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Cauchy sequences

  1. Oct 10, 2012 #1
    I read the proof of the proposition "every cauchy sequence in a metric spaces is bounded" from

    http://www.proofwiki.org/wiki/Every_Cauchy_Sequence_is_Bounded [Broken]

    I don't understand that how we can take m=N[itex]_{1}[/itex] while m>N[itex]_{1}[/itex] ?

    In fact i mean that in a metric space (A,d) can we say that

    [[itex]\forall[/itex]m,n>N[itex]_{1}[/itex][itex]\Rightarrow[/itex] d(x[itex]_{n}[/itex],x[itex]_{m}[/itex])<1][itex]\Rightarrow[/itex][[itex]\forall[/itex]n[itex]\geq[/itex]N[itex]_{1}[/itex][itex]\Rightarrow[/itex] d(x[itex]_{n}[/itex],x_{[itex]_{N_{1}}[/itex]})<1]
     
    Last edited by a moderator: May 6, 2017
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  3. Oct 10, 2012 #2

    Erland

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    You are right. This is an error in the wiki. [itex]m,n>N[/itex] should be changed to [itex]m,n\ge N[/itex] wherever it occurs (this also holds with [itex]N_1[/itex] instead of [itex]N[/itex]). This fits the wiki's definition of Cauchy sequence, which the wiki's proof doesn't.
     
    Last edited by a moderator: May 6, 2017
  4. Oct 11, 2012 #3
    Thank you for the answer i also think like you. This is an error in the wiki. But i saw several functional analysis book which write the proof of proposition same as in wiki. So,

    Who is wrong?
     
  5. Oct 11, 2012 #4

    Erland

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    Well, we can define "Cauchy sequence" with either [itex]>[/itex] or [itex]\ge[/itex], but in the former case, we cannot use [itex]N_1[/itex] the way it is used in the proof in the wiki. Then we also need an [itex]N_2>N_1[/itex] to work with, or something like that.
     
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