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Let $\kappa$ be an uncountable cardinal and $(P(\kappa),\cap,\cup, ^c,\kappa,\emptyset)$ the Boolean Algebra of all subsets of $\kappa$.

Fact: If there exists a countably complete non-principal ultrafilter on $P(\kappa)$, then $\kappa$ is larger than or equal to a measurable.

My question: Is there a Boolean Algebra $\mathcal{A}$ of subsets of $\kappa$, $\mathcal{A}$ of size strictly less than $2^\kappa$, and the existence of a countably complete non-principal ultrafilter on $\mathcal{A}$ would imply that $\kappa$ is (added: larger than or equal to) a measurable/inaccessible/strong limit?

I am interested even in special cases, e.g. $\kappa$ is regular.

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  • $\begingroup$ But then I wonder whether you are really looking for an equivalence ... for example, if $\mathcal{A}$ is any complete atomic Boolean algebra of subsets of $\mathcal{A}$ then your (new) conclusion follows. $\endgroup$ – Nik Weaver Jan 10 '18 at 12:50
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Your question is very interesting in the case $\kappa^+<2^\kappa$.

For this case, recall that a cardinal $\kappa$ is weakly measurable if every family of $\kappa^+$ many subsets of $\kappa$ admits a $\kappa$-complete nonprincipal filter measuring them. This notion was introduced and studied by my student Jason Schanker in his dissertation. There are a variety of equivalent characterizations.

Jason proved that if $2^\kappa$ is larger than $\kappa^+$, then this concept is not necessarily equivalent to measurability, for he constructed models of ZFC with a weakly measurable cardinal, but no measurable cardinal.

In such a model, therefore, where $2^\kappa=\kappa^{++}$, we have measures for all the families of sets of size smaller than $2^\kappa$, but $\kappa$ is not measurable. So this provides a negative answer to your question.

Meanwhile, weakly measurable cardinals are necessarily weakly compact, inaccessible and much more, so there is a positive answer to the version of your question at the end where you ask for less than measurability.

Indeed, the existence of $\kappa$-complete nonprincipal filters measuring families of $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact. In this sense, weak measurability is a generalization of weak compactness from $\kappa$ to $\kappa^+$.

Thus, if $2^\kappa=\kappa^+$, the answer to your question is no, since the existence of such filters for families of at most $\kappa$ many subsets of $\kappa$ is equivalent to $\kappa$ being weakly compact, which is a strictly weaker large cardinal concept than measurability.

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  • $\begingroup$ I'm confused by the second paragraph --- does "family" mean "Boolean subalgebra"? If not, what do you mean by a filter on a family? $\endgroup$ – Nik Weaver Jan 10 '18 at 14:46
  • $\begingroup$ I mean a filter that measures every set in the family---one can without loss move to the Boolean algebra generated by the family. $\endgroup$ – Joel David Hamkins Jan 10 '18 at 14:48
  • $\begingroup$ So we're talking about filters on the Boolean algebra generated by the family? Or maybe the complete Boolean algebra, otherwise I am not sure how to interpret "$\kappa$-complete"? $\endgroup$ – Nik Weaver Jan 10 '18 at 14:56
  • $\begingroup$ It is a filter on $\kappa$, which is to say, a collection of subsets of $\kappa$, and it is $\kappa$-complete as a filter on $\kappa$, and it happens to measure every set in the family, which means for every set in the family, either it or its complement contains a set in the filter. So it isn't necessarily an ultrafilter on $\kappa$, but it is ultra with respect to the family. $\endgroup$ – Joel David Hamkins Jan 10 '18 at 14:58
  • $\begingroup$ If $(\kappa^+)^{<\kappa}=\kappa^+$, one can consider the $\kappa$-complete Boolean algebra generated by the family, and it will also have size at most $\kappa^+$. And then, you can view the filter as a filter on this algebra. $\endgroup$ – Joel David Hamkins Jan 10 '18 at 15:07

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