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seen Dec 15 at 17:11

Dec
8
revised proof that the lebesgue measure of a subspace of lower dimension is 0.
added 108 characters in body
Dec
8
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.
Dec
8
asked proof that the lebesgue measure of a subspace of lower dimension is 0.
Dec
8
awarded  Caucus
Dec
8
answered Is {0} a free module?
Dec
8
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.
Dec
8
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?
Dec
8
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?.
Dec
8
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.
Dec
8
comment Lebesgue measure of a subspace of lower dimension is 0
Yep, It comes way after that.
Dec
8
asked Lebesgue measure of a subspace of lower dimension is 0
Oct
28
awarded  Nice Answer
Jul
2
awarded  Curious
Jun
23
awarded  Commentator
Jun
23
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.
Jun
23
accepted Existence of nontrivial normal subgroups in solvable finite groups
Jun
23
asked Existence of nontrivial normal subgroups in solvable finite groups
Feb
13
awarded  Yearling
Dec
4
comment About some notation of the derivative
TonyPiccolo Look at the formal definition part.
Dec
4
awarded  Promoter