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Intuitive Explanation For Why Continuous functions do not preserve Cauchy sequences

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data
    Why is it that continuous functions do not necessarily preserve cauchy sequences.


    2. Relevant equations
    Epsilon delta definition of continuity
    Sequential Characterisation of continuity


    3. The attempt at a solution
    I can't see why the proof that uniformly continuous functions preserve cauchy sequences doesn't hold for 'normal' continuous functions.
    In particular the example of f(x) = 1/x on (0,1)
    I have worked through the examples
    http://www.mathcs.org/analysis/reals/cont/answers/fcont3.html
    and here
    http://www.mathcs.org/analysis/reals/cont/answers/contuni4.html

    where they address this issue directly, but I can't get my head around it.

    I understand that if we have a cauchy sequence converging to 0, then f(xn) is going to diverge to infinity, but I still can't see what the problem is.

    Any explanation you can offer would be appreciated.

    Kind regards
     
  2. jcsd
  3. Mar 31, 2012 #2

    jgens

    User Avatar
    Gold Member

    Re: Intuitive Explanation For Why Continuous functions do not preserve Cauchy sequenc

    Recall that Cauchy sequences are bounded. So if [itex]\{f(x_n)\}_{n \in \mathbb{N}}[/itex] diverges, then the sequence cannot be Cauchy. In particular, [itex]f[/itex] does not take Cauchy sequences to Cauchy sequences.
     
  4. Mar 31, 2012 #3
    Re: Intuitive Explanation For Why Continuous functions do not preserve Cauchy sequenc

    The reason that we need uniform continuity is that we need to be able to find one [itex]\delta[/itex] for each [itex]\epsilon[/itex] that works for all [itex]x[/itex] in a certain interval. This is because in the proof, we do a "double triangle inequality." So, if [itex]\{f(x_n)\}[/itex] is a sequence of continuous functions that converges to [itex]f(x)[/itex] for each [itex]x[/itex] in the interval [itex](a,b)[/itex] then we want to show that [itex]\forall \epsilon \exists \delta[/itex] such that [itex]|f(x_0) - f(x)| < \epsilon[/itex] whenever [itex]|x_0 - x| < \delta[/itex]. We do this by writting:
    [tex]|f(x_0) - f(x)| = |f(x_0) - f_n(x_0) + f_n(x_0) - f_n(x) + f_n(x_0)-f(x)| \leq
    |f(x_0) - f_n(x_0)| + |f_n(x_0) - f_n(x)| + |f_n(x_0)-f(x)|[/tex]

    Now, since the sequence is Cauchy, we can control the outer two terms with a big enough [itex]n[/itex] and make them both less than [itex]\epsilon / 3[/itex]. So, we need to be able to ensure that [itex]|f_n(x) - f_n(x_0)| \leq \epsilon / 3[/itex] for every [itex]x[/itex] such that [itex]|x_0-x|\leq \delta[/itex]. The only way we can do this is by making [itex]f_n[/itex] uniformly continuous.

    As an example, consider the function [itex]f_n(x) = x^n[/itex] on [itex][0,1)[/itex].
     
  5. Apr 1, 2012 #4
    Re: Intuitive Explanation For Why Continuous functions do not preserve Cauchy sequenc

    Thanks very much to you both.
    I think I can see it more clearly now, (and a good nights sleep always helps too!).
    I will continue to play around with these ideas and if I have any more questions I'll be back.

    Thanks again
     
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