Countably infinite sigma-algebra

[Zaare]

Problem: Is there a countably infinite sigma-algebra?
I can assume there are countably number of disjoint subsets included in the sigma-algebra.

 

[mathman]

Given there are a countable number of disjoint subsets, the number of sets in the sigma algebra has to be non-countable. It is a very simple one to one mapping of all the combinations from the countable number of subsets to all binary numbers between 0 and 1. Specifically, order the countable subsets. For any combination of subsets, the equivalent binary number has zeros for the sets omitted and 1 for sets included.

Example: .001110101... is the image of the set where the 3rd, 4th, 5th, 7th, 9th, etc. sets are include while the 1st, 2nd, 6,th, 8th, etc. sets are omitted (from the union).

 

[tacman]

Yes, there is a countably infinite sigma-algebra. This is because a sigma-algebra is defined as a collection of subsets of a given set that satisfies certain properties, one of which is closure under countable unions. This means that for any countable number of subsets in the sigma-algebra, their union will also be in the sigma-algebra. Therefore, we can construct a countably infinite sigma-algebra by starting with a countable number of subsets and taking their countable unions, which will also be in the sigma-algebra. This process can continue indefinitely, resulting in a countably infinite collection of subsets that form a sigma-algebra.