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A simple question about convergence

  1. Nov 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Let x_n be a convergent sequence with a ≤ x_n for every n, where a is any number. Prove that a ≤ lim x_n when n→∞.

    2. Relevant equations

    Definition of limit. The usual ε, N stuff.

    3. The attempt at a solution

    Let lim x_n = x and choose ε=x_n-a. Hence we have |x_n - x| < x_n - a which shows that -x<-a and thus x>a.

    Is this valid?
     
  2. jcsd
  3. Nov 3, 2012 #2

    micromass

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    You do realize that [itex]\varepsilon[/itex] depends on n?? For a different n, you'll get a different [itex]\epsilon[/itex]. Do you really want that??
    Also, why is [itex]\varepsilon>0[/itex]? Specifically, why is it nonzero?
     
  4. Nov 3, 2012 #3

    tiny-tim

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    hi bedi! :smile:

    (try using the X2 button just above the Reply box :wink:)
    but ε has to be independent of n :confused:

    (ooh, micromass beat me to it! :biggrin:)
     
  5. Nov 3, 2012 #4
    Yes, you are right. However I still can't see the solution :(
     
  6. Nov 3, 2012 #5

    tiny-tim

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    try assuming the contrary :wink:
     
  7. Nov 3, 2012 #6
    Alright, so this will imply that x_n converges to both a and x which is a contradiction. Am I right?
     
  8. Nov 3, 2012 #7

    micromass

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    Take [itex]x_n=2[/itex] for all n and take [itex]a=0[/itex]. Then certainly [itex]x_n[/itex] does not converge to a. So no, you're not right.
     
  9. Nov 3, 2012 #8
    But I assumed that a>x. So -a<-x and x_n-a<x_n-x<ε. Hence x_n-a<ε ?
     
  10. Nov 3, 2012 #9

    micromass

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    Although the idea is there, the proof is still not very nice. For example, what is [itex]\varepsilon[/itex]?? You got to say things like this, not just introduce them without telling anybody what it is.
     
  11. Nov 3, 2012 #10

    tiny-tim

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    bedi, you know this is supposed to be a delta,epsilon proof …

    so you must define delta, and you must define epsilon​
     
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