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Consider, for simplicity, the Borel $\sigma$-algebra of the unit interval $[0,1]$.

Let $\{A_i\}$ be a family of Borel subsets which generate the sigma algebra (can be either countable or uncountable). I was wondering if there is some way of describing all subsets of the sigma-algebra only using $A_i$ and set operations.

Moreover, I am curious if we can linearly order all elements of the Borel sigma algebra in a way that any element in the ordering is either one of the generating elements or can be described as a countable union of elements that have appeared before it.

My motivation is, among other things, to prove following two statements in a straightforward way;

(1) The cardinality of the Borel $\sigma$-algebra equals the cardinality of $\mathbb{R}$.

(2) If a $\pi$ system $K$ is contained in a Dynkin system $D$, then the $\sigma$-algebra generated by $K$ is contained in $D$.

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Yes. The usual construction is the following (from memory, someone please double check me):

Let $\mathcal{C}_0$ be some collection of sets, and for ordinals $\alpha > 0$, define $\mathcal{C}_\alpha$ recursively as the collection of all countable unions and complements of sets from $\bigcup_{\beta < \alpha} \mathcal{C}_\beta$. Then $\sigma(\mathcal{C}_0) = \mathcal{C}_{\omega_1}$, where $\omega_1$ is the first uncountable ordinal.

So in a sense, any Borel set can be produced by taking the sets $A_i$, taking countable unions and complements, and iterating uncountably many times.

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  • $\begingroup$ This is very nice, although I don't think I can prove it. So, the statement $\sigma(C_0)=C_{\omega_1}$ proves (1)? $\endgroup$ – a12345 Dec 10 '13 at 4:43
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    $\begingroup$ @a12345 For the proof, you can take a look at section 4.1 in The joy of sets by Keith Devlin. That is probably the most readable exposition. The book also shows how one can use the construction to show there are as many real numbers as Borel sets. $\endgroup$ – Michael Greinecker Dec 10 '13 at 8:03

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