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There exist a sequence x_n E S s.t. x_n -> sup S

  1. Jan 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Consider sequence of real numbers.
    Theorem: If a= sup S, then there exist a sequence xn E S such that xn ->a

    Take ε = 1/n and find xn E S such that 0 ≤ a - xn < 1/n.
    Now show xn -> a.

    I am very very confused about this proof.
    1) Why are they taking ε = 1/n? What motivates this?

    2) It seems to me that n is simply a "subscript" of the sequence xn and it's a bit weird to talk about ε = 1/n. Is there any relationship between the "n" in ε = 1/n and the "n" in the sequence xn? Are they the SAME "n"?

    3) In the proof, they say "find xn E S such that 0 ≤ a - xn < 1/n", but how do we know that such a thing even EXISTS?

    4) At the end of the proof, they say "show xn -> a", but HOW??

    2. Relevant equations

    3. The attempt at a solution

    Can someone please explain the proof in greater detail?
    Any help is much appreciated! :)
  2. jcsd
  3. Jan 9, 2010 #2
    If there were NOT an element of S so that for all n, [tex]0 \leq a - x_n < 1/n[/tex], is [tex]a[/tex] really the supremum of S?

    Note that this inequality can also be written

    [tex] a- \frac{1}{n} < x_n \leq a [/tex]

    which may be more illustrative. The point of choosing [tex]\epsilon = 1/n[/tex] is because this will shrink the neighborhood around [tex]a[/tex] as [tex]n[/tex] gets larger. This should give you some indication as to the construction of the sequence.
  4. Jan 9, 2010 #3
    OK, and here is my attempt to rewrite and understand the proof in a more presentable manner.

    Take ε = 1/n.
    Now a = sup S => for EACH n=1,2,... there exists an x E S such that a -1/n < x ≤ a.
    For EACH n=1,2,... PICK one such point and CALL it xn,
    so that a -1/n < xn ≤ a for all n=1,2,...
    Now we need to prove that this sequence xn->a.
    By definition, xn->a iff
    for ALL ε>0, there exists an integer N such that n≥N => |xn - a|< ε.
    So for ANY given ε>0, I must be able to choose an N that works.
    But I am stuck here. How can we choose such an N in our case?

    Thank you! :)
  5. Jan 10, 2010 #4
    It looks like you're on the right track. Given [tex] \epsilon >0 [/tex], can you manipulate the quantity [tex] 1/n[/tex] in some manner to help you? Remember, [tex]n[/tex] gets really, really big as you go out in the sequence. What does this say about [tex]1/n[/tex]?

    You have literally constructed the sequence already in the first part of your proof...
  6. Jan 10, 2010 #5
    I've constructed the sequence, but I am not sure how to prove it "rigorously" that the sequence converges to a.

    Definition: xn->a iff
    for ALL ε>0, there exists an integer N such that n≥N => |xn - a|< ε.

    a -1/n < xn ≤ a for all n=1,2,...
    => -1/n < xn - a ≤ 0 < 1/n for all n=1,2,...
    => |xn - a| < 1/n for all n=1,2,...

    But at the beginning of the proof we said that ε = 1/n????? So I am a little confused...why is ε changing for different values of n? This is really werid...shouldn't epsilon be given to be some fixed positive number?
    Also, given ε, how should I choose N such that n≥N => |xn - a|< ε?

    Can someone please help me out?
  7. Jan 10, 2010 #6
    Mmm'kay, I don't really like how they set [tex]\epsilon = 1/n[/tex]. That seems strange.

    Rather, for any given [tex]\epsilon>0[/tex], there always exists an [tex]n[/tex] large enough so that [tex] 1/n < \epsilon[/tex]. But, you've already constructed a sequence such that

    [tex] a - \frac{1}{n} < x_n \leq a [/tex]

    for all [tex]n[/tex]. How is [tex] \left|x_n - a \right| < \epsilon[/tex] related to this inequality given that [tex]1/n < \epsilon[/tex]? What should you choose for your [tex]N[/tex]?

    Try not to get too bogged down in the definitions; i.e., try to look at the big picture before diving into epsilons and stuff.
  8. Jan 16, 2010 #7
    OK, so I think taking N>(1/ε) works.

    At the beginning of the proof, they said that ε = 1/n and ε changing for different values of n? This is werid. I believe that in the definition of limit, ε should be given to be some FIXED arbitrary positive number?

    When they say "ε = 1/n" at the beginning of the proof, is this the "same" ε as the one that appears at the end when we're trying to prove the limit using definition? or are they totally unrelated?

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