Rudin 4.22 Theorem Could you help me understand 
why 1. f(H) = B and 
why 2. $\bar A$ $\cap$ B is empty and
why 3. $\bar G$ $\cap$ H is empty?


*

*Let $x \in f(H)$.
Then, $x = f(h)$ for some $h \in H = E \cap f^{-1}(B) \subset f^{-1}(B)$.
So, $x = f(h) \in B$.
So, $f(H) \subset B$.
Let $x \in B \subset A \cup B = f(E)$.
Then, $x = f(e)$ for some $e \in E = G \cup H$.
If $e \in G = E \cap f^{-1}(A) \subset f^{-1}(A)$, then $x = f(e) \in A$.
Then, $x \in A \cap B$.
But $A \cap B = \emptyset$.
Contradiction.
So, $e \in H$.
So, $x = f(e) \in f(H)$.
So, $B \subset f(H)$.  

*By assumption, $A$ and $B$ are (nonempty) separated subsets of $Y$.
About the definition of "separated", see Definition 2.45 on p.42.
By the definition of "separated", $\overline{A} \cap B = \emptyset.$ 

*Assume $\overline{G} \cap H \neq \emptyset$.
Let $x \in \overline{G} \cap H$.
Then, $f(x) \in f(\overline{G}) \subset \overline{A}$.
Then, $f(x) \in f(H) = B$.
So, $f(x) \in \overline{A} \cap B$.
But $\overline{A} \cap B = \emptyset$.
Contradiction.
So, $\overline{G} \cap H = \emptyset$.  
