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I Proof of convergence & divergence of increasing sequence

  1. Oct 1, 2016 #1
    I'm using the book of Jerome Keisler: Elementary calculus an infinitesimal approach. I have trouble understanding the proof of the following theorem. I'm not sure what it means.

    Theorem: "An increasing sequence <Sn> either converges or diverges to infinity."
    Let T be the set of all real numbers x such that x≤Sn for some n.
    Case 1: T is the whole real line. If H is infinite we have x≤SH for all real numbers x. So SH is positive infinite and <Sn> diverges to ∞.
    Case 2: T is not the whole real line. By the completeness theorem, T is an interval (-∞,b] or (-∞,b). For each real x<b, we have :
    x≤Sn≤Sn+1≤Sn+2 . . . ≤b
    for some n. It follows that for infinite H, SH≤b and SH≈b. therefore, SH converges to b.

    The book states the definition of an interval as the completeness axiom:
    Completeness Axiom:
    "Let A be a set of real numbers such that whenever x and y are in A, then any real number between x and y are in A. Then A is an Interval."

    1.) When it says "Let T be the set of all real numbers x such that x≤Sn for some n". What does it mean? "some n" means not just one n but maybe a few ns. Or does it mean that as long as x is less than some some element of the sequence Sn then it s part of the set T? English isn't my first language.
    2.) If x≤Sk, then x≤Sk≤Sk+1≤Sk+2 . . . because <Sn> is increasing. Then the set T must include all x≤SH where H is infinity. Did I understand it correctly? Again I think it means that as long as x is less than some some element of the sequence Sn then it is part of the set T.
    3.) I think I understand case 1, but please check if I really understood it. My understanding is that:
    Since T is the whole real line then x can be any real number and since x≤Sn for some n, then x≤Sn≤Sn+1≤Sn+2 . . .S. Then x≤S where x is any real number you may think of. S is positive infinite.
    4.)In case 2. If T is not the whole real line, it's easy to visualize why it is an interval (-∞,b] or (-∞,b),but I don't see how it follows from the completeness axiom. It might require several logical steps, but it does not follow immediately, at least for me. But if T is (-∞,b] or (-∞,b), why x<b only why not x≤b?. The rest is just a bit hazy for me, I get it a bit but not clear enough. Please explain case 2. Thanks in advance.
  2. jcsd
  3. Oct 1, 2016 #2


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    What is H?

    For your question (1), the answer is the second of the two alternatives you describe.

    Your point 4 is absolutely correct! The proof in Case (2) is incorrect. It does not follow from the Completeness Theorem that T is either ##(-\infty,b)## or ##(-\infty,b]##, ie that it is bounded above. Additional reasoning is needed to deduce that.

    I believe the proof is poorly written. I believe it would be easier for you, as well as more satisfying, to try to prove the proposition yourself, rather than to make sense of what is written in the book. Why not have a go, and post back here if you get stuck?
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