Is it provable, in ZF (without Choice), that every filter can be extended to one of cardinality continuum?
The extended filter is not requested to be an ultrafilter.
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$\begingroup$ Filter on $\omega$ or filter in general? $\endgroup$– Asaf Karagila ♦Commented Jul 15, 2016 at 8:49
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1 Answer
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If we talk about filters on $\omega$, then the answer is an easy yes.
If $\cal F$ is a filter on $\omega$ such that there is at least one $A\in\cal F$ which is co-infinite, then for every $B\subseteq\omega\setminus A$ we have that $A\cup B\in\cal F$.
But every filter can be extended (perhaps trivially) to include at least one co-infinite set. Otherwise all the co-infinite sets are in the co-ideal, which means that the filter is empty.
If we are talking about general filters, then the answer is negative even with choice, since taking any uniform filter on a large enough set gives a counterexample. Or filters on finite sets.
If we are talking about general filters, but you mean "filters on infinite sets" and "at least continuum", then the answer is consistently negative in the absence of choice. If $A$ is a set with a Dedekind-finite power set, then any filter on $A$ cannot have cardinality greater or equal to the continuum. Such examples include amorphous sets.
answered Jul 15, 2016 at 8:55
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$\begingroup$ Oddly enough, I was thinking about the same question just two days ago. $\endgroup$– Asaf Karagila ♦Commented Jul 15, 2016 at 9:02
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$\begingroup$ Blind spot on my side. Thanks for pointing this out. We indeed cared about filters on $\mathbb{N}$ as you wisely guessed. But I leave the question as is since you answered both possible interpretations. In short, your answer is that you can add all supersets of some coinfinite set (and you can add any coinfinite set you wish if all sets are cofinite). $\endgroup$ Commented Jul 15, 2016 at 9:27
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$\begingroup$ We also arrived at this just now, in a seminar, and this was off the course of the seminar, just a curious question arising from a comment of Piotr Szweczak on selection principles. I never thought about math without choice before, so the coincidence is amazing. $\endgroup$ Commented Jul 15, 2016 at 9:30
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$\begingroup$ Well, I didn't really think about it in the context of choiceless mathematics. Which makes the whole story even weirder now. $\endgroup$– Asaf Karagila ♦Commented Jul 15, 2016 at 9:32