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Given that $A$, $B$ and $C$ are sets, and that $(A ∪ B) \setminus C ⊆ A \setminus B$.

Prove that $(A \setminus C) ∩ B = ∅$.

I tried to prove it this way:

It is given that the containing set $(A \setminus B)$ doesn't contain any $x \in B$ (by the definition of set difference). Therefore, no $x \in B$ & $x \notin C$ exists. Therefore, in no way there is $x \in B$ & $x \notin C$ & $x \in A$ exists (equivalent to $(A \setminus C) ∩ B = ∅ )$.

I'm afraid my proof is incorrect, because $x \notin A$ seems unneccesary.
I'm a new student in the university in Israel, learning parallel to my highschool studies. English isn't my mother tongue. I am sorry for any mistakes and my bad formatting.
Thank you!

http://www.freeimagehosting.net/newuploads/i378c.png


(Added by A.K. translation of the Hebrew in the image)

By the assumption every $x$ which belongs to $A$ or $B$, and does not belong to $C$ is necessarily an element of $A$ and not an element of $B$ (by the definitions of inclusion, union and difference). In the right hand side ($A\setminus B$) no element belongs to $B$, therefore it is impossible that in the left hand side there is an element which belongs to $B$. Therefore there is no $x\in B$ and $x\notin C$. In particular there is no such $x$ for which $$x\in A\land x\in B\land x\notin C$$ Q.E.D

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  • $\begingroup$ I'm not certain what you mean by "containing set" in the first line of your proof. $\endgroup$ – user642796 Nov 11 '12 at 15:02
  • $\begingroup$ I mean A \ B. (the left part of the given expression) $\endgroup$ – Mark Segal Nov 11 '12 at 15:07
  • $\begingroup$ If you are not sure about the correctness of your terminology write it in Hebrew and I'll translate. $\endgroup$ – Asaf Karagila Nov 11 '12 at 15:08
  • $\begingroup$ @Asaf Kargila - I just did. Thanks! $\endgroup$ – Mark Segal Nov 11 '12 at 15:16
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You have to prove that $A\setminus C$ is disjoint from $B$ (i.e. their intersection is empty), when you are given that $(A\cup B)\setminus C\subseteq A\setminus B$.

We need to show that if $x\in A\setminus C$ then $x\notin B$, and if $x\in B$ then $x\notin A\setminus C$. If we have shown both these things then there is no $x$ which is an element of both $A\setminus C$ and $B$, therefore $(A\setminus C)\cap B=\varnothing$.

Given $x\in B$ we know that $x\notin A\setminus B$ by the definition of $\setminus$. In particular this means that $x\notin (A\cup B)\setminus C$ (because the assumption was that $(A\cup B)\setminus C\subseteq A\setminus B$).

Therefore either $x\notin A\cup B$ or $x\in C$. Since we assume $x\in B$ we have to have that $x\in C$ as well. Therefore $x\notin A\setminus C$.

On the other hand, if $x\in A\setminus C$ we have that $x\in A$ and $x\notin C$. In particular this means that $x\in A\cup B$ and $x\notin C$ and therefore $x\in(A\cup B)\setminus C$. The assumption tells us, again, that $x\in A\setminus B$. Therefore $x\notin B$.

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  • $\begingroup$ @amWhy: Thanks. We just finished covering this in our course... (which also happens to be in a university in Israel, but we didn't give this question in any of the sheets.) $\endgroup$ – Asaf Karagila Nov 11 '12 at 15:28
  • $\begingroup$ Thank you very much!!! Great answer. Thanks for the translation also. Took me a minute to grasp the English, but I finally understood the solution. $\endgroup$ – Mark Segal Nov 11 '12 at 15:34

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