Question. Is there a decent classification theorem for linear orders satisfying all three of:

  • Dense. Given a pair of elements $y,x$ with $y>x$, there exists $k$ satisfying $y>k>x$.

  • Complete. Given a non-empty subset $A$, if $A$ is bounded above, then $A$ has a least upper bound.

  • Endless. There is neither a greatest element nor a least element.


  • $\begingroup$ I'm not the best-qualified to answer this, but my understanding is that this is a rather "wild" classification problem and you very quickly run into questions you can't answer in ZFC (a particularly famous example of this is Suslin's problem). By the way, the homogeneity hypothesis you propose is actually extremely restrictive, as it implies every bounded well-ordered sequence in your chain $X$ is countable, which with some work can be used to show $|X|=2^{\aleph_0}$. $\endgroup$ – Eric Wofsey Aug 18 '15 at 16:02
  • $\begingroup$ @EricWofsey, judging by that link, you seem to be right. Perhaps a classification exists under $V=L$ or something like that... $\endgroup$ – goblin Aug 18 '15 at 17:47

On the basis of Eric's comment: no.

In particular, Suslin's problem asks whether or not there exists a non-empty linearly ordered set $R$ that isn't isomorphic to $\mathbb{R}$ satisfying the 3 conditions stated in my question together with the condition: "You can't cram more than countably many disjoint non-empty open intervals into $R$." This is independent of ZFC. Therefore, it seems that no truly useful classification can exist without assuming further axioms.

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