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Given that A, B and C are sets, and that (A ∪ B) \ C ⊆ A \ B
(A cup B) minus C (is contained in) A minus B

Prove that (A \ C) ∩ B = ∅
(A minus C) cap B = Empty set

I tried to prove it this way:

It is given that the containing set (A \ 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 (equivelent to (A \ 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!

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(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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I'm not certain what you mean by "containing set" in the first line of your proof. –  Arthur Fischer Nov 11 '12 at 15:02
    
I mean A \ B. (the left part of the given expression) –  Mark Segal Nov 11 '12 at 15:07
    
If you are not sure about the correctness of your terminology write it in Hebrew and I'll translate. –  Asaf Karagila Nov 11 '12 at 15:08
    
@Asaf Kargila - I just did. Thanks! –  Mark Segal Nov 11 '12 at 15:16

1 Answer 1

up vote 3 down vote accepted

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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+1, quickie!! ;-) –  amWhy Nov 11 '12 at 15:25
    
@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.) –  Asaf Karagila Nov 11 '12 at 15:28
    
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. –  Mark Segal Nov 11 '12 at 15:34

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