On the contrary for DeepSeek you could but not for a non open model.
It says that the OpenAI proof is a different one from the published one in the literature.
Whereas whether the Deepseek proof is the same as the published one, I dont know enough of the math to judge.
That was what I meant.
You could have just rubber-stamped it yourself, for all the mathematical rigor it holds. The devil is in the details, and the smallest problem unravels the whole proof.
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?
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.)
The answer is yes. Assume, for the sake of contradiction, that there exists an \(\epsilon > 0\) such that for every \(k\), there exists a choice of congruence classes \(a_1^{(k)}, \dots, a_k^{(k)}\) for which the set of integers not covered by the first \(k\) congruences has density at least \(\epsilon\).
For each \(k\), let \(F_k\) be the set of all infinite sequences of residues \((a_i)_{i=1}^\infty\) such that the uncovered set from the first \(k\) congruences has density at least \(\epsilon\). Each \(F_k\) is nonempty (by assumption) and closed in the product topology (since it depends only on the first \(k\) coordinates). Moreover, \(F_{k+1} \subseteq F_k\) because adding a congruence can only reduce the uncovered set. By the compactness of the product of finite sets, \(\bigcap_{k \ge 1} F_k\) is nonempty.
Choose an infinite sequence \((a_i) \in \bigcap_{k \ge 1} F_k\). For this sequence, let \(U_k\) be the set of integers not covered by the first \(k\) congruences, and let \(d_k\) be the density of \(U_k\). Then \(d_k \ge \epsilon\) for all \(k\). 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 \(d = \lim_{k \to \infty} d_k \ge \epsilon\). However, by hypothesis, for any choice of residues, the uncovered set has density \(0\), a contradiction.
Therefore, for every \(\epsilon > 0\), there exists a \(k\) such that for every choice of congruence classes \(a_i\), the density of integers not covered by the first \(k\) congruences is less than \(\epsilon\).
\boxed{\text{Yes}}