If $C\subseteq B \subseteq A$ then $A\backslash C = (A\backslash B)\cup (B\backslash C)$ Is it true that if $C\subseteq B \subseteq A$ then $A\backslash C = (A\backslash B)\cup (B\backslash C)$? By drawing Venn diagrams this clearly seems so, but I can't seem to prove it.
Similarly, is it true that if $C\subseteq A$ and $C\subseteq B$ then $A\backslash C \subseteq(A\Delta B)\cup(B\backslash C)$?
If they are, what is the proof? Otherwise, what is a counterexample?
 A: Since $B \subseteq A$, then $A = B \cup ( A \setminus B )$, and since $C \subseteq B$ then $B \cup C = B$.  We then have that:
$$\begin{align}
A \setminus C
&= ( B \cup ( A \setminus B ) ) \setminus C \\
&= ( B \setminus C ) \cup ( ( A \setminus B ) \setminus C )\\
&= ( B \setminus C ) \cup ( A \setminus ( B \cup C ) ) \\
&= ( B \setminus C ) \cup ( A \setminus B )
\end{align}$$
The above relies on two easily proved facts:

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


$A \setminus C \subseteq (A \mathop{\triangle} B) \cup (B \setminus C)$ regardless of the relationships between $A , B , C$.  If $x \in A \setminus C$, then $x \in A$ and $x \notin C$.  Note that either $x \in B$ (whence $x \in B \setminus C$) or $x \notin B$ (whence $x \in A \mathop{\triangle} B$).  In either case, $x \in ( A \mathop{\triangle} B ) \cup ( B \setminus C )$.
A: Hints: 


*

*$X \subseteq Y$ is equivalent to $Y^c \subseteq X^c$.

*$X \setminus Y = X \cap Y^c$.

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


Good luck ;-)
A: First off, is there a way you can pull out $A$ from $(A \backslash B)\cup (A\backslash C)$? In other words, can you prove $(A\backslash B)\cup (A\backslash C)=A\backslash(B \cap C)$ (See DeMorgan's Law http://en.wikipedia.org/wiki/De_Morgan's_laws)? Then since $B\cap C=C$, that will give you $\supseteq$. 
A: Let $x\in A-C$. Then $x\in A$ and $x\not\in C$. If $x\in B$, then $x\in B$ and $x\not\in C$, so $x\in B-C$, while if $x\not\in B$, then $x\in A$ and $x\not\in B$, so $x\in A-B$. This proves that $A-C\subset (A-B)\cup (B-C)$.
Now suppose that $C\subset B\subset A$. Let $y\in (A-B)\cup (B-C)$. If $y\in A-B$, then $y\in A$ and $y\not\in B$, so $y\in A$ and $y\not\in C$, so $y\in A-C$, while if $y\in B-C$, then $y\in B$ and $y\not\in C$, so $y\in A$ and $y\not\in C$, so $y\in A-C$. This proves that $(A-B)\cup (B-C)\subset A-C$.
