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Math Help - prove that the composition gof:A-> C is uniformly continuous

  1. #1
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    prove that the composition gof:A-> C is uniformly continuous

    Let A,B,C be nonempty sets of real numbers, and let f:A->B and g:B->C be uniformly continuous functions. Prove that the composition gof:A->C is also unfiormly continuous,

    Here's what I have. Is it correct?

    Let \epsilon >0 then there is \delta>0 s.t for all x1,y1 \in A if |x_1-y_1|<\delta then | f(x_1)-f(y_1)|<\epsilon so f(x1),f(y1) \inB. Now let \epsilon>0 then there exists \delta>0 s.t for all x2,y2 \inB if |x2-y2|< \delta then |g(x2)-g(y2)|< \epsilon. So |(gof)(x1)-(gof)(y1)|=|g(f(x1))-g(f(y1))|< \epsilon.
    Last edited by Plato; April 9th 2011 at 12:44 PM. Reason: LaTeX fix
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  2. #2
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    No, that won't do. You've given the definitions of uniform continuity for f and g, but you haven't actually shown how they combine to produce the end result.

    Let \epsilon_1>0. The uniform continuity of g gives you a special \delta_1>0. Then let \epsilon_2=\delta_1, and the uniform continuity of f gives you \delta_2>0. Show that if |x-y|<\delta_2 then |(g\circ f)(x)-(g\circ f)(y)|<\epsilon_1.
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  3. #3
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    Quote Originally Posted by alice8675309 View Post
    Let A,B,C be nonempty sets of real numbers, and let f:A->B and g:B->C be uniformly continuous functions. Prove that the composition gof:A->C is also unfiormly continuous
    Start with \epsilon>0 pick a \delta>0 for \mathbf{g}. Then pick a \gamma with respect to \delta for \mathbf{f}

    PS. I did not see reply #2 before posting.
    BUT you need to work on your LaTeX
    I tried to edit your post. Double click on the code to see how it works.
    In particular, how subscripts work.
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