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This question already has an answer here:

For $X$ a set and $A\subset X$ the sigma-algebra generated by $A$ is $$\sigma(A)=\bigcap\{\Sigma: \text{$\Sigma$ $\sigma$-algebra s.t. $A\subset\Sigma$}\}.$$ However why can' t we simply say $$\sigma(A)=\{\text{all countable unions and complements of elements of $A$}\}\cup \{\emptyset,X\}?$$

What would be a good counterexample?

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marked as duplicate by Claude Leibovici, TheSimpliFire, Arnaud Mortier, The Phenotype, Parcly Taxel Feb 17 '18 at 14:27

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When you add all the countable unions and complements the class of sets you get is not closed under countable unions and complements.

Given $F$, a family of subsets of $X$, let's say $C(F)$ is $F$, plus $\emptyset$, $X$, and all countable unions and complements of elements of $F$. Now say $A_1,A_2,\dots\in F$. Then $B=\bigcup A_n\in C(F)$, but the complement of $B$ need not be in $C(F)$. (Similarly, the union of the complements of $A_n$ need not be in $C(F)$.

You could try just repeating it, considering $C(C(F))$. That's not closed under countable unions and complements either. Give up, do it infinitely many times, and take the union of $C(F)$, $(C(C(F))\dots$? That's not enough either, because if $A_n\in C^n(F)$ we never picked up the union of the $A_n$ at any stage.

So apply $C$ infinitely many times, take the union, and then apply $C$ one more time? Not enough. Speaking informally, you have to apply $C$ uncountably many times before you get something that's guaranteed to be closed under countable unions and complements. You can make sense of "apply $C$ uncountably many times" via transfinite recursion; you need to apply $C$ exactly $\omega_1$ times, where $\omega_1$ is the first uncountable ordinal.

(Transfinite recursion is what it actually is - if you look for an explanation of this online you may have better luck searching for "tranfinite induction"...)

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