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Math Help - B={x∈M: rank Df(x)<d} is a closed set,why?

  1. #1
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    B={x∈M: rank Df(x)<d} is a closed set,why?

    M,N smooth manifolds,dim M=m,dim N=n.m>n.
    f:M>N a smooth map.
    d is an integer,0<d<n.
    Then B={x∈M: rank Df(x)<d} is a closed set.
    please help me,thanks!
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  2. #2
    Super Member Rebesques's Avatar
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    Re: B={x∈M: rank Df(x)<d} is a closed set,why?

    f is smooth and the function x\rightarrow \omega(x)=rank(Df)(x) is discrete. So

    B=\{x:rank(Df)(x)<d\}=\{x:rank(DF)(x)=0\}\cup\{x:r  ank(Df)(x)=1\}\cup\ldots\cup\{x:rank(Df)(x)=d-1\}=\omega^{-1}(0)\cup\omega^{-1}(1)\cup\ldots\cup \omega^{-1}(d-1)


    is a finite union of closed sets.
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