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S Sep 16, 2022 at 13:50 history suggested J. W. Tanner CC BY-SA 4.0
corrected spelling in title
Sep 16, 2022 at 12:14 review Suggested edits
S Sep 16, 2022 at 13:50
Sep 15, 2022 at 9:40 history edited user467453 CC BY-SA 4.0
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Sep 15, 2022 at 9:32 answer added Roland Bacher timeline score: 1
Sep 15, 2022 at 9:15 comment added user467453 I am sorry for my earlier mistake. I was confused. Thanks a ton!
Sep 15, 2022 at 8:24 comment added Pietro Majer The approximation you want is very easy then: take the convex hull of a finite $\epsilon$-net $S$ of the compact convex set $K\subset \mathbb R^d$, with the additional property that no $(d+1)$-subset of $S$ is affinely dependent (which is true for a dense open set in $(\mathbb R^d)^{|S|}$ ). Then $\text{co}(S)$ is a simplicial polytope whose Hausdorff distance from $K$ is not larger than $\epsilon$ .
Sep 15, 2022 at 8:13 comment added Pietro Majer Sorry, I still don't get the definition you are referring to, even checking the wikipedia link. Isn't for you 1) A 2D convex polytope just a convex polygon? 2) A facet of a polygon just an edge, which is a 1D simplex?
Sep 15, 2022 at 7:50 comment added user467453 No, only triangles: en.wikipedia.org/wiki/Simplicial_polytope If I had to prove the statement, I would try to prove it for polytopes and then show that simplicial polytopes lie dense in the polytopes, but again that cannot work for $d=2$. I rather not, however, since I assume this should be known.
Sep 15, 2022 at 7:47 comment added Pietro Majer I'd say the answer is yes for any $d\ge0$. Why $d=2$ is special? Aren't then simplicial polytopes just the convex polygons?
S Sep 15, 2022 at 6:24 review First questions
Sep 15, 2022 at 6:43
S Sep 15, 2022 at 6:24 history asked user467453 CC BY-SA 4.0