Hi everyone: Suppose $A\subset \mathbb{R}^{m}$ ($m>1$) is a closed set with empty interior. Which are the necessary and sufficient conditions that $A$ have a neighborhood $V$ such that every bounded component of $ \mathbb{R}^{m}\setminus A$ has a point of $\mathbb{R}^{m}\setminus V$?
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1$\begingroup$ Necessary and sufficient for the existence of such a $V$ is the existence of such a $V$. You need to be more specific. $\endgroup$– user1688Commented Mar 12, 2017 at 7:49
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$\begingroup$ What do you mean more specific? The set $A$ is closed and has no interior points. $\endgroup$– M. RahmatCommented Mar 12, 2017 at 20:00
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$\begingroup$ I added more conditions. $\endgroup$– M. RahmatCommented Mar 12, 2017 at 20:12
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1 Answer
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Suppose $A$ is compact. Then necessary and sufficient is that there is some $\varepsilon > 0$ such that every bounded component of $\mathbb R^n \backslash A$ contains a ball of radius $\varepsilon$.
answered Mar 12, 2017 at 8:40
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$\begingroup$ Correct, but this is not necessary: the union of the spheres $S(n,1/n)$ may have a neighborhood $V$, as explained but the your $\varepsilon$ does not exist. What is the necessary condition? $\endgroup$ Commented Mar 12, 2017 at 18:31
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$\begingroup$ @M.Rahmat Unless I misunderstand your notation, that example is not compact. $\endgroup$ Commented Mar 12, 2017 at 19:11
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$\begingroup$ You are right, this is why I accepted you partial answer. How about the general case, where $A$ is just closed? $\endgroup$ Commented Mar 12, 2017 at 19:57