Can you please explain this simple task with sets to me? I'm solving some problems for practice, and I've come across a something I don't quite understand... So here's the deal:

$A = \{x \in \mathbb{N}: -1 \leq x < 2\}$
$B = \{x \in \mathbb{Z}: -10 < x \leq 0\}$
$C = \{n \in \mathbb{Z}: n = 2k + 1, k \in \mathbb{Z}\}$
a) $C \setminus(A\cap B)$
b) $(B\cup C)\setminus A$

How am I supposed to solve this when $C$ has infinite members?
 A: Let's come up with some more explicit descriptions of these sets:
$$A=\{1\}$$
$$B=\{-9,-8,-7,-6,-5,-4,-3,-2,-1,0\}$$
$$C=\{\ldots,-3,-1,1,3,5\ldots\}=\{\text{odd numbers}\}$$
First, do you understand why the above statements are true? If not I can explain in more detail.
Can you describe $A\cap B$? Then $C\setminus (A\cap B)$ consists of every element of $C$ that isn't in $A\cap B$ (hint: there is a very easy description of this set using $C$).
The second part is a bit trickier. $B\cup C$ consists of every element of either $B$ or $C$, so it is the odd numbers, together with $-8,-6,-4,-2,0$ (the odd numbers $-9,-7,-5,-3,-1$ are in both $B$ and $C$, but they don't get counted twice or anything). Then, $(B\cup C)\setminus A$ consists of every element of $B\cup C$, just throwing away anything in $A$. 
EDIT: We can describe the set explicitly as
$$(B\cup C)\setminus A=\{n\in\mathbb{Z}: n=2k+1,k\in(\mathbb{Z}\setminus\{0\})\}\cup\{-8,-6,-4,-2,0\}$$
The first set is just the odd numbers, except for 1 (which we threw out because it was in $A$), and then the second set is the even numbers we have to add in from $B$.
A: $C$ having infinitely many members should not effect your solution.
Assuming that $0\in \mathbb{N}$ it's easy to see that $C$ is simply the set of odd naturals.
You can explicitly write $A = \{-1, 0\}$ and $B = \{-9, -8, ..., 0\}$ if it makes it easier to think about.
Then, essentially, clause a translates to "What is the set of all odd naturals ($C$), excluding (\) those who are both between -1 and 0 ($A$) and ($\cap$) between -9 and 0?".
It's easy to calculate this to be $C$.
A shorter version would be to notice that, since $A\cap B \subset A$ and $A\cap C = \emptyset$ you get that result immediately...
This kind of lingual breakdown shouldn't be a part of your mathematical handling after some exercise, but it roughly equates to how you should mentally handle such questions...
