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Union of Two Sigma-Algebras

Question:

Let X be a non-empty set and let P and Q be two sigma-algebras on X. Is P U Q(union of P and Q) a sigma-algebra on X?

My approach towards the question:

My intuition was that this is not true in general.

So, the next idea that came to my mind is to disaprove this by using an counterexample.

As the first step, I selected a non-empty set X as follows: X = {a, b, c}, where a, b and c are distinct.

Then I listed out all the possible sigma-algebras that came to mind, which can be derived from X, as follows:
Here, @ denote the null set.

P1 = {@, X}
P2 = { @, X, {a}, {b, c} }
P3 = { @, X, {b}, {a, c} }
P4 = { @, X, {c}, {a, b} }
P5 = { @, X, {a}, {b}, {c}, {a, b}, {a, c}, {b, c} }

Then, I considered P2 U P3 = { @, X, {a, b}, {a, c}, {b, c} }

So, in this case, {c}, the complement of {a, b} is not included in P2 U P3, hence not agreeing with the second property of a sigma-algebra.

Is this a valid counterexample which can prove that the intial statement given in the question is not valid in general?
I would appreciate any feedback on this.
Original post by Ash760
Question:

Let X be a non-empty set and let P and Q be two sigma-algebras on X. Is P U Q(union of P and Q) a sigma-algebra on X?

My approach towards the question:

My intuition was that this is not true in general.

So, the next idea that came to my mind is to disaprove this by using an counterexample.

As the first step, I selected a non-empty set X as follows: X = {a, b, c}, where a, b and c are distinct.

Then I listed out all the possible sigma-algebras that came to mind, which can be derived from X, as follows:
Here, @ denote the null set.

P1 = {@, X}
P2 = { @, X, {a}, {b, c} }
P3 = { @, X, {b}, {a, c} }
P4 = { @, X, {c}, {a, b} }
P5 = { @, X, {a}, {b}, {c}, {a, b}, {a, c}, {b, c} }

Then, I considered P2 U P3 = { @, X, {a, b}, {a, c}, {b, c} }

So, in this case, {c}, the complement of {a, b} is not included in P2 U P3, hence not agreeing with the second property of a sigma-algebra.

Is this a valid counterexample which can prove that the intial statement given in the question is not valid in general?
I would appreciate any feedback on this.


Your general method is along the right lines, except P2P3{@,X,{a,b},{a,c},{b,c}}P2\cup P3 \not = \{ @, X, \{a, b\}, \{a, c\}, \{b, c\} \}

Not got time to look further, as just going out, and not sure if 3 elements are enough, but not got time to check.
(edited 1 year ago)
Reply 2
Original post by ghostwalker
Your general method is along the right lines, except P2P3{@,X,{a,b},{a,c},{b,c}}P2\cup P3 \not = \{ @, X, \{a, b\}, \{a, c\}, \{b, c\} \}

Not got time to look further, as just going out, and not sure if 3 elements are enough, but not got time to check.


Thanks for the quick feedback.
Could you let me know why P2 U P3 is not that, when you are free?
Original post by Ash760
Thanks for the quick feedback.
Could you let me know why P2 U P3 is not that, when you are free?


You seem to be having problems with basic set manipulation, as evidenced by your previous threads as well; probably worth doing a bit of background in elementary set theory.

P2 u P3 is the set of all elements that are either in P2 or in P3.

E.g. What happened to the element "{a}" in P2 - it's not in your version of P2 u P3

And where did the element "{a,b}" come from - it's not in P2 or in P3.

And following my previous comment. 3 elements will be enough, and your counter-example should work out nicely when you've corrected it.
(edited 1 year ago)
Reply 4
Original post by ghostwalker
You seem to be having problems with basic set manipulation, as evidenced by your previous threads as well; probably worth doing a bit of background in elementary set theory.

P2 u P3 is the set of all elements that are either in P2 or in P3.

E.g. What happened to the element "{a}" in P2 - it's not in your version of P2 u P3

And where did the element "{a,b}" come from - it's not in P2 or in P3.

And following my previous comment. 3 elements will be enough, and your counter-example should work out nicely when you've corrected it.


Thank you for the explanation!
So here, P2 U P3 = { @, X, {a}, {b}, {a, c}, {b, c} }

However, here, for each element that belongs to P2 U P3, their complement also do belong to P2 U P3 right?
That is, P2 U P3 is not a valid counterexample in this case?
Original post by Ash760
Thank you for the explanation!
So here, P2 U P3 = { @, X, {a}, {b}, {a, c}, {b, c} }

However, here, for each element that belongs to P2 U P3, their complement also do belong to P2 U P3 right?
That is, P2 U P3 is not a valid counterexample in this case?


You need to check out some of the other axioms/properties of a σ\sigma-algebra.
(edited 1 year ago)
Reply 6
Original post by ghostwalker
You need to check out some of the other axioms of a σ\sigma-algebra.


I believe it must be the 3rd one. But I'm totally clueless on this one?
Could you help me with this one?
Original post by Ash760
I believe it must be the 3rd one. But I'm totally clueless on this one?
Could you help me with this one?


What's the 3rd one?
Reply 8
If E_n belongs to the considered sigma-algebra for all n in natural numbers set, union of E_n from 1 to infinity also belong to the considered sigma-algebra
Original post by Ash760
If E_n belongs to the considered sigma-algebra for all n in natural numbers set, union of E_n from 1 to infinity also belong to the considered sigma-algebra


This looks like part of the definition of a sigma algebra on the natural numbers (but I'm not an expert).

I'll go with the phrasing given in wiki as it's clearer, though the meaning is the same.

"A sigma algebra ... is closed under countable unions ...."

So, no need to go to infinity, you just need to find two elements of P2 u P3, whose union, as sets, is not in P2 u P3. You've actaully mentioned it already. :smile:
(edited 1 year ago)
Reply 10
Original post by ghostwalker
This looks like part of the definition of a sigma algebra on the natural numbers (but I'm not an expert).

I'll go with the phrasing given in wiki as it's clearer, though the meaning is the same.

"A sigma algebra ... is closed under countable unions ...."

So, no need to go to infinity, you just need to find two elements of P2 u P3, whose union, as sets, is not in P2 u P3. You've actaully mentioned it already. :smile:


In this case, the union of {a} and {b}, {a, b} should work since it's not in P2 U P3 right?
Original post by Ash760
In this case, the union of {a} and {b}, {a, b} should work since it's not in P2 U P3 right?


Yep, that's it.
Reply 12
Original post by ghostwalker
Yep, that's it.


Great! Finally figured this out. Thank you so much for the help!

Btw, if you don't mind me asking, are you a Math major?
Original post by Ash760
Great! Finally figured this out. Thank you so much for the help!

Btw, if you don't mind me asking, are you a Math major?


Sounds like an American term. I did mathematics at Warwick (England) a very long time ago.
Reply 14
Original post by ghostwalker
Sounds like an American term. I did mathematics at Warwick (England) a very long time ago.

That's great!

Thank you very much for helping me figure out many concepts and errors in my work!

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