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I'm not a set-theorist, but I hope this question is appropriate. This is just a question about names:

Fix a cardinal $\lambda$. I'd like to know if there is a name for regular cardinals $\kappa$ such that for all $\mu <\kappa, \mu^\lambda < \kappa$. For now, let me call them $\lambda$-good (I hope that's not the official name)

I'm also wondering if there is a reference for the following elementary fact:

Fact: For any $\kappa_0$, there exists a $\lambda$-good cardinal $\kappa \geq \kappa_0$.

I believe I know how to prove this (just pick $\kappa = (2^{\max(\lambda, \kappa_0)})^+$) but I would like to know a reference for this.

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  • $\begingroup$ When $\lambda<\kappa$, there is a related notion of $\lambda^+$-closed cardinal that perhaps fits you (at least it is implied by what you want). It's Definition 2.2 i) in the paper "Internal sizes for $\mu$-abstract elementary classes" by Lieberman-Rosický-Vasey. For the second fact I would just quickly mention the reason you gave instead of a reference, since it seems to be a kind of a folklore fact. $\endgroup$
    – godelian
    Commented Mar 13 at 15:34
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    $\begingroup$ In fact $\lambda$-good and $\lambda^+$-closed coincide (when $\lambda<\kappa$, which it has to be if $\mu>1$). $\endgroup$
    – godelian
    Commented Mar 13 at 15:50
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    $\begingroup$ Godelian : if you write this as an answer I can accept it :) $\endgroup$
    – Maxime Ramzi
    Commented Mar 14 at 13:00
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    $\begingroup$ I usually just write out the statement, $(\forall\alpha<\kappa)\alpha^\lambda<\kappa$. $\endgroup$
    – Monroe Eskew
    Commented Mar 14 at 14:02

2 Answers 2

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As suggested by the Maxime, I am turning my comments into an answer.

The notion of $\lambda$-good cardinal has appeared in the literature under a different terminology and in an equivalent form. In their paper "Internal sizes for $\mu$-abstract elementary classes", Lieberman, Rosický and Vasey introduce the Definition 2.2, whose first item defines a $\lambda^+$-closed cardinal $\kappa>\lambda$ as a a cardinal such that for each $\mu<\kappa$ one has:$$\mu^{<\lambda^+}:=\Sigma_{\theta<\lambda^+}\mu^{\theta}<\kappa$$. Clearly, a $\lambda^+$-closed cardinal is $\lambda$-good, but also, if $\kappa$ is $\lambda$-good, since $\lambda<\kappa$ we necessarily have: $$\mu^{<\lambda^+} \leq \lambda \mu^{\lambda}=\max(\lambda, \mu^{\lambda})<\kappa$$

and so $\kappa$ is $\lambda^+$-closed. So the two notions are equivalent, as one can always assume $\lambda<\kappa$ in the definition of a $\lambda$-good cardinal $\kappa$, as one is essentially interested in infinite cardinals.

The second fact is likely folklore, but since it is just an easy computation one could just include it in a parenthetical remark. The cited paper in fact discusses $\mu$-closed cardinals in the context of the Singular Cardinal Hypothesis and weakenings of it, and contains similar folklore remarks like e.g. the first two sentences of Remark 2.3.

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  • $\begingroup$ In general, $\mu^{<\lambda^+}=\mu^\lambda$ whenever $\lambda$ is non-zero. $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 14 at 14:33
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    $\begingroup$ Thanks ! I like "closed" a bit better than "strong" especially in the context I want to use this for :) $\endgroup$
    – Maxime Ramzi
    Commented Mar 14 at 15:55
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In their book, "Introduction to Cardinal Arithmetic", Holz, Steffens, and Weitz define (on p.71 of the second edition) as follows.

Assume that $\kappa$ is an infinite and $\lambda$ is an uncountable cardinal number. We say that $\lambda$ is $\kappa$-strong iff $\rho^\kappa<\lambda$ for every cardinal number $\rho$ less than $\lambda$.

The term "strong" is also appearing in the large cardinal hierarchy and can be a bit confusing in that sense. But it is certainly a term that appears in a relevant book and some people might use.

Note that this property does not imply regularity. Namely, $\lambda$ (in the above definition) need not be regular. Indeed, $\lambda$ is a strong limit cardinal if it is $\kappa$-strong for all $\kappa<\lambda$ (and the books defines exactly that in the same paragraph). And it is easy to see that being a strong limit is equivalent to $2^\kappa<\lambda$ for all $\kappa<\lambda$. And $\sf ZFC$ proves that there is a proper class of singular strong limit cardinals (and it does not prove that there are any regular strong limit cardinals, of course).

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  • $\begingroup$ (I've marked this as CW since it's really just a glorified comment. But the book is worth your effort in searching for these kind of propositions that you're asking about.) $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 14 at 14:42
  • $\begingroup$ Thanks ! As I said below godelian's answer, I think the terminology "closed" will be more convenient for my purposes so I'll stick to that, especially if "strong" can be confusing. And indeed, it does not imply regularity but again for my purposes I need to restrict to regular cardinals $\endgroup$
    – Maxime Ramzi
    Commented Mar 14 at 15:56

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