# Is it true that, if $A\setminus B \subseteq C$, then $A\setminus C \subseteq B$?

Prove or provide counterexample for :

If $A\setminus B \subseteq C$ then $A\setminus C \subseteq B$

My approach was, supposed $A\setminus B \subseteq C$, then let $x \in A\setminus B$, since $A\setminus B \subseteq C$, then $x \in C$. From there on I don't know what to do, please point me in the right direction. Thanks

• I mean no disrespect by this, but this website is not a place where people will do your homework for you. Please edit the question to show your own attempt at the problem and where you get lost/what's confusing to you. There will be no judgements for not knowing how to do something but plenty of judgment for not trying! – user52969 May 31 '16 at 21:53
• @user52969 You are absolutely correct, OP could have at least dignified the question with whether or not they thought it was true or not! – Inazuma May 31 '16 at 21:56
• Yeah sorry, this is my first time posting a question. I'll keep what you guys said in mind. – Iamlegend1996 May 31 '16 at 22:00
• One approach to problems where we try to show the roles of two terms can be swapped, as is the case here with $B,C$, is to look for a way to rewrite the condition in a way that makes the symmetry of their roles evident. – hardmath May 31 '16 at 23:30
• Did any of the answers help, or are you still having trouble with this? If the former, $\color{green}\checkmark$ the answer. If the later, edit the question to include further thoughts. – Graham Kemp Jun 2 '16 at 1:29

S'pose $A/B \subset C$.
Let $x \in A/C$. Then $x \not \in C$. Then $x \not \in A/B \subset C$. But $x \in A \subset A/C$ while $x \not \in A/B \subset A$. So $x \not: \not \in B$. So $x \in B$. So $A/C \subset B$.
$$x\in A\setminus C\implies x\notin C\;\text{but}\;x\in A\implies x\notin A\setminus B\;\text{but}\;x\in A\implies x\in A\setminus B$$
Begin: $A\setminus B\subseteq C$ means that any element which is in $A$ is either in $B$ or else it is in $C$.   That is $A\subseteq B\cup C$.
$$A\setminus B\subseteq C ~\implies~ A\subseteq B\cup C$$