Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
show 3 more comments

5 Answers

up vote 1 down vote accepted

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 $3 \not \in A \setminus C$.

share|improve this answer
add comment

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).

share|improve this answer
add comment

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

share|improve this answer
add comment

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 $$(A\setminus B) \cup (B\setminus C) =A\cup B \ne A\setminus C=A.$$

share|improve this answer
add comment

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)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.