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.

  • $\begingroup$ Consider an open annulus, which is a counterexample to the claim in the title... $\endgroup$ – Mariano Suárez-Álvarez Nov 2 '16 at 4:09
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    $\begingroup$ Then it's complement isn't connected which contradicts our hypothesis. So that example doesn't work. $\endgroup$ – Enigma Nov 2 '16 at 14:38
  • $\begingroup$ Are you familiar with Poincare/Alexander duality? $\endgroup$ – Moishe Kohan Nov 2 '16 at 15:22
  • $\begingroup$ Yes I am familiar with Poincare/ Alexander duality. How does this fit into establishing the result above? I don't quite see it. $\endgroup$ – Enigma Nov 2 '16 at 16:11
  • $\begingroup$ @enigma, it is a counterexample to the claim made in your title (as I wrote). It is a really bad idea to pick a title which claims something different than the actual question and which, to boot, is false! $\endgroup$ – Mariano Suárez-Álvarez Nov 2 '16 at 18:27

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


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