# 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...