Let $(P,\leq)$ be a totally ordered set such that
- for all $p\in P$ the set $\{q\in P: q > p\}$ has a smallest element (that is, $p$ has a successor), and
- for all $p\in P$ the set $\{q\in P: q < p\}$ has a largest element (that is, $p$ has a predecessor).
It seems fairly intuitive to me that this forces that $P\cong\mathbb{Z}$ - but how does one prove this?
How about $\mathbb{Z}^2$ with the lexicographic order?
$$(a,b) < (a', b') \text{ if } \begin{cases} a< a' \text{ or} \\ a=a', b<b'\end{cases}$$
The predecessor of $(a,b)$ is $(a, b-1)$ and the successor of $(a,b)$ is $(a, b+1)$.
In the usual order on $\mathbb{Z}$, every interval $[p,q]$ is finite. But the interval $[(0,0), (1,1)]$ is infinite.