Let $\Omega$ be an ordered set. We say that it is order-isomorphic to another totally-ordered set $E$ whenever there is a bijection $m:\Omega\to E$ satisfying $x<y$ in $\Omega$ if and only if $m(x)<m(y)$ in $E$.

Clearly, if $\Omega$ contains a strictly monotone sequence $(x_\alpha)_{\alpha\in\omega_1}$ (where $\omega_1$ is the first uncountable ordinal) then it is not order-isomorphic to any subset of $\mathbb{R}$. I was thinking the converse might be true. That is:

Question 1. If $\Omega$ is a totally-ordered set which does not contain any strictly monotone sequence $(x_\alpha)_{\alpha\in\omega_1}$, is $\Omega$ order-isomorphic to a subset of $\mathbb{R}$?

If yes, then that gives us a characterization of subsets of $\mathbb{R}$ in terms of their order structures. More generally, I was wondering about the following.

Question 2. Are there any well-known characterizations of totally-ordered sets which are order-isomorphic to subsets of $\mathbb{R}$?


  • $\begingroup$ (@Asaf I added the set theory tag back in light of your future edits...) $\endgroup$ Feb 28, 2019 at 14:55
  • $\begingroup$ @Andrés: Yes I agree. Or maybe, I will have agreed on the future. $\endgroup$
    – Asaf Karagila
    Feb 28, 2019 at 14:56

1 Answer 1


Well, consider $\Bbb R^2$ with the lexicographic ordering. This is not isomorphic to any subset of $\Bbb R$, since it can be partitioned into $2^{\aleph_0}$ nontrivial intervals.

However, if $(x_\alpha,y_\alpha)$ is a family of size $\omega_1$ in $\Bbb R^2$, then on at least one of the coordinates it has to violate monotonicity.

In general, the real numbers is the unique linear order which is:

  1. Complete.
  2. Dense.
  3. Separable.

So it means that a linear order can be mapped into $\Bbb R$ if and only if its completion is separable. Which holds, by the definition of completion, if the linear ordering is separable.

To see why this is indeed the case, pick a countable subset which is dense in the order, then that countable subset is mapped into the rationals, and by completeness of $\Bbb R$, we can now realize the rest of the linear ordering as well.

It follows from separability that every family of pairwise disjoint open intervals is countable. Call this property the countable chain condition.

Suslin asked in 1920 whether or not one can replace "separable" by "ccc" (although the terminology wasn't developed until much later). It turned out that the answer is independent of the usual axioms, i.e. ZFC. In other words, it is consistent that there is a linearly ordered space which is complete, dense, and ccc but not separable, and it is consistent that there is none.

And since we mention Suslin lines, let us also mention Aronszajn lines. Which are linear orders of size $\aleph_1$ but contain no uncountable subset which can be embedded in the real numbers, and no uncountable monotone sequence. The existence of these, unlike Suslin lines, is provable from ZFC.

  • $\begingroup$ Perfect answer, thanks! $\endgroup$
    – Ben W
    Feb 28, 2019 at 14:27
  • $\begingroup$ (Suslin lines...) $\endgroup$ Feb 28, 2019 at 14:31
  • $\begingroup$ @Andrés: I started by writing about those. But figured that since I have an example already in ZFC... :) $\endgroup$
    – Asaf Karagila
    Feb 28, 2019 at 14:31
  • $\begingroup$ Yes, but the point is that you get a different characterization if they don't exist. :-) $\endgroup$ Feb 28, 2019 at 14:46
  • $\begingroup$ Right. Good point. I'll edit later... $\endgroup$
    – Asaf Karagila
    Feb 28, 2019 at 14:48

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