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Homework Help: Content math problem

  1. Dec 10, 2007 #1
    1. The problem statement, all variables and given/known data
    If we let J be in R^p so J=J1x....xJp the cartesian product of p cells in R. Then we define the content of J as c(J)=(b1-a1)....(bp-ap) so in like R it would be length area in R^2 etc. Then define content zero by Z in R^p has content zero if for each epsilon > 0 there exists a finite set J1,...,Jn of cells whose union contains Z and such that c(J1)+.....c(Jn) < epsilon. In an example we did we let the space in R^2 S=((x,y): |x|+|y|=1) and we proved that S has content zero by letting n natural number introducing squares with diagonals on the line in S. then S can be in enclosed in 4n squares each with content 1/n^2 so the total content is 4/n which can be made small so S has content zero. Where i'm confused is S is in R^2 so the cells of S are (-1,1) and (-1,1) so c(S)=(-1-1(-1-1)=4 but we showed S has content zero. This is confusing to me can anyone help clarify?
    Thanks


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 10, 2007 #2

    Dick

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    I have no idea what you are talking about. S is contained in [-1,1]x[-1,1] which has area 4. Sure. It's also contained in many collections of cells having smaller area.
     
  4. Dec 10, 2007 #3
    in the space in R^2, S is diamond shaped and S={(x.y): |x|+|y|=1}. take n to be a natural number. then if you introduce squares with diagonals along S you can enclose S in 4n squares. The content of each square is 1/n^2. so the total content is 4/n which can be made arbitrarily small. So by definition S has content zero. My question is S is in R^2 and the x values range from -1 to 1 and same with the y values. So the content of S is c(S)=(1-(-1))(1-(-1))=4. I'm confused about this. Because by definition a set has zero content if for every epsilon > 0 there exists a finite set J1, ...,Jn of cells whose union contain the set and s.t. c(J1)+...+c(Jn)<epsilon. The squares in the example contain S and all contents added together can be made smaller than any epsilon by increasing epsilon. So it makes sense that S has content zero but I also see how the content could be 4. Also, it's not a square it's a diamond. I don't know if that makes a difference though. thanks
     
  5. Dec 10, 2007 #4

    Dick

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    Area 4 is ONE way to cover the set with rectangles. 'Content' is defined as the limit of the MINIMUM area needed to cover the set with a FINITE number of rectangles, isn't it? The area 4 cover is hardly minimum.
     
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