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Homework Help: Prove sequence converges to sup

  1. Apr 29, 2013 #1


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    1. The problem statement, all variables and given/known data
    Suppose that E is contained in ##\mathbb{R}## is a nonempty bounded set and that ##\sup E## is not in E. Prove that there exists a strictly increasing sequence ##\left\{x_n\right\}## that converges to ##\sup E## such that ##x_n \in E## for all n in ##\mathbb{N}##.

    2. Relevant equations
    More like Relevant theorems:
    Monotone convergence.

    3. The attempt at a solution

    E is contained in R and nonempty so by Completeness, sup E exists. E is bounded so -a < e < a for all e in E and let supE = a which is not in E. I suppose this might be the essence of the proof, but how exactly do we know such a sequence in E exists? If there existed a function ##f: \mathbb{N} \rightarrow \mathbb{R}## with all the ##x_n## in E, then provided I can show such a sequence is strictly increasing (and that it exists) then the rest of the problem becomes trivial. I thought about using nested intervals but I didn't get very far. Any ideas?

    Many thanks.
  2. jcsd
  3. Apr 29, 2013 #2
    Here's an outline/suggestion on how to proceed: let [itex]sup(E)=s_E[/itex].
    1) Can you show that given any [itex]\epsilon>0[/itex], there exists an [itex]x\in E[/itex] such that [itex]s_E-\epsilon<x[/itex]?
    2) Consider the sequence [itex]\epsilon_n=\frac{1}{n}[/itex]; apply this sequence of epsilons to part 1 to show that you can generate a sequence in [itex]E[/itex], call it [itex]x_n[/itex].
    3) Show that this sequence converges to [itex]sup(E)=s_E[/itex].

    Hint for part 1): this is a good exercise in proof by contradiction which teaches you how to negate quantifiers; "for all"'s become "there exists" and vice-versa, and statements get negated.
  4. Apr 29, 2013 #3


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    Yes, this is the Approximation property for suprema

    I can make the following statements: supE - 1 < x for some x, supE - 1/2 < x for some x, etc.. So supE-1, supE-1/2,...supE-1/n.. would make a sequence and indeed it would be strictly increasing. My only worry is that how can I be sure that the elements would be in E?
  5. Apr 29, 2013 #4
    The numbers [itex]Sup(E)-1, Sup(E)-1/2, Sup(E)-1/3,...[/itex] aren't the numbers you're interested in; it's the x's that bound them that you want to look at.

    Let's start at n=1. You said that there exists an [itex]x\in E[/itex] such that [itex]Sup(E)-1<x[/itex]. Let's call this x we found [itex]x_1[/itex].
    Now, n=2. We know that there exists an [itex]x\in S[/itex] such that [itex]Sup(E)-1/2<x[/itex]. Let's call this second x we've found [itex]x_2[/itex].
    Then we could do the same for n=3, n=4, etc.

    The important part thing is we can do this for ANY n; so in particular, there exists a sequence [itex]x_n\in E[/itex] with the property that [itex]Sup(E)-1/n<x_n[/itex]. Can you show that this sequence converges? Hint below if you can't get it.

    You also know that, because [itex]Sup(E)[/itex] is the supremum, that each [itex]x_n[/itex] is bounded by [itex]Sup(E)[/itex]. So you have [itex]Sup(E)-1/n<x_n\leq Sup(E)[/itex].
    Last edited: Apr 29, 2013
  6. Apr 29, 2013 #5


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    Ok so the last statement implies $$\lim_{n \rightarrow \infty} \sup E - \lim_{n \rightarrow \infty} \frac{1}{n} \leq \lim_{n \rightarrow \infty} x_n$$ From this ##x_n## must converge to ##\sup E##. It caanot be greater because the ##x_n## are bounded above by ##\sup E##. Is this all I need to do? I haven't checked your hint yet, I prefer to have feedback first.
  7. Apr 29, 2013 #6
    Yes, that is basically the idea. Since the sequence [itex]sup(E)-1/n[/itex] converges to [itex]sup(E)[/itex], and [itex]x_n[/itex] is bounded above by the [itex]sup(R)[/itex] and below by this [itex]sup(E)-1/n[/itex], it must converge to [itex]sup(E)[/itex].
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