# Thread: disjointlize an arbitrary sequence in a ring?

1. ## 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 $\mathbb N$) $$ in R can be disjointlized to a disjoint sequence $$ such that $\bigcup E_i=\bigcup F_i$ by traditional induction using the equation $F_i=E_i-\bigcup \limits_{j . But for arbitrary uncountable sequence $$ in R, I either do not know if it is still possible to turn $$ 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!

2. 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 $$ is an indexed class of the set being unioned. First $$ 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 $$ can be disjointliazed to be a uncountable sequence $$ in R satisfying $\cup F_\alpha=\cup E_\alpha$ and $$ disjoint, since every term $E_\alpha$, as an element of R, contains a point of rational coordinate, it would follow that there are uncountable distinct rational coordinates. However, the set of all points of rational coordinate in $\mathbb R^2$ has countable cardinality. So, the disjointlization for sequence $$ is not possible. The same argument applies when we are to extend this impossibility to $\sigma$-ring.