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For algebras generated by a family of sets there is a way to explicitly describe how this algebra looks like. But my measure-theory lecture notes tell me, that for $\sigma$-algebras this is not the case anymore since one can prove that

the $\sigma$-algebra generated by some family of sets $\mathcal{S}$ is strictly greater that the family of sets that can be constructed by a sequence of countable infinite set-theoretic operation applied to $\mathcal{S}$.

My questions are:

  • Since this last assertion to me has a little bit of a meta-mathematical flavour (since "countable infinite many set-theoretic operations" is not a mathematical object) I would like to know how do you formulate it mathematically (if necessary in predicate logic) ?
    I believe this may be rather complicated since I can't think of an efficient way to describe all sets one can obtain this way (we have all countable unions of sets from $\mathcal{S}$ in it -- but also the set where we successively intersect two sets and then take the complement, than intersect with some other two sets and take the complement again and so on...this is already pretty cumbersome to describe)

  • Can you give me a reference for its proof ?

share|cite|improve this question
Are you familiar with transfinite recursion? – Michael Greinecker Aug 14 '13 at 5:33
@MichaelGreinecker No, I'm not, but I've heard that this is the tool of choice in these situations (do you maybe mean transfinition induction?). But isnt't it possible to answer my question without appealing to transfinite recursion/induction ? – MyCatsHat Aug 14 '13 at 19:15

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