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Math Help - If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

  1. #1
    Senior Member Pinkk's Avatar
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    If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

    Here's my attempt:

    Assume to the contrary that f is a uniformly continuous function on a bounded set S but f is unbounded on S. Then there exists a sequence (s_{n})\in S such that |f(s_{n})|\ge n for all n\in \mathbb{N}. Since (s_{n}) is a bounded sequence, there exists a subsequence (s_{n_{k}}) that is a Cauchy sequence. Since f is uniformly continuous and (s_{n_{k}}) is a Cauchy sequence in S, then (f(s_{n_{k}})) is a Cauchy sequence. So (f(s_{n_{k}})) converges to some real number. But |f(s_{n_{k}})|\ge n_{k} for all n_{k}\in \mathbb{N} as well, which is a contradiction. Q.E.D.

    Is this correct? Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Here's my attempt:

    Assume to the contrary that f is a uniformly continuous function on a bounded set S but f is unbounded on S. Then there exists a sequence (s_{n})\in S such that |f(s_{n})|\ge n for all n\in \mathbb{N}. Since (s_{n}) is a bounded sequence, there exists a subsequence (s_{n_{k}}) that is a Cauchy sequence. Since f is uniformly continuous and (s_{n_{k}}) is a Cauchy sequence in S, then (f(s_{n_{k}})) is a Cauchy sequence. So (f(s_{n_{k}})) converges to some real number. But |f(s_{n_{k}})|\ge n_{k} for all n_{k}\in \mathbb{N} as well, which is a contradiction. Q.E.D.

    Is this correct? Thanks.
    Looks good to me.

    Another way to look at it might be as follows:

    Since f is unif. cont. we may choose a \delta such that |x-y|<\delta\implies |f(x)-f(y)|<1.

    Now notice that since S is bounded we have that S\subseteq [-M,M] for some M>0. But, this latter set is compact and thus we may cover it with finitely many intervals of the form I_1,\cdots,I_n such that \text{diam }I_k<\frac{\delta}{2}. Let I_{\alpha_1},\cdots,I_{\alpha_m} be those that intersect S. Choose x_k\in I_{\alpha_k}\cap S then I_{\alpha_k}\subseteq B_{\frac{\delta}{2}}(x_k). It follows that B_{\frac{\delta}{2}}(x_1),\cdots,B_{\frac{\delta}{  2}}(x_m) cover S.

    So, let x\in S. Then x\in B_{\frac{\delta}{2}}(x_k) for some k and so |f(x)|=|f(x)-f(x_k)+f(x_k)|\leqslant |f(x)-f(x_k)|+f(x_k)\leqslant 1+\max_{1\leqslant j\leqslant m}f(x_j)
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  3. #3
    Senior Member Pinkk's Avatar
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    Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all). Thanks for the feedback!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Hmm, thanks. The teacher's assistant for my real analysis class talks about using compactness in proofs for uniform continuity but our professor and curriculum doesn't cover compactness (at least we haven't touched on it at all).
    That's a shame. Compactness is one of the most important parts of analysis. A lot of important theorems are just statements about compactness and continuous maps (e.g. the extreme value theorem)

    Thanks for the feedback!
    Any time
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