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Math Help - Uniform Continuity

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    Uniform Continuity

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    Last edited by lovesmath; November 17th 2012 at 05:13 PM.
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    Re: Uniform Continuity

    Quote Originally Posted by lovesmath View Post
    Prove that the function is uniformly continuous on the set of real numbers.
    f(x)=|x|
    I think I have to use the reverse triangle inequality, but I am not sure how to do it.
    There you have it.

    \left| {\left| x \right| - \left| {x_0 } \right|} \right| \leqslant \left| {x - x_0 } \right| < \delta  = \varepsilon
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    Re: Uniform Continuity

    Here is an additional question.

    Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.
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    Re: Uniform Continuity

    Quote Originally Posted by lovesmath View Post
    Here is an additional question.

    Suppose f is a uniformly continuous function on a bounded set D contained in the set of real numbers. Prove that f(D) is a bounded set.
    What have you tried so far?

    -Dan
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    Re: Uniform Continuity

    I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.

    Not sure what to do next.
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    Re: Uniform Continuity

    Quote Originally Posted by lovesmath View Post
    I know I can set Epsilon=1. There exists a delta>0 such that x,y are elements in D and |x-y|<delta implies that |f(x)-f(y)|<E. Then |f(x)-f(y)|<1.
    The key to this is the fact that D is bounded.
    That means that \exists M>0 such that D\subset [-M,M].

    In your text material you should have a proof that a continuous function is bounded on a closed interval of real numbers. This new proof follows that proof very closely.
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