TITLE: A question about well ordered subsets of totally ordered countable sets
QUESTION [6 upvotes]: 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")

REPLY [4 votes]: The converse is a theorem of Đ. Kurepa in:
Sur les ensembles ordonnés dénombrables, Hrvatsko Prirodoslovno Društvo. Glasnik Mat.-Fiz. Astr. Ser. II. 3, (1948). 145–151.