Dense and complete totally ordered sets The totally ordered set $\mathbb{R}$ has the following properties.


*

*[Density]. Given any two $x,y \in \mathbb{R}$, we can find $\eta \in \mathbb{R}$ such that $x < \eta < y$.

*[Dedekind Completeness]. Every non-empty subset that admits an upper bound also admits a supremum.
Adjoining a greatest and/or least element to $\mathbb{R}$ preserves the above properties. This gives us a total of four non-isomorphic totally ordered sets satisfying properties 1 and 2, namely $$(-\infty,\infty),\quad[-\infty,\infty),\quad (-\infty,\infty],\quad [-\infty,\infty].$$
This is probably a silly question, but is the above list exhaustive (up to isomorphism)?
 A: No.
A simple example that doesn't quite work is the total order on $\mathbb{R}\times\mathbb{R}$ defined such that $(a,b) < (c,d)$ if $a<c$ or if $a=c$ and $b<d$.  (In other words, uncountably many copies of $\mathbb{R}$, lined up end-to-end.)  This is dense and clearly not order-isomorphic to the real line (extended or not), because it contains uncountably many disjoint intervals that are order-isomorphic to the real line (unlike the real line itself, which can contain only countably many of these).  However, it is not Dedekind-complete per your definition, because $\{0\}\times\mathbb{R}$ is bounded above (say, by $(1,0)$), but has no least upper bound.  This can be repaired by considering $\mathbb{R}\times (\mathbb{R} + \{-\infty,\infty\})$ instead, in which case the supremum of $\{0\}\times \mathbb{R}$ is just $(0,\infty)$; and you are guaranteed greatest lower bounds as well.
A: Additional information for the interested reader:
Joel David Hamkins writes here that

The real line $\langle \mathbb{R},<\rangle$ is (up to isomorphism) the unique nonempty, separable, complete, dense, endless total order. 

