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Thread: Uniformly continuous functions

  1. March 18th 2011, 06:30 AM #1
    worc3247
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    Uniformly continuous functions

    The question:
    Suppose g(a,b] \rightarrow \mathbb{R} is uniformly continuous. Show that if { x_n} is a Cauchy sequence in (a,b], then { g(x_n)} is always a Cauchy sequence.

    So I need to show that \forall \epsilon >0 \exists N \in \mathbb{N} s.t. \forall m,n>N |g(x_m)-g(x_n)|<\epsilon. I have then written out the definition for g being uniformly continuous on (a,b] but I can't see where to go from this definition to show the Cauchy sequence. Help?
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  2. March 18th 2011, 06:55 AM #2
    girdav
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    Applying the definition of uniformly continuous, for \varepsilon >0 given we can find a \delta such that if x,y\in\left(a,b\right] and |x-y|<\delta then |g(x)-g(y)|<\delta.
    We can find N_0 such that if m,n\geq N_0, then |x_m-x_n|< \delta.
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  3. March 18th 2011, 07:26 AM #3
    worc3247
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    Ok, so then I guess my question is how do we go about finding an N_0 that satisfies this?
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  4. March 18th 2011, 07:32 AM #4
    girdav
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    What about |g(x_n)-g(x_m)| if n,m\geq N_0?
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  5. March 18th 2011, 09:04 AM #5
    worc3247
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    Ok, I think i've got it thanks.
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