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Nov 17, 2020 at 0:54 vote accept Lostsoul
Nov 9, 2020 at 9:20 answer added mlk timeline score: 0
Nov 3, 2020 at 21:13 comment added mlk By drawing some circles I think I am able to show that the tangent space of $\partial X$ in any point exists and is equal to the $x_n$-hyperplane. This should imply the form that Pietro is suggesting, but there might be some weirdness that I am missing, e.g. if F is something like the Cantor set.
Nov 3, 2020 at 12:36 comment added Lostsoul Right, I agree with that claim. The proof or some hints about X being a union of hyperplanes would be very helpful.
Nov 3, 2020 at 7:08 comment added Pietro Majer I'm saying $X$ is a product $F\times\mathbb{R}^{n-1}$ for a closed $F\subset\mathbb R$, without assuming $X$ is a $d$-dimensional regular set. I'll try to write a proof later.
Nov 3, 2020 at 2:35 comment added Lostsoul The closest point is not always unique. There could be two, the $z_{+}$ and $z_{-}$. A zone you have mentioned won't be $d$-regular though unless I am missing something. Any suggestions why this condition necessarily implies that $X$ is a union of parallel hyperplanes? It seems intuitive but can't come up with a proper argument.
Nov 3, 2020 at 0:17 comment added Pietro Majer And if the closest point is assumed to be always unique, $X$ is a zone $\{a\le x_n\le b\}$ for some $a\le b$
Nov 3, 2020 at 0:07 comment added Pietro Majer It seems to me that the only assumption on the form of the closest points already implies that $X$ is a union of parallel hyperplanes orthogonal to $e_n$.
Nov 2, 2020 at 23:56 comment added Pietro Majer It is not clear to me if the closest point is assumed to be always unique.
Nov 2, 2020 at 1:12 review First posts
Nov 2, 2020 at 4:04
Nov 2, 2020 at 1:05 history asked Lostsoul CC BY-SA 4.0