Let $U$ be a set. I denote $\mathfrak{A}$ the lattice of filters on $U$ ordered *reverse* to set theoretic inclusion of filters. I denote $\bigvee$ and $\bigwedge$ correspondingly the supremum and infimum on $\mathfrak{A}$.


Let $N$ be a (possibly infinite) index set. Let $f$ be an $N$-ary relation.


I denote $\langle f\rangle\_k L = \{ x \mid L\cup\{(k,x)\}\in f \}$ for $k\in N$.


I denote $\langle \upuparrows f\rangle\_k \mathcal{L} = \bigwedge\_{L\in\prod \mathcal{L}} \uparrow \langle f\rangle\_k L$ for every $(N\setminus\{k\})$-indexed family $\mathcal{L}$ of filters on $U$ (here $\uparrow X$ is the principal filter corresponding to $X$).


**Conjecture** $$\langle \upuparrows f\rangle\_k \mathcal{L} = \bigvee\_{a\in \prod\_{i\in N\setminus\{k\}} \operatorname{atoms} \mathcal{L}\_i} \langle \upuparrows f\rangle\_k a,$$


where $\operatorname{atoms} \mathcal{X}$ is the set of ultrafilters over $\mathcal{X}$.



Expressing a value related to an infinitary relation through ultrafilters
There is a counter-example againt my conjecture.


I will denote $a\not\asymp b$ iff there is a non-least element which is below both $a$ and $b$.


Take $\mathcal{P} \in F$ from [the previous counter-example](https://mathoverflow.net/a/190653/4086). We have
$$ \forall a \in \prod\_{i \in \operatorname{dom} F} \operatorname{atoms} \mathcal{P}\_i : a
 \notin \mathcal{P} . $$
Take $k = 1$.


Let $\mathcal{L} = \mathcal{P} |\_{(\operatorname{dom} F) \setminus \{ k \}}$. Then $a
\notin \mathrel{\upuparrows F}$ and thus $a\_k \asymp \langle
\upuparrows F \rangle\_k a|\_{(\operatorname{dom} F) \setminus \{ k \}}$.


Consequently $\mathcal{P}\_k \asymp \langle \upuparrows F \rangle\_k
a|\_{(\operatorname{dom} F) \setminus \{ k \}}$ and thus $\mathcal{P}\_k \asymp
\bigvee^{\mathfrak{F}}\_{a \in \prod\_{i \in (\operatorname{dom} F) \setminus \{ k \}}
\operatorname{atoms} \mathcal{L\_{}}\_i} \langle \upuparrows F \rangle\_k a$
because $\mathcal{P}\_k$ is principal.


But $\mathcal{P}\_k \not\asymp \langle \upuparrows F \rangle\_k 
\mathcal{L}$. Thus follows $\langle \upuparrows F \rangle\_k 
\mathcal{L} \neq \bigvee^{\mathfrak{F}}\_{a \in \prod\_{i \in (\operatorname{dom} F)
\setminus \{ k \}} \operatorname{atoms} \mathcal{L}\_i} \langle \upuparrows
F \rangle\_k a$.


