# Showing that if the boundary of a set is connected, then the closure is connected.

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?

• 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. Jul 12 '20 at 6:05
• 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.
– bof
Jul 12 '20 at 7:53
• 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. – bof Jul 12 '20 at 8:01 ## 2 Answers 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. • +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...? Jul 13 '20 at 0:50 I found the counterexample of an annulus with a radial line removed • What is the counterexample? What is the set$A\$ whose boundary is connected but whose closure is disconnected?
– bof
Jul 13 '20 at 7:33