analytic sets define a closure operator

For a paving on a set X, we denote the collectionMathworldPlanetmath of all -analytic setsMathworldPlanetmath ( by a(). Then, a() is a closure operatorPlanetmathPlanetmathPlanetmath on the subsets of X. That is,

  1. 1.


  2. 2.

    If 𝒢 then a()a(𝒢).

  3. 3.


For example, if 𝒢 is a collection of -analytic sets then 𝒢a() gives a(𝒢)a(a())=a() and so all 𝒢-analytic sets are also -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 Aa(a()) we show that Aa(). First, there is a compactPlanetmathPlanetmath paved space ( (K,𝒦) and S(a()×𝒦)σδ such that A is equal to the projection πX(S). Write


for Am,na() and Bm,n𝒦. It is clear that Am,n×Bm,n is ×𝒦-analytic and, as countable unions and intersections of analytic sets are analytic, S is also ×𝒦-analytic. Finally, since projections of analytic sets are analytic, A=πX(S) must be -analytic as required.

Title analytic sets define a closure operator
Canonical name AnalyticSetsDefineAClosureOperator
Date of creation 2013-03-22 18:46:30
Last modified on 2013-03-22 18:46:30
Owner gel (22282)
Last modified by gel (22282)
Numerical id 4
Author gel (22282)
Entry type Theorem
Classification msc 28A05