# Prove question $(A\setminus B) \cup (B\setminus C) = A\setminus C$ , $(A\setminus B)\setminus C= A\setminus(B\cup C)$

I want to prove the following statements but for do it I need some hint.

\begin{align} \tag{1} (A\setminus B) \cup (B\setminus C) &= A\setminus C\\ \tag{2} (A\setminus B)\setminus C&= A\setminus(B\cup C) \end{align} Thanks!

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Geometrical hint: Draw the sets $A, B$ and $C$. Identify what the left-hand side and right-hand side of the equation corresponds to on your drawing. Confirm that they are equal. Try to figure out why they are equal no matter how the sets intersect. –  Arthur Jul 6 '13 at 12:47
MathJax/TeX tip: \setminus creates $\setminus$. –  Lord_Farin Jul 6 '13 at 12:49
$1)$ is incorrect, I guess: $A=[0,1]$, $B=[0.5,1]$, $C=[0,0.5]$ gives : $(A\backslash B)\bigcup(B\backslash C)=[0,1]$ while $A\backslash C=[0.5,1]$. –  zuggg Jul 6 '13 at 12:50
if I draw it its ok but there is way to write it? not by drawing –  Ofir Attia Jul 6 '13 at 12:51
Axiomatic hint: Try to put into symbols what it means that $x$ is an element in the left-hand side and of the right-hand side of the equation. See if you can logically make each of them imply the other. –  Arthur Jul 6 '13 at 12:51

For the first one, suppose that $(A \setminus B) \cup (B \setminus C)$ is not empty. Take any $x \in (A \setminus B) \cup (B \setminus C)$. Then either $x \in A \setminus B$ or $x \in B \setminus C$. Note that in this particular case, both cannot be true (why?). If $x \in A \setminus B$, then $x \in A$ and $x \not \in B$. If $x \in B \setminus C$, then $x \in B$ and $x \not \in C$. This does not imply that $x \in A \setminus C$. If $x \in A \setminus B$, one of the possibilities above, then this does not give us any information about whether $x \in C$.

For example, suppose $A = \{1,2,3\},\ B = \{1,2\}$, and $C = \{3\}$. Then $3 \in A \setminus B$ and so $3 \in (A \setminus B) \cup (B \setminus C)$, but $A \setminus C = \{1,2\}$ and so $3 \not \in A \setminus C$.

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Recall that $x\in A\setminus B\iff x\in A\wedge x\notin B.$

1)Let $x\in A\setminus C$ so $x\in A\wedge x\notin C$ then there's two cases

• if $x\in B$ then $x\in B\setminus C$
• if $x\notin B$ then $x\in A\setminus B$ hence we find $$A\setminus C\subset (A\setminus B)\cup (B\setminus C)$$

The second inclusion is false: an element in $B$ isn't necessary in $A$.

Do the same method to show 2)

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The first one is not true if you don't require something extra. Consider the case where $A\cap B=\varnothing$, but neither $A$ nor $B$ is empty; and $C=\varnothing$.

For the second one, pick $x\in(A\setminus B)\setminus C$. Then $x\notin C$ and $x\in A\setminus B$, by the definition of the set. It follows that $x\notin C$ and $x\notin B$ and $x\in A$. Continue with the manipulation until you prove that $x\in A\setminus(B\cup C)$. Then prove the other inclusion in a similar fashion (by going in reverse, for example).

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Use the fact that $A\setminus B=A\cap \overline{B}$

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We write $$x\in (A\backslash B)\backslash C\iff x\in(A\backslash B)\& x\notin C\iff (x\in A\&x\notin B)\&x\notin C$$ $$\iff x\in A\&x\notin B \&x\notin C\iff x\in A\&(x\notin B \&x\notin C )\iff x\in A\&(x\notin B\cup C )$$ $$\iff x\in A\backslash(B\cup C).$$

The first one is false; let's take $B\ne \emptyset$, $A\cap B=\emptyset$, $A\cap C=\emptyset$, $B\cap C=\emptyset$, then $$(A\setminus B) \cup (B\setminus C) =A\cup B \ne A\setminus C=A.$$

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