# analytic sets define a closure operator

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

1. 1.

$\mathcal{F}\subseteq a(\mathcal{F})$.

2. 2.

If $\mathcal{F}\subseteq\mathcal{G}$ then $a(\mathcal{F})\subseteq a(\mathcal{G})$.

3. 3.

$a(a(\mathcal{F}))=a(\mathcal{F})$.

For example, if $\mathcal{G}$ is a collection of $\mathcal{F}$-analytic sets then $\mathcal{G}\subseteq a(\mathcal{F})$ gives $a(\mathcal{G})\subseteq a(a(\mathcal{F}))=a(\mathcal{F})$ and so all $\mathcal{G}$-analytic sets are also $\mathcal{F}$-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 $A\in a(a(\mathcal{F}))$ we show that $A\in a(\mathcal{F})$. First, there is a compact paved space (http://planetmath.org/PavedSpace) $(K,\mathcal{K})$ and $S\in\left(a(\mathcal{F})\times\mathcal{K}\right)_{\sigma\delta}$ such that $A$ is equal to the projection $\pi_{X}(S)$. Write

 $S=\bigcap_{m=1}^{\infty}\bigcup_{n=1}^{\infty}A_{m,n}\times B_{m,n}$

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

Title analytic sets define a closure operator AnalyticSetsDefineAClosureOperator 2013-03-22 18:46:30 2013-03-22 18:46:30 gel (22282) gel (22282) 4 gel (22282) Theorem msc 28A05