TITLE: Connectedness of the boundary
QUESTION [28 upvotes]: My question is about the following claim:

For $n \geq 2$, let $A\subset \mathbb R^n$ be a non-empty, open, bounded set. Assume $A$ and its complement are connected and $\text{int}(\text{cl}(A)) = A$. Then $\partial A$ is connected.

Without the assumption $\text{int}(\text{cl}(A)) = A$, the statement is false (just take any open set and remove an interior point). This condition is not necessary but I think it is sufficient to get the claim.
This seems a trivial matter, but I cant's find a proof using only basic topological tools. Does anyone know something about this ? Thank you very much.

REPLY [9 votes]: This is true in general: a connected open set $A$ in $\mathbb{R}^n$ has disconnected complement if and only if it has disconnected boundary.
The "only if" follows from $\mathbb{R}^n$ being normal.
As for the "if", consider that $\partial A$ has two components, $X$ and $Y$. Neither of them can lie in the interior of closure of $A$ (it is then a clopen set in the complement of $A$, and we assumed it to be connected). So assume to the contrary, and this actually means that both $X$ and $Y$ have points from the closure of interior of complement of $A$ (don't worry, that was the longest:), say $x$ and $y$. Now you can connect $x$ to $y$ by a curve through $A$ (we can do such things in $\mathbb{R}^n$), and then connect $y$ to $x$ with a curve through the complement of $A$ (we need that nonempty interior of complement for that). This gives a loop that cannot be homotoped to constant, yet $\mathbb{R}^n$ is simply connected, a contradiction.
This cannot be done non-homotopically. The equivalence as I stated it characterises (is equivalent) to the ambient space having trivial first cohomology. The map $f$ from Deian Govc's answer cannot exist precisely because it would give a nontrivial cohomology class.