In my research the following problem appeared (and if it is true, this solves positively several my conjectures):

Let $U$ be a fixed set (usually $U$ is infinite). Let $n$ be a fixed index set (usually $n$ is infinite).

I call a staroid such an $n$-ary relation $f$ between sets (which are subsets of some fixed set $U$) such that:

  1. $X\notin f$ if any component $X_i$ of $X$ is empty set.
  2. $\{(k,I\cup J)\} \cup L\in f\Leftrightarrow \{(k,I)\} \cup L\in f \vee \{(k,J)\} \cup L\in f$ for every index $k\in n$, sets $I,J\in\mathscr{P}U$ and an indexed family $L$ of subsets f $U$ such that $\operatorname{dom} L=n\setminus\{k\}$.
  3. If $X\in f$ and $\forall i\in n:Y_i\supseteq X_i$ then $Y\in f$ for every $n$-indexed family $Y$ of subsets of $U$.

Let $f$ be a staroid.

Let $X$ be an $n$-indexed family of subsets of $U$ and $Y\in f$. Let also $\forall L \in f, i \in n : L_i \cap X_i \ne \varnothing$.

Conjecture: $(\lambda i\in n: X_i\cap Y_i) \in f$.

Note that $(\lambda i\in n: X_i\cap Y_i) = \{ (i,X_i\cap Y_i) | i\in n \}$ (and it is a function defined on the set $n$).


The answer is no. Let $\mathscr{U}$ be an ultrafilter on $\mathbb{N}$. Let $A,B \in \mathscr{U}$ be such that $A \cap B$ is a proper subset of both $A$ and $B$. Let $\mathscr{F}$ be the collection of functions $f : \mathbb{N} \to \mathscr{U}$ such that either $f(n) \supseteq A$ for all but finitely many $n$, or $f(n) \supseteq B$ for all but finitely many $n$.

$\mathscr{F}$ satisfies your definition of staroid, and for any $f,g \in \mathscr{F}$, $f(n) \cap g(n) \not= \emptyset$ for all $n \in \mathbb{N}$. But let $f$ be the constant function with value $A$ and $g$ the constant function with value $B$. Then $\{ (n,f(n) \cap g(n)) : n \in \mathbb{N} \} \notin \mathscr{F}$.


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