Let $A\subset \mathbb{R}^n$ such that $\pi_i(A)$ is a Lebesgue null set for $1\leq i\leq n$. Is the complement $\mathbb{R}^n\!\setminus\! A$ pathwise connected? If so, is it always possible to construct a rectifiable path from $x$ to $y$ in the complement with length near $|x-y|$ ?
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1This is closely related: http://math.stackexchange.com/questions/77791/separation-of-two-points-with-null-sets/77800#77800. I'm not marking as a duplicate because I assume you mean $|x-y|2$ and that question asked for $|x-y|\infty$. It so happens that I had an idea for getting close to $|x-y|_2$ while answering that other question; I'll think about it and answer here if I come up with something. But it does answer the first part of your question, since the construction generalizes to higher dimensions. – joriki Nov 15 '11 at 08:45
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4You do want to specify that $n>1$. – Brian M. Scott Nov 15 '11 at 08:53
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What is $\pi_i$? The projection to the $i$th component? – Willie Wong Nov 15 '11 at 11:52
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$\pi_i$ is the projection to the ith component and n has to be >1, yes. – juffo Nov 16 '11 at 02:09
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The solution to the other problem was very helpful. It would be great to also see how path length $|x-y|_2$ (or $\leq |x-y|_2+\varepsilon$ for arbitrary $\varepsilon>0$) could be achieved. – juffo Nov 16 '11 at 02:17