text: # How to construct a disjoint sequence from an infinite sigma-algebra?
  
  1. Sep 18, 2010
  
  ### zzzhhh
  
  Suppose $$\mathcal A$$ is an infinite $$\sigma$$-algebra, how to construct a disjoint sequence in $$\mathcal A$$ such that each term is nonempty? Thanks!
  
  2. Sep 19, 2010
  
  ### zzzhhh
  
  Last edited by a moderator: Apr 25, 2017
  3. Jul 2, 2011
  
  ### stat22
  
  Let X1, X2, X3, ..., Xn, ... are the elements of the said $\sigma$-algebra.
  
  Then, for any positive integer n>1, let En = Xn - $[$Xn$\bigcap$$($Xn-1$\bigcup$ ... $\bigcup$X1$)$$]$
  
  The result is a disjoint sequence {En}, where E1= X1
  
  4. Jul 2, 2011
  
  ### micromass
  
  And who says that the terms are nonempty?? For example, take X1=[0,2] and X2=[0,1]. Then
  
  $$X_2\setminus (X_2\cap X_1)=[0,1]\setminus [0,1]=\emptyset$$
  
  5. Jul 2, 2011
  
  ### stat22
  
  micromass,
  
  in your example X1={0,2} and X2={0,1}, the intersection is {0} not {0,1}. Thus X2-(X2$\bigcap$X1)={0,1}-{0}={1} not empty!
  
  6. Jul 2, 2011
  
  ### micromass
  
  I'm talking about the interval [0,1], not {0,1}.
  
  But, if you don't like intervals, consider X1={0,1}, X2={0}. Then
  
  $$X_2\setminus(X_2\cap X_1)=\emptyset$$
  
  7. Jul 3, 2011
  
  ### Bacle
  
  I think the standard way, given a collection A:={A_1,...,A_n,..} (I think this works only for countable collections ) to define B:={B_1,...,B_n,...} by:
  
  B_1:=A_1
  B_2:=A_2-A_1
  ......
  ......
  B_n:=A_n-[A_1\/A_2\/......\/A_(n-1)]
  
  8. Jul 3, 2011
  
  ### micromass
  
  Yes, these are certainly disjoint, but perhaps empty!
  
  9. Jul 3, 2011
  
  ### Bacle
  
  Then I imagine one can ignore the empty sets, but that depends on what zzzhhh wants.
  
  10. Jul 3, 2011
  
  ### micromass
  
  Yes, but you still need to end up with an infinite number of sets. If you throw away too many sets, then you might end up with a finite number of sets...