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Prove function continuous at only one point

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the function defined as f(x)= x when x is rational and -x when x is irrational is only continuous at 0.


    2. Relevant equations



    3. The attempt at a solution
    I have been looking at this website which proves this:
    http://planetmath.org/encyclopedia/FunctionContinuousAtOnlyOnePoint.html [Broken]

    I don't understand how the writer proves that the function is not continuous when a[tex]\neq[/tex]0.
    Why do they take xk= to something? How does this help them show that the limits equal a and -a.
    How does showing that prove that f is not continuous at a?

    I know those are a lot of question, but maybe an answer to even just the first ones would help me understand this proof.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 30, 2009 #2

    CompuChip

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    Homework Helper

    Do you agree that
    "f is continuous at a, if for all sequences xk that converge to a, f(xk) converges to f(a)" ?

    So then to show that f is not continuous at a, all you need is to find one sequence that does converge to a but for which the limit of the f(xk) is not f(a). (The negation of "for all sequences S, P(S)" is "there exists a sequence S such that not P(S)").

    So what they do is take two different sequences which both have limit a, and show that when you apply f first and take the limit, you get two different results. Since f(a) cannot be equal to a and -a at the same time, at least one of this sequence forms a counter-example to the statement of continuity they start out with.
     
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