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Infinite intersection of indexed sets

  1. Jul 16, 2011 #1
    Every element of a set [itex]A[/itex] can be written [itex]a=w.a_1a_2a_3\ldots{a_n}\ldots[/itex] with [itex]w, a_n\in\mathbb{Z}[/itex] and [itex]0\leq a_n\leq9[/itex] for every [itex]n\in\mathbb{N}.[/itex] If [itex]A[/itex] is bounded, there exists a greatest whole part [itex]\overline{w}[/itex] of the elements of [itex]A,[/itex] and because any set [itex]S[/itex] of elements [itex]a_n[/itex] is bounded, for every [itex]n,[/itex] there exists a greatest element [itex]\overline{a_n}[/itex] of [itex]S.[/itex]

    Let [itex]A_0\subset{A}[/itex] be the set of elements [itex]a=\overline{w}.a_1a_2a_3\ldots{a_n}\ldots[/itex] and [itex]A_1\subset{A_0}[/itex] the set of elements [itex]a=\overline{w}.\overline{a_1}a_2a_3\ldots{a_n}\ldots,[/itex] and define [itex]A_{n+1}\subset{A_n}[/itex] as the set of elements [itex]a=\overline{w}.\overline{a_1}\overline{a_2}\overline{a_3}\ldots\overline{a_n}\overline{a_{n+1}}a_{n+2}\ldots.[/itex]

    It would appear as though [itex]\alpha\in\bigcap_{n=1}^\infty{A_n}[/itex] contains a single element--namely, [itex]\sup(A)[/itex]. However, this seems to lead to the absurd conclusion that for every set [itex]A,\sup(A)\in{A}[/itex]. Perhaps we would say simply that [itex]\bigcap_{n=1}^\infty{A_n}=\emptyset[/itex] in case [itex]\sup(A)\not\in{A}[/itex]?
     
    Last edited: Jul 16, 2011
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  3. Jul 16, 2011 #2

    micromass

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    Hi math771! :smile:

    Here you went a bit too fast. It isn't necessary that [itex]a=\overline{w}.a_1a_2a_3\ldots{a_n}...[/itex] or [itex]a=\overline{w}.\overline{a_1}a_2a_3\ldots{a_n}...,[/itex] are in A. So it isn't necessary that [itex]A_1\subseteq A[/itex].

    For example, take [itex]A=\{222.000000..., 0.3333.....\}[/itex], then it isn't true that [itex]222.3333.....[/itex] is in A, but that is what you are claiming!! Furthermore, 222.3333.... isn't even the supremum of A! So you'll need a different method to actually obtain the supremum.
     
  4. Jul 16, 2011 #3
    I understand what you're trying to say. However, I meant to imply that A_0, for example, would be defined as the set of precisely those elements of A whose whole part is equal to [itex]\overline{w}[/itex] by stating that A_0 is a subset of A.
     
  5. Jul 16, 2011 #4

    micromass

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    Ah yes, I see now. Well, then I see no real reason why

    [tex]\bigcap_{n}{A_n}[/tex]

    should be nonempty. If you use completeness, then this intersection is nonempty if the An are closed and nonempty. But the An aren't necessarily closed here!
     
  6. Jul 16, 2011 #5

    disregardthat

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    This is how I understand you:

    A is a bounded subset of the reals. Let w be the largest integer such that w.a1a2a3.... for some sequence a_i is in A, and let A_0 be the subset of A such that w.a1a2... is in A for some sequence a_i. Let b0 be the largest integer in A_0 such that w.b0a1a2... is in A_0 for some sequence a_i. Recursively define A_n as such: let b_n be the largest integer such that w.b0b1b2...b_(n-1)a1a2a3... is in A_(n-1) for some sequence a_i, and let A_n be the subset of A_(n-1) of the elements on the form w.b0b1...bna1a2... for some sequence a_i.

    Now what you seem to want to conclude is that w.b0b1b2... is in the intersection of the A_n's. But this is not necessarily so. For every n, there is a sequence a_i such that w.b0b1...bna1a2... is in A_n for some sequence a_i, but you don't know whether w.b0b1b2.... is in A_n for any n.
     
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