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The totally ordered set $\mathbb{R}$ has the following properties.

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

  2. [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)?

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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.

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  • $\begingroup$ Fine, but why not just call it $\mathbb R\times[0,1]$ or for that matter use $[0,1]\times[0,1]$. $\endgroup$ – bof Oct 23 '13 at 23:46
  • $\begingroup$ $[0,1]$ is order-isomorphic to $\mathbb{R} + \{-\infty,\infty\}$, so it doesn't matter... I just wanted to use the same sets that OP was already referring to. $\endgroup$ – mjqxxxx Oct 24 '13 at 2:39
  • $\begingroup$ Okay, I think I get it now. However I'm finding your counterexample a bit hard to comprehend at the moment though (my fault, not yours); so can you let me know whether the following analysis is correct? Let $\alpha$ denote an arbitrary Dedekind-complete totally ordered set, like an ordinal number. Then if we stick $\alpha$-many copies of $[0,1)$ end-on-end so to speak, the resulting total order will be both dense and complete. Thus, taking $\alpha$ to denote any uncountable Dedekind-complete totally ordered set (such as $\omega_1$) yields the desired counterexample. $\endgroup$ – goblin Oct 24 '13 at 6:14
  • $\begingroup$ In particular, I can't figure out why the example you give has the density property. I mean, if we just stuck [0,1] end-on-end with itself once, the result wouldn't be dense. Why does your example give a dense order when just two copies of $[0,1]$ stuck end-on-end isn't? $\endgroup$ – goblin Oct 24 '13 at 7:36
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    $\begingroup$ $X \times [0,1]$ is dense in lexicographic order iff $X$ is dense. If it isn't, then there are $x,y\in X$ such that $x<y$ and no $z\in X$ satisfies $x<z<y$; and in that case, $(x,1)$ is adjacent to $(y,0)$. (So it's not the uncountability of $X$ that matters, it's the denseness. $\mathbb{Q}\times[0,1]$ is dense, too, just not Dedekind-complete.) $\endgroup$ – mjqxxxx Oct 24 '13 at 13:50
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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.

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  • $\begingroup$ Do you mean to say "there exists $t \in A$"? $\endgroup$ – mjqxxxx Oct 24 '13 at 13:44
  • $\begingroup$ @mjqxxxx, good catch. $\endgroup$ – goblin Oct 24 '13 at 21:11
  • $\begingroup$ It must also be dense. Otherwise consider the integers. It looks like this got corrected in the source already. $\endgroup$ – Anguepa Jul 13 '17 at 9:09

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