Assume that $\omega<\kappa_1<\dotsb< \kappa_n$ are infinite cardinals such that for each $1\le i\le n$ there is a $\kappa_i$-complete, $\kappa_i^+$-saturated ideal $\mathcal I_i\subset \mathcal P(\kappa_i)$. Can you obtain a ZFC model which contains $n$-many measurable cardinals? The natural candidate is $L[\mathcal I_1,\dotsc, \mathcal I_n]$.

It is well known that the answer is yes for $n=1$ (see Kunen: Some applications of iterated ultrapowers in set theory. Ann. Math. Logic 1 (1970), 179–227.)


1 Answer 1


Yes. And if the $\mathscr{I}_k$s are normal, then the suggested candidate works (I haven't really thought about whether the candidate still works without normality). Here is the argument, which is a typical one:

First, we may assume that each $\mathscr{I}_k$ is normal, by Jech Lemma 22.28.

Case 1. There is an proper class transitive inner model of ZFC with $n+1$ measurable cardinals (not just $n$).

Let $M$ be the minimal proper class transitive inner model satisfying ZFC with $n$ (not $n+1$) measurable cardinals $\mu_1<\ldots<\mu_n$, as witnessed by (unique) normal measures $U_1,\ldots,U_n$. Then (this uses the case hypothesis) $\mu_1,\ldots,\mu_n$ are countable. So we can iterate $M$ out at its measurables (using the $U_k$s and their images), eventually sending each $\mu_k$ to $\kappa_k$. Let $\mathcal{T}$ be this iteration (consisting of the sequence of iterates $M^{\mathcal{T}}_\alpha$, iteration maps $i_{\alpha\beta}^{\mathcal{T}}$, etc), and $M'=M^{\mathcal{T}}_{\kappa_n}$ be the final iterate and $i:M\to M'$ the iteration map. Then I claim that $i(U_k)\subseteq\mathscr{F}_k$, the filter dual to $\mathscr{I}_k$. For let $j:V\to M$ be a generic embedding given by $\mathscr{I}_k$. So $j(\mathcal{T})$ is an iteration with last model $j(M')$, and $j(\mathcal{T})\upharpoonright(\kappa_k+1)=\mathcal{T}\upharpoonright(\kappa_k+1)$, but the $\kappa_k$th measure used in $j(\mathcal{T})$ is $i(U_k)$, and $j(\mathcal{T})$ eventually sends $\kappa_k$ further out to $j(\kappa_k)$. Now standard calculations show that the iteration map $i^{\mathcal{j(\mathcal{T})}}_{\kappa_kj(\kappa_k)}$ agrees with $j$ over $\mathcal{P}(\kappa_k)\cap M^{\mathcal{T}}_{\kappa_k}$, which implies that $j(U_k)\subseteq G$, the generic filter, and since this is independent of $G$, therefore $j(U_k)\subseteq\mathscr{F}_k$, as desired.

Since $i(U_k)$ is an ultrafilter in $M'$ and $i(U_k)\subseteq\mathscr{F}_k$, it follows that $$L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[i(U_1),\ldots,i(U_n)]\subseteq M',$$ and that $i(U_k)=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$; let $\bar{\mathscr{F}}_k$ denote this measure. Therefore $L[\mathscr{F}_1,\ldots,\mathscr{F}_k]\models$"$\bar{\mathscr{F}_k}$ is a $\kappa_k$-complete normal measure on $\kappa_k$". Of course $L[\mathscr{I}_1,\ldots,\mathscr{I}_n]=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, so we are done. (Using the minimality of $M$, can also show that $M'=L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$.)

Case 2: Otherwise.

Then the core model $K$ exists. Let $j:V\to M$ be a generic embedding given by $\mathscr{I}_k$. Then (using core model theory for this level) $j(K)$ is an iterate of $K$ and $j\upharpoonright K$ is the iteration map. But $\mathrm{crit}(j)=\kappa_k$. Therefore $\kappa_k$ is measurable in $K$ (as witnessed by its extender sequence), and because $j\upharpoonright K$ is the iteration map, letting $D_k$ be the unique normal measure on $\kappa_k$ in $K$ (uniqueness because otherwise we get a measure of Mitchell order 1), we have $D_k\subseteq\mathscr{F}_k$. It follows that $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]=L[D_1,\ldots,D_n]\subseteq K$, and that letting $\bar{\mathscr{F}}_k=\mathscr{F}_k\cap L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, then $\bar{\mathscr{F}}_k$ is a normal, $\kappa_k$-complete measure on $\kappa_k$ in $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$, as desired. (With the case hypotheses as they are, it might be that $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]\subsetneq K$. I could have made the case hypothesis to case 1 be that the sharp for an inner model with $n$ measurables exists, and case 2 its negation. Then we would get that $K$ and $L[\mathscr{F}_1,\ldots,\mathscr{F}_n]$ have the same universe.)

  • $\begingroup$ Thank you very much. I should admit that I can not follow all the technival details, but i am convinced that your solution is correct. $\endgroup$ Mar 14, 2022 at 20:07

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