MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


I am reading Rosenstein's "Linear Orderings" and I am not sure if I am missing something, or if there is an error.

He gives the definition of a $\beta$-limit ordinal inductively, as follows (his exact wording is given):

  1. $L_0 = $ all ordinals
  2. $\alpha\in L_{\beta+1}$ if and only if $\alpha=\lim \{\alpha_m | m<\omega\}$ where each $\alpha_m\in L_\beta$
  3. If $\lambda$ is a limit ordinal, $\alpha\in L_\lambda$ if and only if $\alpha\in L_\beta$ for all $\beta<\lambda$

He then remarks that the 1-limit ordinals (for example) are the limit ordinals, the 2-limit ordinals are precisely the limit ordinals which are limits of limit ordinals, and so on.

But, what about $\omega_1$? It has uncountable cofinality, so no strictly increasing $\omega$-sequence of ordinals will limit to it. So $\omega_1\not\in L_1$, unless we all constant sequences, in which case $L_\alpha$ is the class of all ordinals, for all $\alpha$.

One obvious modification is to allow any cofinality of a sequence in the definition (2). However, the chapter is especially concerning the finite condensation, which naturally collapses $\omega$-sequences. So, I'm not sure if this is the right answer, or if he really only means ordinals with countable cofinality, or what.

Unfortunately, I have no idea if this terminology is standard, or just useful for this book, so I don't know if people here will have an answer.

To summarize: either the definition above, or the comments proceeding it, or both, are wrong. I would like to know the correct definition, if it is known.Blockquote

share|cite|improve this question
Another issue: you haven't ruled out $\alpha_m=\alpha$ in item 2. But clearly allowing that would trivialize the hierarchy, since one could use constant sequences. – Joel David Hamkins Oct 17 '11 at 14:14
I don't have the book in front of me, but it seems likely that only countable ordinals are being considered. But are there any particular problems with the subsequent arguments if $\omega_1$ is not in $L_1$? – Clinton Conley Oct 17 '11 at 15:59
up vote 4 down vote accepted

Not only the remark right after the definition, but also Exercise 5.2 ($\alpha$ is a $\beta$-limit ordinal iff the last exponent in the Cantor normal form of $\alpha$ is at least $\beta$) suggest that Rosenstein meant to allow any cofinality, as you suggested. This definition seems very natural, but I do not know if it is used elsewhere.

It seems to me that Rosenstein does not use the notion of $\beta$-limit points a lot; the last usage seems to be in Proposition 5.7, which among other things states:

  • If $x$ and $y$ are distinct $\beta$-limits of a well-order, then they will not be identified by a $\beta$-fold application of the condensation operator. (The condensation operator maps an order to a quotient, identifying two points if they span a finite interval.)

This theorem is true with "your" definition of $\beta$-limits; it is of course also true (but weaker) with Rosenstein's definition read literally, since in this case ordinals of uncountable cofinality are not $\beta$-limits.

share|cite|improve this answer

If one replaces the cofinality $\omega$ requirement by an arbitrary cofinality (and if one also insists that $\alpha_m\lt\alpha$, as seems intended), then the construction is the same as that of the Cantor–Bendixson derivative. This is an idea that applies in any topological space, and ordinals are topological spaces under the order topology.

One begins with $X_0$ as the whole space, and then defines $X_{\alpha+1}$ to be all the limit points of $X_\alpha$, and for limit ordinals $\lambda$, one defines $X_\lambda=\bigcap_{\alpha\lt\lambda}X_\alpha$. Thus, the process proceeds by throwing out all the isolated points, and then throwing out the isolated points among the remaining points, and so on, iterating transfinitely, taking intersections at limit stages. The Cantor–Bendixson rank of a point $x$ is the first ordinal stage $\alpha$ at which it becomes isolated among the remaining points at that stage; in other words, the least $\alpha$, such that $x\notin L_{\alpha+1}$.

If the initial space is all the ordinals, then the Cantor–Bendixson construction is identical to the $L_\beta$ construction (corrected as indicated above). In particular, the $\beta$-limit ordinals (as corrected) are precisely the ordinals having Cantor–Bendixson rank at least $\beta$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.