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Sard's theorem's vector space

  1. Jul 5, 2012 #1
    There is a part in the proof of sard's theorem where we restrict our discussion to a point x such that Df(x)=0, and then declare that f ' (x) is a proper (n-1) subspace (f is n-dim). What I don't understand is, the argument then goes by considering any two points in a sub-rectangle around this point and stating that all such points "lie within ε√n (l/N) of the (n-1)-plane V+f(x)." Where √n (l/N) is the length of the longest "diagonal" of our rectangles used in the proof and epsilon pops up from a "continuity" argument. Anyway, my question is about the V+f(x) plane.

    I don't really see how its a plane. I believe its "centered" around f(x),.... is this correct?

    Also, it says that {Df(x)(y-x): y in rectangle} lies in an (n-1)-dim subspace V of R^n. Is it saying that {Df(x)(y-x)} constitutes all of V ? or just that it is locally approximated by this vector space..... I feel like V is a plane tangent to this point, if that's the case then this all makes sense, but I'm not really sure..... All help appreciated. Are there any other suggestions on how I can "see" this.
     
  2. jcsd
  3. Jul 5, 2012 #2
    Which proof of Sard's theorem? Without being able to look at these things in context, it's very difficult to answer your questions.

    You say that V is an (n-1)-dimensional subspace of R^n, so I guess that's the sense in which V+f(x) is an (affine hyper-) plane. I don't know that "centered" is the right way to say it, but V+f(x) does contain f(x) (because V, being a subspace, contains zero).
     
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