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

  1. Mar 4, 2012 #1
    I'm trying to prove the following and all I've got is like one line worth of proof.

    sfdsdffsd.png

    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?
     
  2. jcsd
  3. Mar 4, 2012 #2

    Office_Shredder

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    Re: sigma-rings are closed under countable intersections (sigma-rings are delta-rings

    Instead of complementing take relative complements in A1.
     
  4. Mar 4, 2012 #3
    Re: sigma-rings are closed under countable intersections (sigma-rings are delta-rings

    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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