A filter $\mathcal{F}$ over subsets of $\Omega$ is a set such that:

  1. $\Omega \in \mathcal{F}$.
  2. For every $F_1$ and $F_2$ in $\mathcal{F}$, $F_1 \cap F_2 \in \mathcal{F}$.
  3. If $F_1 \in \mathcal{F}$ and $F_1 \subset F_0 \in \Omega$, then $F_0 \in \mathcal{F}$.

Furthermore, $\mathcal{F}$ is closed under $\lambda$-intersection if,

  1. If $(F_i)_{i \in I}$ is such that $|I| \leq \lambda$ and for every $i \in I, F_i \in \mathcal{F}$, then $\cap_{i \in I}{F_i} \in \mathcal{F}$.

It is possible to show that, if $\mathcal{F}$ is a filter over subsets of $\Omega$ such that $\mathcal{F}$ is closed under $|\Omega|$-intersections, then $\mathcal{F}$ is a principal filter, that is, there exists $R \subset \Omega$, such that $F \in \mathcal{F}$ if and only if $R \subset F$. My question:

Do filters closed under $\lambda$-intersection, where $\lambda$ is an arbitrary infinite cardinal, admit similar characterizations as filters closed under $|\Omega|$-intersections? If $|\Omega|$ is an accessible cardinal, are filters closed by $\lambda$-intersections essentially a principal filter?

Disclaimer: This question might be related to this other one that I asked about ultrafilters: Existence of non-trivial ultrafilter closed under countable intersection . If this is the case, it is ok to assume that $|\Omega|$ is small. For example, $\Omega = \mathbb{R}^{\mathbb{R}}$.

  • $\begingroup$ The term is principal filter, not trivial. $\endgroup$
    – Asaf Karagila
    Dec 5, 2017 at 20:19
  • 1
    $\begingroup$ Either there is a typo in your penultimate paragraph or I don't understand the question. However, ignoring that, if you ask whether $\lambda$-closed filters must essentially be trivial, the answer is 'no'. See for example measurable cardinals. $\endgroup$ Dec 5, 2017 at 20:52
  • $\begingroup$ I changed the name from trivial to principal. Also, I clarified that I am interested in the case in which the cardinal is not measurable. $\endgroup$
    – madprob
    Dec 6, 2017 at 1:02

1 Answer 1


Let $\Omega$ be an uncountable set and let $\mathcal F$ be the filter consisting of those subsets of $\Omega$ whose complement is countable. Then $\mathcal F$ is closed under $\aleph_0$-intersection and is not principal.

You can get lots of similar examples: For any set $\Omega$ and any cardinal $\lambda<|\Omega|$, the filter of subsets of $\Omega$ whose complements have cardinality at most $\lambda$ is closed under $\lambda$-intersection and not principal.


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