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Proof: Every convergent sequence is Cauchy

  1. Nov 16, 2015 #1
    Hi,

    I am trying to prove that every convergent sequence is Cauchy - just wanted to see if my reasoning is valid and that the proof is correct.

    Thanks!

    1. The problem statement, all variables and given/known data


    Prove that every convergent sequence is Cauchy

    2. Relevant equations / Theorems

    Theorem 1: Every convergent set is bounded

    Theorem 2: Every non-empty bounded set has a supremum (through the completeness axiom)

    Theorem 3: Limit of sequence with above properties = Sup S (proved elsewhere) Incorrect - not taken as true in second attempt of proof

    3. The attempt at a solution

    Suppose (sn) is a convergent sequence with limit L. Let S = {sn: n∈ℕ}

    By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

    Since (sn) is convergent, for every ε > 0, ∃N such that ∀n > N, | sn - L | < ε, where L = M

    Now, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

    But then this means ∃sn such that M - φ < sn < M.

    i.e. sn + α = M for some α > 0

    So M - ε < M - φ < sn < M.

    This is the same as M - ε < M - φ < M - α < M

    But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

    I.e. the sequence is Cauchy.
     
    Last edited: Nov 16, 2015
  2. jcsd
  3. Nov 16, 2015 #2

    LCKurtz

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    There is no reason to suppose L = M. And your theorem 3 above, whatever it means, is false. And you have not even stated what a Cauchy sequence is, let alone proved that property.
     
  4. Nov 16, 2015 #3
    Thanks LCKurtz,

    I think i implcitly assumed that the sequence is monotonically increasing. But the Theorem 3 is not true as a a sequence denoted by the sine function would be bounded, yet not convergent I suppose.

    I will give it another crack
     
  5. Nov 16, 2015 #4

    LCKurtz

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    Start by writing the definition of a Cauchy sequence. Then use:

    (sn) is convergent [to L if], for every ε > 0, ∃N such that ∀n > N, | sn - L | < ε, where L = M

    without the incorrect part in red, to prove it. It has nothing to do with sup.
     
  6. Nov 16, 2015 #5
    3. The attempt at a solution

    Suppose (sn) is a convergent sequence with limit L. Let S = {sn: n∈ℕ}

    By Theorem 1, every convergent set is bounded. By Theorem 2, sup S exists, let Sup S = M.

    Now, since M = Sup S, consider some M - φ < M for some 0 < φ < ε.

    But then this means ∃sn such that M - φ < sn < M.

    i.e. sn + α = M for some α > 0

    So M - ε < M - φ < sn < M.

    This is the same as M - ε < M - φ < M - α < M

    But then | (M - ε) - (M - φ) | > | (M - φ) - (M - α) |.

    I.e. the sequence is Cauchy.
     
  7. Nov 16, 2015 #6

    LCKurtz

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    No. See my post #4 which I apparently posted the same time you were posting #5.
     
  8. Nov 16, 2015 #7
    Attempt

    A sequence is called a Cauchy sequence if the terms of the sequence eventually all become arbitrarily close to one another

    That is, given ε > 0 there exists N such that if m, n > N then | am - an | < ε

    (sn) is convergent to L if for every ε > 0, ∃N such that ∀n > N, | sn - L | < ε

    which implies sn < ε + L

    Now consider three consecutive numbers in the sequence sn, so, sp. It follows that

    sn < ε + L

    so < α + L

    sp < β + L

    where β ≤ α ≤ ε by notion of convergence.

    it then follows that the distance between sn and so; | sn - so | = | ε - α | ≥ | so - sp | = | α - β |.

    Hence the terms get closer and closer, so the sequence is Cauchy.
     
  9. Nov 16, 2015 #8
    Thanks,

    Would the argument I posted before be incorrect? I essentially just used the supremum as a point of reference to guage distance between points.
     
  10. Nov 16, 2015 #9

    LCKurtz

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    None of your arguments look good, sorry to say. You have the definitions you have given earlier (I have edited them slightly for clarity):

    The sequence ##\{s_n\}## is convergent to L if for every ε > 0, ∃N such that ∀n > N, ##|s_n-L|<\epsilon##. That is what you are given to work with.

    And the sequence ##\{s_n\}## is Cauchy means that given ε > 0 there exists N such that if m, n > N then ##|s_m-s_n|<\epsilon##. That is what you have to prove.

    Now if you think about that, you are informally saying that ##s_n## gets close to L for large n because the sequence converges to it. So for large n and m, both ##s_n## and ##s_m## get close to L. That is what you are given. Now, if they both get close to L, wouldn't they have to get close to each other? Intuitively it makes sense. Your problem is to write down the appropriate inequalities to prove it. I hope that's enough of a hint to get you going.
     
  11. Nov 17, 2015 #10
    OK so from the above, I would do as follows:

    Third Attempt

    Since ##|s_n-L| < \epsilon##

    ##|s_n| < |L + \epsilon|##

    ##|s_n - s_m| < |L + \epsilon - s_m|##

    ##| (s_n - L) - (s_m - L) | < |L + \epsilon - s_m|##

    But ##|s_n-L| < \epsilon## and ##|s_m-L| < \epsilon## as the both ##s_m## and ##s_n## converge to L, for ##∀m,n##

    So ##| (s_n - L) - (s_m - L) | < 2\epsilon## (by triangle inequality)

    i.e. ##| s_n - s_m | < 2\epsilon##, that is ##s_n## and ##s_m## get arbitrarily close. The sequence is therefore Cauchy.
     
    Last edited: Nov 17, 2015
  12. Nov 17, 2015 #11

    Samy_A

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    This last inequality isn't necessarily true.
    Take ##s_n=L-\frac{9}{10}\epsilon## and ##s_m=L+\frac{9}{10}\epsilon##
    Then ##|s_n - s_m|=\frac{18}{10}\epsilon##, and ##|L + \epsilon - s_m|=\frac{1}{10}\epsilon##

    But it doesn't really matter because you don't use that inequality.

    The core argument is here:
    Now you have to word this correctly (for example, writing "both ##s_m## and ##s_n## converge to L" is sloppy).
     
  13. Nov 17, 2015 #12
    Thanks Samy,

    I did not realise a sequence can converge to a Limit from both the left and the right - will keep that handy fact in mind.

    So in terms of wording I would approach as follows;

    ##∀\epsilon > 0, ∃N ∋## ##∀n,m > N, | s_n - L | < \epsilon## and ##| s_m - L | < \epsilon##.

    By the trianle inequality, ## | s_n - L + L - s_m | ≤ | s_n - L | + | s_m - L | < \epsilon + \epsilon = 2\epsilon##

    Since ##2\epsilon## is arbitrarily large, we can make both ##s_n## and ##s_m## as close together as we like, for ##∀n,m > N##.

    Therefore, the sequence is Cauchy.

    Would this be a more rigid argument?
     
  14. Nov 17, 2015 #13

    Samy_A

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    There is a typo, as you write "arbitrarily large", while you clearly mean "arbitrarily small".
    One can nitpick about details, but I think this proof is correct.
     
  15. Nov 17, 2015 #14
    Thank you for the help
     
  16. Nov 17, 2015 #15

    LCKurtz

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    And if you want to spiff it up a little, pick N so that if n,m > N then ##|s_n-L|<\frac \epsilon 2## and ##|s_m-L|<\frac \epsilon 2## in the first place, so ##|s_m-s_n|<\epsilon##.
     
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