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If $X$ is a connected metric space and $A\subseteq X$ has a connected boundary, then $\overline{A}$ is connected and $\overline{A^c}$ are both connected.

My attempted proof is as follows. As $\partial{A}=\overline{A}\cap\overline{A^c}$, then $\partial{A}\subseteq\overline{A}$. If $\overline{A}=C\cup D$, with $\overline{C}\cap D=\overline{D}\cap C=\emptyset$ with $C,D$ not empty.

Then $$\partial{A}=\partial{A}\cap\overline{A}=\partial{A}\cap(C\cup D)=(\partial{A}\cap C)\cup(\partial{A}\cap D)$$. But $\overline{(\partial{A}\cap C)}\cap(\partial{A}\cap D)\subseteq\overline{C}\cap D=\emptyset$ and similarly $\overline{(\partial{A}\cap d)}\cup(\partial{A}\cap C)=\emptyset$.

This means that $\partial{A}$ is disconnected, a contradiction, thus $\overline{A}$ is connected.

An analogous reasoning also shows that $\overline{A^c}$ is connected.

Is this correct, or am I missing something?

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  • $\begingroup$ Thanks, I added the hypothesis of the space being connected. I am struggling to prpove that $\partial{A}\cap C$ and $\partial{A}\cap D$ are non empty. $\endgroup$
    – user489562
    Jul 12 '20 at 6:05
  • $\begingroup$ I'm not sure but it seems to me you don't need a metric, it should work in a topological space. But maybe you haven't had topological spaces yet. $\endgroup$
    – bof
    Jul 12 '20 at 7:53
  • $\begingroup$ Here's an idea that might work. Assume for a contradiction that $\overline A$ is disconnected. So $\overline A$ is the union of two nonempty separated sets $C$ and $D$; "separated" means that each is disjoint from the closure of the other. Since $partial A\subseteq\overline X$ and since $\partial A$ is connected, $\partial A$ is a subset of either $C$ or $D$; say $\partial A\subset $C$, so $D\cap\partial A=\emptyset$. Now, if you can show that $D$ and $X\setminus D$ are separated, then you've got a contradiction, since $X$ is connected. $\endgroup$
    – bof
    Jul 12 '20 at 8:01
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I usually work with the definition that a set $S$ is connected if it cannot be written as a union of two disjoint nonempty sets which are both open (and closed) in $S$.

Assume that $C$ and $D$ are two disjoint nonempty sets which are both open (and closed) in $\overline{A}$. Since they are open in $\overline{A}$, there exist open sets $U,V \subseteq X$ such that $$C = U \cap \overline{A}, \quad D = V \cap \overline{A}.$$

We have that $$C \cap \partial A = U \cap \partial A \quad\text{ and }\quad D \cap \partial A = V \cap \partial A$$ are disjoint sets which are both open in $\partial A$ and whose union is $\partial A$ so they cannot both be nonempty.

WLOG assume that $D \cap \partial A = \emptyset$. Then also $V \cap \partial A = \emptyset$ so we have $$D = D \cap \overline{A} = V \cap (\operatorname{Int} A \cup \partial A) = (V \cap \operatorname{Int} A) \cup \underbrace{(V \cap \partial A)}_{=\emptyset} = V \cap \operatorname{Int} A$$ so we conclude that $D$ is open in $X$. Furthermore, $D$ is closed in $\overline{A}$ and $\overline{A}$ is closed in $X$ so $D$ is closed in $X$.

Hence $D$ and $X\setminus D$ (which contains $C\ne\emptyset$) are nonempty disjoint sets which are both open in $X$. This is a contradiction since $X$ is connected.

We conclude that $\overline{A}$ is connected.

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  • $\begingroup$ +1. The proposer's attempt has 2 flaws. One is the assumption that neither $C\cap \partial A$ nor $D\cap \partial A$ is empty. The other is failing to use the (necessary) connectedness of $X$... E.g. if $X$ is disconneced and $A=X$ then...? $\endgroup$ Jul 13 '20 at 0:50
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I found the counterexample of an annulus with a radial line removedenter image description here

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  • $\begingroup$ What is the counterexample? What is the set $A$ whose boundary is connected but whose closure is disconnected? $\endgroup$
    – bof
    Jul 13 '20 at 7:33

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