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While working on a project in operator algebras with a collaborator (and fellow MO user), we are able to successfully complete a transfinite induction assuming that the following has an affirmative answer.

Question: Let $\delta$ be a cardinal, considered as an initial ordinal, so that it is equal to the set of all ordinals of cardinality strictly less than $\delta$. Does there exist a nondecreasing function $\phi \colon \delta \to \delta$ such that, for all ordinals $1 \leq \lambda < \delta$, there exists $\gamma < \lambda$ such that $\phi(\gamma) \geq \lambda$?

If it happens to matter, we are only concerned with the case where $\delta$ is a limit cardinal. Any reasonably (transfinitely) constructive approach to writing down such $\phi$ seems to quickly run into issues of ordinal notation that are beyond our expertise. If an abstract existence argument is available, we will certainly still be happy. On the other hand, we fear that this may somehow depend on large cardinal issues.

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Your property must fail for $\lambda=0$, since there are no $\gamma$ less than $0$. – Joel David Hamkins Jun 7 '12 at 21:08
As Nik pointed out such functions exist only when $cf(\delta) = \omega$ (and trivially when $\delta$ is a successor ordinal). However, it often happens that the $cf(\delta) = \omega$ and $cf(\delta) \gt \omega$ cases must be dealt with using different techniques. Perhaps you want to follow up this question with another on how to handle the $cf(\delta) \gt \omega$ case... – François G. Dorais Jun 7 '12 at 21:27
@Joel, oops! Fixed, thank you. – Manny Reyes Jun 7 '12 at 22:57
@Francois, would you kindly be able to point me to a (reasonably simple) instance where the $cf(\delta)>\omega$ case is successfully handled by separate means? A concrete example may help jog some new ideas for me. – Manny Reyes Jun 7 '12 at 23:03
One of the key things that happens with ordinals of uncountable cofinality is the existence of the club filter and stationary sets, which have tons of wonderful properties. This is often the key to handling that case - – François G. Dorais Jun 7 '12 at 23:37
up vote 15 down vote accepted

Not unless $\delta$ has countable cofinality (e.g., $\delta = \aleph_\omega$). This will fail for $\delta = \aleph_1$, for example. Let $\phi: \delta \to \delta$ be any increasing function and recursively define $\lambda_0 = 0$ and $\lambda_{n+1} = \phi(\lambda_n)+1$. Since $\phi$ is increasing, the sequence $(\lambda_n)$ is increasing, and since $\delta$ has uncountable cofinality we have $\lambda = \sup \lambda_n < \delta$. However, for any $\gamma < \lambda$ we must have $\gamma < \lambda_n$ for some $n$, so that $\phi(\gamma) \leq \lambda_{n+1} < \lambda$.

(If $\delta$ has countable cofinality it's easy. For instance, if $\delta = \aleph_\omega$ then we define $\phi$ by letting $\phi(\lambda) = \aleph_{n+1} + \lambda$ for $\aleph_n \leq \lambda < \aleph_{n+1}$.)

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In the cofinality $\omega$ case, one can also use the natural step function with $\omega$ many values. For $\delta=\aleph_\omega$, let $f(\alpha)=|\alpha|^+$ for infinite $\alpha$. In the general case, where $\delta=\sup_n\lambda_n$, let $f(\alpha)=\lambda_{n+1}$ for $\alpha\in[\lambda_n,\lambda_{n+1})$. – Joel David Hamkins Jun 7 '12 at 22:56
Well, that was much simpler than I expected! Thanks for setting me straight. – Manny Reyes Jun 7 '12 at 23:04

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