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Let $C$ be a compact subset of $\mathbb R^n$ and suppose that for every $\varepsilon >0$ there exists a finite family of open disks $B_i$ s.t. $C \subset \bigcup_{i} B_i$ and $\sum_i r_i \le \varepsilon$ (where $r_i$ denotes the radius of $B_i$).

(a) Suppose $n\ge 2$. Is $\mathbb R^n \setminus C$ connected?

If the answer is yes, then

(b) For which $n \ge 2$ is $\mathbb R^n \setminus C$ path-connected? And simply-connected?

Similar questions can be found here (and also on MathOverflow, with a beautiful but difficult answer). Anyway, I do not know the answer to the questions. What do you think? Do you have any references? Do you know any elementary proofs?

Thanks in advance.

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    What do you mean by "and totally *bounded*"? If $C$ is compact, then it is necessarily totally bounded according to the usual definition, so maybe you mean something different from this?2012-08-07
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    If $C$ is just compact, there is a simple counter-example to a): The unit sphere $S^{n-1}$ is obviously compact, as it is a closed set in $\mathbb R^n$ and bounded, but it clearly divides $\mathbb R^n$ into an inside and outside part.2012-08-07
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    Thanks for the comment, I edited the problem. Hope now is clear.2012-08-07
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    I think you're asking about sets of [Hausdorff dimension](http://en.wikipedia.org/wiki/Hausdorff_dimension) smaller than 1.2012-08-08

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