Souslin operation

I'm reading J. Neveu's "Mathematical Foundations of the Calculus of Probability". There is a question regarding the Souslin operation and proving it is idempotent. In essence, we start with a collection $\mathcal{F}$ of subsets of $\Omega$.

Here is my line of thought. Also, note that the notation I'm using is from the book. We denote the collection of sets resulting from the operation as $\mathcal{F}_S$. We have to show that $\mathcal{F}_{SS}=\mathcal{F}_S$. It is easy to show that $\mathcal{F}_{S} \subset{F}_{SS}$. Assume now that $A\in \mathcal{F}_{SS}$. Therefore, 

$$ A=\bigcup_{v \in \mathrm{N^N}} \bigcap_{k \ge 1} \overline{F}_(v_1,...,v_k), $$ where $\overline{F} \in \mathcal{F}_{S}$. But, 

$$\overline{F}_(v_1,\ldots,v_k)=\bigcup_{u \in \mathrm{N^N}}\bigcap_{l\ge 1}F_{(u_1,\ldots,u_l);(v_1,\ldots,v_k)},$$
where $F \in \mathcal{F}.$

This implies that we can write $A$ as:

$$ A=\bigcup_{v \in \mathrm{N^N}} \bigcap_{k \ge 1}\bigcup_{u \in \mathrm{N^N}}\bigcap_{l\ge 1}F_{(u_1,\ldots,u_l);(v_1,\ldots,v_k)}.$$


Neveu provides the following hint: use the following distributivity formula:
$$\bigcup_{j \in J} \bigcap_{i\in i_j} F_i^j=\bigcap_{\{i_j\}\in K}\bigcup_{j\in J} F_{i_j}^j.$$

We can translate the notation I have thus far to the hint in the following way:

$$F_i^j:=F_{(v_1,\ldots,v_i)}^v$$.

It is then easy to apply the hint to change the order of inner unions and intersections, but I'm not sure how that will help me here..It looks like I need to "eliminate" somehow some of the unions and intersections.

Note: I am looking for a solution that involves the notation I just used and the hint!

  • What do you mean by F_{(u_1,...,u_l);(v_1,...,v_k)}. The ; symbol confuses me.

  • I understand now, you fix the origin of (v_1,...,v_k)

  • Yes..the whole notation is pretty confusing, sorry

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  • Thanks! I am going through the solution. I have a question regarding the following equality: Give a family $(B_{k,m})$ where $k$ is in $N$ and $m$ is in $N^N$ we have that $\bigcap_{k\in N} \bigcup_{m\in N^N^}B_{k,m}=\bigcup_{f: N\rightarrow N^N}\bigcap_{k \in N} B_{k,f(k)}$ I dont' understand the RHS. What is the index running on in $f: N \rightarrow N^N$?

  • f itself is the index: you're taking the union over all the possible functions f : N -> N^N.

  • It might help to notice that the set of such functions is isomorphic to N^N^N, that's how you get it.

  • Maybe I should parenthesize. I mean (N^N)^N.

  • Ohh..interesting. First time I encounter such indexing. So these f take a natural number as input and have an infinite sequence as output..and obviously there are uncountably many of these functions.

  • Can you point me in the direction of the solution involves the hint by Neveu and how it relates to the indexing on f?

  • The way you swap intersection and union when that indexing appears is essentially the same as the distributivity formula, up to swapping intersection and union: your I_j are all N^N, J is N, and so the product over J of I_j is just (N^N)^N, which is exactly your collection of functions.

  • I guess it is tricky that Neveu states the distributivity formula that way, when you really need the statement with union and intersections swapped, but they are equivalent, and you can easily see it by taking the complements of the sets F_i^j, noting that the complement of the union is the intersection of the complements, and the complement of the intersections is the union of the complements.

  • If you really want to use the hint in the way it is given, you can prove the statement on the complements, saying you take A such that Omega \ A is in F_{SS}, which flips all the unions and the intersections, and then use the same argument to show that Omega \ A is in F_{S} too, but it seems an unnecessary complication to me.

  • It's probably easier to take the complement on both sides on the distributivity formula, which yields the same formula but with unions and intersections swapped, and with F_i^j replaced by Omega \ F_i^j. The distributivity formula does not use the fact that F_i^j is in \mathcal{F}, so it holds all the same. I hope it is clear, otherwise I'll just type it more formally in the body of the answer.

  • Thanks! Got it. The whole notation confused me.

  • Cool, glad that everything was fine!

The answer is accepted.