Does $A\setminus B = A\setminus C$ imply $B=C$? Let $A, B, C$ be sets with $B \subset  C$ and $C \subset  A$. Does $A\setminus B = A\setminus C$ imply $B=C$?
I am not sure what the "\" means, so I don't know how to solve this. 
 A: Yes it does, since, if $B\subset A$, then $A\setminus (A\setminus B)=B$. So $A\setminus B = A\setminus C$ implies $A\setminus (A\setminus B) = A\setminus (A\setminus C)$, i.e.  $\,\,B=C$.
A: We have $A\setminus B=A\setminus (B\cap A)$ and $A\setminus C=A\setminus (C\cap A)$. Given $B\subset C\subset A$.
$$\begin{align}A\setminus B&=A\setminus C\\\implies A\setminus (B\cap A)&=A\setminus (C\cap A)\\\implies B\cap A&=C\cap A\\\implies B&=C{[\text {using} B\subset C\subset A]}.\end{align}$$
A: see here, for what "∖" means. for other part: 
you have $B \subset  C$. I prove $C \subset B$  
Let $x\in C$. so $$ x\in A\ and\   x\notin (A\setminus C)$$ which means $$x\in A\ and\   x\notin (A\setminus B)$$ therefore $x\in B$
A: Without loss of generality you can assume that $A$ is the universal set. Then your assumption means $B^c=C^c$; so $B=C$
A: As a complement to good the answers already proposed here. I'd like to emphasize the fact that the assumption: $B\subset A$ and $C \subset A$ is crucial in order to conclude $B=C$.
Example: $A=\{1,2,3\}, B=\{1\},C=\{1,4\}\subset \Bbb N$ then $B \subset A$ but $\require{cancel} C \cancel{\subset} A$. Moreover, $A\setminus B = \{2,3\} = A\setminus C$, however, $B \neq C$.
