proof of equivalent definitions of analytic sets for Polish spaces
is -analytic (http://planetmath.org/AnalyticSet2).
is the projection (http://planetmath.org/GeneralizedCartesianProduct) of a closed subset of onto .
is the image (http://planetmath.org/DirectImage) of a continuous function for some Polish space .
is the image of a continuous function .
is the image of a Borel measurable function .
is the projection of a Borel subset of onto .
(1) implies (2): Let be the paving consisting of closed subsets of . The collection of -analytic sets contains the Borel -algebra of (see countable unions and intersections of analytic sets are analytic) and, as the analytic sets are given by a closure operator (http://planetmath.org/AnalyticSetsDefineAClosureOperator) it follows that it contains all analytic subsets of . So, any analytic subset of is -analytic. Then, there is a closed and a function such that
(see proof of equivalent definitions of analytic sets for paved spaces). For each let denote the closed subset of with . Then, we can rearrange the above expression to get where is the projection map and
It is easily seen that is a closed subset of for each , and therefore is closed, as required.
(2) implies (3): Suppose that for a closed subset of , where is the projection map. As the product of Polish spaces is Polish, and every closed subset of a Polish space is Polish, then will be a Polish space under the subspace topology. So, we can take and let be the restriction of to .
(3) implies (4): Suppose that is the image of a continuous function , for a Polish space . As Baire space is universal for Polish spaces, there exists a continuous and onto (http://planetmath.org/Surjective) function . The result follows by taking .
(4) implies (5): Suppose that is the image of a continuous function . Since uncountable Polish spaces are all Borel isomorphic (see Polish spaces up to Borel isomorphism), there is a Borel isomorphism . The result follows by taking .
The projection of onto is equal to , so the result will follow once it is shown that is a Borel set.
Choose a countable and dense subset of , and let be a metric generating the topology on . Then, for integers , denote the open ball about of radius by . Since the form a dense set, for each . Let us define
which contains . Furthermore, since are Borel, are Borel sets. Suppose that . Then, for each , there is an such that and . So,
This holds for all , showing that and so . We have shown that is Borel, as required.
|Title||proof of equivalent definitions of analytic sets for Polish spaces|
|Date of creation||2013-03-22 18:48:48|
|Last modified on||2013-03-22 18:48:48|
|Last modified by||gel (22282)|