Borel sets as the countable intersection of open sets

Question

" The intersection of every sequence of open subsets of R is a borel set.However set of all such intersections is not the borel set because it is not closed under countable unions"

I cannot see why the set is not closed under countable unions. Any countable union of such sets can be expressed as the countable intersection of opensets using the fact that countable union of countable sets is countable. More specifically if the first set is the intersection of all O_1n Second set O_2n and so on... Then union of all such sets can be written as intersection of the sets where each set is the union of a possible combination of O _1(n_1),O_2(n_2) and so on...certainly number of such possibilities is countable. I dont understand where i'm going wrong.

Answer 1

As it stands, your argument is all hand-waving; if, as was suggested in the comments, you try to make it precise, you’ll find yourself running into difficulties. In the meantime, here is an actual example of a countable union of $G_\delta$ sets that is not itself a $G_\delta$ set.

Let $P=\Bbb R^+\setminus\Bbb Q$, the set of positive irrationals, and let $Q=\Bbb Q\setminus\Bbb Q^+$, the set of non-positive rationals. Then $P$ is a $G_\delta$ in $\Bbb R$, and so is $\{q\}$ for each $q\in Q$, so $P\cup Q$ is a countable union of $G_\delta$ sets. It is not, however, a $G_\delta$ set itself: if it were one, $Q$ would also be one, and the Baire category theorem shows that $Q$ is not a $G_\delta$.

Answer 2

Let $ K = \mathbb N \times \mathbb N \times \mathbb N \times \dots $. $$ \bigcup _ { j \in \mathbb N } \left( \bigcap _ { i \in \mathbb N } U ^ j _ i \right) = \bigcap _ { ( n _ 1 , n _ 2 , \dots ) \in K } \left( \bigcup _ { j \in \mathbb N } U ^ j _ { n _ j } \right) \text . $$ Whats wrong with the notation used here?