MHB Disconnected Sets .... Palka, Lemma 3.1 ....

[Math Amateur]

I am reading Bruce P. Palka's book: An Introduction to Complex Function Theory ...

I am focused on Chapter 2: The Rudiments of Plane Topology ...

I need help with some aspects of the proof of Lemma 3.1 ...

Lemma 3.1 and its proof read as follows:View attachment 7373
View attachment 7374

In the above text from Palka Section 2.2 we read the following:" ... ... The sets $$S$$ and $$T$$ are non-empty - $$S$$ contains $$A \cap U^*$$ and $$T$$ contains $$A \cap V^* $$ - and disjoint. ... ... "I am trying to show/demonstrate rigorously that the sets $$S$$ and $$T$$ are non-empty - $$S$$ contains $$A \cap U^*$$ and $$T$$ contains $$A \cap V^*$$ - and disjoint ... can someone please help ...?
Help will be much appreciated ...

Peter===================================================================================Readers of the above post will be assisted by having access to Palka's introduction to disconnected sets which includes the key definition ... so I am providing the same ... as follows ... :https://www.physicsforums.com/attachments/7375

 

[Opalg]

I am trying to show/demonstrate rigorously that the sets $$S$$ and $$T$$ are non-empty - $$S$$ contains $$A \cap U^*$$ and $$T$$ contains $$A \cap V^*$$ - and disjoint ... can someone please help ...?

If $z\in A\cap U^*$ (which is a nonempty set) then $z\notin A \cap U^*\cap V^*$ (because $A \cap U^*\cap V^*$ is empty). In particular, $z\notin V^*$. Therefore $z \in (A \cap U^*)\sim V^* \subseteq U^*\sim V^* = S.$ That shows that $A\cap U^*\subseteq S$, so that in particular $S$ is nonempty. The proof that $A\cap V^*\subseteq T$ is similar.

The fact that $S$ and $T$ are disjoint follows from their definitions: $S = U^*\sim V^*$, which is contained in $U^*$. But $T = V^*\sim U^*$, which is contained in the complement of $U^*$.

 

[Math Amateur]

If $z\in A\cap U^*$ (which is a nonempty set) then $z\notin A \cap U^*\cap V^*$ (because $A \cap U^*\cap V^*$ is empty). In particular, $z\notin V^*$. Therefore $z \in (A \cap U^*)\sim V^* \subseteq U^*\sim V^* = S.$ That shows that $A\cap U^*\subseteq S$, so that in particular $S$ is nonempty. The proof that $A\cap V^*\subseteq T$ is similar.

The fact that $S$ and $T$ are disjoint follows from their definitions: $S = U^*\sim V^*$, which is contained in $U^*$. But $T = V^*\sim U^*$, which is contained in the complement of $U^*$.

Thanks Opalg ... most helpful post!

Peter