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Math Help - Uniformly continuous #3

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Uniformly continuous #3

    Prove that if f : (a,b)-->R is is uniformly continuous function in (a,b) so f is bounded function.
    Last edited by mr fantastic; January 3rd 2010 at 09:59 PM. Reason: Re-titled
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    Assume it's not bounded then pick a (unbounded) sequence (x_n) \subset (a,b) such that f(x_n)<f(x_{n+1}) for all n then since this sequence is bounded we get a Cauchy subsequence (x_{n_k}) by Bolzano-Weierstrass. Since f is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about f(x_{n_k}) by construction?.
    Last edited by Jose27; January 3rd 2010 at 01:55 PM.
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    Quote Originally Posted by Jose27 View Post
    Assume it's not bounded then pick a (unbounded) sequence (x_n) \subset (a,b) such that x_n<x_{n+1} for all n then since this sequence is bounded we get a Cauchy subsequence (x_{n_k}) by Bolzano-Weierstrass. Since f is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about f(x_{n_k}) by construction?.
    Jose27, I think you want too 'clean up' that proof.
    I know what you mean but you did not do it.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    hmmm... I don't understand...
    What is Cauchy subsequence ?
    "Since f is uniformly cont. it sends Cauchy seq..." why is that?
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    Quote Originally Posted by Jose27 View Post
    Assume it's not bounded then pick a (unbounded) sequence (x_n) \subset (a,b) such that f(x_n)<f(x_{n+1}) for all n. Since \left(x_n\right) is bounded we get a Cauchy subsequence (x_{n_k}) by Bolzano-Weierstrass. Since f is uniformly cont. it sends Cauchy seq. into Cauchy seq. but what can you say about f(x_{n_k}) by construction?.
    I believe that now it is easier to understand it.

    Quote Originally Posted by Also sprach Zarathustra View Post
    hmmm... I don't understand...
    What is Cauchy subsequence ?
    "Since f is uniformly cont. it sends Cauchy seq..." why is that?
    If f is unif. cont and \left(x_{n_k}\right) is Cauchy, then f\left(x_{n_k}\right) is Cauchy too.

    Let \epsilon>0. Since f is unif cont, there exists \delta>0 such as |x-y|<\delta\Rightarrow |f(x)-f(y)|<\epsilon. Since \left(x_{n_k}\right) is Cauchy, there existe K such as |x_{n_i}-x_{n_j}|<\delta for all i,j\geq K.
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    Quote Originally Posted by Plato View Post
    Jose27, I think you want too 'clean up' that proof.
    I know what you mean but you did not do it.
    Yeah, already did. Sorry, it made no sense as it was.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Dear Jose! Can you please write me why the existence of such K ==> boundedness of f?
    So you know a Cauchy sequence is bounded, and Abu-Khalil proved that if f is a unif. cont. function and x_n is a Cauchy seq. then f(x_n) is also a Cauchy seq. but by how we constructed them f(x_n) \rightarrow \infty (or -\infty in which case invert all inequalities in my first post) clearly contradicting the fact that f is unif. cont. so we conclude that our initial hypothesis is false ie. f is bounded.
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