Closure and interior of a connected set $A$

Question

Let $\overline{A}$ be the closure of $A$, and $A^{\circ}$ be the interior of $A$.

We assume that $A$ be connected.

1. Is $\overline{A}$ connected?

My answer is yes. Because, if $\overline{A}$ is not connected, we can write $$\overline{A}=B\cup C$$ such that $B,C$ are separated. i.e., no limit point of $B$ is in $C$ and vice versa. Also, both $B,C$ are clopen in $\overline{A}$. And note that $B,C$ are non-empty.

WLOG, suppose $A\subset B$. Since $$\overline{A}=A\cup A'$$ where $A'$ is the set of all limit points of $A$, $C$ must contain a limit point $l$ of $A$. But then, since $C$ is open, $C$ is a neighborhood of $l$ that doesn't intersect $A$; which makes $l$ not a limit point of $A$. A contradiction. So $A\not\subset B$ and by the same process, $A\not \subset C$. Hence $$A=(A\cap B)\cup(A\cap C)$$ separated by two non-empty clopen (in $A$) sets, thus $A$ is not connected.

So we have the contrapositive statement "$\overline{A}$ not connected $\implies$ $A$ not connected", which proves "$A$ connected $\implies$ $\overline{A}$ connected".

2. Is $A^{\circ}$ connected?

My answer is no. Consider $$A=\{(x,y)\in \mathbb{R}^2:x^2+y^2\leq 1\}\cup\{(x,y)\in\mathbb{R}^2:(x-2)^2+y^2\leq 1\}$$

two circles tangent to each other at $(1,0)$. Clearly this $A$ is connected. Then the interior would be

$$A^{\circ}=\{(x,y)\in \mathbb{R}^2:x^2+y^2< 1\}\cup\{(x,y)\in\mathbb{R}^2:(x-2)^2+y^2< 1\}$$

a union of two disjoint open sets. Which is separated. Thus in some cases $A^{\circ}$ is not connected even if $A$ is connected.

Does this argument make correct proof?

Answer 1

Your proof of the first assertion is incomplete. How do you know that $A\subset B$ or $A\subset C$? Yes, it is true, but you did not justify it.

Your counterexample for the second assertion is correct. Another possibility would be $\{(x,y)\in\mathbb R^2\,|\,xy\geqslant0\}$, for instance.

Answer 2

For $1.$

For brevity let $\overline A=D.$

Let $B'=B\cap A$ and $C'=C \cap A.$ Then $B',C'$ are open disjoint subsets of the sub$space $ $A$ (as a subspace of $D$), and $$B'\cup C'=(B\cap A)\cup (C\cap A)=(B\cup C)\cap A =D\cap A=A.$$ So at least one of $B',C'$ is empty. Suppose $B'=\emptyset.$ Then $$D=\overline A = \overline {B'\cup C'}=\overline {B'} \cup \overline {C'}=$$ $$=\overline \emptyset \cup \overline {C'}=\overline {C'}\subset$$ $$\subset \overline C\subset \overline D=D$$ so $D=\overline C.$ Since $C$ is a closed subset of the closed sub$space $ $D $, we also have $\overline C=C, $ so $D=C.$ And $B=D \setminus C=D\setminus D=\emptyset.$