How to construct a disjoint sequence from an infinite sigma-algebra?

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Suppose \mathcal A is an infinite \sigma-algebra, how to construct a disjoint sequence in \mathcal A such that each term is nonempty? Thanks!
 
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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 ... \bigcupX1)]

The result is a disjoint sequence {En}, where E1= X1
 
stat22 said:
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 ... \bigcupX1)]

The result is a disjoint sequence {En}, where E1= X1

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
 
micromass,

in your example X1={0,2} and X2={0,1}, the intersection is {0} not {0,1}. Thus X2-(X2\bigcapX1)={0,1}-{0}={1} not empty!
 
stat22 said:
micromass,

in your example X1={0,2} and X2={0,1}, the intersection is {0} not {0,1}. Thus X2-(X2\bigcapX1)={0,1}-{0}={1} not empty!

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
 
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)]
 
Bacle said:
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)]

Yes, these are certainly disjoint, but perhaps empty!
 
Then I imagine one can ignore the empty sets, but that depends on what zzzhhh wants.
 
  • #10
Bacle said:
Then I imagine one can ignore the empty sets, but that depends on what zzzhhh wants.

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...
 
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