Proving Sigma-Rings Are Closed under Countable Intersections

[jdinatale]

I'm trying to prove the following and all I've got is like one line worth of proof.

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?

 

[Office_Shredder]

Instead of complementing take relative complements in A₁.

 

[jdinatale]

Instead of complementing take relative complements in A₁.

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

I'm not seeing where this is leading.