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Thread: Help with Continuity Proof

  1. #1
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    Help with Continuity Proof

    I think I kinda know the direction this proof needs to go but I'm not sure how to word it all. Any help would be appreciated.

    Let $\displaystyle f:[0,1]\rightarrow [0,1]$ be continuous. Prove that there is a $\displaystyle c\in [0,1]$, such that $\displaystyle f(c)=c.$ [Hint: consider $\displaystyle g(x)=x-g(x)$

    I'm thinking a proof by contradiction might be apprpriate here. Assume that there is no c that works and then find a contradiction by using the hint somehow????
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    For which x in [0,1] g(x)>0, for which x in [0,1] g(x)<0 ?

    Now use Intermediate value theorem - Wikipedia, the free encyclopedia
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Suppose that $\displaystyle f(x)\ne x$ for all $\displaystyle x\in[0,1]$ and consider $\displaystyle \displaystyle g:[0,1]\to\{-1,1\}:\frac{f(x)-x}{|f(x)-x|}$. Since the denominator is non-zero this is continuous and surjective. But, this implies that $\displaystyle \{-1,1\}$ is connected....
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