I have to prove that for my discrete class, and I know it's true, but I'm stuck. I stopped to see my professor but I am still confused. Any help is greatly appreciated.

I started the proof going left to right:

pf: Let x

f(S U T). Then f(x)

S U T and there exists a z

S U T such that f(z) = x

$\displaystyle \color{red}\mbox{This doesn't make sense}:$

$\displaystyle \color{red}x \in f(S \cup T) \Longrightarrow \exists y \in S \cup T\,\,s.t.\,\, x = f(y).\,If \,\,y \in S \Longrightarrow $ $\displaystyle \color{red}x \in f(S)\,,\,\,if\,\,y \in T \Longrightarrow x \in f(t)$ $\displaystyle \color{red}\,,\,so\,\,anyway\,\,x \in f(s) \cup f(T)$

Since z

S U T then z

S or z

T.

Assume z

S. Since f(z) = x, then x

f(S) and x

f(S) U f(T).

Assume z

T. Since f(z) = x, then x

f(T) and x

f(S) U f(T).

Therefore, since x

f(S) U f(T), and f(S U T) is a subset of f(S) U f(T).

I know I need to go back the other way, and as long as the part shown is correct then the other way is too. Any criticism is welcome, or tell me where I went wrong, etc.

Thanks again in advance for any help!!