Give an example of an infinite disjoint sequence of subsets of the real numbers, each of which is of second category in the real numbers. The imperative in the title is Exercise 21.12 from Elementary Analysis: The Theory of Calculus, Second Edition, written by Kenneth Ross. (page 178)
I have searched on this website for "elementary analysis 21.12" and "infinite disjoint sequence second category" but didn't find anything that seemed relevant.
For context, this question follows a section that stated and proved the Baire Category Theorem.
My problem:
I wanted to give the example $(0, 1), (1, 2), (2, 3), \dots$ but I am not quite sure if these sets are of second category in $\mathbb{R}$. I think they are, because none of them be written as the union of a sequence of nowhere dense subsets of $\mathbb{R}$. (Informally, you cannot union a countable number of single points to create an interval.)
I found what purports to be a solution at the link https://www.slader.com/textbook/9781461462705-elementary-analysis-the-theory-of-calculus-2nd-edition/178/exercises/12/#, but the answer given is $U_n = \mathbb{Q} \setminus [n, n + 1)$ for $n = 0, 1, -1, 2, -2, \dots$. This seems overly elaborate, and I'm not even sure if it meets the criteria since it doesn't look to me like the sets $U_n$ are disjoint.
Am I missing something obvious?
Thanks.
 A: The sets $(0, 1), (1, 2), (2, 3), \dots$ are clearly disjoint. We want to prove that they are of second category in $\mathbb{R}$.
We will assume that the set $(n, n + 1)$---where $n$ is a non-negative integer---is of first category in $\mathbb{R}$ and find a contradiction. This proof uses the Baire Category Theorem in the formulation "the union of a sequence of nowhere dense subsets of $\mathbb{R}$ has dense complement." It also uses the definition of a set of first category, which is that a set is of first category in $\mathbb{R}$ if can be written as the union of a sequence of nowhere dense subsets of $\mathbb{R}$. (Category 2 sets are all sets not in Category 1.)
Assume $(n, n + 1)$ is of first category in $\mathbb{R}$. Using the definition of a set of first category, we see that $(n, n + 1)$ can be written as the union of a sequence of nowhere dense subsets of $\mathbb{R}$. By the aforementioned formulation of the BCT, the complement $\mathbb{R} \setminus (n,n+1)$ is dense in $\mathbb{R}$. This is clearly false, so we conclude that $(n,n+1)$ is of second category.
Postscript: This proof seems to show that any non-degenerate interval of finite length is of second category in $\mathbb{R}$.
