Lowering from filters to ultrafilters for an infinitary relation

Question

Let $U$ be a set. Let $N$ be a (possibly infinite) index set. Let $f$ be an $N$-ary relation on $U$ (that is $f$ is a set of functions $N\rightarrow U$).

I denote $\mathcal{L}\in \upuparrows f \Leftrightarrow \forall L\in\prod_{i\in N}\mathcal{L}_i: f \cap \prod_{i\in N} L_i\ne\emptyset$ for every $N$-indexed family $\mathcal{L}$ of filters on $U$.

Does $\mathcal{L}\in \upuparrows f$ imply existence of $a\in \prod_{i\in N} \operatorname{atoms} \mathcal{L}_i$ (where $\operatorname{atoms}\mathcal{X}$ is the set of ultrafilters over filter $\mathcal{X}$) such that $\forall A\in\prod_{i\in N} a_i: f\cap \prod_{i\in N} A_i\ne\emptyset$?

This question is probably related with my other question.

Answer 1

Counter-example:

Example There exists such an (infinite) set $N$ and $N$-ary relation $f$ that $\mathcal{P} \in \upuparrows f$ but there are no indexed family $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ of atomic filter such that $\forall A \in \operatorname{up} a : f \cap \prod A \ne \emptyset$.

Proof Take $\mathcal{L}_0$, $\mathcal{L}_1$ and $f$ from the the answer to this question. Take $\mathcal{P} = \operatorname{up}\mathcal{L}_0 \cap \operatorname{up}\mathcal{L}_1$. If $a \in \prod_{i \in N} \operatorname{atoms} \mathcal{P}_i$ then there exists $c \in \{ 0, 1 \}^N$ such that $a_i \sqsubseteq \mathcal{L}_{c (i)} (i)$ (because $\mathcal{L}_{c (i)} (i) \ne 0$). Then from that example it follows that $(\lambda i \in N : \mathcal{L}_{c (i)} (i)) \not\in \upuparrows f$ and thus $a \not\in \upuparrows f$.