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Thread: HEINE iff CAUCHY

  1. #1
    Jun 2013


    So I uploeded a picture of a theorem and it's proof, I understood the first part, however the second part I didn't understand why he just chose 1/n and wrote
    |xn-a|<1/n, I mean from what I understand he did it so he could use the sandwich theorem and show that 0<|xn-a|<1/n , 1/n converges to 0 , zero converges to 0 and there fore xn-a converges to zero and we get xn-->a, but because we assumed the that it is not continuous at a we got the sequence {f(xn)} doesn't converge to f(a).

    But was it necessary to just assume |xn-a|<1/n ?
    Firstly, if we just assume that for every xn that converges to a (in the range of the domain of the function), and we have the contrapositive of the continuity and therefore we get that f(xn) doesn't converge to f(a) and that's enough to prove without assuming |xn-a|<1/n... Isn't it? Or am I missing something here?

    Secondly, I think assuming that |xn-a|<1/n is problematic because in this case we are talking about a specific situation that is when
    |xn-a|<1/n . When we prove don't we have to talk about all sequences that converges to a, In other words, we need to prove for every sequence in general and not a specific situation when the sequence is smaller than 1/n. Because we are talking about the sequences only that are smaller than 1/n.

    Appreciate if someone can give me feedback, I am clearly missing something here...
    Thank you in advance.
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  2. #2
    MHF Contributor

    Aug 2006

    Re: HEINE iff CAUCHY

    Frankly, I don't think that the proof can be improved upon. One must fully understands what it means for $\displaystyle f$ not to be continuous at $\displaystyle a$.

    Please note there is one $\displaystyle \varepsilon > 0$. But corresponding to any $\displaystyle \delta > 0$ there is at least one

    That is the reason that we can be assured that $\displaystyle x_n$ exists for each $\displaystyle \frac{1}{n}$.

    EDIT: it should be
    Last edited by Plato; Feb 8th 2014 at 07:02 AM.
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