Let A then A c = ∪ n = i n A i , where A 1 , A 2 , . . . , A n are disjoint sets in R.Since A , A i ∈ R and R is closed under finite intersection then A ∩ A i ∈ R.Hence ϕ = A ∩ A c = ∪ n = i n ( A ∩ A i ), where A ∩ A i are disjoint sets in R.Since ϕ is a finite union of disjoint sets in R, then ϕ c = Ω ∈ R.But now how to prove that ϕ ∈ R?

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Let   A then       A    c    =      ∪          n      =      i              n            A    i  , where       A    1    ,      A    2    ,  .  .  .  ,      A    n   are disjoint sets in   R.Since   A  ,      A    i    ∈  R and   R is closed under finite intersection then   A  ∩      A    i    ∈  R.Hence   ϕ  =  A  ∩      A    c    =      ∪          n      =      i              n        (  A  ∩      A    i    ), where   A  ∩      A    i   are disjoint sets in   R.Since   ϕ is a finite union of disjoint sets in   R, then       ϕ    c    =  Ω  ∈  R.But now how to prove that   ϕ  ∈  R?

Bruno Dixon · Accepted Answer

Let   A be a member of the semi-algebra   R  ≠  ∅. Then       A    ∁   is a disjoint union of some       B    1    ,  …  ,      B    n   for some finite   n and all       B    i    ∈  R. If   n  =  1, then       A    ∁   is in   R so   A  ∩      A    ∁    =  ∅ is in   R too. If   n  ≥  2 then       B    1    ∩      B    2    =  ∅  ∈  R as well, by closedness under finite intersections.So   Ω  =      ∅    ∁   is also a finite disjoint union of sets       C    i    ∈  R for   i  =  1  ,  …  ,  m  ∈      N  . Again if   m  =  1 we&#039;re done and   Ω  ∈  R. If   m  ≥  1 I don&#039;t see how to make progress towards   Ω  ∈  R. Applying intersections to the       C    i   is pointless and the complement property yields nothing new for them either.And   {  ∅  ,  {  1  }  ,  {  2  }  } is a semi-algebra on   Ω  =  {  1  ,  2  } without   Ω in it...

Let A then A c </msup> = <msubsup> ∪ <mrow

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