Linked Questions

Possible Duplicate: Prove $f(S \cup T) = f(S) \cup f(T)$ I'm revisiting set theory and am troubled by this question. Let $f:A \rightarrow B$, and $C \subset A$, $D \subset A$. Prove that $f(C \...

I have a question. How do I prove the following identity? $$ f(S\cup T) = f(S) \cup f(T) $$

I need to prove this and I'm pretty sure this is straight forward. $$f[A\cup{B}]=f\{x|x\in{A} \text{ or } x\in{B}\}=\{f(x)|x\in{A} \text{ or } x\in{B}\}=f[A]\cup{f[B]}$$ I'm writing this because of ...

$$f(A\cup B)=f(A)\cup f(B)$$ My attempt: Suppose that $f(A\cup B) \neq f(A) \cup f(B)$, then $f(A\cup B)$ has a variable $x$ that is neither in $A$ nor $B$; however, $A\cup B$ implies that $x$ must ...

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 ...

If $S_1$ and $S_2$ are both subsets of some arbitrary set $A$, then how do I prove that $f(S_1\cup S_2) = f(S_1) \cup f(S_2)$ for ALL cases I understand that it is true, but I don't know how to prove ...

Are there some good overviews of basic facts about images and inverse images of sets under functions?

$f(S \cap T) \subseteq f(S) \cap f(T)$ $x$ lies in ($S \cap T$), which means the domain has fewer elements than the domain of $S$ and $T$, since $x$ must be in $S$ and $T$. All $f(x)$ values of $x$, ...

I have an example is like this Let f be a function from the set A to the set B. Let S and T be subsets of A. Show that $f(S \cup T) = f(S) \cup f(T)$ Answer: $y\in f(S\cup T) \rightarrow \exists x \...

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