How do I construct this isomorphism?

Question

Let $X$ be a completely ordered set, with the following properties:

a)$X$ does not have neither a maximum nor minimum element.

b)Any bounded subset of $X$ is finite.

Then $X$ is isomorphic to the set of al integers.

I was thinking on using a partition for $\mathbb{Z}$ which consists of only finite sets, such as intervals. However I'm struggling to see how can the isomorphism be constructed from there.

Answer 1

SKETCH: Let $x\in X$; $X$ has no maximum element, so there is a $y\in X$ such that $x<y$. Use the fact that the interval $[x,y]$ is finite to show that $x$ has an immediate successor. Then use a similar argument to show that $x$ has an immediate predecessor. For each $x\in X$ let $x^+$ be the immediate successor of $x$ and $x^-$ the immediate predecessor of $x$.

Now define a function $f:\Bbb Z\to X$ as follows. Let $f(0)$ be any element of $X$. Given $f(n)$ for some $n\ge 0$, let $f(n+1)=f(n)^+$. Given $f(n)$ for some $n\le 0$, let $f(n-1)=f(n)^-$. Show that $f$ is an order-isomorphism.

Answer 2

Hints: Choose arbitrary $x_0\in X$ and send to $0$.
We can define $\varphi(x):=|[x_0,x)|$ if $x\ge x_0$ and $\varphi(x):=-|[x,x_0)|$ if $x<x_0$.