proof of countable unions and intersections of analytic sets are analytic

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

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

$$A_n=\{x\in X:(x,y)\in S_n\text{ for some }y\in K_n\}.$$

We start by showing that ${\bigcap }_n{A}_n$ is analytic. Let $K={\prod }_n{K}_n$ and $𝒦={\prod }_n{𝒦}_n$ be the product paving, and ${\pi }_n:K\to K_n$ 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 $y=(y_1,y_2,\mathrm{\cdots })$, then $(x,y)\in {\bigcap }_n{\pi }_n^{-1}(S_n)$. However, this is in ${(ℱ\times {𝒦}_n)}_{\sigma \delta }$ and we can write,

$$\bigcap _n{A}_n={\pi }_X\left(\bigcap _n{\pi }_n^{-1}(S_n)\right),$$

where ${\pi }_X:X\times K\to X$ is the projection map.As products of compact pavings are compact, $(K,𝒦)$ is compact and it follows from the definition that ${\bigcap }_n{A}_n$ is $ℱ$-analytic.

We now show that ${\bigcup }_n{A}_n$ is analytic. Let $K={\sum }_n{K}_n$ and $𝒦={\sum }_n{𝒦}_n$ be the direct sum paving, which is compact (http://planetmath.org/SumsOfCompactPavingsAreCompact). Also, write $S_n={\bigcap }_{m=1}^{\mathrm{\infty }}{T}_{m,n}$ for $T_{m,n}\in {(ℱ\times {𝒦}_n)}_{\sigma }$.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 $n_1\ne n_2$ says that $T_{m,n_1},T_{m,n_2}$ are disjoint and, therefore,

$$\bigcup _n{S}_n=\bigcup _n\bigcap _m{T}_{m,n}=\bigcap _m\bigcup _n{T}_{m,n}\in {\left(ℱ\times 𝒦\right)}_{\sigma \delta }.$$

So ${\bigcup }_n{A}_n$ is $ℱ$-analytic.