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)
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)
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 $C = \emptyset$.
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 $C=\emptyset$.
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 $C=\varnothing$.
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