7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Henning Makholm Aug 5 '16 at 19:38
  • 1
    $\begingroup$ A principal ultrafilter can be easily generated by $\kappa$ sets :-) $\endgroup$ – Henning Makholm Aug 5 '16 at 19:39
  • 2
    $\begingroup$ @Henning: Yes. $\endgroup$ – Brian M. Scott Aug 5 '16 at 19:39
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ This is Exercise III.1.36 in Kunen's new set theory book. $\endgroup$ – hot_queen Aug 6 '16 at 2:00
4
$\begingroup$

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 $L$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.