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.
Best Answer
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 $\mathbb{R}$.