Given $f : X {\to} Y$; $ C, D \subset X$. Is it true that $f(A-B) = f(C)-f(D)$. Prove it if so or give an example if its false I can't seem to prove this. I have tried many ways but haven't come close to finishing it. I know that $C-D$ is everything in $A$ that isn't in $B$. The function is what is confusing me, as im not entirely sure the difference between $f(C-D)$ and $f(C)-f(D)$. Any clarification would help.
 A: Regarding your question: presumably, $A - B$ is the elements of $A$ that do not appear in $B$.  Thus, we have
$$
\begin{align*}
f(A - B) &= \{y: y = f(x) \text{ for some } x \text{ for which } x \in A \text{ and } x \notin B\},\\
f(A) - f(B) &= \{y: y = f(x) \text{ for some } x \in A\}
- \{y: y = f(x) \text{ for some } x \in B\}
\\
&= \{y: y = f(x) \text{ for some } x \in A, \text{ but } y \text{ is not equal to }f(x) \text{ for any } x \in B\}.
\end{align*}
$$
With this in mind,
Counterexample (following rogerl's hint): If $X = \{1,2\}$ (or is any set that contains at least two elements) and $Y = \{1\}$ (or any set containing $1$), then the statement will not be true for the function defined by $f(x) = 1$ for all $x \in X$.
Consider in particular what happens when $A = \{1,2\}$ and $B = \{1\}$.
A: Just like in elementary algebra, you evaluate expressions like this by starting with the innermost sub-expressions and working your way outward. 
So, $f(A-B)$ is evaluated by starting with the set $A$ and the set $B$, then removing from $A$ all of its elements that are also in $B$ to get the set $A-B$, then taking the image of the set $A-B$ under the function $f$.
On the other hand $f(A) - f(B)$ is evaluated like this. Start with the set $A$, then taking its image under the function $f$ to get the set $f(A)$, and then set that aside for the moment. Next, take the set $B$, and take its image under the function $f$ to get the set $f(B)$. Finally, take the set $f(A)$, and remove from it all of its elements that are also in the set $f(B)$, to get the set $f(A)-f(B)$.
You could produce your own counterexample by starting with a function $f$ that is not one-to-one, and taking $A$ and $B$ to be disjoint subsets of the domain, but doing this carefully so that there is an element $a \in A$ and $b \in B$ such that $f(a)=f(b)$.
