23
$\begingroup$

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.

  • 10
    Your counterexample is already excluded by the requirement that the complement of $A$ be connected.2012-07-13

2 Answers 2