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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.

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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. – Arturo Magidin Apr 21 '11 at 19:51
@Arturo, quite right, thanks for pointing that out. – yunone Apr 21 '11 at 19:52
up vote 12 down vote accepted

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).

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And equally simple, the decomposition is unique. – Andrés E. Caicedo Apr 21 '11 at 19:59
@Andres: Yes, of course. – Asaf Karagila Apr 21 '11 at 20:00
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. – yunone Apr 21 '11 at 20:09
@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): – Asaf Karagila Apr 21 '11 at 20:12
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 $0<n<\omega$, 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. – Andrés E. Caicedo Apr 21 '11 at 21:28

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