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](https://mathoverflow.net/questions/190585/expressing-a-value-related-to-an-infinitary-relation-through-ultrafilters).



Lowering from filters to ultrafilters for an infinitary relation
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](https://mathoverflow.net/questions/189424/interweaving-two-indexed-families-of-filters).
 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 $a \not\in \upuparrows f$.


