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Im trying to find a some conditions on the sets $B,C$ so the following will be correct:

for all $A$, $(A\setminus B)\setminus C= A\setminus (B\setminus C)$ if and only if ** condition **

(set theory)

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  • $\begingroup$ Draw some Venn diagrams and see which regions the two sets represent. $\endgroup$ – Matthew Daly Nov 9 at 14:00
  • $\begingroup$ I did, still coudn't understand what I should define for B,C $\endgroup$ – Shiran Shaharabani Nov 9 at 14:02
  • $\begingroup$ I think the condition you are searching for should be on A and C not on B and C . $\endgroup$ – lessili Nov 9 at 14:12
  • $\begingroup$ Any region(s) where those Venn diagrams disagree must represent intersections of sets that would have to be empty in order for the "identity" to hold. $\endgroup$ – Matthew Daly Nov 9 at 14:13
  • $\begingroup$ the instruction says "find a conditions on C and B". I guess if it was on A,C I would choose to have their intersection the empty set $\endgroup$ – Shiran Shaharabani Nov 9 at 14:14
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Break things into logical statements. Note that $$ x \in (A \setminus B) \setminus C \iff (x \in A \text{ and } x \notin B) \text{ and } x \notin C\\ x \in A \setminus (B \setminus C) \iff x \in A \text{ and } x \notin B \setminus C\\ \iff x \in A \text{ and not}(x \in B \text{ and } x \notin C)\\ \iff x \in A \text{ and } (x \notin B \text{ or }x \in C) $$ In other words, we want conditions that will guarantee that $$ x \in A \text{ and } x \notin B \text{ and } x \notin C \iff x \in A \text{ and } (x \notin B \text{ or }x \in C). $$ So, the sets coincide iff given that $x \in A$, it must be the case that $x$ satisfies $$ x \notin B \text{ and } x \notin C \iff x \notin B \text{ or }x \in C. $$ Note that the statement on the left implies the statement on the right. So really, what we need is that for all $x \in A$, $$ x \notin B \text{ or }x \in C \implies x \notin B \text{ and } x \notin C. $$ In other words, the sets coincide if the possibilities $$ x \notin B \text{ and } x \in C, \quad x\in B \text{ and } x\in C $$ never occur when $x$ is in $A$. In other words, the sets coincide iff $x \notin C$ for all $x \in A$. In other words, the sets coincide iff $$ A \cap C = \emptyset. $$ This will only be true for every $A$ if $C = \emptyset$.

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Another way:

Since $(A\setminus B)\setminus C = A \setminus (B\cup C)$, the associativity equation can equivalently be written as $$A\setminus (B\cup C) = A\setminus (B\setminus C)$$ Now obviously that is true for all $A$ exactly if $$B\cup C = B\setminus C$$ But those two sets differ exactly by the elements of $C$, thus they are equal if and only if $C=\emptyset$.

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First observe that: $$\left(A\setminus B\right)\setminus C=A\cap B^{\complement}\cap C^{\complement}$$

and: $$A\setminus\left(B\setminus C\right)=A\cap\left(B\cap C^{\complement}\right)^{\complement}=A\cap\left(B^{\complement}\cup C\right)=\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)$$

So we have: $$\left(A\setminus B\right)\setminus C=A\cap B^{\complement}\cap C^{\complement}\subseteq A\cap B^{\complement}\subseteq\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)=A\setminus\left(B\setminus C\right)\tag1$$

And what we need is: $$A\cap B^{\complement}\cap C^{\complement}=A\cap B^{\complement}=\left(A\cap B^{\complement}\right)\cup\left(A\cap C\right)\tag2$$

For the second equality it is necessary and sufficient that $A\cap C=\varnothing$

If that is satisfied then $A\cap C^{\complement}=A$ so that also $A\cap B^{\complement}\cap C^{\complement}=A\cap B^{\complement}$.

So the associativity is accomplished if and only if $A\cap C=\varnothing$.

If this must be the case for every set $A$ then it is unescapable that $C=\varnothing$.

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  • $\begingroup$ Hi, can you explian the line after where you wrote "so what we need is:" ? I mean I understand the line before it but why it is a reuqirement? could it be another requirment? $\endgroup$ – Shiran Shaharabani Nov 9 at 17:30
  • $\begingroup$ Do you agree with statement $(1)$? In that statement you find two inclusion signs. Associativity will be there if LHS and RHS of $(1)$ are equal. For this the inclusion signs must become equality signs. This is what we need and it is stated in $(2)$. $\endgroup$ – drhab Nov 9 at 18:27
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If for all A , (A \ B)\ C=A \ (B \ C) , then take A=C , it gives (C \ B)\C=C \ (B\C) , thus $\emptyset=C$ is the condition you are searching for

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  • $\begingroup$ I thought about that when I draw the diagrams, can you try to give me some intuition to why would you take A=C ? thank you! $\endgroup$ – Shiran Shaharabani Nov 9 at 14:24

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