proof of countable unions and intersections of analytic sets are analytic

Let $(X,\mathcal{F})$ be a paved space and $(A_{n})_{n\in\mathbb{N}}$ be a sequence of $\mathcal{F}$ -analytic sets (http://planetmath.org/AnalyticSet2). We show that their union and intersection is also analytic.

From the definition of $\mathcal{F}$ -analytic sets, there exist compact paved spaces $(K_{n},\mathcal{K}_{n})$ and $S_{n}\in(\mathcal{F}\times\mathcal{K}_{n})_{\sigma\delta}$ such that

$A_{n}=\left\{x\in X\colon(x,y)\in S_{n}\text{ for some }y\in K_{n}\right\}.$

We start by showing that $\bigcap_{n}A_{n}$ is analytic. Let $K=\prod_{n}K_{n}$ and $\mathcal{K}=\prod_{n}\mathcal{K}_{n}$ be the product paving, and $\pi_{n}\colon K\rightarrow K_{n}$ be the projection map. Then $x\in\bigcap_{n}A_{n}$ if and only if for each $n$ there is a $y_{n}\in K_{n}$ with $(x,y_{n})\in S_{n}$ . Equivalently, setting $y=(y_{1},y_{2},\cdots)$ , then $(x,y)\in\bigcap_{n}\pi^{-1}_{n}(S_{n})$ . However, this is in $(\mathcal{F}\times\mathcal{K}_{n})_{\sigma\delta}$ and we can write,

$\bigcap_{n}A_{n}=\pi_{X}\left(\bigcap_{n}\pi^{-1}_{n}(S_{n})\right),$

where $\pi_{X}\colon X\times K\rightarrow X$ is the projection map. As products of compact pavings are compact, $(K,\mathcal{K})$ is compact and it follows from the definition that $\bigcap_{n}A_{n}$ is $\mathcal{F}$ -analytic.

We now show that $\bigcup_{n}A_{n}$ is analytic. Let $K=\sum_{n}K_{n}$ and $\mathcal{K}=\sum_{n}\mathcal{K}_{n}$ be the direct sum paving, which is compact (http://planetmath.org/SumsOfCompactPavingsAreCompact). Also, write $S_{n}=\bigcap_{m=1}^{\infty}T_{m,n}$ for $T_{m,n}\in(\mathcal{F}\times\mathcal{K}_{n})_{\sigma}$ . We identify $K_{n}$ with a subset of $K$ , so that $K$ is the union of the disjoint sets $K_{n}$ . Then $x\in\bigcup_{n}A_{n}$ if and only if $(x,y)\in S_{n}$ for some $n$ and some $y\in K$ ,

$\bigcup_{n}A_{n}=\pi_{X}\left(\bigcup_{n}S_{n}\right).$

However, the fact that $K_{n_{1}},K_{n_{2}}$ are disjoint for $n_{1}\not=n_{2}$ says that $T_{m,n_{1}},T_{m,n_{2}}$ are disjoint and, therefore,

$\bigcup_{n}S_{n}=\bigcup_{n}\bigcap_{m}T_{m,n}=\bigcap_{m}\bigcup_{n}T_{m,n}% \in\left(\mathcal{F}\times\mathcal{K}\right)_{\sigma\delta}.$

So $\bigcup_{n}A_{n}$ is $\mathcal{F}$ -analytic.

Title	proof of countable unions and intersections of analytic sets are analytic
Canonical name	ProofOfCountableUnionsAndIntersectionsOfAnalyticSetsAreAnalytic
Date of creation	2013-03-22 18:46:19
Last modified on	2013-03-22 18:46:19
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	4
Author	gel (22282)
Entry type	Proof
Classification	msc 28A05