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I have really no idea how to prove this question, any help?

Question: Show that $X\backslash Y=X\cap Y^c$ (where the complement may be taken with respect to $X\cup Y:$ that is, you may take the universal set $U$ to be $X\cup Y$).

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    $\begingroup$ Perhaps working it out element-wise would make things easier? $\endgroup$ – user66354 Mar 22 '13 at 6:17
  • $\begingroup$ I have always taken the expression on the right to be the definition of the notation on the left. $\endgroup$ – Sammy Black Mar 22 '13 at 6:23
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    $\begingroup$ By the way, Isn't it better to replace number-theory tag with set theory? $\endgroup$ – Zeta.Investigator Mar 22 '13 at 6:26
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To prove an equality of sets we must show that each set contains the other. Begin by taking $a\in X\backslash Y$. Clearly we have $a\in X$ and the definition of set difference implies that $a\not\in Y$. However, this implies that $a\in U\backslash Y$ and $a \in Y^c$ by the definition of the set compliment. So $a\in X$ and $a \in Y^c$, thus we have $a\in X\cap Y^c$. This shows that $X\backslash Y\subseteq X\cap Y^c$.

Next, take $b\in X\cap Y^c$. So we have $b\in X$ and $b\in Y^c$. Since $b\in Y^c$ we know that $b\in U\backslash Y=(X\cup Y)\backslash Y$. Hence $b\in X\cup Y$ and $b\not\in Y$ so $b\in X\backslash Y$. So we have $X\cap Y^c\subseteq X\backslash Y$ and also $X\cap Y^c= X\backslash Y$.

Really all we need is that $z\not\in Z\Leftrightarrow z\in Z^c$.

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