Let $A, B ,C$ be sets. Prove that $(A-B) - C = (A - C) - (B - C)$

Question

Prove that

1. $(A-B) - C = (A - C) - (B - C)$
2. $(A \cup B) \cap C \subseteq A \cup (B \cap C)$

For the first part, I tried picking a general object in the set $(A - B) - C$, showed that it's an element of $A$ and not an element of $B$ and not an element of $C$. I didn't know how to carry on though to show that $(A-B) - C$ is a subset of $(A-C) - (B-C)$.

Answer 1

So, these seem to be pretty standard exercises in basic set theory. A standard method for showing containment $A\subseteq B$ is to take an $x \in A$ and show $x\in B$ by restating your statement in the metatheory (i.e. $x\in A\cup B$ becomes $x$ is in $A$ or $x$ is in $B$).

To show $A=B$ when dealing with sets, we show $A\subseteq B$ and $B\subseteq A$.

Answer 2

1. \begin{align} (A−C)−(B-C)&=(A\cap C^c)\cap (B\cap C^c)^c \\ &=(A\cap C^c)\cap (B^c\cup C) \\ &=((A\cap C^c)\cap B^c)\cup ((A\cap C^c)\cap C) \\ &=((A\cap B^c)\cap C^c)\cup (A\cap (C^c\cap C)) \\ &=((A\cap B^c)\cap C^c)\cup (A\cap \varnothing) \\ &=((A-B)- C)\cup \varnothing \\ &=(A-B)- C \end{align} 2. $$ (A∪B)∩C=(A\cap C)\cup (B\cap C)\subset A\cup (B\cap C) $$ for $A\cap C\subset A$.