Hi everyone: Let $A$ be a closed subset of $\mathbb{R}^{k}$, $k\geq2$, with empty interior, with no isolated point and such that if $W$ is an open set in $\mathbb{R}^{k}\setminus A$ then the interior of the boundary of $W$ does not meet $A$. Is there a sequence of closed and disjoint sets $B_{n}$ in $\mathbb{R}^{k}\setminus A$ such that for all $a$ in $ A$ there is a sequence $(b_{n})$ in $B_{n}$ that converges to $a$, as $n\rightarrow\infty$?
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2$\begingroup$ This does not seem to be research level. Try $B_n=\lbrace x\in\mathbb R^k:$ dist$(x,A)\ge 1/n\rbrace$. $\endgroup$– Jochen WengenrothMar 29, 2017 at 6:17
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$\begingroup$ Thanks for your answer. The sets $B_{n}$ must be disjoint and I forgot to mention. I tried $B_{n}=\lbrace x: d(x,A)=1/n\rbrace$. But how would you prove that for each $a$ in $A$ there is $(b_{n})$ in $B_{n}$ that converges to $a$? $\endgroup$– M. RahmatMar 29, 2017 at 16:47
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$\begingroup$ I corrected the question. $\endgroup$– M. RahmatMar 29, 2017 at 17:00
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1$\begingroup$ Taking $B_{n}=\lbrace x: d(x,A)=1/n\rbrace$ does not work if $A$ is the union of all circles centered at the origin with radius of the form $1/n$ together with the origin. $\endgroup$– Ramiro de la VegaMar 31, 2017 at 13:35
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1$\begingroup$ Work in the Hausdorff compactification of ${\bf R}^k$ to get compactness. Note that $A^c$ is totally bounded. Build the $B_n$ as disjoint finite subsets of $A^c$ recursively using total boundedness so that every point of $A^c$ is at distance less than $1/n$ from $B_n$. $\endgroup$– coudyMar 31, 2017 at 19:22
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