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I'm working on this problem. I found a solution but I don't really get it..

Let $A_1 \in \mathcal{B}$ be neither the empty set nor the whole set. Then for every other $W \in \mathcal{B}, W = (W \bigcap A_1) \bigcup (W \bigcap A_1^c)$. Let $B_{A_1}$ denote the σ-algebra restricted to $A_1$ and $\mathcal{B}_{A^c_1}$ the one restricted to $A^c_1$.

Then $\mathcal{B} = \mathcal {B}_{A_1} × \mathcal{B}_{A^c_1}$ as sets. (1)

Since $\mathcal{B}$ is infinite, either $B_{A_1}$ or $\mathcal{B}_{A^c_1}$ (or both) must be infinite. Without loss of generality (replacing $A_1$ with $A^c_1$) let us suppose that $\mathcal{B}_{A^c}$ is infinite. Then let $X_1 = A_1$ and repeat the process with $\mathcal{B}_{A^c_1}$ . By induction we have a sequence $X_1, X_2, . . .$ such that $X_k \bigcap X_j = \emptyset$ for $k\neq j$ and $X_k \in \mathcal{B}$, as required.

My problem is the point (1), why can I write $\mathcal{B}$ in that way? Thanks

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  • $\begingroup$ Do you meant them to be non-empty as well? Because it seems like you can just take a collection of empty sets. As for (1), what does $\times$ means, because it sure doesn't appear to be a Cartesian product. $\endgroup$
    – Gina
    Commented Jan 5, 2014 at 8:07
  • $\begingroup$ If I understand it correctly assuming that $A_i$ aren't empty then my sequence is a sequence of non-empty sets. Yes the cross product I really don't understand but that's the proof that I have for the statement... $\endgroup$
    – sky90
    Commented Jan 5, 2014 at 8:26
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    $\begingroup$ I have a suspicion that they meant to say $|\mathcal{B}|=|\mathcal{B}_{A_{1}}\times\mathcal{B}_{A_{1}^{c}}|$ which make more sense. Since you can do the bijection $f:\mathcal{B}_{A_{1}}\times\mathcal{B}_{A_{1}^{c}}\rightarrow\mathcal{B}$ as $f(X,Y)=X\bigcup Y$. Then that $\times$ would be the Cartesian product. After all, you only need the argument that one of them is infinite, so cardinality argument is fine. $\endgroup$
    – Gina
    Commented Jan 5, 2014 at 8:30
  • $\begingroup$ I see.. thanks for your help :) $\endgroup$
    – sky90
    Commented Jan 5, 2014 at 8:40

1 Answer 1

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Take $X_{1}$ a set in $\mathbb{B}$ including only $1$ element; then $\mathbb{B}/X_{1} $ is non-empty because $\mathbb{B}$ is infinite.

So take $X_{2}$ a set in $\mathbb{B}/X_{1}$ including only $1$ element and so on. In this way you get a sequence of pairwise disjoint sets.

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  • $\begingroup$ ok but why can I write this cross product? $\endgroup$
    – sky90
    Commented Jan 5, 2014 at 8:18
  • $\begingroup$ @sky90: What do you mean by cross product ? $\endgroup$
    – WLOG
    Commented Jan 5, 2014 at 10:58
  • $\begingroup$ Cartesian product :-) but I think that Gina has already helped me. Thanks $\endgroup$
    – sky90
    Commented Jan 9, 2014 at 10:39

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