← Back to Problems
Logic and Order TheoryResearchAI-Generated

Does every uncountable meet-continuous lattice that satisfies a natural completeness condition admit a Scott sober topology without requiring the axiom of choice?

Related: Hofmann-Mislove theorem, Scott topology sobriety conjecture for dcpos, independence of sobriety results from choice axioms

The recent paper establishes that every countable meet-continuous lattice is Scott sober, a result connecting order theory and topology in a clean and foundational way. The open problem is whether this result extends to uncountable meet-continuous lattices, and moreover whether such an extension can be achieved without invoking the axiom of choice. The countable case may admit proofs that exploit enumerability in ways that break down entirely when the lattice is uncountable, leaving the general case unresolved.

View Source Paper →