每个可数交连续格都是 Scott 素朴的
Every countable meet-continuous lattice is Scott sober
- Nanchang Institute of Technology（南昌工程学院）
- Guilin University of Technology（桂林理工大学）
机构由 AI 辅助整理，请以论文原文为准。
AI总结：
本文证明每个可数交连续格是Scott素朴的，并肯定回答Xu的问题7.1和7.2，通过对角线论证和并平移构造Scott开矩形，还刻画了非素朴格的基数谱。
AI中文摘要：
我们证明了每个可数交连续格都是 Scott 素朴的。对于任何这样的格的族，乘积上的 Scott 拓扑等于因子 Scott 拓扑的乘积，并且乘积 Scott 空间是素朴的。这些结果肯定地回答了 Xu ( arXiv:2609.18032v1 ) 的问题 7.1 和 7.2，将可数框架的结果扩展到有限分配性之外。对交坐标的对角线论证从非主理想中提取有向穿孔区间。这些区间的并平移形成一个可数族，用于检测 Scott 开集，并通过逐次有限选择产生 Scott 开矩形。我们还证明了 Scott 非素朴交连续格的基数谱是向上封闭的，并且在连续统假设下，由所有不可数基数组成。
英文摘要：
We prove that every countable meet-continuous lattice is Scott sober. For any family of such lattices, the Scott topology on the product equals the product of the factor Scott topologies, and the product Scott space is sober. These results answer affirmatively Questions 7.1 and 7.2 of Xu (arXiv:2609.18032v1), extending the countable-frame results beyond finite distributivity. A diagonal argument on meet coordinates extracts directed punctured intervals from nonprincipal ideals. Join translates of these intervals form a countable family detecting Scott openness and yield Scott open rectangles by successive finite choices. We also show that the cardinal spectrum of Scott non-sober meet-continuous lattices is upward closed and, under the Continuum Hypothesis, consists of all uncountable cardinals.
