Frankl for infinite set

Question

Let $A$ be a set and $F\subset \mathcal P(A)$, and for any $a\in A$, let's define $F_a:= \left\{X\in F, a\in X\right\}$

Suppose that :

1. $F$ is not a finite union of chains

2. for any $E\subset F$, $\bigcap E\in F$

Does there necessarilly exists $a\in A$ such that for any $a\in A$, $card(F_a)\leq card(F-F_a)$

Answer 1

Let $(F_n,<_n)_{n\in \mathbb N}$ be a sequence of disjoint finite lattices .

We consider the lattice $(\bigcup F_i,<<)$ where we keep the original computation on each $F_i$ (for any interger $i$, and any $x<_iy$ we say $x<<y$) and such that for any $i<j$ and any $(x_i,x_j)\in F_i\times F_j$, $x_i<<x_j$

For any element $x$ of the lattice, we considere $M(x):=\left\{y\in \bigcup F_i, y<<x\right\}$

Then $\left\{M(x), x\in \bigcup F_i\right\}$ does the job as soon as $i\mapsto |F_i|$ is unbounded and well chosen (for example one can take for the $F_n$ distributive lattices that cardinality increase with $n$).

Answer 2

Let $(A_i)_{i\in \mathbb N}$ be a sequence of disjoint finite set and for any $i\in\mathbb N$ let $F_i\subset \mathcal P(A_i)$ closed under intersection.

For any $i>0$ we then define $F^*_i:=\bigcup_{j<i} A_j\cup F_i$, and we state $F_0=:F^*_0$

Then $\left\{F^*_i, i\in \mathbb N\right\}$ is intersection closed on $A:=\bigcup A_i$ and does not satisfies the condition as soon as the $F_i$ are well choosen.

For example, for all $i$, $card (A_i)=i+1$ and the element of $F_i$ are $n$ distinct singleton and the emptyset.