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  1. #1
    mus
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    Uniformly continuous

    Wazzup!

    How do I get around proving (by an indirect argument) that a continuos mapping f: X -> Y between metric-spaces (X,d_X) and (Y,d_y) is uniformly continuous in every sequentially compact subset K of X.
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  2. #2
    Senior Member roninpro's Avatar
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    My suggestion is to suppose that f is not uniformly continuous. Then, there exists \varepsilon> 0 such that for every \delta>0 there exist x,y\in X such that d(x,y)<\delta but d(f(x),f(y))\geq \varepsilon.

    Try to construct a sequence that behaves badly. Keep in mind that the space is sequentially compact.
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  3. #3
    mus
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    How do I use tha fact that the space is sequentially compact.
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  4. #4
    Senior Member roninpro's Avatar
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    You first need to construct some kind of sequence out of the failure of uniform continuity. For example, for \delta=1 we can find points x_1,y_1\in X such that d(x_1,y_1)<1 but d(f(x_1),f(y_1))\geq \varepsilon. For \delta=1/2, we can find x_2,y_2\in X such that d(x_2,y_2)<1/2 but d(f(x_2),f(y_2))\geq \varepsilon. Continuing the process, we can find \delta=1/n such that d(x_n,y_n)<1/n but d(f(x_n),f(y_n))\geq \varepsilon.

    This gives you two sequences \{x_n\} and \{y_n\} such that d(x_n,y_n)<1/n and d(f(x_n),f(y_n))\geq \varepsilon for every n. Use sequential compactness here to relate them to the continuity of f.
    Last edited by roninpro; January 12th 2011 at 10:14 AM. Reason: Included another condition for the sequences.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mus View Post
    Wazzup!

    How do I get around proving (by an indirect argument) that a continuos mapping f: X -> Y between metric-spaces (X,d_X) and (Y,d_y) is uniformly continuous in every sequentially compact subset K of X.
    Note: This is more high-powered for than roninpro's suggestion. If it's gobbeldygook to you just ignore it.


    It's trivial that for metric spaces sequential compactness and compactness are synonymous. So, let \varepsilon>0 be arbitrary and cover f(K) with the open balls \left\{B_{\varepsilon}(f(x))\right\}_{x\in K}. Clearly then we have that \left\{f^{-1}\left(B_{\varepsilon}(f(x))\right)\right\}_{x\in K} is an open cover for K. Then, since K is compact we know that the open cover \left\{f^{-1}\left(B_{\varepsilon}\left(f(x)\right)\right)\ri  ght\}_{x\in K} admits a Lebesgue number \delta. Note then that if d_X(x,y)<\delta then we have that \text{diam}\{x,y\}<\delta (this last part was pedantic, but just in case) and so by definition we have there is some f^{-1}\left(B_{\varepsilon}(f(x_0))\right) such that x,y\in f^{-1}\left(B_{\varepsilon}(f(x_0))\right) and so f(x),f(y)\in B_{\varepsilon}(f(x)) and thus d_Y(f(x),f(y))<2\varepsilon.
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