Consider the following statement (P):



> 
> For every subset $\emptyset \not = A \subset P(\omega)$, there exists a subset $B\subset A$ such that $B \subset A$ such that :
> 
> 
> 1) $B$ is countable (i.e. is either finite or has cardinal $\aleph\_0$)
> 
> 
> 2) $\bigcap B = \bigcap A$
> 
> 
> 3) $\bigcup B = \bigcup A$
> 
> 
> 


Is (P) true? Asaf Karagila have given me [an answer on MSE](https://math.stackexchange.com/questions/2713031/countable-and-uncountable-subset-of-p-omega/2713111#2713111), namely that (P) is true under the Axiom of Choice (AC).


My question is : is (P) true regardless of AC ?!


If not could you please post the best way to use AC in this question ?



countable and uncountable subset of $P(\omega)$
It is relatively consistent with ZF that this is impossible. 


Suppose that there is an infinite Dedekind finite set $A\subset P(\omega)$. Since ZF proves that the continuum is bijective with the set of branches through the binary tree $2^{<\omega}$, we may assume by labeling the nodes of this tree that 
no finite subset of $A$ has the same union as $A$. But since $A$ is Dedekind finite, it has no countably infinite subsets. So in this case, there is no countable $B\subset A$ with the same union as $A$. So it is a counterexample to your requested property. 


So you cannot prove the existence of $B$ in ZF.


