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

[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!

 

[zzzhhh]

solved.
There are mainly three methods to this question:
1)by concept of "atom" (http://at.yorku.ca/cgi-bin/bbqa?forum=calculus;task=show_msg;msg=0125.0001")
2)by continuously dividing into two subsets (http://www.mathkb.com/Uwe/Threads/S...2e-6e8fb10de953@f36g2000hsa.googlegroups.com")
3)by equivalence class (http://planetmath.org/?op=getmsg&id=5849")
The latter two are explicit construction; the first one proves existence of such sequence by contradiction.

 

[stat22]

Let X₁, X₂, X₃, ..., X_n, ... are the elements of the said $\sigma$-algebra.

Then, for any positive integer n>1, let E_n = X_n - $[$X_n$\bigcap$$($X_n-1$\bigcup$ ... $\bigcup$X₁$)$$]$

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

 

[micromass]

Let X₁, X₂, X₃, ..., X_n, ... are the elements of the said $\sigma$-algebra.

Then, for any positive integer n>1, let E_n = X_n - $[$X_n$\bigcap$$($X_n-1$\bigcup$ ... $\bigcup$X₁$)$$]$

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

And who says that the terms are nonempty?? For example, take X₁=[0,2] and X₂=[0,1]. Then

$$X_2\setminus (X_2\cap X_1)=[0,1]\setminus [0,1]=\emptyset$$

 

[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!

 

[micromass]

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!

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

$$X_2\setminus(X_2\cap X_1)=\emptyset$$

 

[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)]

 

[micromass]

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!

 

[Bacle]

Then I imagine one can ignore the empty sets, but that depends on what zzzhhh wants.

 

[micromass]

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...