Theorem 4.22 from baby Rudin

Question

Can anyone explain me what would be if one of them is empty?

Answer 1

If I understand correctly, then the argument is as follows: since $A$ and $B$ are nonempty and $f(E)=A\cup B$, there exist $a,b\in E$ such that $f(a)\in A$ and $f(b)\in B$. Therefore, $a\in G$ and $b\in H$, so $G$ and $H$ are nonempty.

Hope this helps!

Answer 2

The reason $G$ is not empty is that $A$ is a non-empty subset of $f(E)$. That means there exists $x\in E$ so that $f(x)\in A$. that same $x$ belongs to $f^{-1}(A)$. so $x$ belongs to $f^{-1}(A)\cap E$ proving $G=A\cap E$ is not empty.

The same argument proves $H$ is not empty.

Answer 3

If one of them is empty, then we would have no contradiction on the definition of $E$.

We know that $E$ is connected, so it can't be written as union of non-empty separated sets: from this at the last line we derive a contradiction.

Neither $G$ nor $H$ are empty basically because of the definition of $f(E)$. Is it clear to you why?