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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.

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

2 Answers 2

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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.

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

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    $\begingroup$ You probably have to add "countable" in your theorem. $\endgroup$
    – GEdgar
    Commented Aug 8, 2013 at 18:55
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    $\begingroup$ Relevant earlier discussion; it agrees with you that the theorem is due to Cantor. Also relevant. $\endgroup$
    – MJD
    Commented Aug 8, 2013 at 19:00
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    $\begingroup$ @Michael Hardy. Thanks for your recommendation and your interesting comment. $\endgroup$
    – gamma
    Commented Aug 8, 2013 at 19:16
  • $\begingroup$ @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. $\endgroup$ Commented Aug 9, 2013 at 17:48

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