proof of countable unions and intersections of analytic sets are analytic
From the definition of -analytic sets, there exist compact paved spaces and such that
We start by showing that is analytic. Let and be the product paving, and be the projection map. Then if and only if for each there is a with . Equivalently, setting , then . However, this is in and we can write,
where is the projection map. As products of compact pavings are compact, is compact and it follows from the definition that is -analytic.
We now show that is analytic. Let and be the direct sum paving, which is compact (http://planetmath.org/SumsOfCompactPavingsAreCompact). Also, write for . We identify with a subset of , so that is the union of the disjoint sets . Then if and only if for some and some ,
However, the fact that are disjoint for says that are disjoint and, therefore,
So is -analytic.
|Title||proof of countable unions and intersections of analytic sets are analytic|
|Date of creation||2013-03-22 18:46:19|
|Last modified on||2013-03-22 18:46:19|
|Last modified by||gel (22282)|