If $|X| = \kappa$, can it be the case that every ultrafilter $\mathcal{F}$ on $X$ is generated by a set $\mathcal{F}_0 \subseteq PX$ of size $<2^\kappa$? Here I say that $\mathcal{F}$ is generated by $\mathcal{F}_0$ if $\mathcal{F}$ is the smallest filter containing $\mathcal{F}_0$. Let's say that $\mathcal{F}$ is small-generated if there exists $\mathcal{F}_0$ which generates $\mathcal{F}$ and has size $|\mathcal{F}_0| <2^\kappa$. So the question is whether every ultrafilter is small-generated. I'd like to show that the answer is no. If it helps to assume that $\kappa$ is regular or something, that would be fine.

Since there are $2^{2^\kappa}$-many ultrafilters on $X$, an easy counting argument shows that not every ultrafilter is generated by a set of size $\kappa$ (there can be at most $2^\kappa$ of these). Nor can every ultrafilter be generated by a set of size $\lambda$ if $2^\lambda < 2^{2^{\kappa}}$. So under GCH, the answer is in fact no. But without GCH, we could have $\lambda < 2^\kappa$ but $2^\lambda = 2^{2^\kappa}$. So counting only gets us so far.

It's also easy to show that if $\mathcal{F}$ is not small-generated, then for any $A \subseteq X$, either the filter generated by $\mathcal{F} \cup \{A\}$ is not small-generated, or else the filter generated by $\mathcal{F} \cup \{\neg A\}$ is not small generated, where $\neg A$ is the complement of $A$. This suggests trying to build a non-small-generated ultrafilter by starting with a non-small-generated filter and expanding it inductively. But I think the union of a small set of non-small-generated filters can be small-generated, so I don't know what to do at limit steps. And anyway, I don't even know of a base case to use.

  • $\begingroup$ Right. Btw is it possible for an ultrafilter on $X$ with $|X| = \kappa$ to be generated by a set of size $\kappa$? $\endgroup$ – tcamps Aug 5 '16 at 19:36
  • $\begingroup$ It is clear that there are at most $2^{2^\kappa}$ ultrafilters, but do we know that there are necessarily that many? $\endgroup$ – hmakholm left over Monica Aug 5 '16 at 19:38
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    $\begingroup$ A principal ultrafilter can be easily generated by $\kappa$ sets :-) $\endgroup$ – hmakholm left over Monica Aug 5 '16 at 19:39
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    $\begingroup$ @Henning: Yes. $\endgroup$ – Brian M. Scott Aug 5 '16 at 19:39

Let $F \subseteq [\kappa]^{\kappa}$ be an independent family (every finite boolean combination has size $\kappa$) of size $2^{\kappa}$. Put $E = \{\kappa \setminus \bigcap A: A \in [F]^{\aleph_0}\}$. Let $U$ be an ultrafilter containing $F \cup E$. Then $U$ is not generated by fewer than $2^{\kappa}$ sets. You can easily generalize this construction to get $2^{2^{\kappa}}$ such ultrafilters.

It is, however, unknown if it is consistent to have a uniform ultrafilter on $\omega_1$ which can be generated by fewer than $2^{\omega_1}$ sets.

  • $\begingroup$ Excellent, thanks! I'll think about verifying this. Is there a canonical reference for a fact like this, or is it more folklore? $\endgroup$ – tcamps Aug 6 '16 at 1:57
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    $\begingroup$ This is Exercise III.1.36 in Kunen's new set theory book. $\endgroup$ – hot_queen Aug 6 '16 at 2:00

In fact it is a more tricky problem to get a model with a "small-generated" non-principal uf over $\omega$. See Kunen p 289 (old book) p 345 (new book), where an uf base of cardinality $\aleph_1$ in a model with $\mathfrak c >\aleph_1$ is generically defined. Or Baumgartner/Laver in Annals of Mathematical Logic 17 (1979) 271-288 where it is shown that some ufs in $L$ remain uf bases in $\omega_2$-long Sacks-forcing iterated extensions of $L$.


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