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Proof about uniformly continuous functions

  1. Mar 8, 2009 #1
    Let f be a uniformly continuous function on Q... Prove that there is a continuous function
    g on R extending f (that is, g(x) = f(x), for all x∈Q


    I think I am supposed to somehow use the denseness of Q and the continuity of a function to prove this, but I am not quite sure where I should start. Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 8, 2009 #2
    Pick an irrational t, what can you do with rationals to estimate it?
     
  4. Mar 8, 2009 #3

    Dick

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    First you have to define what g(x) is. Yes, use the denseness of Q. Can you do that?
     
  5. Mar 9, 2009 #4
    so would i be right to start out defining g as uniformly continuous on R?

    then somehow using the denseness of Q to show that all the Qs are in R so g extends f?

    ... or am i not understanding what this means by extending?
     
  6. Mar 9, 2009 #5

    Dick

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    For every irrational t there are an infinite number of sequences q_i of rationals such that limit q_i->t. So you can define g(t)=limit f(q_i). Now you have to show that g(t) is independent of the sequence q_i. Use the continuity of f on the rationals. To show it's uniformly continuous, you can't define it to be. You have to show it is. Again use the uniform continuity of f.
     
  7. Mar 10, 2009 #6
    ok I think I can do it now. thanks :) I'll let you know how it goes in a little bit!
     
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