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Math Help - disjointlize an arbitrary sequence in a ring?

  1. #1
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    disjointlize an arbitrary sequence in a ring?

    In a ring R (nonempty class of sets closed under difference and finite union), any sequence (here means a function on natural numbers \mathbb N) <E_i> in R can be disjointlized to a disjoint sequence <F_i> such that \bigcup E_i=\bigcup F_i by traditional induction using the equation F_i=E_i-\bigcup \limits_{j <i} E_j. But for arbitrary uncountable sequence <E_\alpha> in R, I either do not know if it is still possible to turn <E_\alpha> into a disjoint sequence with the same union or have no idea how to use transfinite induction to prove it if disjointlization holds, can you help me with this problem? Thanks!
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  2. #2
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    Thank fedja for constructing a brilliant counterexample. I write it as follows.
    Define R to be a class of finite unions of half-open rectangles [a_1,b_1)\times[a_2,b_2) in \mathbb R^2. It can be verified that R is a ring. Consider point set A=\{(x,y)|x+y\geq 0\} in \mathbb R^2. A can be expressed as a union \bigcup\{E\in{\bf R}|E\subseteq A\}=\cup E_\alpha where sequence <E_\alpha> is an indexed class of the set being unioned. First <E_\alpha> can not be reduced to a countable sequence with the same union, for otherwise the line x+y=0 would be obtained by a countable union of points, which contradicts the fact that any line is a continuum. Second, If uncountable sequence <E_\alpha> can be disjointliazed to be a uncountable sequence <F_\alpha> in R satisfying \cup F_\alpha=\cup E_\alpha and <F_\alpha> disjoint, since every term E_\alpha, as an element of R, contains a point of rational coordinate, it would follow that there are uncountable distinct rational coordinates. However, the set of all points of rational coordinate in \mathbb R^2 has countable cardinality. So, the disjointlization for sequence <E_\alpha> is not possible. The same argument applies when we are to extend this impossibility to \sigma-ring.
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