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Math Help - Pathwise Connected

  1. #1
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    Pathwise Connected

    Problem: Show that the set S={u=(x,y,z) in R^3 | 2x^2 + y^2 + z^2 = 1} is pathwise-connected.

    I plotted the function in MATLAB and attached it. In a way, this looks very similar to the unit sphere in R^3, which is pathwise connected where the parametric path is
    f(t) = costu+sintw

    where w is a point in the sphere that is perpendicular to u and v = -u. However, I am having a hard time to find a parametric path for this set.

    The book says to use two previous exercises 1.) The union of two pathwise-connected subsets of R^n is pathwise-connected and 2.)the graph of the mapping F:A -> R^m is continuous then if A is pathwise-connected, then the set
    G = {(u,v) in R^(n+m)|u in A, v = F(u))} is pathwise-connected.

    Image:


    Any help is greatly appreciated. Thank you.
    Last edited by Paperwings; July 23rd 2008 at 05:20 PM.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by Paperwings View Post
    Problem: Show that the set S={u=(x,y,z) in R^3 | 2x^2 + y^2 + z^2 = 1} is pathwise-connected. (...)

    The book says to use two previous exercises 1.) The union of two pathwise-connected subsets of R^n is pathwise-connected and 2.)the graph of the mapping F:A -> R^m is continuous then if A is pathwise-connected, then the set
    G = {(u,v) in R^(n+m)|u in A, v = F(u))} is pathwise-connected.
    I suggest you use the two previous exercises. Let . Let A be the domain of this two functions. (A is a subset of R^2) The set S can be described as the union of two sets :

    Using , one can write

    Is A pathwise connected ? Are F+ and F- continuous ?
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  3. #3
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    Is A pathwise connected ? Are F+ and F- continuous ?

    1. A is pathwise connected since there exists a

    2. F+ and F- are continuous. Let , then the polynomial is continuous since polynomials are continuous.
    Then let be the boundary of the the region.

    For such that
    then . Thus, F+ is continuous.
    Similarly, F- is continuous.

    Odd... latex is not working so I attached images.
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Paperwings View Post
    Is A pathwise connected ? Are F+ and F- continuous ?

    1. A is pathwise connected since there exists a
    If gamma is supposed to be a path between two points, why does it exist ?

    I think it's easier to show that A={(x,y) in R^2 | 2x^2+y^2 <= 1} is pathwise connected because it is homeomorphic to {(x,y) in R^2 | x^2+y^2 <= 1} which is pathwise connected. This can be done using f(x,y)=(x/sqrt(2),y) for example.
    2. F+ and F- are continuous. Let , then the polynomial is continuous since polynomials are continuous.
    Then let be the boundary of the the region.

    For such that
    then . Thus, F+ is continuous.
    Instead of using the epsilon-delta definition of continuity, you can simply say that g(x)=sqrt(x) is continuous on [0,infinity[ hence the composite function F+ is continuous on A.

    Using these results you can conclude and say that S is pathwise connected.
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  5. #5
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    Thank you for help. I haven't learned what a homeomorphic is nor is it in my book, but I'll look it up.
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