$\bigcup_{I} \bigcap_{n=1}^{\infty} A_{i_{n}} \in \mathcal{A}$

Asked Oct 01 '20 at 09:16

Active Oct 01 '20 at 10:53

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Suppose $A_{n} \in \mathcal{A}$, $\mathcal{A} - \sigma$-algebra. And $I:=\{ i_{1},i_{2},...,i_{n},...\}$ - some infinite sequence of natural numbers. I need to prove that $\bigcup_{I} \bigcap_{n=1}^{\infty} A_{i_{n}} \in \mathcal{A}$.

It's easy to see that $ \bigcap_{n=1}^{\infty} A_{i_{n}} \in \mathcal{A}$ for every $I$. However, the number of infinite sequances is uncountable and I don't know how to solve this problem. Please, give me any hints.

Thank you!!!

edited Oct 01 '20 at 10:53

asked Oct 01 '20 at 09:16

Ingrid

Your notation is confusing. Is each $i_j$ a sequence? – Oct 01 '20 at 09:27
No, each $I$ is a sequence – Ingrid Oct 01 '20 at 09:28
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This isn't true. This is the Suslin $\mathcal A$-operation, which when applied to the Borel sets (in a complete separable metric space with no isolated points) can produce non-Borel sets. The collection of sets obtainable by applying the $A$-operation to closed sets produces all Borel sets and other sets (collectively called the analytic sets). The collection of analytic sets is NOT a $\sigma$-algebra. See also What are some good intuitions for understanding Souslin's operation $\mathcal{A}$? – Dave L. Renfro Oct 01 '20 at 09:29
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Incidentally, the term "$C$ sets" is often used to name the smallest $\sigma$-algebra containing the analytic sets. It can be formed by a transfinite construction of $\omega_1$ many steps that alternate between applying the Suslin operation and the operation of taking taking complements, and taking the union of all previous collections at the limit ordinal steps. For a little more about $C$ sets, see this answer. – Dave L. Renfro Oct 01 '20 at 09:37
OK. So $I\in ?$ – Oct 01 '20 at 10:07

$\bigcup_{I} \bigcap_{n=1}^{\infty} A_{i_{n}} \in \mathcal{A}$

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