projections of analytic sets are analytic
Projections along compact paved spaces
If is -analytic then is -analytic.
The proof of this follows easily from the definition of analytic sets. First, there is a compact paved space and a set such that . Then,
However, is a compact paved space (see products of compact pavings are compact (http://planetmath.org/ProductsOfCompactPavingsAreCompact)), which shows that satisfies the definition of -analytic sets.
Projections along Polish spaces
Theorem 1 above can be used to prove the following result for projections from the product of a measurable space and a Polish space. For -algebras (http://planetmath.org/SigmaAlgebra) and , we use the notation for the product -algebra (http://planetmath.org/ProductSigmaAlgebra), in order to distinguish it from the product paving .
Let be a measurable space and be a Polish space with Borel -algebra (http://planetmath.org/BorelSigmaAlgebra) .
If is -analytic, then its projection onto is -analytic.
Although Theorem 2 applies to arbitrary Polish spaces, it is enough to just consider the case where is the space of real numbers with the standard topology. Indeed, all Polish spaces are Borel isomorphic to either the real numbers or a discrete subset of the reals (see Polish spaces up to Borel isomorphism), so the general case follows from this.
If , then the Borel -algebra is generated by the compact paving of closed and bounded intervals. The collection of analytic sets is closed under countable unions and countable intersections so, by the monotone class theorem, includes the product -algebra . Then, as the analytic sets define a closure operator,
Thus every -analytic set is -analytic, and the result follows from Theorem 1.
|Title||projections of analytic sets are analytic|
|Date of creation||2013-03-22 18:46:21|
|Last modified on||2013-03-22 18:46:21|
|Last modified by||gel (22282)|