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◇ arXiv2026-09-17· math.LO

Every countable meet-continuous lattice is Scott sober

Xiaoquan Xu, Wei Ji

原始摘要(英文原文)· Original abstract
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.
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Every countable meet-continuous lattice is Scott sober — 科研速览 Science Skim