# A variant of selectivity for ultrafilters on $\omega$

Question. Does there exist a name for the the class (or a subclass) of free ultrafilters $$\mathscr{F}$$ on $$\omega$$ with the following property?

Property: There exists a partition $$\{A_n: n \in \omega\}$$ of $$\omega$$ into consecutive finite intervals such that $$\max A_n/\min A_n\to \infty$$ and, if $$\{I_k: k \in \omega\}$$ is a family of disjoint finite intervals with $$I:=\bigcup_k I_k \in \mathscr{F}$$ and each $$I_k$$ contains at least one $$A_n$$, then also $$\bigcup_{n: A_n\subseteq I} A_n \in \mathscr{F}$$.

Ps. As proved by bof below, no such ultrafilters exist. Hence I remove the tag "reference-request".

• Doesnt this trivially hold for all ultrafilters with $A_n=\{n\}$? – Andreas Lietz Jul 25 at 3:02
• Why aren't you asking "are there any ultrafilters with this property" instead of "is there a name for ultrafilters with this property"? – bof Jul 25 at 7:47
• The above property is a sufficient to imply that $\{a\in \{0,1\}^\omega: \mathscr{F}-\lim \mu_n(a)=0\}$ is not analytic, where $(\mu_n)$ is a sequence of "well-behaved" lower semicontinuous submeasures. I could prove that at least one $\mathscr{F}_0$ of this type exists. However, I was not able to show that $\mathscr{F}_0$ satisfies the above property. And your answer explains why, good job. – Paolo Leonetti Jul 25 at 8:07

I'll try to prove that no such ultrafilter can exist. Assume for a contradiction that the ultrafilter $$\mathscr F$$ and the partition $$\{A_n:n\in\omega\}$$ have the stated properties.

Choose $$N$$ so that $$\#A_n\ge2$$ for all $$n\ge N$$. For each $$n\ge N$$ choose nonempty intervals $$B_n$$ and $$C_n$$ so that $$A_n=B_n\cup C_n$$ and $$\max B_n\lt\min C_n$$.

One of the sets $$A_{N+1}\cup A_{N+3}\cup A_{N+5}\cup A_{N+7}\cup\cdots$$ and $$A_{N+2}\cup A_{N+4}\cup A_{N+6}\cup A_{N+8}\cup\cdots$$ belongs to $$\mathscr F$$; without loss of generality we may assume that $$A_{N+1}\cup A_{N+3}\cup A_{N+5}\cup A_{N+7}\cup\cdots\in\mathscr F.$$ Now one of the following cases holds:

Case 1. $$B_{N+1}\cup B_{N+3}\cup B_{N+5}\cup B_{N+7}\cup\cdots\in\mathscr F$$.

Let $$I_1=A_N\cup B_{N+1}$$, $$I_2=A_{N+2}\cup B_{N+3}$$, $$I_3=A_{N+4}\cup B_{N+5}$$, $$I_4=A_{N+6}\cup B_{N+7}$$, etc. Now

$$I=I_1\cup I_2\cup I_3\cup I_4\cup\cdots\supseteq B_{N+1}\cup B_{N+3}\cup B_{N+5}\cup B_{N+7}\cup\cdots\in\mathscr F$$, while
$$\bigcup_{n:A_n\subseteq I}A_n=A_N\cup A_{N+2}\cup A_{N+4}\cup A_{N+6}\cup\cdots\notin\mathscr F.$$

Case 2. $$C_{N+1}\cup C_{N+3}\cup C_{N+5}\cup C_{N+7}\cup\cdots\in\mathscr F$$.

Let $$I_1=C_{N+1}\cup A_{N+2}$$, $$I_2=C_{N+3}\cup A_{N+4}$$, $$I_3=C_{N+5}\cup A_{N+6}$$, $$I_4=C_{N+7}\cup A_{N+8}$$, etc. Now

$$I=I_1\cup I_2\cup I_3\cup I_4\cup\cdots\supseteq C_{N+1}\cup C_{N+3}\cup C_{N+5}\cup C_{N+7}\cup\cdots\in\mathscr F$$, while $$\bigcup_{n:A_n\subseteq I}A_n=A_{N+2}\cup A_{N+4}\cup A_{N+6}\cup A_{N+8}\cup\cdots\notin\mathscr F.$$