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I have just read that a selective ultrafilter must necessary be an ultrafilter. Is this also true for q-filters?

Im not sure if using CH for instance, we can follow the construction of a q-point but this time by fixing a set and ensuring neither the fixed set nor its complement should be in the filter.

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2 Answers 2

Yes, exactly what you suggest works. Constructing by transfinite induction an increasing family of countably generated filters $F_\xi$ such that the evens and the odds are both $F_\xi$ positive for all $\xi$ you are confronted at stage $\xi$ with a partition $P$ into finite sets. Choosing by finite induction a selector for $P$ you notice that given any infinite set $X$ it is possible to extend your partial selector to intersect $X$ (this is not possible in the selective construcution). So you choose your selector meeting all sets in $F_\xi$ and their intersections with either the evens or the odds and let $F_{\xi+1}$ be generated by $F_\xi$ and this new selector.

I think you can also produce a counterexample from the assumption that there are two non RK equivalent selectives, simply by putting one of these selectives on the evens and the other on the odds.

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Another way to get a Q-filter that is not an ultrafilter is to destroy a Q-point using an $\omega^\omega$-bounding forcing, e.g., Grigorieff forcing with a selective ultrafilter.

Grigorieff forcing destroys the selective ultrafilter, but because Grigorieff forcing with a P-filter is $\omega^\omega$-bounding, the (formerly selective) filter will remain a Q-filter.

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