Is the Intersection of a Sequence in a Sigma-Algebra Also in the Sigma-Algebra?

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The intersection of a sequence of elements in a sigma-algebra is indeed contained within the sigma-algebra itself. Given a sigma-algebra F, it is established that the empty set is included, complements of elements in F are also included, and the union of a sequence of elements in F is part of F. By demonstrating that the intersection of elements can be expressed as the union of their complements, which are in F, it follows that the intersection is also in F.

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Given the definition of a sigma-algebra, I need to show that the intersection of a sequence of elements in a sigma-algebra is in the sigma-algebra:

Given:
Let F be a sigma-algebra, then:
1) The empty set is in F.
2) If A is in F, then so is the complement of A.
3) The union of a sequence of elements in F is also in F.

To prove:
The intersection of a sequnce of elements in F is also in F.

I'm quite stuck and seem to go around in circles. Any help on how to attack this problem would be appreciated.
 
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Take an intersection AnBnC... what is its complement?
 
Right, the intersection is the union of the complements. And since the complements, and therefore the union, are in F, the intersection must also be in F.
 

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