11
$\begingroup$

I've been browsing through Jech's and Levy's texts on set theory, and the ideas of ordinals come up fairly quickly. The idea of a limit ordinal is introduced, which is an ordinal with no maximum element. My question is, can any infinite ordinal be written as the sum of a limit ordinal and a finite ordinal, possibly unique?

My thinking was, if $\alpha$ is an infinite ordinal with no maximum, it is a limit ordinal, so $\alpha=\alpha+0$. Otherwise, suppose $\alpha$ has some order type $\{a_0,a_1,\ldots, b\}$, so $\alpha=\omega+1$. Similarly, if $\alpha$ has order type $\{a_0,a_1,\ldots,b,c\}$, we could write it as $\omega+2$.


(Thanks to Arturo Magidin, for pointing out that the following example I gave is not an ordinal.) But what about an order type like $\{a_0,a_1,a_2,\ldots, b_2,b_1,b_0\}$, this has order type $\omega+\omega^*$, would it still be possible to write is a sum of a limiting ordinal and a finite ordinal? Thanks.

  • 2
    The last order type is not an ordinal: it is not well ordered (the set of $b$s has no first element). Ordinals are always well-ordered.2011-04-21
  • 0
    @Arturo, quite right, thanks for pointing that out.2011-04-21

1 Answers 1

15

This is true by a simple (complete) induction:

Case I: $\alpha$ is limit, vacuously true (as you observed)

Case II: $\alpha=\beta+1$, then $\beta=\beta'+n$ and thus $\alpha=\beta'+(n+1)$.

This of course can be expanded to inversed ordinals as well, resulting that every order of the form $\alpha^*+\beta$ can be written as a sum of two limit ordinals (one which is the inverse of an ordinal, to be accurate) and two finite ordinals (again, one is inverse of a finite number).

  • 2
    And equally simple, the decomposition is unique.2011-04-21
  • 0
    @Andres: Yes, of course.2011-04-21
  • 0
    Thank you Asaf. I've never inducted on ordinals, so can you tell me if my understanding is correct? Case I is the base case. So assume now $\alpha$ is not a limit, and for all ordinals less than $\alpha$, the result holds. Since $\alpha$ is not a limit, it has a maximum, so $\alpha=\beta+1$ for some $\beta<\alpha$, and then Case II follows like you said.2011-04-21
  • 0
    @yunone: True. You might want to refer to this answer of mine on the most general case of induction (which is easily applied to ordinals): http://math.stackexchange.com/questions/22357/22363#223632011-04-21
  • 0
    Great, thank you for the link.2011-04-21
  • 0
    @Andres, do you mind explaining why the decomposition is unique, please? I suppose $\alpha=\beta+n=\beta'+m$. I guess it is fairly easy to see that $m=n$, for if $m, then 'chopping off' the last $m$ elements of $\beta+n$ and $\beta'+m$ would leave no maximum in the new $\beta'+m$, but would leave a maximum in $\beta+n$. I'm not sure how to say this formally. And how would one conclude $\beta=\beta'$ from that? Sorry I don't see such a simple thing.2011-04-21
  • 3
    Yunone: What you are saying is correct, and can be formalized. Or you can just argue by transfinite induction: Clearly, a limit does not have the form $\lambda+n$ for any limit $\lambda$ and $0, so the result is immediate for $\alpha$ limit. If $\alpha$ is not limit, say $\alpha=\beta+1$, and we have $\alpha=\lambda+n=\lambda'+n'$ for some $\lambda,\lambda'$ limit, $n,n'<\omega$, then $n,n'>0$ (or else $\alpha$ is limit), so $n=m+1$, $n'=m'+1$ for some $m,m'$, and then $\beta=\lambda+m=\lambda'+m'$, and the inductive assumption completes the argument.2011-04-21
  • 0
    @Andres, makes sense now, thank you.2011-04-21
  • 0
    @yunone: What book are you using to study these things? Sounds like you could use a better book on the topic :-)2011-04-21
  • 0
    @Asaf, I've been going through most of Herbert Enderton's Elements of Set Theory, I'm nearing the end where he talks about ordinals but I've been trying to branch out to Thomas Jech's Set Theory and Azriel Levy's basic set theory, since I feel they might give a better treatment. Do you have other suggestions?2011-04-22
  • 0
    @yunone: Depends on what your goals are. I studied cardinal (and ordinal) arithmetics from "Introduction to Cardinal Arithmetic" by Holz, Steffends and Weitz. I think that the first few chapters discussing these topics can be very helpful for these two topics.2011-04-22
  • 0
    Thank you Asaf, the specificity of that book sounds helpful.2011-04-22
  • 0
    @yunone: I got the book from one of my teachers who gave me a guided reading course on cardinal arithmetic which carried over to PCF theory. The former topic should be well taught in the book, the latter - as I was informed by him - can be found written better in other texts. I did not read the later parts, though and I cannot tell.2011-04-22