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I'm studying forcing from Kunen's Set Theory (ed. 1983), and I came across this lemma

Lemma 6.10. Fn$(I,J,\lambda)$ has the $\left(|J|^{<\lambda}\right)^+$-cc.

proof. [...] First assume $\lambda$ regular. Then $\left(|J|^{<\lambda}\right)^{<\lambda} = |J|^{<\lambda}$ [...]

In this lemma, we are not assuming GCH and no assumption is made on $|J|$. I tried to prove the cardinal arithmetic fact that appears in the proof, but I only succeeded in proving it in specific cases, not in generality (i.e. for all regular cardinals $\lambda$).

In fact it is trivial in case $\lambda$ is a successor cardinal, since then we have $|J|^{<\mu^+}=|J|^\mu$. It also follows if we assume that $\lambda$ is limit (hence weakly inaccessible) and $\text{cof}\left(|J|^{<\lambda}\right)\neq \lambda$, since then we'd have that the the $\lambda$-sequence $\left(|J|^\kappa\right)_{\kappa < \lambda}$ cannot be cofinal in $|J|^{<\lambda}$, hence it is eventually constant. But if we were to deal with a weakly inaccessible cardinal $\lambda$ s.t. $\text{cof}\left(|J|^{<\lambda}\right) = \lambda$, then my attempts fail.

For what I have seen afterward the problematic case does not appear since mostly we are dealing with successor cardinals or we are assuming some form of CH. But still, I wonder, how it can be proved in the general case?

Thanks

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2 Answers 2

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Note that since $\lambda$ is regular, for any $\mu<\lambda$, $f\colon\mu\to\lambda$ is bounded.

Now think about $g\in\left(\kappa^{<\lambda}\right)^{<\lambda}$ as some $g\colon\mu\to\kappa^{<\lambda}$. Then there is some $\nu<\lambda$ such that $g\colon\mu\to\kappa^\nu$. So we get the wanted result, since clearly $\left(\kappa^{<\lambda}\right)^\mu=\kappa^{<\lambda}$ for any $\mu<\lambda$.

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  • $\begingroup$ Thank you! Why do you say that $(\kappa^{<\lambda})^\mu = \kappa^{<\lambda}$? I would say, in order to finish, that $(\kappa^\nu)^{<\lambda} = \kappa^{<\lambda}$... $\endgroup$
    – Lorenzo
    Apr 28, 2020 at 14:47
  • $\begingroup$ To quote Shelah, "think!" $\endgroup$
    – Asaf Karagila ♦
    Apr 28, 2020 at 14:47
  • $\begingroup$ Let me give you a hint, $\mu<\lambda$, and $\kappa^{<\lambda}=\sup\{\kappa^\nu\mid\nu<\lambda\}$. $\endgroup$
    – Asaf Karagila ♦
    Apr 28, 2020 at 16:06
  • $\begingroup$ Ok, so I'd like to say that $(\sup\{\kappa^\nu \mid \nu < \lambda\})^\mu = \sup\{\kappa^{\nu \ \mu} \mid \nu < \lambda\} = \sup\{\kappa^\nu \mid \nu < \lambda\}$, right? So $\sup\{\kappa^\nu \mid \nu < \lambda\}$ should be a continuity point of $\alpha \mapsto \alpha^\mu$. The last question I have posted regards the continuity of such function, and what I got from the responses is that it is not obvious to check whether a certain cardinal is a continuity point or not. Probably I'm missing something trivial.. $\endgroup$
    – Lorenzo
    Apr 28, 2020 at 16:54
  • $\begingroup$ Yes, you can assume that $\nu\geq\mu$. What's $\nu\mu$? $\endgroup$
    – Asaf Karagila ♦
    Apr 28, 2020 at 17:33
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I want to add some details to Asaf's aswer and slightly modify its final argument:

Suppose $\lambda$ weakly inacessible and $\text{cof}(k^{<\lambda})=\lambda$ (the other cases are dealt in the body of the question), then, if we have $g \in \left(k^{<\lambda}\right)^{<\lambda}$ with $g:\mu \longrightarrow k^{<\lambda}$, $g$ must be bounded in $k^{<\lambda}$ (because of its cofinality),
hence $\exists \nu < \lambda$ s.t. $g: \mu \longrightarrow k^\nu$

So we have $$\left(k^{<\lambda}\right)^{<\lambda} = \left|\bigcup_{\mu,\nu<\lambda}\left(\kappa^\nu\right)^\mu\right| = \kappa^{<\lambda} $$

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