Proof that $((A\setminus B)\setminus (B\setminus C))^{c} = A^{c} \cup B$

Question

I'm lost in this question, how can I deal with multiple complements and differences?

Answer 1

Well first of all, notice that there are no points in common between $A\setminus B$ and $B\setminus C$. So $$(A\setminus B) \setminus (B\setminus C)=A\setminus B \quad (*)$$

Next, notice that in general for sets, $$ X \setminus Y = X \cap Y^{c} \quad (**)$$

Finally, we have a basic fact of set theory (draw a picture and you can see it), that the compliment of the intersection is the union of the compliments: $$ (X \cap Y)^{c}=X^{c} \cup Y^{c} \quad (***)$$

So to answer your question, $$ ((A\setminus B) \setminus (B\setminus C))^{c} \overset{(*)}{=} (A\setminus B)^{c} \overset{(**)}{=} (A\cap B^{c})^{c} \overset{(***)}{=}A^{c} \cup B$$