(Joint question with Piotr Szewczak.)

Definitions and notation. By filter we mean a filter on $\omega$ containing the cofinite sets at least.

For a filter $\mathcal{F}$, let $\mathcal{F}^+:=\{A\subseteq\omega : A^c\notin \mathcal{F}\}$.

For an infinite set $A\subseteq\omega$ and a natural number $n$, define $\mathrm{next}(A,n)=\min\{k\in A:n<k\}$.

For natural numbers $m<n$, let $[m,n]:=\{m,m+1,\dots,n\}$.

The question. Which filters $\mathcal{F}$ we have the following property?

  • For each $A\in \mathcal{F}^+$ and all $B\in \mathcal{F}$, the set $C := \bigcup\{[n,\mathrm{next}(A,n)] : n \in A\cap B\}$ is in $\mathcal{F}$.

The Frechet filter (of all cofinite sets), and every ultrafilter, have this proeprty.

In particular, are there concrete examples of filters that are not Frechet and not uf (and better not the above under a finite-to-one map) but have this property?

Update. Blass and Brian provide below examples that are ultrafilters under finite-to-one maps. Our research developed in the meanwhile to see that we need examples that cannot be mapped to the Frechet filter or an ultrafilter by a finite-to-one function. Consistently, as Blass points out, there are no such examples (Filter Dichotomy), so the question is, for example, what can be said if CH holds? More concretely, how prevalent are such examples under CH? Are there "important" examples (i.e., not ones cooked up by transfinite induction for the purpose of the question).

Any other axioms relevant for a positive answer?

  • $\begingroup$ Does "not the above under a finite-to-one map" mean that the filter you seek should not have the Frechet filter or an ultrafilter as a finite-to-one image? So, in particular, you would require Filter Dichotomy to fail? $\endgroup$ Mar 26, 2015 at 23:49
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    $\begingroup$ You probably already have examples of filters lacking your property, but, just in case, here's one: For $i=0,1,2$, let $A_i=\{n\in\omega:n\equiv i\pmod3\}$. Let $\mathcal U$ and $\mathcal V$ be non-principal ultrafilters containing $A_0$ and $A_2$, respectively, and let $\mathcal F=\mathcal U\cap\mathcal V$, so $\mathcal F^+=\mathcal U\cup\mathcal V$. I claim this $\mathcal F$ does not have the property you described. The counterexample is $A=A_0\cup A_1$ and $B=A_0\cup A_2$. Then $A\cap B=A_0$ and $\bigcup\{[n,\text{next}(A,n)]:n\in A\cap B\}=A_0\cup A_1\notin\mathcal F$. $\endgroup$ Mar 27, 2015 at 0:56
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    $\begingroup$ If $L$ is a minimal left ideal of $(\omega^*,+)$ and if $\mathcal F$ is the Stone dual of $L$, then $\mathcal F$ has the property you describe. $\mathcal F$ is neither an ultrafilter nor the cofinite filter, but it can be sent to an ultrafilter under a finite-to-one map. So I don't know if this is really the kind of answer you're looking for. If you want to see the details, let me know and I'll be happy to write them out. $\endgroup$
    – Will Brian
    May 18, 2015 at 19:24

1 Answer 1


The filters that you're looking for don't exist. Every filter with your property, other than the Frechet filter, maps to an ultrafilter via a finite-to-one map. Let's call the filters with your property Szewczak-Tsaban filters, or S-T filters for short.

Theorem: If $\mathcal F$ is an S-T filter other than the Frechet filter, then there is a finite-to-one map that sends $\mathcal F$ to an ultrafilter.

Lemma: Suppose $\mathcal F$ is an S-T filter and $B \in \mathcal F$. If $\mathcal F$ is not an ultrafilter, then $B-1 \in \mathcal F$.

Proof: Suppose $B \in \mathcal F$ and $B-1 \notin \mathcal F$, and suppose $\mathcal F$ is not an ultrafilter. We will show that $\mathcal F$ is not S-T. To do this, notice that $B \setminus (B-1) \in \mathcal F^+$. (In plain English, $B \setminus (B-1)$ is the set of right-hand endpoints of maximal subintervals of $B$). Using that $\mathcal F$ is not an ultrafilter, we can find some $A_0 \subseteq B \setminus (B-1)$ with $A_0 \in \mathcal F^+$ but $A_0 \notin \mathcal F$. Let $$A = A_0 \cup (A_0+1).$$ Now look at the definition of an S-T filter using these particular values of $A$ and $B$. The set $C$ that we get is precisely equal to $A_0$, which is not in $\mathcal F$ by design. Thus $\mathcal F$ is not S-T. QED(lemma).

Proof of theorem: Fix a non-principal filter $\mathcal F$ other than the Frechet filter, and assume that $\mathcal F$ does not map to an ultrafilter by a finite-to-one map. We'll show $\mathcal F$ is not S-T. Because $\mathcal F$ is not Frechet, there is some infinite $D \subseteq \omega$ such that $\omega \setminus D \in \mathcal F$.

Partition $\omega$ into intervals as follows: if $n \in D$, then $[n,n]$ is in our partition, and we partition $\omega \setminus D$ into maximal intervals (which are all finite because $D$ is infinite). Let $\{I_n : n \in \omega\}$ be an enumeration of this partition. Let $E_0$ be the set of $n$ such that $I_n$ is a maximal interval of $\omega \setminus D$. Notice that $\bigcup_{n \in E_0}I_n = \omega \setminus D \in \mathcal F$.

Consider the finite-to-one map induced by this partition (i.e., the map that sends each element of $I_n$ to $n$). By assumption, this map does not send $\mathcal F$ to an ultrafilter. Therefore there is some $E \subseteq E_0$ such that $\bigcup_{n \in E_0}I_n$ is in $\mathcal F^+$ but not $\mathcal F$.

Let $A = \bigcup_{n \in E}I_n$ and let $$B = \left(\omega \setminus D\right) \cap \left(\left(\omega \setminus D\right)-1\right).$$ By our choice of $E$, we have $A \in \mathcal F^+$. By our lemma and our choice of $D$, we have $B \in \mathcal F$. Now plug this $A$ and $B$ into your definition of an S-T filter. You'll find that what pops out is precisely the set $A$. But by design, we have $A \notin \mathcal F$, so $\mathcal F$ is not S-T. QED


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