If $A, B, C$ are sets prove that $(A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)$

If $A, B, C$ are sets prove that $(A\setminus B)\setminus C\subseteq A\setminus(B\setminus C)$. Find a description of when it is that we have equality, and give an example where the inclusion is strict.

I'm not sure how to do this one I know that to prove that two sets $X$ and $Y$ are equal, we need to show that if $x$ is any element of $X$, then $x\in Y$, and that if $x$ is any element of $Y$, then $x\in X$: the first of these means that $X\subseteq Y$ and the second that $Y\subseteq X$, so the two together establish that $X=Y$. I'm not sure how to use this for a question and how to finish it.

Hint:

$(A\setminus B)\setminus C=A\setminus(B\cup C)$

$A\setminus V =A \setminus U$ iff $V\cap U\cap A=U\cap A=V\cap A$.

• What does this strict means? – Nikola Sep 13 '13 at 1:28
• U is strictly contained in V if U is contained in V and there exists $x\in V$ such that $x\notin U$. – Pocho la pantera Sep 13 '13 at 1:47

We have $(A \setminus B) \setminus C = A \setminus (B \cup C)$, and $A \setminus (B \cup C) \subseteq A \setminus (B \setminus C)$ because $B \setminus C \subseteq B \cup C$. The inclusion is strict whenever $A \cap C$ is nonempty.

A different approach would be using DeMorgans law and basic set operations

$$(A \setminus B) \setminus C = A \cap B^c \cap C^c$$

and

$$A \setminus (B \setminus C) = A \setminus (B \cap C^c) = A \cap (B^c \cup C) = (A \cap B^c) \cup (A \cap C)$$

Now it should be clear that

$$(A \setminus B) \setminus C = A \cap B^c \cap C^c \subseteq A \cap B^c \subseteq (A \cap B^c) \cup (A \cap C) = A \setminus (B \setminus C)$$