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

cardinalslogicset-theory

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

Best Answer

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 $\mu<\lambda$.