A Besicovitch-type Covering Theorem

Question

In the book The Geometry of Domains in Spaces by Krantz and Parks, the authors proved the weak $(1,1)$-type estimate of the maximal function $M_\mu f$, where $\mu$ is a Radon measure, using their version of the Besicovitch covering theorem.

Let $d$ be a positive integer. Then there exists a constant $C=C(d)$ such that for any finite collection of balls $\mathcal B = \{B_i\}_{i=1}^m$ in $\Bbb R^d$ with the property that no ball contains the center of any other ball, we can partition the family $\mathcal B$ into $$ \mathcal B = \mathcal B_1 \cup\mathcal B_2 \cup \dots \cup \mathcal B_C, $$ where each subfamily $\mathcal B_j$ consists of disjoint balls.

This version of the covering theorem seems pretty restrictive, especially the part about no ball contains the center of any other balls. Indeed, in theor proof of the weak $(1,1)$-type estimate, they relied on a certain claim that they did not prove.

Edit As Skeeve mentioned in the comment, this claim is not explicitly stated in the book but more of a paraphrasing of the part the authors left out in a proof.

Claim: Let $K\subset \Bbb R^d$ be a compact set such that each $x\in K$ is associated with a real number $r_x>0$. Then $K$ can be covered by a family of balls $$ \mathcal B = \{ B(x_i,r_i) : i=1,\dots,k\ \}, $$ where $r_i := r_{x_i}$, such that for any distinct $i,j \le k$, we have $$ x_i\notin B(x_j,r_j) \quad\text{and}\quad x_j\notin B(x_i,r_i). $$

I don't find this claim to be trivial at all. In fact, I tried many different methods but failed to prove it. Note that the mapping $x\mapsto r_x$ doesn't enjoy any nice property like continuity of any kind.

While the usual version of Besicovitch covering theorem circumvents this problem, I still would like to know how to prove the above claim (or a counter example if it is actually false).

Answer 1

This trivial counterexample in $\mathbb R^2$ should have taken me five minutes. Instead, I spent almost two days. The moral is the usual one: after 50 you'd better give up on mathematics.

Let $y,z$ be 2 points at distance $1$ from each other. We shall construct by induction a sequence of points $x_j$ and radii $r_j>\max(d(x_j,y),d(x_j,z))$ such that $x_j\to y$ when $j$ is odd, $x_j\to z$ when $j$ is even, $x_j$ do not lie on the line $yz$, $\max(d(x_j,y),d(x_j,z))<1$, the disk $D(x_j,\max(d(x_j,y),d(x_j,z)))$ contains $x_1,\dots, x_j$ but $x_{j+1}\notin D(x_j,r_j)$ . If you choose the radius $\rho$ for $y$ and $z$ small enough so that the corresponding disks do not contain $x_1$, you'll get a bad configuration.

Indeed, an attempt to choose $y$ or $z$ as one of the centers results in the exclusion of all the centers $x_j$, after which covering $x_1$ gets impossible.

Out of $x_i$, we can choose only one (if $i<j$, then $x_i\in D(x_j,r_j)$). But then, if we choose $x_i$, the point $x_{i+1}$ is not covered.

Now the sequences. Start with any $x_1$ very close to $y$ and not on the line $yz$ so that $d(x_1,z)<1$. Assume that $x_1,x_2,\dots x_j$ and $r_1,\dots,r_{j-1}$ are already constructed and, say $j$ is odd, so $x_j$ is close to $y$. Then the circles centered at $x_j$ and $y$ containing $z$ cross at an angle, so $D(y,1)\setminus \bar D(x_j,d(x_j,z))$ is an open set containing points arbitrarily close to $z$. Choose $x_{j+1}$ to be any point in that difference that doesn't lie on the line $yz$ and satisfies $d(y,x_{j+1})\ge 1-d(z,x_{j+1})>\max_{i\le j}d(z,x_i)+d(z,x_{j+1})\ge \max_{i\le j}d(x_i,x_{j+1})$ and choose $r_j$ anywhere between $d(x_j,z)$ and $d(x_j,x_{j+1})$.

Clearly, we can keep $x_j$ with odd indices at the distance $<1/3$ to $y$ and converging to $y$ and similarly for even indices and $z$.