# find a condition on B,C so set difference is associative (A\B)\C=A\(B\C)

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)

• Draw some Venn diagrams and see which regions the two sets represent. – Matthew Daly Nov 9 at 14:00
• I did, still coudn't understand what I should define for B,C – Shiran Shaharabani Nov 9 at 14:02
• I think the condition you are searching for should be on A and C not on B and C . – lessili Nov 9 at 14:12
• 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. – Matthew Daly Nov 9 at 14:13
• 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 – Shiran Shaharabani Nov 9 at 14:14

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 $$C=\varnothing$$.

• 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? – Shiran Shaharabani Nov 9 at 17:30
• 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 $(2)$. – drhab Nov 9 at 18:27

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

• 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! – Shiran Shaharabani Nov 9 at 14:24