## set problem

I have the case

$
we\ have\ a\ function\ g:\textsl{P(Q)}\rightarrow\textsl{P(Q)}$

$
With\ \forall A,B \in \textsl{P(Q)}, A\subset B\ implies\ g\left(A\right) \subset g\left(B\right)$

$
prove\ that\ \exists \ an\ A\in {P(Q)}\ such\ that\ g\left(A\right)=A.$

$
Now\ show\ also\ that\ \underline{Q} = \cap A \subset A
$

So P(Q) is the power set of Q and Q underline is the fixed point of Q.
But my main question is; If I take infinite intersections of sets do I get the Null set?