# countable unions and intersections of analytic sets are analytic

###### Theorem 1.

Let $(X,\mathcal{F})$ be a paved space and $(A_{n})_{n\in\mathbb{N}}$ be a sequence  of $\mathcal{F}$-analytic sets. Then, $\bigcup_{n}A_{n}$ and $\bigcap_{n}A_{n}$ are $\mathcal{F}$-analytic.

###### Corollary.

Let $\mathcal{F}$ be a nonempty paving on a set $X$ such that the complement (http://planetmath.org/Complement) of any $S\in\mathcal{F}$ is a union of countably many sets in $\mathcal{F}$.

Then, every set $A$ in the $\sigma$-algebra generated by $\mathcal{F}$ is $\mathcal{F}$-analytic.

For example, every closed subset of a metric space $X$ is a union of countably many open sets. Therefore, the corollary shows that all Borel sets are analytic with respect to the open subsets of $X$.

That the corollary does indeed follow from Theorem 1 is a simple application of the monotone class theorem. First, as the collection  $a(\mathcal{F})$ of $\mathcal{F}$-analytic sets is closed under countable unions and finite intersections, it will contain all finite unions of finite intersections of sets in $\mathcal{F}$ and their complements, which is an algebra (http://planetmath.org/RingOfSets). Then, Theorem 1 says that $a(\mathcal{F})$ is closed under taking limits of increasing and decreasing sequences of sets. So, by the monotone class theorem, it contains the $\sigma$-algebra generated by $\mathcal{F}$.

Title countable unions and intersections of analytic sets are analytic CountableUnionsAndIntersectionsOfAnalyticSetsAreAnalytic 2013-03-22 18:45:18 2013-03-22 18:45:18 gel (22282) gel (22282) 4 gel (22282) Theorem msc 28A05