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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
  2. jcsd
  3. Jul 16, 2011 #2
    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
    Ah yes, I see now. Well, then I see no real reason why


    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


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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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