I’m learning about Borel sets and I’m struggling to understand the following sentence.
The intersection of every sequence of open subsets of $\mathbb{R}$ is a Borel set. However, the set of all such intersections is not the set of Borel sets (because it is not closed under countable unions).
Why isn’t the set of all countable intersections closed under countable unions?
Let
$$(A_{1,1}\cap…\cap A_{1,n}\cap…)\cup(A_{2,1}\cap…\cap A_{2,n}\cap…)\cup…$$
be a countable union of intersections of open sets
$A_{i,j}$, with $i,j \in \mathbb{Z}^+$. If we were to expand this, we would get an intersection of unions like
$$(A_{1,1}\cup A_{2,1}\cup A_{3,1}\cup…)\cap(A_{1,2}\cup A_{2,1}\cup A_{3,1}\cup…)\cap…$$
Since each union above is a countable union of open sets, and the countable union of open sets is also an open set, we thus have the overall quantity written as an intersection of open sets.
Where is the flaw in my reasoning?
Best Answer
If you actually fully expand out, you'll realize you actually have an uncountable intersection, because the index set of your intersection is actually the set of all sequences $\mathbb{N} \to \mathbb{N}$. For an explicit example, $\mathbb{Q}$ is a countable union of countable intersections of open sets, but it is not a countable intersection of open sets itself. See the example section of https://en.wikipedia.org/wiki/G%CE%B4_set