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".

  • $\begingroup$ Doesnt this trivially hold for all ultrafilters with $A_n=\{n\}$? $\endgroup$ – Andreas Lietz Jul 25 '19 at 3:02
  • $\begingroup$ Why aren't you asking "are there any ultrafilters with this property" instead of "is there a name for ultrafilters with this property"? $\endgroup$ – bof Jul 25 '19 at 7:47
  • $\begingroup$ 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. $\endgroup$ – Paolo Leonetti Jul 25 '19 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 $$\bigcup_{n:A_n\subseteq I}A_n=A_{N+2}\cup A_{N+4}\cup A_{N+6}\cup A_{N+8}\cup\cdots\notin\mathscr F.$$


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