Boundary of bounded connected open subset of $\mathbb{R}^n$, whose complement is also connected, is connected

Question

So I know that there are other threads available on this topic, but there seems to be proofs that miss some details and I was wondering if someone could extrapolate or give a more detailed proof. Namely, if $U\subseteq\mathbb{R}^n$ is a bounded connected open subset and $U^c$ is connected then $\partial A$ is connected as well.

Answer 1

Here is an argument using homology. Suppose that $K=\partial U$ is not connected. Then there exists a pair of disjoint open subsets $A, B\subset R^n$ each of which has nonempty intersection with $K$. Without loss of generality we may assume that each component of $A$ and of $B$ has nonempty intersection with $K$ (otherwise we discard these components). It follows therefore that the open sets $X=A\cup U\cup B, Y= A\cup B\cup U^c$ are both connected (since $U, U^c$ are connected and each component of $A, B$ has nonempty intersection with both $U$ and $U^c$). We have an open cover of $R^n$ by $X, Y$. Furthermore, $Z=X\cap Y= A\cup B$ is not connected. Now, apply Mayer-Vietoris sequence (integer coefficients) for the reduced homology for the open cover $\{X, Y\}$ of $R^n$. We obtain: $$ 0=H_1(R^n)\to \tilde{H}_0(Z)\to\tilde{H}_0(X)\oplus \tilde{H}_0(Y)= 0 $$ Hence, $\tilde{H}_0(Z)=0$, which contradicts the fact that $Z$ is not path connected. qed