# Given $\lambda$ regular cardinal, $\left(\kappa^{<\lambda}\right)^{<\lambda}=\kappa^{<\lambda}$?

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

• Use \left(\right) when you have some more complex terms, like fractions, powers, etc., so that all can fit inside the parentheses. (: – Invisible Apr 28 '20 at 11:06
• @ms._VerkhovtsevaKatya Thanks! – Lorenzo Apr 28 '20 at 11:07

## 2 Answers

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 $$\mu<\lambda$$.

• Thank you! Why do you say that $(\kappa^{<\lambda})^\mu = \kappa^{<\lambda}$? I would say, in order to finish, that $(\kappa^\nu)^{<\lambda} = \kappa^{<\lambda}$... – Lorenzo Apr 28 '20 at 14:47
• To quote Shelah, "think!" – Asaf Karagila Apr 28 '20 at 14:47
• Let me give you a hint, $\mu<\lambda$, and $\kappa^{<\lambda}=\sup\{\kappa^\nu\mid\nu<\lambda\}$. – Asaf Karagila Apr 28 '20 at 16:06
• 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.. – Lorenzo Apr 28 '20 at 16:54
• Yes, you can assume that $\nu\geq\mu$. What's $\nu\mu$? – Asaf Karagila Apr 28 '20 at 17:33

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