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Definition A fine measure on $P_\kappa(\lambda)$ is a non-principal ultrafilter on $P_\kappa(\lambda)$ which contains all upper cones $\uparrow{x}=\{y\in P_\kappa(\lambda)|x\subset y\}$, for all $x\in P_\kappa(\lambda)$.

Bagaria and Magidor defined a cardinal $\kappa$ to be $\aleph_1$-strongly compact if there exists a non-principal $\aleph_1$-complete fine measure on $P_\kappa(\lambda)$, for all $\lambda\ge\kappa$. This is a refinement of the usual definition of strongly compacts where $\kappa$-completeness has been relaxed to $\aleph_1$-completeness.

I want to further relax the assumption that all upper cones belong to $U$.

Definition Let $F$ be a subset of $\lambda$ (which maybe empty). An ultrafilter on $P_\kappa(\lambda)$ is $F$-fine if for every $a\in \lambda$, $\uparrow{\{a\}}\in U$ if and only if $a\in F$. If $F=\lambda$ then we recover the definition of a fine measure.

Question Assume $\kappa$ is an $\aleph_1$-strongly compact cardinal. Let $F\subset \lambda$ be of size $<\lambda$. Can we prove there is a non-principal $F$-fine $\aleph_1$-complete ultrafilter on $P_\kappa(\lambda)$?

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  • $\begingroup$ The definition of $F$-fineness seems restrictive with the "if and only if"... $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 17, 2022 at 19:19
  • $\begingroup$ @AsafKaragila Yes, it is very restrictive. We want "total control"! $\endgroup$
    – Ioannis Souldatos
    Commented Mar 18, 2022 at 6:51

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Yes, if $F$ is large enough. Let $j : V \to M$ be the embedding derived from an $\aleph_1$-strongly compact ultrafilter $U$ on $P_\kappa(\lambda)$. Let $[\mathrm{id}]$ be the set in $M$ represented by the identity function. Define $W \subseteq P(P_\kappa(\lambda))$ by: $$A \in W \Leftrightarrow [\mathrm{id}] \cap j(F) \in j(A).$$

Then $W$ is a countably complete ultrafilter. If $\alpha \in F$, then $j(\alpha) \in [\mathrm{id}] \cap j(F)$, so $[\mathrm{id}] \cap j(F) \in j({\uparrow}\{\alpha\})$. If $\alpha\notin F$, then $j(\alpha) \notin j(F)$, so $[\mathrm{id}] \cap j(F) \notin j({\uparrow}\{\alpha\})$. Thus $W$ is $F$-fine.

It remains to check that $W$ is nonprincipal. This holds when $|F| \geq \kappa$. To see this, suppose on the contrary that there is $z_0$ such that $A \in W$ iff $z_0 \in A$. Then $[\mathrm{id}] \cap j(F) = j(z_0)$. But $|[\mathrm{id}] \cap j(F)| \geq |j[\lambda] \cap j(F)| \geq j(\kappa)$, which means $|z_0| \geq \kappa$, which is false.

On the other hand, if $|F|<\mathrm{crit}(j)$, then $j(F) = j[F] = [\mathrm{id}] \cap j[F]$, so $A \in W$ iff $j(F) \in j(A)$, thus $A \in W$ iff $F \in A$. In this case, $W$ is principal.

I'm not sure what happens when $\mathrm{crit}(j) \leq |F| < \kappa$.

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    $\begingroup$ You can make $W$ nonprincipal by deriving instead from $\sigma = [\text{id}]\cap j(F)\cup \{\alpha\}$ where $\alpha < j(\lambda)$ does not belong to $j(A)$ for any $A$ with $|A| < \kappa$. The $F$-fineness argument is the same. To see $W$ is nonprincipal, we need that $\sigma\notin \text{ran}(j)$. If $|\sigma|\geq \kappa$, $\sigma\notin \text{ran}(j)$ since otherwise $|\sigma|$ would be in $\text{ran}(j)$ but no cardinal between $\kappa$ and $j(\kappa)$ is. If $|\sigma| < \kappa$, assume towards a contradiction that $\sigma = j(A)$. Then $|A| < \kappa$ but $\alpha\in j(A)$, contradiction. $\endgroup$
    – Gabe Goldberg
    Commented Mar 17, 2022 at 23:18
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    $\begingroup$ If $\kappa < \lambda$, then such an ordinal $\alpha$ exists, namely $\alpha = \sup j[\kappa^+]$. If $\kappa = \lambda$, there is an issue when $\kappa$ is singular. Then one must be more careful about the choice of $U$. But (assuming now $\kappa$ is the least $\aleph_1$-strongly compact without loss of generality) one can choose $U$ to be a uniform $\aleph_1$-complete ultrafilter on $\kappa$, viewed an ultrafilter on $P_\kappa(\kappa)$ concentrating on ordinals. Take $\alpha = [\text{id}]_U$. The existence of $U$ is not too hard to prove, although it is not immediate from the definition. $\endgroup$
    – Gabe Goldberg
    Commented Mar 17, 2022 at 23:23
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    $\begingroup$ @MonroeEskew This is equivalent to the fact that the ultrafilter on $\kappa^+$ derived from $j$ using $\sup j[\kappa^+]$ is uniform. Concretely, fix $A$ with $|A| \leq \kappa$. Note that $j(\kappa^+) > \sup j[\kappa^+]$ since [id] witnesses $\text{cf}^M(\sup j[\kappa^+]) < j(\kappa)$ whereas $j(\kappa^+)$ is regular in $M$. So to show $\sup j[\kappa^+]\notin j(A)$, it suffices to show that $\sup j[\kappa^+]\notin j(A\cap \kappa^+)$, which is obvious since $A\cap \kappa^+$ is bounded below some $\beta < \kappa^+$ and so $j(A\cap \kappa^+)\subseteq j(\beta) < \sup j[\kappa^+]$. $\endgroup$
    – Gabe Goldberg
    Commented Mar 18, 2022 at 18:29
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    $\begingroup$ I just noticed that someone downvoted this answer. I do not see a reason for this, but who downvoted can explain what is wrong with this answer. $\endgroup$
    – Ioannis Souldatos
    Commented Mar 21, 2022 at 6:29
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    $\begingroup$ I think Eskew's answer together with Goldberg's comments answer my question. I will accept. $\endgroup$
    – Ioannis Souldatos
    Commented Mar 21, 2022 at 6:30

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