analytic sets define a closure operator

For a paving $ℱ$ on a set $X$, we denote the collection of all $ℱ$-analytic sets (http://planetmath.org/AnalyticSet2) by $a(ℱ)$.Then, $ℱ\mapsto a(ℱ)$ is a closure operator on the subsets of $X$.That is,

1. $ℱ\subseteq a(ℱ)$.

2. If $ℱ\subseteq 𝒢$ then $a(ℱ)\subseteq a(𝒢)$.

3. $a(a(ℱ))=a(ℱ)$.

For example, if $𝒢$ is a collection of $ℱ$-analytic sets then $𝒢\subseteq a(ℱ)$ gives $a(𝒢)\subseteq a(a(ℱ))=a(ℱ)$ and so all $𝒢$-analytic sets are also $ℱ$-analytic. In particular, for a metric space, the analytic sets are the same regardless of whether they are defined with respect to the collection of open, closed or Borel sets.

Properties 1 and 2 follow directly from the definition of analytic sets. We just need to prove 3. So, for any $A\in a(a(ℱ))$ we show that $A\in a(ℱ)$. First, there is a compact paved space (http://planetmath.org/PavedSpace) $(K,𝒦)$ and $S\in {\left(a(ℱ)\times 𝒦\right)}_{\sigma \delta }$ such that $A$ is equal to the projection ${\pi }_X(S)$.Write

$$S=\bigcap _{m=1}^{\mathrm{\infty }}\bigcup _{n=1}^{\mathrm{\infty }}{A}_{m,n}\times B_{m,n}$$

for $A_{m,n}\in a(ℱ)$ and $B_{m,n}\in 𝒦$. It is clear that $A_{m,n}\times B_{m,n}$ is $ℱ\times 𝒦$-analytic and, as countable unions and intersections of analytic sets are analytic, $S$ is also $ℱ\times 𝒦$-analytic. Finally, since projections of analytic sets are analytic, $A={\pi }_X(S)$ must be $ℱ$-analytic as required.