Prove that for all $U\subseteq A$ and $V\subseteq A$, that $f(U\cup V)=f(U)\cup f(V)$.

Question

Prove that for all $U\subseteq A$ and $V\subseteq A$, that $f(U\cup V)=f(U)\cup f(V)$.

Hello everyone, I am having some trouble trying to prove this problem. Could I get some hints on starting the proof? Thank you.

Answer 1

If $x \in U \cup V$, then $x \in U$ or $x \in V$. So $f(x) \in f(U)$ or $f(x) \in f(V)$, i.e., $f(x) \in f(U)\cup f(V)$, thus $f(U \cup V) \subset f(U) \cup f(V)$.
If $y \in f(U) \cup f(V)$, then $y \in f(U)$ or $y\in f(V)$. Then there is $u \in U$ s.t. $f(u)=y$, or there is $v \in V$ s.t. $f(v)=y$, i.e., there is $x \in U\cup V$ s.t. $f(x)=y$. Hence $y \in f(U \cup V)$. Thus $f(V) \cup f(U) \subset f(U \cup V)$.
Hence $f(V) \cup f(U) = f(U \cup V)$.