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Math Help - Help with Intro to Real Analysis proof with sequence and limit.

  1. #1
    Member ilikedmath's Avatar
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    Exclamation Help with Intro to Real Analysis proof with sequence and limit.

    Problem states:
    Let be a sequence in with = a.
    If a > 0, prove (the set of Natural numbers) such that .

    My work so far:
    I don't have a proof written up for this problem yet, but I have jotted down some ideas that I think might help me:

    I know that the sequence { a_n} converges to a and a > 0.
    So by definition of convergence, I know that there exists an n_0 in the Naturals such that
    | a_n - a| < ε for all n ≥ n_0.

    There is also the theorem that every convergent sequence is bounded, so by definition of bounded: there exists an M > 0 such that
    | a_n| ≤ M for all n in the Naturals.

    So for this proof do I need cases? Would I do proof by contradiction?
    I worked with a study group on this, and my classmate drew a picture and talked us through the thinking. It kind of made sense to me that if a sequence converges to zero, then eventually the sequence has to be greater than zero at some point.

    Any help, suggestions, and/or tips are greatly appreciated.
    Thank you for your time! =)
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  2. #2
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    From the fact that \left( {a_n } \right) \to a>0 you use the definition of sequence convergence.
    But in place of just \varepsilon >0 you are free to use \frac{a}{2} >0.
    \begin{gathered}  \left| {a_n  - a} \right| < \frac{a}{2} \hfill \\<br />
   - \frac{a}{2} < a_n  - a < \frac{a}{2} \hfill \\  0 < \frac{a}{2} < a_n  < \frac{{3a}}{2} \hfill \\ \end{gathered}
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  3. #3
    Member ilikedmath's Avatar
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    Okay, so can I just choose any value for epsilon? Like could I have chosen a/3 or a/7 > 0?

    Here's my proof so far:

    QED.

    is that okay?
    Last edited by ilikedmath; November 1st 2008 at 12:18 PM.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    You could do this proof by contradiction. Assume that there does not exist a N such that N\leqslant n\implies a_n>0

    By definition of \lim_{n\to\infty}a_n=a we must have that for every \varepsilon>0 there exists a corresponding N such that N\leqslant n\implies|a_n-a|<\varepsilon. So if we take \varepsilon=\frac{a}{2} we must be able to find a N such that |a_n-a|<\frac{a}{2}, but now by assumption a\leqslant|a_n-a| which gives a contradiction.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by Mathstud28 View Post
    You could do this proof by contradiction. Assume that there does not exist a N such that N\leqslant n\implies a_n>0

    By definition of \lim_{n\to\infty}a_n=a we must have that for every \varepsilon>0 there exists a corresponding N such that N\leqslant n\implies|a_n-a|<\varepsilon. So if we take \varepsilon=\frac{a}{2} we must be able to find a N such that |a_n-a|<\frac{a}{2}, but now by assumption a\leqslant|a_n-a| which gives a contradiction.
    Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last n is not the same as the n introduced earlier).

    CB
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    Confused, it can be done this way (though I can't see why one would want to) and this is probably OK when all the missing detail is put in but in this form it is near incomprehensible (and the last n is not the same as the n introduced earlier).

    CB
    Yes you are right, that is unclear. Let me fix it up.

    Suppositions:
    1. a>0
    2. \lim a_n=a
    3. There exists a N such that N\leqslant n\implies a_n\leqslant 0

    A fact we will use, if x>0 and y<0 then x\leqslant|x-y|

    Now by the definition of a sequence \left\{a_n\right\} converging to a we must have that for every \varepsilon>0 there exists a corresponding N_1 such that N_1\leqslant n\implies |a-a_n|<\frac{a}{2}. So now let \varepsilon=\frac{a}{2}, so we must be able to find a N_1 such that N_1<n\implies |a_n-a|<\varepsilon. So it must be true that if we take N_2=\max\left\{N,N_1\right\} that N_2\leqslant n\implies |a_n-a|<\frac{a}{2}. But here we arrive at a contradiction. The reason being that because of supposition three we must have that N_2\leqslant n\implies |a-a_n|\geqslant a\implies |a_n-a|\not<\frac{a}{2}
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