Unions and Functions on Sets Given these conditions, I seek a proof.
Let $f: A \rightarrow B$ be a function, and let $X$ and $Y$ be subsets of $A$.
Prove that $f(X \cup Y) = f(X) \cup f(Y)$.
I can't seem to figure it out. It appears obvious, but materializing a proof is troubling me. What is the best method of proof for this?

To get started, write out the definitions of $f(X\cup Y)$, $f(X)$, and $f(Y)$:
$$f(X\cup Y)=\{b\in B\mid \text{there exists an }a\in X\cup Y\text{ such that }b=f(a)\}$$
$$f(X)=\{b\in B\mid \text{there exists an }a\in X\text{ such that }b=f(a)\}$$
$$f(Y)=\{b\in B\mid  \text{there exists an }a\in Y\text{ such that }b=f(a)\}$$
Also, remember that for two sets $C$ and $D$, 
$$z\in C\cup D\iff z\in C\;\text{ or }\,z\in D.$$
Do you see how to go the rest of the way?
