In Terence Tao's paper Exploring the toolkit of Jean Bourgain is stated:

Theorem 3.1 (Furstenberg–Katznelson–Weiss theorem, qualitative version). Let $A\subset\Bbb R^2$ be a measurable set whose upper density $$δ∶=\limsup_{R→∞}\frac{|A∩\mathrm B(0,R)|}{|\mathrm B(0,R)|}$$is positive. Then there exists $l_0$ such that, for all $l≥l_0$, there exist $x, y∈A$ with $|x−y|\geqslant l$.

In this paper, in addition to the usual meaning, $|\centerdot|$ denotes the Lebesgue measure of a subset of $\Bbb R^2$, and I interpret $\mathrm B(0,R)$ as the (open) ball of radius $R$ and centered at the origin in $\Bbb R^2$.

A minor point is that the conclusion of the theorem is trivial unless $l_0$ is required to be positive. That granted, though, it seems that $l_0$ is entirely redundant. Thus, why does the last sentence of the theorem not stay simply “Then, for all $l$, there exist $x, y∈A$ with $|x−y|\geqslant l$.”? Moreover, I don't even see the need for $\delta$ to be positive. All that is necessary is for $A$ to be unbounded (e.g. $A=\Bbb N\times\{0\}$), in which case the theorem's conclusion is pretty well tautologous.

I guess that there is a crucial typo somewhere, whose correction would make the theorem nontrivial. But I can't see what that could be.

  • 2
    $\begingroup$ I believe it's $|x-y| = l$ that might make it the correct statement. $\endgroup$ – asahay Nov 13 '20 at 11:55
  • $\begingroup$ @asahay : Thanks, that would make sense. If you posted that as your answer, I could accept it. $\endgroup$ – John Bentin Nov 13 '20 at 12:16

I looked up the paper in which Furstenberg-Katznelson-Weiss proved their result. The correct statement has $|x-y| = \ell$ instead of $|x-y| \geq \ell$.


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