Let $X$ be a connected topological space, $U$, $V\in X$ two non-disjoint open subsets none of which contains the other one. Prove that if their boundaries $Fr(U)$ and $Fr(V)$ are connected, then $Fr(U)\cap Fr(V)\neq\emptyset$.

I've tried that since $X$ is connected, and $U, V$ are proper subsets of $X$, then $Fr(U)$ and $Fr(V)$ are not empty. And also since $Fr(U)$ and $Fr(V)$ are connected, then $Cl(U)$ and $Cl(V)$ are connected. Then $Cl(U)\cap V\neq\emptyset$, and $Cl(U)\cap (X-V)\neq\emptyset$, then $Cl(U)\cap Fr(V)\neq\emptyset$, but I can not see this goes anywhere, any help, thanks!

  • $\begingroup$ what are $A$ and $B$? $\endgroup$ – ΑΘΩ Dec 21 '19 at 1:01
  • $\begingroup$ Should be U and V, typed wrong $\endgroup$ – Cathy Dec 21 '19 at 1:02

After edit: Still false. Consider two points $x,y\in X=\Bbb R^n$, and then $U=X\setminus\{x\}$ and $V=X\setminus\{y\}$.

Before edit: False. Consider $X=\Bbb R^2$, $U=\{(x,y)\in\Bbb R^2\,:\, x^2+y^2<1\}$ and $V=\{(x,y)\in\Bbb R^2\,:\, (x-1)^2+y^2<1\}$.

  • $\begingroup$ typed wrong, sorry, the question should be $Fr(U)\cap Fr(V)\neq\emptyset$ $\endgroup$ – Cathy Dec 21 '19 at 1:08

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