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Let $U$ be a connected open set in $\mathbb{R}^n$ with $n>1$. If the boundary of $U$ is disconnected, can we say that the complement of $U$ in $\mathbb{R}^n$ is disconnected?

Here is what I did. Let $F$ be the complement of $U$ in $\mathbb{R}^n$. Since $F$ and $U$ have the same boundary, suppose he boundary of $F$ is the disjoint union of two closed sets: $\partial F=F_1\cup F_2$. We take $x_1$ in $F_1$ and $x_2$ in $F_2$, and since every ball of center $x_j$ ($j=1,2$) contains a point of $F$, we can find two points $y_1$ and $y_2$ in $F$, and we must prove that every connected component of $F$ containing $y_j$ must intersect $U$.

Sumanta
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M. Rahmat
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    Have you tried anything? Do you have any ideas or examples to mention and ask about? – Steve Kass Oct 22 '21 at 15:35
  • Let $U$ be punctured disk (in the plane). What happens then? – Reveillark Oct 23 '21 at 05:36
  • The boundary of $U$ is disconnected and so is the complement of $U$. At least in this case the answer to my question is yes, but how about the general case? – M. Rahmat Oct 23 '21 at 17:37
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    This seems like a question that will require some advanced tools, and the answer is yes. For example , see the "if" part of lemon314's answer here. The answer is slightly unrigorous, but clear enough for me. Having said that : we might , just might be able to write that proof in a manner which you might be able to understand a little more clearly even if you did not have the required background. I could be wrong : but I wouldn't mind trying. – Sarvesh Ravichandran Iyer Oct 25 '21 at 02:55
  • Very good. Please write down your solution. – M. Rahmat Oct 27 '21 at 20:57

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