# A question about well ordered subsets of totally ordered countable sets

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")

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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.

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Linear orders that do not contain a copy of $\Bbb{Q}$ are known in the literature as scattered; the Hausdorff theorem on scattered orders mentioned in the nice solution by Andreas Blass can be found in Rosenstein's text on Linear Orders (among other places). Indeed, Exercise 5.19.2 of Rosenstein's text is precisely what Blass shows in his outline. It is also worth pointing out that Hausdorff's result from 1908 was re-discovered in a 1962-paper of Erdős and Hajnal. –  Ali Enayat Jan 16 '13 at 2:52