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HELP: Set theory Question

  1. Sep 19, 2006 #1
    Let F be the label of an non-empty set and let [tex](B_m)_{m \geq 1}[/tex] be elements in [itex]2^F[/itex]

    Then I need to prove the following:

    [tex]\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m} = \mathrm{lim}_{m} \ \mathrm{inf} \ \mathrm{B_m} = \cup _{m= 1} ^{\infty} B_{m}[/tex]

    if [tex]B_{m} \uparrow[/tex] which implies that [tex]B_{m} \subseteq B_{m+1}[/tex] for all [tex]m \geq 1[/tex] and

    [tex]\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m} = \mathrm{lim}_{m} \ \mathrm{inf} \ \mathrm{B_m} = \cap _{m= 1} ^{\infty} B_{m}[/tex]

    if [tex]B_{m} \downarrow[/tex] which means that [tex]B_m \supseteq B_{m+1}[/tex] for all [tex]m \geq 1[/tex]

    how do I go about proving this? Do I need to show the infimum of F first?

    Sincerely
    mb20
     
    Last edited: Sep 19, 2006
  2. jcsd
  3. Sep 19, 2006 #2

    HallsofIvy

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    Staff Emeritus
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    What you have makes no sense. I guess that you really mean 2F rather than 2E (or you mean E rather than F in the first line), the power set of F. But then (Bm) is just some collection of subsets of F. Don't you have some conditions on (Bm)? Finally, you say you need to show
    [tex]\mathrm{lim}_{m} \ \mathrm{sup} \ B_{m}[/tex]
    but that's not a statement you can "show", that's an expression. What is supposed to be true about it?
     
  4. Sep 19, 2006 #3
    My solution

    here is my solution

    1.
    I say B_m "uparrow" if B_m is a subset of B_{m+1} for all m, so they are all nested upward. In this case I want to prove that

    limsup B_m = liminf B_m = union of B_m over all m.

    Let B be the union over all m of B_m. Since B_m is a subset of F for every m, B is a subset of F. Furthermore, every B_m is contained in B so B is an upper bound for the sequence (B_m). Recall that
    liminf <= limsup for abstract reasons, so I show liminf = B then as limsup <= B. liminf {B_m} = sup{inf{B_k: k>= m}: m> 0}.

    As B_{k+1} >= B_{k} for every k, so inf{B_k:k>= m}=B_m. Hence
    sup{inf{B_k:k>= m}: m>0}=sup{B_m:m>0}. But B_{m+1}>= B_m so
    sup{B_m:m>0}=union of all B_m, which is B.

    Thus liminf{B_m} = B.

    2. I say B_m "downarrow" if B_{m+1} is a subset of B_{m} for all m, so they are all nested downward. In this case I want to prove that

    limsup B_m = liminf B_m = intersection of B_m over all m.

    As B_{k+1} <= B_k I know that sup{B_k:k>=m}=B_m.

    Thus limsup B_m=inf{B_m:m>0}=B -- the intersection of all B_m's.

    Does this sound right?

    Sincerely Mathman20
     
    Last edited: Sep 20, 2006
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