Proving Infinite Sigma-Algebra's Countable Disjoint Subsets

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Discussion Overview

The discussion revolves around the question of whether a sigma-algebra can be infinite and countable, specifically focusing on the implications of having countable disjoint subsets within the context of sigma-algebras. Participants explore theoretical aspects and potential proofs related to this concept.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant suggests that if a sigma-algebra has a countable number of disjoint subsets, it cannot be countable due to the possible combinations of those subsets.
  • Another participant challenges this assertion, indicating that the original claim may not hold true and suggests a clarification regarding the term "countably infinite."
  • There is a proposal to use proof by contradiction to demonstrate that if a sigma-algebra consists of an infinite number of subsets, it must have a countably infinite number of disjoint subsets.

Areas of Agreement / Disagreement

Participants express differing views on the implications of countable disjoint subsets in relation to the countability of sigma-algebras, indicating that the discussion remains unresolved with competing interpretations.

Contextual Notes

Participants have not reached a consensus on the definitions and implications of countable versus countably infinite subsets within sigma-algebras, and there are indications of potential misunderstandings in terminology.

Zaare
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I'm supposed to answer the question "Can a sigma-algebra be infinite and countable?"
I think I can show that if it has a countable number of disjoint subsets, then it can't be countable considering the possible combinations of the subsets.
Now I need to show that if a sigma-algebra consists of an infinite number of subsets, then it has a countable number of disjoint subsets.
Any ideas on how I can do this?
 
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I think I can show that if it has a countable number of disjoint subsets, then it can't be countable considering the possible combinations of the subsets.

No you can't.

Now, if you instead said countably infinite... :smile:


Now I need to show that if a sigma-algebra consists of an infinite number of subsets, then it has a [countably infinite] number of disjoint subsets.

(I edited it)

Proof by contradiction, maybe?
 
Hurkyl said:
No you can't.

Now, if you instead said countably infinite... :smile:

That's what I meant. I was sloppy. :redface:
Thank you both for the help.
 

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