I’m reading a paper and I’m unable to verify arguments. I try to avoid technical terms in the paper by using simple set-theoretic notation.
(a) Right now, I’m having $\partial A\subset \overline{X}$ and $X\subset A$. Similarly, $\partial B\subset \overline{Y}$ and $Y\subset B$. Additionally, $A$ and $B$ are compact. The complement of A is connected, likewise the complement of B is connected.
Prove that if $X=Y$ then $A=B$.
Here $\partial A$ means the boundary of set A and $\overline{A}$ denote the closure of set $A$. All sets are subsets in the complex plane.
(b) Suppose that the complement of $A$ and $B$ in the complex plane are connected (i.e., $A$ and $B$ are simply connected) and $A$ and $B$ are compact.
Show that if $\partial A=\partial B$, then $A=B$.
My issue is that I have no idea how to use the condition that the complement of $A$ and $B$ are connected in $\mathbb{C}$.
From part (a), I’ve got $\partial A\subset A$ and $\partial A\subset B$. Similarly, $\partial B\subset A$ and $\partial B\subset B$.
The important trick is to instead show the equality of the complements: $\mathbb{C} \setminus A = \mathbb{C} \setminus B$.
For $(a)$ we will show that the inclusion $X \subseteq Y$ implies the inclusion $A \subseteq B$ or, equivalently, $\mathbb{C} \setminus B \subseteq \mathbb{C} \setminus A$ - the symmetry of the given situation then yields the claim.
Since $A$ is compact and in particular closed in $\mathbb{C}$, $\mathbb{C} \setminus A$ is open in $\mathbb{C}$. Thus $$U := (\mathbb{C} \setminus B) \cap (\mathbb{C} \setminus A)$$ is open in $\mathbb{C} \setminus B$. Furthermore $$ U = \mathbb{C} \setminus (A \cup B) \neq \emptyset$$ because $A \cup B$ is again compact and $\mathbb{C}$ is not. Now the given inclusion comes into play: We have $$\partial A \subseteq \overline{X} \subseteq \overline{Y} \subseteq \overline{B} = B$$ and therefore $\mathbb{C} \setminus \partial A \supseteq \mathbb{C} \setminus B$. With $A = \overline{A} = \partial A \cup \operatorname{int}(A)$ we hence obtain $$U = (\mathbb{C} \setminus B) \cap (\mathbb{C} \setminus A) = (\mathbb{C} \setminus B) \cap (\mathbb{C} \setminus \partial A) \cap (\mathbb{C} \setminus \operatorname{int}(A)) = (\mathbb{C} \setminus B) \cap (\mathbb{C} \setminus \operatorname{int}(A)),$$ so $U$ is closed in $\mathbb{C} \setminus B$.
Altogether $U$ is a nonempty clopen subset of the connected space $\mathbb{C} \setminus B$ and we conclude that $$\mathbb{C} \setminus B = U = (\mathbb{C} \setminus B) \cap (\mathbb{C} \setminus A) \subseteq \mathbb{C} \setminus A.$$
For $(b)$ one just applies $(a)$ with $X := \partial A$ and $Y := \partial B$.