Munkres-Analysis on Manifolds: Theorem 20.1

  • #1
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Hello. I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20. It states that:

Let A be an n by n matrix. Let h:R^n->R^n be the linear transformation h(x)=A x. Let S be a rectifiable set (the boundary of S BdS has measure 0) in R^n. Then v(h(S))=|detA|v(S) (v=volume).

The author starts his proof by considering tha case of A being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that h(intS)=int h(S) and h(S) is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).

He proceeds by considering the case where A is singular, so detA=0. He tries to show now that v(T)=0. He states that since S is bounded so is h(S) (I think thats true because |h(x)-h(a)|<=n|A|*|x-a| for each x in S and fixed a in S, if there is again a better explanation please tell me).

Then he says that h(R^n)=V with dimV=p<n and that V has measure 0 (for each ε>0 it can be covered by countably many open rectangles of total volume less than ε), a statemant that I have no clue how to prove. Then he says that the closure of h(S)=cl h(S) is closed and bounded and has neasure 0 (of course cl h(S) is closed but why is it bounded with measure 0?). Then makes an addition step (which I understand) and proves the theorem for that case too.

Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!
 

Answers and Replies

  • #2
mathwonk
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try showing that the x axis in R^2 can be covered by a sequence of rectangles of total area < 1. e.g. take them all of base 1, and make the height of the one whose base goes from n to n+1 something like 1/2^(n+2). or whatever works.

by the way have you had a calculus course at the level of spivak? i ask because it seems this analysis on manifolds course is a little beyond your training. of course one can always catch up in time.
 
  • #3
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You have a linear transformation between finitedimensional spaces. Such transformation is always continuous, or equivalently, bounded. This explains the boundedness part in your question.
 
  • #4
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try showing that the x axis in R^2 can be covered by a sequence of rectangles of total area < 1. e.g. take them all of base 1, and make the height of the one whose base goes from n to n+1 something like 1/2^(n+2). or whatever works.

by the way have you had a calculus course at the level of spivak? i ask because it seems this analysis on manifolds course is a little beyond your training. of course one can always catch up in time.
I have had a course in one variable at a level a bit higher than Spivak and a (slightly less rigorous) course in multivariable course. I am studying this book to proceed to Spivak's Differential Geometry books but I am having second thoughts. Seeing how much topology Munkres' proofs have I am thinking of studying his topology book instead.
 
  • #5
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You have a linear transformation between finitedimensional spaces. Such transformation is always continuous, or equivalently, bounded. This explains the boundedness part in your question.
I thought boundedness is something I can claim for a range if I have fa compact domain.
 
  • #6
mathwonk
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for a linear transformation T the meaning of "bounded" is that the image of the unit sphere is a bounded set. equivalently there is a finite bound C such that for every vector v we have |Tv| ≤ C|v|, where | | denotes the length of a vector
 
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