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bio website location Santa Cruz, Bolivia age 21 member for 2 years, 10 months seen Dec 15 at 17:11 profile views 58

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 Dec8 revised proof that the lebesgue measure of a subspace of lower dimension is 0. added 108 characters in body Dec8 comment proof that the lebesgue measure of a subspace of lower dimension is 0. I wanna see if it's right, I would also like alternative solutions. Dec8 asked proof that the lebesgue measure of a subspace of lower dimension is 0. Dec8 awarded Caucus Dec8 answered Is {0} a free module? Dec8 comment Lebesgue measure of a subspace of lower dimension is 0 I kinda get where you are going, you want to regard $Y$ as $0\times...\times 0\times R^m$, that has measure 0 cause it is written as a countable union of hypercubes $[0,1]^m$ that have measure 0 (from this logic chain it is clear cause page 52 shows that $m([0,1]^m)=vol([0,1]^m)=0$ cause it has a side 0. But I think the problem is when you regard $Y$ as $0\times ...\times 0\times R^m$. They are homeomorphic, yes, but $(0,1)$ and $R$ are also isomorphic but their measure isn't the same. I'm writing a possible fix using determinants and the like tomorrow,there is no other option, thanks anyways. Dec8 comment Lebesgue measure of a subspace of lower dimension is 0 Ok, so we write Y as a countable union of some kind of subsets of Y that are translates of each others and we now have to prove that any of these subsets have measure 0. How can we build those open sets of arbitrary small measure in the simplest way? Dec8 comment Lebesgue measure of a subspace of lower dimension is 0 I don't get the meaning of "dimension of an hypercube". By an hypercube you mean a subset of Y?. Dec8 comment Lebesgue measure of a subspace of lower dimension is 0 What do you mean by "An hypercube of the same dimension as the subspace". I got a more or less similar idea but it seems kinda lenghty. Dec8 comment Lebesgue measure of a subspace of lower dimension is 0 Yep, It comes way after that. Dec8 asked Lebesgue measure of a subspace of lower dimension is 0 Oct28 awarded Nice Answer Jul2 awarded Curious Jun23 awarded Commentator Jun23 comment Existence of nontrivial normal subgroups in solvable finite groups The problem talked about solvable groups, so I supposed it had to be used somewhere, idk how i didn't come up with this, thanks anyways. Jun23 accepted Existence of nontrivial normal subgroups in solvable finite groups Jun23 asked Existence of nontrivial normal subgroups in solvable finite groups Feb13 awarded Yearling Dec4 comment About some notation of the derivative TonyPiccolo Look at the formal definition part. Dec4 awarded Promoter