Not every countable complete lattice is sober

TL;DR

This paper investigates conditions under which Scott spaces of posets and lattices are sober, proving that certain countability conditions ensure sobriety and providing a counterexample to a longstanding open question.

Contribution

It establishes that the Scott space of a product of posets is topologically the product of their Scott spaces under countability conditions and constructs a countable complete lattice with a non-sober Scott space, answering Jung's problem.

Findings

The topology of the product Scott space coincides with the Scott topology on the product poset when ideal sets are countable.

A directed complete poset with countable ideals and certain topological properties has a sober Scott space.

A constructed countable complete lattice with a non-sober Scott space disproves Jung's conjecture.

Abstract

The study of the sobriety of Scott spaces has got an relative long history in domain theory. Lawson and Hoffmann independently proved that the Scott space of every continuous directed complete poset (usually called domain) is sober. Johnstone constructed the first directed complete poset whose Scott space is non-sober. Not long after, Isbell gave a complete lattice with non-sober Scott space. Based on Isbell's example, Xu, Xi and Zhao showed that there is even a complete Heyting algebra whose Scott space is non-sober. Achim Jung then asked whether every countable complete lattice has a sober Scott space. Let $\Sigma P$ be the Scott space of poset $P$. In this paper, we first prove that the topology of the product space $\Sigma P\times \Sigma Q$ coincides with the Scott topology on the product poset $P\times Q$ if the set $Id(P)$ and $Id(Q)$ of all non-trivial ideals of posets $P$ and $Q$ are both countable. Based on this result, we deduce that a directed complete poset $P$ has a sober Scott space, if $Id(P)$ is countable and the space $\Sigma P$ is coherent and well-filtered. Thus a complete lattice $L$ with $Id(L)$ countable has a sober Scott space. Making use the obtained results, we then construct a countable complete lattice whose Scott space is non-sober and thus give a negative answer to Jung's problem.