Proving $A\subset B\implies (C\backslash B)\subset(C\backslash A)$ I am attempting to prove the set inclusion above, my construction is:
If $A\subset C\implies A\cup B=B$, so set $C\backslash B = C\backslash(A\cup B)$ and $C\backslash A = C\backslash(A\cap B)$, so set $C\backslash B$ is a smaller set which doesn't include set $B$ while set $C\backslash A = C\cup(B\backslash A)$, hence $(C\backslash B)\subset(C\backslash A)$.
Is this solution correct? It seems a little shabby to me but im not sure about other ways of proving it.
Thanks!
 A: The standard way of proving an inclusion like $S\subset T$ is to prove that every element of $S$ is also an element of $T$.  It can be easier than trying to manipulate the sets themselves, as you seem to be discovering.  Plus, checking your work is much less of a headache.
Here's how that proof would look.
Let $x\in C\setminus B$ be given.  Therefore, $x\in C$ and $x\notin B$.  Since we know that $A\subset B$, it follows that $x\notin A$ as well.  Therefore, $x\in C\setminus A$.  Since $x$ was an arbitrary member of $C\setminus B$, we conclude that $(C\setminus B)\subset (C\setminus A$).
A: $A \subset B$ implies every element that is in $A$ is in $B$.
Contrapositively that means every element not in $B$ is not in $A$.
$C\setminus B$ are precisely the elements that are in $C$ but not $B$, so these elements are among the elements that are in $C$ and not in $A$, or in other words among the elements of $C\setminus A$.  
So $C\setminus B\subset C\setminus A$.
A: If you intend the symbol '$\subset$' to denote the 'strictly proper' subset relation (i.e every element of $A$ is in $B$ but there are elements of $B$ not in $A$), then your conclusion is false.  Consider the simple case of $C$ = $A$; the two differences are both empty, hence equal.  More generally, $C$ may contain elements from $A$ together with elements in neither $A$ nor $B$.
If you intend the symbol to allow set equality, then the conclusion is true (and @fleablood has given a proof).
