# The order type of the rationals.

Herewith another mind-numbingly naive question from a reader of philosophy.

My question concerns the order type of the rational numbers.

Omega squared seems a natural first choice, but obviously this does not look anything like the natural ordering of the rationals.

Is it known where the order type of Q occurs in the hierarchy of ordinal numbers? Is there a known ordinal-arithmetic expression describing it a function of omega?

Finally, I really must buy a textbook on the subject of Set Theory. Wiki is a fantastic resource and the maths pages are of exceptionally high quality, but I don't want to get into bed at night with my laptop. Is there a standard, undergraduate text that could be recommended.

• The order type of the rationals, which as Asaf points out is not an ordinal, is often denoted with $\eta$.
– MJD
Commented Aug 8, 2013 at 18:39
• Note we are all assuming that when you say "the rationals", you are referring to the rationals along with their usual ordering.
– user14972
Commented Aug 8, 2013 at 18:43
• @Hurkyl That's a very good point. I really need to take a course on this subject! Commented Aug 8, 2013 at 18:59
• @MJD Somehow I missed your comment earlier. That's very helpful. Not all order types are represented as ordinal numbers. Commented Aug 9, 2013 at 0:06

The ordinals are order types of well-ordered partial orders. The rational numbers are not well-ordered, therefore their order type does not occur within the ordinal hierarchy.

• Thanks again Asaf. I really must do a course on this subject! Commented Aug 8, 2013 at 18:57
• Yes. That is a good idea. Commented Aug 8, 2013 at 19:31

E. Kampke's book on set theory, which I think has a Dover edition, has some material on the order type of the rationals. It's not found among the ordinals because it's not well-ordered. However, there's a proof, which I seem to recall goes back to Cantor, proving that any two countable linearly ordered sets without endpoints that are densely ordered (i.e. between any two points there's another) are order-isomorphic.

• You probably have to add "countable" in your theorem. Commented Aug 8, 2013 at 18:55
• Relevant earlier discussion; it agrees with you that the theorem is due to Cantor. Also relevant.
– MJD
Commented Aug 8, 2013 at 19:00
• @Michael Hardy. Thanks for your recommendation and your interesting comment. Commented Aug 8, 2013 at 19:16
• @GEdgar : Done. I'm surprised I forgot that. BTW, the same method of proof show that up to isomorphism there is only one countably infinite atomless Boolean algebra. Commented Aug 9, 2013 at 17:48