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Cauchy sequence

  1. Jan 15, 2010 #1
    "Definition: A sequence of real numbers (an) is Cauchy iff
    for all ε>0, there exists N s.t. n≥N and m≥N => |an-am|<ε.

    An equivalent definition is:
    for all ε>0, there exists N s.t. n≥N => |an-aN|<ε. "
    =============================================

    I don't exactly see why these definitions are equivalent.

    One direction (from 1st one to 2nd one) is clear, we can just take m=N which is clearly ≥N.

    But how can we prove the converse (i.e. starting with the 2nd definition, prove the 1st)?

    Any help is much appreciated!
     
    Last edited: Jan 16, 2010
  2. jcsd
  3. Jan 15, 2010 #2

    D H

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    |an-aN|<ε means for all n≥N, an is a member of the open interval (aN-ε,aN+ε). Given this, what can deduce whether some bounds (if any) exist for |an-am|, where n,m≥N?
     
  4. Jan 15, 2010 #3
    I agree with you that:
    for all n≥N, an is a member of the open interval (aN-ε,aN+ε)

    But I still don't see the connection, how can we get the "m" part in definition 1? (definition 2 doesn't contain any statements about "m")

    Thanks!
     
  5. Jan 16, 2010 #4
    For all ε≥0, there exists N s.t. n≥N => |an-aN|<ε/2. Fix this N. Suppose m≥N. By the triangle inequality, [itex]|a_n - a_m| \leq |a_n - a_N| + |a_N - a_m| < \varepsilon.[/itex]
     
  6. Jan 16, 2010 #5
    We know that n≥N => |an-aN|<ε/2 (*)

    So m≥N means that also |am-aN|<ε/2, right? (i.e. in (*) can we replace n by m?)
    Why are we labelling the subscript of the subsequence as "m" now? (i.e. why {am} not {an}?)

    thanks.
     
    Last edited: Jan 16, 2010
  7. Jan 16, 2010 #6
    Yes, you can replace n by m if m≥N (you can use any symbol you want as long as it fits the criteria)
     
  8. Jan 17, 2010 #7

    You have:

    For all ε>0.there exists a natural No N ,such that:

    For all n: [tex] n\geq N\Longrightarrow |a_{n}-a_{N}|<\epsilon[/tex]...............................................................................................1


    For all ε>0,there exists a natural No N such that:

    For all n ,for all m: [tex]n\geq N,m\geq N\Longrightarrow |a_{n}-a_{m}|<\epsilon[/tex]................................................................................................2

    Notice the "for all" expression that is missing in your statements and which will be the central issue in the proof that will follow


    And you want to show that (1) implies (2)


    LET ε>0,then from (1) we have :

    There exists a natural No N such that:

    For all n: [tex] n\geq N\Longrightarrow |a_{n}-a_{N}|<\epsilon[/tex]..................................................................................................3

    Since (3) holds for all ,n it will hold for :

    a) n=n and b) n=m

    The law of logic that allows to do that is called:

    Universal Elimination

    Hence we have:

    [tex] n\geq N\Longrightarrow |a_{n}-a_{N}|<\frac{\epsilon}{2}[/tex]...........................................................................................................4


    [tex] m\geq N\Longrightarrow |a_{m}-a_{N}|<\frac{\epsilon}{2}[/tex].......................................................................................................5


    Now ,let : [tex] n\geq N ,m\geq N[/tex] ,and by the law of logic called : M,Ponens and using (4) and (5) we have :


    [tex] |a_{n}-a_{N}|<\frac{\epsilon}{2}[/tex]..........................................................................................................6

    AND

    [tex]|a_{m}-a_{N}|<\frac{\epsilon}{2}[/tex]............................................................................................................7


    HENCE:

    [tex] |a_{n}-a_{m}| = |a_{n}-a_{N}+a_{N}-a_{m}|\leq |a_{n}-a{N}| + |a_{N} -a_{m}|<\frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon[/tex]


    So we have proved (2)


    BOTTOM LINE IS:

    To get into the mysteries of Analysis ,one need to do proof analysis or to become more formal in the proofs of Analysis
     
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