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Math Help - D bounded => f is bounded on D

  1. #1
    Sep 2009

    D bounded => f is bounded on D

    I need some help with a proof
    Assume \to \mathbb{R}" alt="f\to \mathbb{R}" /> is uniformly continous on D.
    Show that D being bounded implies f is bounded on D.

    All I can think of is to assume f is unbounded and show that it leads to a contradiction, but not sure how to proceed with the proof.
    any suggestions? tHanks
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  2. #2
    Jun 2009
    Precompactness is the key. Let f be uniformly continuous on D\subseteq  \mathbb{R}^n. You get d>0 such that the oscilation of fin each ball of radius d is smaller than 1. Every bounded subset in \mathbb{R}^n is precompact, hence you can find x_1,\ldots, x_n\in D such that D\subset\cup_{i=!}^{n}B(x_i,d). Now you can easily get a bound for |f|.
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