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F continuous at x[sub]0[/sub], prove g is continuous atx[sub]0[/sub]

  1. Nov 15, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f: E--> R is cont at x0 and x0 is an element of F contained in E. Define g:F--->R by g(x)=f(x) for all x elemts of F. Prove g is continuous at x0. Show by example that the continuity of g at x0 need not imply the continuity of f at x0.

    2. Relevant equations
    lx-x0l<delta
    lf(x)-f(x0)l<epsilon


    3. The attempt at a solution
    lx-x0l<delta
    lg(x)-g(x0)l<epsilon
    lf(x)-f(x0l<epsilon
    Ok, then it's continuous because g(x)=f(x)?
     
  2. jcsd
  3. Nov 15, 2008 #2

    morphism

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    Your proof isn't very convincing. Write it out in words.
     
  4. Nov 15, 2008 #3

    HallsofIvy

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    Morphism's point is that "sketched" the proof but you need to say exactly why those statements prove the theorem.
     
  5. Nov 15, 2008 #4
    Ok, but how would I do the part with showing by example f doesn't need to be continuous. Wouldn't it ahve to be continuous since g(x)=f(x)?
     
  6. Nov 15, 2008 #5

    morphism

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    Do g and f have the same domain...?
     
  7. Nov 15, 2008 #6
    No, so if f is in E, then g could still be continuous?
     
  8. Nov 15, 2008 #7

    Office_Shredder

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    F is a subset of E, so we know f is continuous. Hence, for all e>0, there exists d>0 such that
    |x-x0<d and x in E implies |f(x)-f(x0|<e

    Hence
    |x-x0<d and x in F implies |f(x)-f(x0|<e

    But f(x)=g(x) and f(x0)=g(x0)

    Hence |x-x0<d and x in F implies |g(x)-g(x0|<e

    This is a much clearer way of writing the proof

    For a counterexample, I would suggest looking at step functions
     
  9. Nov 15, 2008 #8
    Thanks!
     
  10. Nov 19, 2008 #9
    Ok, i'm confused on finding an example. Like what if I choose the function [x]?
     
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