Let us assume ZFC and let Q be the set of rational numbers ordered according to size. There is a
well known theorem which implies that if S is any totally ordered countable set containing a subset
ordinally similar to Q, then S contains well ordered subsets having arbitrarily large countable
ordinal numbers. Is there a converse to this theorem which implies that if S has the second of these
properties, then it has the first? I have been unable to find any mention of such a converse theorem
or to come up with any obvious counter-examples. (I apologize if my question is not considered
appropriate for "mathoverflow.net")



A question about well ordered subsets of totally ordered countable sets
If a linear order doesn't contain a copy of $\mathbb Q$, then, by a theorem of Hausdorff, it can be obtained by a transfinite sequence of steps, starting with singletons, and at each step forming well-ordered or reverse-well-ordered sums of previously constructed orderings. If the final result is to be a countable set, then the transfinite sequence will be only countably long, and each of the well-ordered or reverse-well-ordered index sets used along the way will also be countable. I believe this will allow a proof, by induction along the transfinite sequence, that at no stage does it become possible to embed arbitrarily large countable ordinals. Unfortunately, I don't have time to work out the details right now. I'll come back to it later if no one else does it first.


