I am reading about the Baire Category Theorem in Jech's book on set theory.
4.8: Baire Category Theorem: Let $D_0,D_1,\dots,D_n,\dots$, $n \in \mathbb{N}$, be open dense subsets of $\mathbb{R}$. Then the intersection, $D=\bigcap_{n=1}^{\infty} D_n\ $ is dense in $\mathbb{R}$.
My question:
Isn't any dense open set in $\mathbb{R}$ equal to $\mathbb{R}$? I mean, that $U$ is dense means that, for every open interval $(a,b)$, $U\cap(a,b)\neq\emptyset$. Furthermore, that $U$ is open means that every point is an interior point. So, isn't $U$ just a union of open intervals which equals $\mathbb{R}$? I can't figure out what I am missing here..
Thank you!
Enumerate the rational numbers $q_1, q_2, q_3, \dots$. Then place an open interval with length $1/2^n$ centered at $q_n$. The union of these intervals is open and dense, yet has finite total measure (length) and thus cannot be all of $\mathbb{R}$.