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(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ between subsets of $U$ such that for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)

(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ between subsets of $U$ such that

1. for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

2. for every $k \in I$, and family $L = L_{i \in I}$ we have $L_k=\emptyset \Rightarrow \neg f (L)$.

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)

(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ such that for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)

(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ between subsets of $U$ such that for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)

(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ such that for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)

Addition: I think that the following condition may be necessary: $f(\{(i;A_i\cup B_i) | i\in I\}) \Leftrightarrow f(\{(i;A_i) | i\in I\}) \vee f(\{(i;B_i) | i\in I\})$ for each families $A=A_{i\in I}$ and $B=B_{i\in I}$ of subsets of $U$.

(This problem appeared in face of me trying to generalize my theory of (binary) funcoids to the theory of $n$-ary funcoids (I call them "multifuncoids") for arbitrary $n$.)

Let $I$ is some indexing set.

By filters I will mean (not necessarily proper) filters on some fixed set $U$.

I will call a multifuncoid a $I$-ary relation $f$ such that for every $k \in I$, subsets $A$ and $B$ of $U$, and family $L = L_{i \in I \setminus \{ k \}}$ of subsets of $U$ we have $$ f ( L \cup \{ (k ; A \cup B) \} ) \Leftrightarrow f ( L \cup \{ (k ; A) \} ) \vee f ( L \cup \{ (k ; B) \} ) . $$

Let $a = a_{i \in I}$ is some family of filters.

I will call funcoidal product $\prod a$ of a family $a = a_{i \in I}$ of filters an $I$-ary relation between subsets of $U$ such that for every family $R = R_{i \in I}$ of sets we have $$ \left( \prod a \right) R \Leftrightarrow \forall i \in I \forall A \in a_i : A \cap R_i \neq \emptyset . $$ It simple to show that funcoidal product is a multifuncoid.

Conjecture For every non-empty multifuncoid $f$ there exist a family $a = a_{i \in I}$ of ultrafilters such that $f \supseteq \prod a$.

If this conjecture is false, under which additional conditions it will be true? (I know that it is true for finite set $I$, but am interested also in the infinite case.)