baire category and the union of dyadic balls of rational center

Question

Suppose that $\{r_n\}_{n=0}^\infty$ is an enumeration of $\mathbb{Q}^N$ and $U = \bigcup_{n=0}^\infty B(r_n,2^{-n})$. We can use a trivial measure theory argument to prove that $U \neq \mathbb{R}^N$.

I was poking around in Real Analysis by Carothers in the section on Baire category (before measure theory has been developed in the book), and I noticed that there's a problem there that asks to prove that $U \neq \mathbb{R}^N$. There are no hints, but given the placement in the book, the goal is presumably to use Baire category.

I have never seen a non-measure based proof of this result, so I gave it a try. I'm a bit stuck, though. It's clear that Baire tells us that one of the sets $\bigcup_{n=k}^\infty B(r_n,2^{-n})$ for $k \in \mathbb{N}$ must not be dense, but I'm not sure how to use this information to prove the result. Any thoughts would be greatly appreciated.

EDIT: Here are the details on my above claim. Set $U_k = \bigcup_{n=k}^\infty B(r_n,2^{-n})$, which is clearly open. Suppose, BWOC, that each $U_k$ is dense in $\mathbb{R}^N$. Then Baire says that $\bigcap_{k=0}^\infty U_k$ is dense in $\mathbb{R}^N$, but on the other hand by construction $\bigcap_{k=0}^\infty U_k = \varnothing$. This contradiction means we must have $E_k$ not dense for some $k$.

EDIT 2: As pointed out in the comments below, this argument doesn't seem to fly. So, back to square one...

Answer 1

You can get inspiration from measure theory without actually using it, as follows.

Let $D$ be any straight line in ${\mathbb R}^N$ (which we will identify with $\mathbb R$). For any $n\geq 0$, $I_n=B(r_n,2^{-n}) \cap D$ is a (possibly empty) open line segment of length at most $2\times 2^{-n}$. So the sum of the lengths of the $I_n$ is at most $4$. It is then easy to construct by induction a decreasing sequence of compact subsets $(K_n)_{n\geq 1}$ of $D$, such that for each $n$, $K_n$ is disjoint from $I_n$, and $K_n$ is a finite union of closed line segments whose lengths sum up to at least $2015-\sum_{k=1}^{n}{\textsf{length}}(I_k)$.

By the Heine-Borel property, $\bigcap_{n\geq 1}K_n$ is nonempty and any $x$ in it will be outside your $U$.

Answer 2

Carothers (2000, p. 136) notes that this problem was adapted from a short paper by Wilansky (1953). The said paper, in turn, uses the classic measure-theoretic argument to show that the complement of $U$ is not empty. The proofs of the other statements, namely that

• $U$ is open,
• $U$ is dense, and
• $U^c$ is nowhere dense,

do not require Baire's category theorem, either.

Thus, it looks like Carothers never intended to have the reader come up with a purely topological argument excluding measure theory, despite the fact that measure theory is discussed way later in the book.

Answer 3

@EwanDelanoy alrady posted the main idea, but since I worked quite a bit for this, I will post my version of the argument as well.

First of all, there is no Baire-category argument, or any purely topological (non-metric) argument, that can give the result. The statement is a metric one, and depends intimately on the diameters of the covering sets. Here is an exercise: take the dimension to be $1$. If $a_n=1$ is the sequence of radii, show that for any enumeration of the rationals, the balls $B(r_n,a_n)$ cover $\mathbb{R}$. If $a_n=\frac{1}{n}$ , show that for some enumerations the balls cover, for some others they don't. Then classify precisely those sequences $a_n$ so that the balls cover for every enumeration, those that cover for some enumerations, and those that do not cover for any enumeration.

Now for the problem at hand, let's assume that $U=\mathbb{R}^n$ and proceed to derive a contradiction. In $\mathbb{R}^n$ take $\overline{0}$ to be the origin and $\overline{N} = (N,0,\cdots,0)$. Let $I$ be the closed line segment joining them. $U$ is an open cover of the compact set $I$, so there exist finitely many balls centered at say $q_0,\cdots, q_M$ that cover $I$.

By strict convexity (this is not necessary, but it simplifies arguments), the intersections of the open balls with $\textrm{R}\times 0 \times\cdots\times 0$ are open intervals denoted $V_i = (l_i,r_i)$. Let $V_0$ be an open interval containing $0$ by relabeling if necessary.

Perform the following iterative procedure: start with $V_0$ and suppose we have constructed $V_{i-1}$. If $V_{i-1}$ contains $N$, then stop.
If not, there exists an interval, call it $V_i$, so that $l_i<r_{i-1}<r_i$.
If it did not exist, then the $V$ would be partitioned into two sets: those with $r_i\leq r_{i-1}$ and those with $r_{i-1}\leq l_i$. The first (open) set is contained in some interval $[-1,r_{i-1}-\epsilon]$ and the second one in some $[r_{i-1}+\epsilon, N+1]$; thus the collection $V_i$ does not cover $[0,N]$, a contradiction.

Since we have a finite number of intervals, this procedure will terminate with an interval $V_m$ ($0\leq m\leq M$) that contains $N$. So now we have a chain of intervals $V_0,\cdots, V_m$ so that $$V_i\cap V_{i+1}\neq \emptyset,$$ $$0\in V_0$$ and $$N\in V_m.$$

Let $t_0=0,t_m=N$ and pick $$t_i\in V_{i-1}\cap V_i$$ for $1\leq i \leq m$ (if $m=0$ skip to the contradiction immediately). Then $$N-0 = t_m-t_{m-1}+t_{m-1}-t_{m-2}+\cdots +t_1-t_0\leq \sum |t_i-t_{i-1}|.$$

But the segment* $[t_0,t_1]$ is in $V_0$; the segment $[t_1,t_2]$ is in $V_1$; the segment $[t_2,t_3]$ is in $V_2$ and so on. Thus the length of each such segment is at most the length of the $V_i$ it is contained in, which in turn is at most the diameter of the ball it came from, and the sum of these diameters is at most $3$, so $N\leq 1$ giving a contradiction for large $N$.

*(Note it is not necessarily true that $t_i<t_{i+1}$, i.e. the path $t_i$ might oscillate, so when I write $[t_i,t_{i+1}]$ I mean the smallest of the two is the leftmost endpoint and then the largest of the two the rightmost one).