Sigma-rings are closed under countable intersections (sigma-rings are delta-rings)

  • Thread starter jdinatale
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  • #1
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I'm trying to prove the following and all I've got is like one line worth of proof.

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If we had that sigma-rings were closed under complementation, this would be easier, but we only know that if A in R and B in R, then A \ B in R and B \ A in R (symmetric difference). Is there a way to approach this using the symmetric difference?
 

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  • #2
Office_Shredder
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Instead of complementing take relative complements in A1.
 
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Instead of complementing take relative complements in A1.

Ok. So assume that [itex]A_1, A_2, A_3 ... \in R[/itex]. Then [itex]A_1 \backslash A_1, A_1 \backslash A_2, A_1 \backslash A_3 ... \in R[/itex]. Since [itex]R[/itex] is a [itex]\sigma[/itex]-ring, [itex]X = \cup_{n = 1}^\infty A_1 \backslash A_n\in R[/itex]. Also [itex]X \backslash A_1 \in R[/itex].

I'm not seeing where this is leading.
 

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