Can any subset of $\mathbb{R}$ be generated by taking countable unions, countable intersections and complements of open intervals?
Clearly, singletons can be generated from the complement of the union of half-rays, e.g.: $$a=((-\infty, a) \cup (a, \infty))^C .$$
Closed intervals can also be generated by a countable intersection of open sets of the form $(a- \frac1n, b+\frac1n)$ and similarly for half-open intervals.
From this, it seems obvious that any countable union/intersection of intervals can be generated.
How about for example uncountable unions/intersections of intervals? It is unclear to me whether this is enough to generate all subsets of $\mathbb{R}$. Can this be done?
Best Answer
Not every subset of $\mathbb{R}$ can be "created" by this process. If it were possible, we would get that the Borel $\sigma$-algebra would contain all subsets of $\mathbb{R}$, which it does not.