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Not every countable complete distributive lattice is sober

Published online by Cambridge University Press:  28 July 2023

Hualin Miao ,
Xiaoyong Xi Open the ORCID record for Xiaoyong Xi [Opens in a new window] ,
Qingguo Li Open the ORCID record for Qingguo Li [Opens in a new window]  and
Dongsheng Zhao
Hualin Miao
Affiliation:
School of Mathematics, Hunan University, Changsha, Hunan, China
Xiaoyong Xi*
Affiliation:
School of Mathematics and Statistics, Yancheng Teachers University, Yancheng, Jiangsu, China
Qingguo Li
Affiliation:
School of Mathematics, Hunan University, Changsha, Hunan, China
Dongsheng Zhao
Affiliation:
Mathematics and Mathematics Education, National Institute of Education, Nanyang Technological University, Singapore
*
Corresponding author: Xiaoyong Xi; Email: littlebrook@jsnu.edu.cn
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Abstract

The study of the sobriety of Scott spaces has got a relatively 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. Soon after, Isbell gave a complete lattice with a 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. The main aim of this paper is to answer Jung’s problem by constructing a countable complete lattice whose Scott space is non-sober. This lattice is then modified to obtain a countable distributive complete lattice with a non-sober Scott space. In addition, we prove that the topology of the product space $\Sigma P\times \Sigma Q$ coincides with the Scott topology of the product poset $P\times Q$ if the set Id(P) and Id(Q) of all incremental ideals of posets P and Q are both countable. Based on this, it is deduced that a directed complete poset P has a sober Scott space, if Id(P) is countable and $\Sigma P$ is coherent and well filtered. In particular, every complete lattice L with Id(L) countable has a sober Scott space.

Keywords

CountabilitySobrietyScott topologyComplete lattices

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Type
Paper
Information
Mathematical Structures in Computer Science , Volume 33 , Issue 9 , October 2023 , pp. 809 - 831
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Copyright
© The Author(s), 2023. Published by Cambridge University Press

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