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Construction of a sequence

  1. Nov 19, 2011 #1
    Hi,

    1. The problem statement, all variables and given/known data
    V is a non-empty, upper bounded subset of R. Show that a sequence [tex](v_n)_{n \geq 0}[/tex] in V exists such that: 1)[tex]v_0 \leq v_1 \leq ...[/tex] and 2) the limit of the sequence is sup V. (Hint: use a recursive construction)


    2. Relevant equations



    3. The attempt at a solution
    V is non-empty and upper bounded so sup V exists.
    Suppose [tex]sup V \in V[/tex] Construct the sequence: sup V = v0 = v1...
    And I know how to go from there.

    I'm having problems with the case: sup V is not in V.
    I was thinking of constructing the sequence:
    [tex]v_0 \in V[/tex]
    [tex] (v_n, supV) \subset (v_{n-1}, supV)[/tex]

    But I'm not sure if this is right or how to proof that the limit of this sequence is sup V.
    Any ideas?
     
  2. jcsd
  3. Nov 19, 2011 #2

    HallsofIvy

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    Let [itex]v_0[/itex] be any member of V. We can do this because V is non-empty.

    The recursion step: for any [itex]v_i[/itex], [itex](sup V+ v_i)/2< sup(V)[/itex] and so is not an upper bound on V/. There must exist some [itex]v_{i+1}[/itex] in V such that [itex](v_i+ sup(V))/2< v_{i+1}[/itex]. That cuts the distance from the previous term to sup(V) in half on each step.
     
  4. Nov 19, 2011 #3
    Thanks!
     
  5. Nov 19, 2011 #4
    Sorry for the double post, but I came across another problem when writing it down. v(i+1) doesn't have to exist right? And there doesn't have to be a smallest v(i+1). So how would one formulate the sequence?

    [tex] v_{i+1} \in [v_1, sup V) [/tex] perhaps?

    Or is it enough to state v(i+1) exists and v(i+1) ? vi and v(i+1) is in V, to formulate a sequence?
     
    Last edited: Nov 19, 2011
  6. Nov 20, 2011 #5

    HallsofIvy

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    No, [itex]v_{i+1}[/itex] by this method must exist: Since in this case we are assuming that sup(V) is not in V itself, our [itex](v_i+ sup(V))/2[/itex] is NOT sup(V) and so is not an upper bound for V. There must exist at least one member of V larger than [itex](v_i+ sup(V))/2[/itex]. I said nothing about [itex]v_{i+1}[/itex] being the smallest such member of V- choose any of them.

    But it is not sufficient just to say that [itex]v_{i+1}[/itex] is some number in V larger than [itex]v_i[/itex]. That sequence would not necessarily converge to sup(V).
     
  7. Nov 20, 2011 #6
    I understand. I have one final question.
    Nevermind, thanks :)
     
    Last edited: Nov 20, 2011
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