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Thread: Borsuk-Ulam Theorem 1 dimensional

  1. #1
    MHF Contributor
    Mar 2010

    Borsuk-Ulam Theorem 1 dimensional

    Let $\displaystyle f:S^1\to \mathbb{R}$ be a continuous map. Show there exists a point x of $\displaystyle S^1$ such that $\displaystyle f(x)=f(-x)$.

    It says to define $\displaystyle g:S^1\to \mathbb{R}$ by $\displaystyle g(x)=f(x)-f(-x)$.

    $\displaystyle g(x)=f(x)-f(-x)=-(f(-x)-f(x))=-g(x)$

    By the Intermediate Value Theorem, there exist $\displaystyle x_0\in S^1$ such that $\displaystyle g(x_0)=0\Rightarrow f(x)=f(-x)$.

    Here is my question: Why did we need to define a new function g?
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  2. #2
    Senior Member roninpro's Avatar
    Nov 2009

    Re: Borsuk-Ulam Theorem 1 dimensional

    You defined the new function $\displaystyle g$ so that you could apply the Intermediate Value Theorem in a straightforward manner.

    Did you have something else in mind?
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