functions - Prove $f(S cup T) = f(S) cup f(T)$

$f(S \cup T) = f(S) \cup f(T)$

$f(S)$ encompasses all $x$ that is in $S$
$f(T)$ encompasses all $x$ that is in $T$

Thus the domain being the same, both the LHS and RHS map to the same $y$, since the function $f$ is the same for both.

Can you post the solution?

Answer

$$y\in f(S\cup T)\Longrightarrow \exists\,x\in S\cup T\,\,s.t.\,\,f(x)=y$$

and now:

$$x\in S\Longrightarrow\,y=f(x)\in f(S)\;\;;\;\;x\in T\Longrightarrow\,y=f(x)\in T$$

so that anyway $\,y=f(x)\in f(S)\cup f(T)\,\Longrightarrow f(S\cup T)\subset f(S)\cup f(T)$

Now you try to do the other way around: $\,f(S)\cup f(T)\subset f(S\cup T)$