A filter $\mathcal{F}$ over subsets of $\Omega$ is a set such that:
- $\Omega \in \mathcal{F}$.
- For every $F_1$ and $F_2$ in $\mathcal{F}$, $F_1 \cap F_2 \in \mathcal{F}$.
- 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,
- 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, $\Omega = \mathbb{R}^{\mathbb{R}}$.