# Thread: Proof of f(S U T) = f(S) U f(T) ?

1. ## Proof of f(S U T) = f(S) U f(T) ?

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.
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!!

2. Originally Posted by mathgirl13
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!!
The other direction: $\displaystyle x \in f(S) \cup f(T) \Longrightarrow x \in f(S) \vee x \in F(T) \Longrightarrow$ $\displaystyle x=f(y)\,,\,\,y \in S\,\,\vee \,x=f(y)\,,\,\,y \in T\,\Longrightarrow$ $\displaystyle \,anyway\,,\,\,y \in S \cup T\,\,\,and\,\,\,x \in f(S \cup T)$

Tonio

3. Thank you!! I didn't think it made sense either? but I thought that was what my professor said so I put it, thank you for correcting it and your help!!

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