ve done this, you can prove by induction both that $\varphi$ is a surjection and that it is an injection. After that it only remains to show that $\varphi$ is order-preserving, which is pretty straightforward given the definition of $\varphi$.

Answer 2

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 $\{ x \in X \, | \, x<x_0 \}$.