alt Hacker News"Since \(U_{k+1} \subseteq U_k\), the sets \(U_k\) are decreasing and periodic, and their intersection \(U = \bigcap_{k \ge 1} U_k\) has density \(d = \lim_{k \to \infty} d_k \ge \epsilon\)."
Is this enough? Let $U_k$ be the set of integers such that their remainder mod 6^n is greater or equal to 2^n for all 1<n<k. Density of each $U_k$ is more than 1/2 I think but not the intersection (empty) right?
Indeed. Your sets are decreasing periodic of density always greater than the product from k=1 to infinity of (1-(1/3)^k), which is about 0.56, yet their intersection is null.
This would all be a fairly trivial exercise in diagonalization if such a lemma as implied by Deepseek existed.
(Edit: The bounding I suggested may not be precise at each level, but it is asymptotically the limit of the sequence of densities, so up to some epsilon it demonstrates the desired counterexample.)