PreprintCC BY 4.0arXiv
The product of two Scott sober complete lattices is not always Scott sober
Abstract
In this paper, we obtain two complete lattices that are sober with respect to the Scott topology, whereas the Scott space of their product is not a sober space. Thus, we answer a question posed by Miao, Xi, Jia, Li and Zhao in [8].
Related
- A co-sober Alexandrov space whose Scott power space is not co-sober
- Every countable meet-continuous lattice is Scott sober
- A Lean 4 Formalization of Scott's Continuous Lattices (1972)
- Scott topologies on products of countable complete Heyting algebras
- The noncommutative topological factor theorem for rank-one product lattices
BibTeX
@misc{he2026product,
title = {{The product of two Scott sober complete lattices is not always Scott sober}},
author = {Zhengmao He},
howpublished = {arXiv},
year = {2026},
month = oct,
doi = {10.48550/arxiv.2609.39246},
eprint = {2609.39246},
archivePrefix = {arXiv},
url = {https://arxiv.org/abs/2609.39246},
note = {CC BY 4.0},
}
RIS
TY - UNPB
TI - The product of two Scott sober complete lattices is not always Scott sober
AU - Zhengmao He
PY - 2026
DA - 2026/10/13
PB - arXiv
DO - 10.48550/arxiv.2609.39246
UR - https://arxiv.org/abs/2609.39246
AB - In this paper, we obtain two complete lattices that are sober with respect to the Scott topology, whereas the Scott space of their product is not a sober space. Thus, we answer a question posed by Miao, Xi, Jia, Li and Zhao in [8].
ER -