Mathematics Asked by Nothingone on December 5, 2021
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 $partial Asubset overline{X}$ and $Xsubset A$.
Similarly, $partial Bsubset overline{Y}$ and $Ysubset B$. 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 $partial Asubset A$ and $partial Asubset B$. Similarly, $partial Bsubset A$ and $partial Bsubset B$.
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 $Y := partial B$.
Answered by Sebastian Spindler on December 5, 2021
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