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 is possible, can you help me with this problem? Thanks!
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$\begingroup$ Exactly the same formula works. $\endgroup$– Mariano Suárez-ÁlvarezCommented Apr 27, 2010 at 4:40
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1$\begingroup$ @Mariano: It does? Don't I need to form some infinite unions, which might take me outside of R? $\endgroup$– Reid BartonCommented Apr 27, 2010 at 4:44
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1$\begingroup$ Well, $\bigcup E_i$ must make sense, at least, for the question to make sense, so assuming the existence of unions involving less terms is not a far stretch :P $\endgroup$– Mariano Suárez-ÁlvarezCommented Apr 27, 2010 at 5:14
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$\begingroup$ My first impression is that Mariano is correct, as long as you have a well-ordering on your indexing set. However, given the importance of countable additivity in measure theory, I'm not sure which examples benefit from having "disjointlization" in the more general sense. $\endgroup$– Yemon ChoiCommented Apr 27, 2010 at 5:30
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$\begingroup$ The unions do exist. They just aren't in the ring (and there is no reason to suppose them to be there). $\endgroup$– fedjaCommented Apr 27, 2010 at 20:53
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2 Answers
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Let's take the usual ring of finite unions $E$ of half-open rectangles $[a,b)\times [c,d)$ on the plane. The closed half-plane $x+y\ge 0$ is a union of continuum of such rectangles (all possible rectangles contained in that closed half-plane) but, since each $E$ contained in this half-plane can intersect the boundary line by only finitely many points, you cannot get the half-plane as a union of countably many ring elements. On the other hand, any disjoint family of ring elements is at most countable.
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$\begingroup$ Thank you. I got the example but do not understand the last sentence:"On the other hand, any disjoint family of ring elements is at most countable." could you please explain in detail why any disjoint family of ring elements is at most countable? $\endgroup$– zzzhhhCommented Apr 27, 2010 at 22:01
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$\begingroup$ Any family of ring elements is the countable union of the ring elements of area at least 1/N, for every positive integer N. If a family of ring elements is uncountable, then one of these sets is uncountable, so there are uncountably many ring elements with area at least 1/N for some N. But the plane can only fit countably many such ring elements. (This is not hard to see by examining how many of these ring elements can be within some distance of the origin.) $\endgroup$ Commented Apr 27, 2010 at 22:40
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$\begingroup$ I cannot get it. Why "Any family of ring elements is the countable union of the ring elements of ..."? Above all, a family of ring elements is a set of subsets of X, while union of the ring elements is just a subset of X, here X means the whole space. $\endgroup$– zzzhhhCommented Apr 28, 2010 at 6:04
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2$\begingroup$ The correct English phrase would be "any family $F$ of ring elements is the countable union of its subfamilies
$F_N$consisting of all elements in $F$ having the area $\frac 1N$ or greater". Anyway, if Qiaochu's approach seems too hard, just notice that every rectangle contains a point with rational coordinates. $\endgroup$– fedjaCommented Apr 28, 2010 at 13:47 -
$\begingroup$ Thank you very much, fedja, for your brilliant counterexample and rational-coordinate explanation. So disjointlization is only possible for countable sequence in a ring, and even in a sigma-ring by applying the same argument. $\endgroup$– zzzhhhCommented Apr 29, 2010 at 2:21
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"Disjointlization" ... the world would be a better place without this word!
answered Apr 28, 2010 at 0:31
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1$\begingroup$ Agreed. The alternatives "disjointify" or "disjointize" are awkward but still much better. $\endgroup$ Commented Apr 28, 2010 at 0:40