Characterizations of simple connectivity of subsets of $\mathbb{C}$ 
A nonempty open and connected set $G\subset \mathbb{C}$ is said to be simply connected if its complement in the extended complex plane $\overline{\mathbb{C}}= \mathbb{C}\cup \left \{ \infty  \right \}$ is connected.

How can one show that the previous condition is equivalent to any one of the following conditions?


*

*If  $\mathbb{C}\setminus G=K\cup A$ where $K$ is compact, $A$ is closed and $K\cap A=\varnothing $,  then $K=\varnothing$;

*$\mathbb{C}\setminus G$ has no bounded components.
 A: Let $G$ be nonempty, open and connected.
(1) Assume $G$ is simply connected.
Assume $\mathbb C\setminus G=K\cup A$, $K$ compact, $A$ closed, $K\cap A=\emptyset$.
Note that $A\cup\{\infty\}$ is closed in $\overline{\mathbb C}$ and by compactness $K$ is closed in $\overline{\mathbb C}$. Thus via $\overline{\mathbb C}\setminus G=(A\cup\{\infty\})\cup K$ we have written the complement of $G$ as disjoint union of two closed sets. By connectednes of $\overline{\mathbb C}\setminus G$, one of them must be empty, and that can only mean $K=\emptyset$.
(2) Assume property 1 holds.
Assume $B$ is a (nonempty) bounded component of $\mathbb C\setminus G$.
Then the closure $K$ of $B$ is compact and is still disjoint to $G$.
Since $B$ is a component, $A:=\mathbb C\setminus G\setminus K$ is closed and we have $A\cap K=\emptyset$. By property 1, $K=\emptyset$, contradiction.
(3) Assume property 2 holds.
Write $\overline{\mathbb C}\setminus G=A\cap A'$ with $A,A'$ closed and disjoint. Wlog. $\infty\in A'$. Then $A$ is bounded. As $\mathbb C\setminus G$ has no bounded components, $A$ must be empty. Hence $\overline{\mathbb C}\setminus G$ is connected.
A: Connected components are always closed, so indeed I think that $B=K$. On the other hand they need not be open, so I don't see either why $A$ should be closed. See however this closely relation question, where the equivalence between (2) and simple connectivity is proved (Lemma 1 in the accepted answer).
