Let $\mathscr{A} $ be a set of sets. Let's denote $\{A 
\setminus B : A,B \in \mathscr{A}\}$ by $\mathscr{A} \setminus \mathscr{A} $. 


The *Marica-Schönheim* theorem says that $|\mathscr{A} \setminus \mathscr{A}| \geq |\mathscr{A}|$ for every **finite** $\mathscr{A}$. 


This immediately implies the result for countably-infinite $\mathscr{A}$, since if we had $|\mathscr{A} \setminus \mathscr{A}|=n $ is finite, then taking a subset of size $n+1$ out of $\mathscr{A}$ gives a contradiction.


There seems to be no natural one-to-one mapping $\mathscr{A} \to \mathscr{A} \setminus \mathscr{A} $, so this raises the question: 



> 
> Do we have $|\mathscr{A} \setminus \mathscr{A}| \geq |\mathscr{A}|$ for families $\mathscr{A}$ of arbitrary cardinality? 
> 
> 
> 


I [asked](https://math.stackexchange.com/questions/1835849/every-family-mathscra-of-sets-satisfies-mathscra-setminus-mathscra) it on math.SE a while ago.



For every family $\mathscr A$ of sets, there are at least $|\mathscr{A}| $ sets of the form $A_1 \setminus A_2$
The answer is yes, because from $A$ and $A-B$ and $B-A$, you can reconstruct $B$ via $$B=(B-A)\cup(A-(A-B)).$$ So if we fix $A$, we get a map from $(\mathscr{A}\setminus\mathscr{A})^2$ onto $\mathscr{A}$. So $\mathscr{A}\setminus\mathscr{A}$ must be at least as large as $\mathscr{A}$. 


(Note that this argument used $\kappa^2=\kappa$ for infinite cardinals, which requires AC.)


