Is this a condition to be order isomorphic to a subset of the integers? Let $S$ be a totally ordered set such that any nonempty subset of $S$ bounded above has a greatest element and any nonempty subset of $S$ bounded below has a least element.  Is $S$ order isomorphic to a subset of the integers?
 A: Yes.
Assume without loss of generality that $S$ has a least element. (Otherwise pick an element and handle the part of $S$ above and below the chosen element separately -- for those below mutatis mutandis with the order swapped).
Now we can keep assigning the "smallest element of $S$ not yet used" to integers starting with $0$ and counting upwards. (After the first step, the set of integers not yet used is always bounded below by the original least element).
This can either end when we've used up all of the elements, or when we've assigned elements to all the natural numbers. But in the latter case, the set of elements that we have assigned yet cannot be upwards bounded in $S$ -- because then they would have a largest element, which they haven't -- so also in that case there cannot be any elements left.
A: Pick any element $a\in S$, and consider $T=\{x\in S:a\leq x\}$ and $U=S\setminus T$.
Every subset of $T$ has least element, so $T$ is well ordered, and thus order isomorphic to some ordinal number. Now for every $\alpha > \omega$, take $\omega \subset \alpha$ to be subset bounded above with no greatest element. This imply, that $T$ is order isomorphic to some subset of $\mathbb{N}$. Dual argumentation shows that $U$ is order isomorphic to subset of $\mathbb{Z}\setminus\mathbb{N}$.
