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Continuous functions on dense subsets

  1. Apr 15, 2012 #1
    Hi, can someone give me pointers on this question
    1. The problem statement, all variables and given/known data

    Prove or provide a counterexample: If f : E -> Y is continuous on a
    dense subset E of a metric space X, then there is a continuous function
    g: X -> Y such that g(z) = f(z) for all z element of E.

    3. The attempt at a solution
    I'm not sure if the statement is true or not. I have tried to find counter-examples using continuous functions on the rationals or irrationals. For example f: Q -> R , f(x) = x. This is continuous for every x in Q. However it is easy to find a mapping g: R -> R which is continuous and g(z) = f(z) for all z element of Q. ie given by g(x) = x. I am yet to find a counter-example (in R anyway). However if the statement holds I'm not to sure how I would begin to prove it.

    Thanks!
     
  2. jcsd
  3. Apr 15, 2012 #2

    HallsofIvy

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    There is a reason you can't find a counter-example!

    Let a be a point in X that is not in E. Since E is dense in X, there exist a sequence of points, [itex]\{x_n\}[/itex] in E that converges to a. Show that, since f is continuous in E, the sequence [itex]\{f(x_n}\}[/itex] converges and define that limit to be f(a).
     
  4. Apr 15, 2012 #3

    micromass

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    There is a counterexample.

    Take E=]0,1] and X=[0,1].

    Can you find a function on E whose limit to 0 does not exist?
     
  5. Apr 15, 2012 #4
    Thanks guys, makes a lot more sense now!
     
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