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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 $\displaystyle f$ is not uniformly continuous. Then, there exists $\displaystyle \varepsilon> 0$ such that for every $\displaystyle \delta>0$ there exist $\displaystyle x,y\in X$ such that $\displaystyle d(x,y)<\delta$ but $\displaystyle d(f(x),f(y))\geq \varepsilon$.

    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 $\displaystyle \delta=1$ we can find points $\displaystyle x_1,y_1\in X$ such that $\displaystyle d(x_1,y_1)<1$ but $\displaystyle d(f(x_1),f(y_1))\geq \varepsilon$. For $\displaystyle \delta=1/2$, we can find $\displaystyle x_2,y_2\in X$ such that $\displaystyle d(x_2,y_2)<1/2$ but $\displaystyle d(f(x_2),f(y_2))\geq \varepsilon$. Continuing the process, we can find $\displaystyle \delta=1/n$ such that $\displaystyle d(x_n,y_n)<1/n$ but $\displaystyle d(f(x_n),f(y_n))\geq \varepsilon$.

    This gives you two sequences $\displaystyle \{x_n\}$ and $\displaystyle \{y_n\}$ such that $\displaystyle d(x_n,y_n)<1/n$ and $\displaystyle d(f(x_n),f(y_n))\geq \varepsilon$ for every $\displaystyle n$. Use sequential compactness here to relate them to the continuity of $\displaystyle f$.
    Last edited by roninpro; Jan 12th 2011 at 09:14 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 $\displaystyle \varepsilon>0$ be arbitrary and cover $\displaystyle f(K)$ with the open balls $\displaystyle \left\{B_{\varepsilon}(f(x))\right\}_{x\in K}$. Clearly then we have that $\displaystyle \left\{f^{-1}\left(B_{\varepsilon}(f(x))\right)\right\}_{x\in K}$ is an open cover for $\displaystyle K$. Then, since $\displaystyle K$ is compact we know that the open cover $\displaystyle \left\{f^{-1}\left(B_{\varepsilon}\left(f(x)\right)\right)\ri ght\}_{x\in K}$ admits a Lebesgue number $\displaystyle \delta$. Note then that if $\displaystyle d_X(x,y)<\delta$ then we have that $\displaystyle \text{diam}\{x,y\}<\delta$ (this last part was pedantic, but just in case) and so by definition we have there is some $\displaystyle f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ such that $\displaystyle x,y\in f^{-1}\left(B_{\varepsilon}(f(x_0))\right)$ and so $\displaystyle f(x),f(y)\in B_{\varepsilon}(f(x))$ and thus $\displaystyle d_Y(f(x),f(y))<2\varepsilon$.
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