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Hi everyone: Suppose $A$ is a closed set with empty interior in $\mathbb{R}^{m}$, $m\geq2$. Does there exist a neighborhood $V$ of $A$ such that each bounded component of $\mathbb{R}^{m}\setminus A$ has a point of $\mathbb{R}^{m}\setminus V$?

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Letting $m=2$, consider a sequence $C_1,C_2,C_3,...$ of concentric circles in $\mathbb{R}^2$, centered at the origin, with strictly decreasing radii $r_1,r_2,r_3, ...$ approaching $0$.

Then $\mathbb{R}^2 \setminus A$ has infinitely many components.

Let $S$ be the union of the circles, and let $A = S \cup \{(0,0)\}$. It's easily seen that $A$ is closed, and has empty interior.

But any open set containing $A$ contains an open disc centered at $(0,0)$, hence covers all but finitely many circles, and in the process, covers (and therefore removes) all but finitely many components of $R^2 \setminus A$.

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