Involution that brings sets to disjoint sets Let $A$ be a collection of subsets of $\{1,2,\dots,n\}$ that is closed under taking subsets (that is, if $U\in A$ and $V\subseteq U$ then $V\in A$). Is there always an involution $f:A\to A$ such that $f(V)\cap V=\emptyset$ for all $V\in A$? I'm guessing yes.
Note that if $A=\mathcal P(\{1,2,\dots,n\})$, then taking the complement works. Also note that if $|A|$ is odd, we can send the empty set to itself (the empty set is disjoint from itself, isn't that weird?).
I tried working through a few small examples. I haven't found a counterexample, but I also haven't found a proof.

For any $A$ start with the smallest powerset that includes $A$ (e.g. if $A=\{\emptyset,\{1\},\{2\}\}$ start with $\mathcal P(\{1,2\})$). Use the trivial involution you suggested for this power set. Then transform this involution to a new involution by eliminating elements one by one to reach $A$ (start with bigger elements). In doing so alternate between reassigning $\emptyset$ to itself and the element that is mapped to the eliminated element. This way you can construct an involution for any $A$. You need to prove that the required conditions are preserved under these transformations. Note that each transformation is applied on the previous transformed involution.
