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?
 A: 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?
A: The simplest method two prove two sets are equal is to show that each one is contained in the other.
The simplest method to show that one set is contained in the other is to show that any element in the one set is also an element in the other.
Here you will want to show that
$$f(X\cup Y)\subseteq f(X)\cup f(Y)\quad\text{and}\quad f(X)\cup f(Y)\subseteq f(X\cup Y)$$
both hold.
For example, to show that $f(X\cup Y)\subseteq f(X)\cup f(Y)$, let $b\in f(X\cup Y)$. We need to show that $b\in f(X)\cup f(Y)$; that is, we need to show that either $b\in f(X)$, or $b\in f(Y)$.
Since $b\in f(X\cup Y) = \{f(a)\mid a\in X\cup Y\}$, there exists $a\in X\cup Y$ such that $b=f(a)$. Since $a\in X\cup Y$, either $a\in X$ or $a\in Y$. If $a\in X$, then $b=f(a)\in f(X)\subseteq f(X)\cup f(Y)$, and we are done. If $a\in Y$, then $b=f(a)\in f(Y)\subseteq f(X)\cup f(Y)$, and we are done. Since these are the only cases, we see that if $b\in f(X\cup Y)$, then $b\in f(X)\cup f(Y)$. This proves the first inclusion.
I'll let you work out the second inclusion.
A: I suppose the best way to do this is to show the two inclusions (namely $f(X\cup Y)\subseteq f(X)\cup f(Y)$ and $f(X)\cup f(Y)\subseteq f(X\cup Y)$), and to use definitions. For the first inclusion, this gives :
Let $x\in f(X\cup Y)$. This means that there exists a $y\in X\cup Y$ such that $f(y)=x$. Now $y\in X\cup Y$ means that either $y\in X$ or $y\in Y$. Now if $y\in X$, then $x\in f(X)$, and if $y\in Y$, then $x\in f(Y)$. In any instance, $x\in f(X)\cup f(Y)$. We proved that any $x$ in $f(X\cup Y)$ is also in $f(X)\cup f(Y)$, that is precisely $f(X\cup Y)\subseteq f(X)\cup f(Y)$. The other inclusions proceeds similarly.
