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Math Help - Completeness

  1. #1
    MHF Contributor Drexel28's Avatar
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    Completeness

    I am supposed to prove that the set of all bounded real functions on a metric space X with metric d(x,y)=\left\|f-g\right\|_{\infty}=\sup_{x\in X}\left|f(x)-g(x)\right| is complete.

    So we have some Cauchy sequence \left\{f_n\right\}_{n\in\mathbb{N}}. If we fix \xi\in X we can clearly see that \left\{f_n(\xi)\right\}_{n\in\mathbb{N}} is a Cauchy sequence in \mathbb{R} and thus there exists some \alpha\in\mathbb{R} such that f_n(\xi)\to\alpha. So, define \bar{f}:X\mapsto\mathbb{R} by \bar{f}(\xi)=\lim\text{ }f_n(\xi). It is an elementary concept from analysis that a Cauchy sequence of functions converges uniformly. Thus, we can make \left|f_n(x)-\bar{f}(x)\right|<\varepsilon by making n sufficiently large regardless of what value x is. Thus, for every \varepsilon>0 there exists some N\in\mathbb{N} such that n\leqslant N\implies \left|f_n(x)-\bar{f}(x)\right|<\varepsilon,\text{ }\forall x\in X and so n\leqslant N\implies \left\|f_n(x)-\bar{f}(x)\right\|_{\infty}<\varepsilon. So, it remains to show that \bar{f}(x) is bounded. But, this is trivial, since there exists some n\in\mathbb{N} such that \left|\bar{f}(x)\right|-\left|f_n(x)\right|\leqslant\left|\bar{f}(x)-f_n(x)\right|\leqslant 1 for every x\in X and thus \left|\bar{f}(x)\right|\leqslant\left|f_n(x)\right  |+1 but since f_n(x) is bounded for every n we see that there exists some M\in\mathbb{R} such that \left|\bar{f}(x)\right|\leqslant M+1 and thus \bar{f}(x) is bounded and thus it is in our metric space.

    So, we may conclude that this space is complete.


    How's that look?
    Last edited by Drexel28; January 11th 2010 at 11:29 PM. Reason: Clarity
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  2. #2
    Senior Member Shanks's Avatar
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    It is correct except some typo.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Shanks View Post
    It is correct except some typo.
    Thanks, Shanks! What specific typos?
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