Explore BrainMass

Explore BrainMass

    Fixed Point Theorem and Closed Unit Ball in Euclidean Space

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    The Brouwer Fixed-Point Theorem

    Let denote the closed unit ball in Euclidean space :
    .
    Any continuous map from onto itself has at least one fixed point, i.e. a point such that .

    Proof Suppose has no fixed points, i.e. for .
    Define a map , , by letting be the point of intersection of and the ray starting at the point and going through . For see figure below:

    We have
    with , (1)
    Then
    (2)
    and so is continuous. Could you please explain, in as much detail as possible, how (1) and (2) were derived?

    © BrainMass Inc. brainmass.com March 4, 2021, 6:55 pm ad1c9bdddf
    https://brainmass.com/math/geometry-and-topology/fixed-point-theorem-closed-unit-ball-euclidean-space-69927

    Attachments

    Solution Preview

    Please see the attached file for the complete solution.
    Thanks for using BrainMass.

    The Brouwer Fixed-Point Theorem

    Let denote the closed unit ball in Euclidean space :
    .
    Any continuous map from onto itself has at least one fixed point, i.e. a point such that .

    Proof Suppose has no fixed points, i.e. for .
    Define a map , , by letting be the point of intersection of and the ray starting at the point and going through . For see figure below:

    We have
    with , (1)
    Then
    (2)
    and so is continuous. Could you please explain, in as much detail as possible, how (1) and (2) were derived?

    EXPLANATION: Since be the point of intersection of and the ray starting at the point and going through .
    So r(x) is a point on the surface of Hollow sphere. Which is the point of intersection.

    Since it is a point of intersection . Therefore r(x) is a point. Also given that r(x ) lies on
    Sn-1(x) . And it is also specific that

    so we can easily imagine that

    Any point on Sn-1 (x) is in such a way that x belongs to R n such that mod (x )= 1.

    So clearly any point on Sn-1 (x) is in such a way that

    with ,

    Also it is great that r(x) is a function in such a way it generates the points which are on the Hollow ball(sphere)(because mod (x) = 1) with the help of m and t.

    Where these m and t are just defined functions or expressions or standard expressions

    and

    Here m is taken in order that r(x) will lie on Sn-1(x).

    Also t is taken in such a way that totally r(x) will lie on Sn-1(x).

    So here Sn-1(x) is a hollow ball without any material in it.It just contains the over surface to cover like a ball.

    These are the explanations according to my point of view.

    CLARIFICATION PART:

    As with ,

    Also x + mt = 1 implies x + mt = +1 or - 1.

    Implies x + mt = 1 or x + mt = -1

    Implies mt = 1 - x or t = ( 1 - x )/m

    ANOTHER Equation ...

    Solution Summary

    Fixed Point Theorem and Closed Unit Ball in Euclidean Space are investigated. The solution is detailed and well presented.

    $2.49

    ADVERTISEMENT