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A complete Heyting algebra whose Scott space is non-sober

Volume 252 / 2021

Xiaoquan Xu, Xiaoyong Xi, Dongsheng Zhao Fundamenta Mathematicae 252 (2021), 315-323 MSC: Primary 06B35; Secondary 54A05. DOI: 10.4064/fm704-4-2020 Published online: 31 August 2020

Abstract

We prove that (1) for any complete lattice $L$, the set $\mathcal {D}(L)$ of all non-empty saturated compact subsets of the Scott space of $L$ is a complete Heyting algebra (with the reverse inclusion order); and (2) if the Scott space of a complete lattice $L$ is non-sober, then the Scott space of $\mathcal {D}(L)$ is non-sober. Using these results and Isbell’s example of a non-sober complete lattice, we deduce that there is a complete Heyting algebra whose Scott space is non-sober, thus giving an affirmative answer to a problem posed by Achim Jung. We also prove that a $T_0$ space is well-filtered iff its upper space (the set $\mathcal {D}(X)$ of all non-empty saturated compact subsets of $X$ equipped with the upper Vietoris topology) is well-filtered, which answers another open problem.

Authors

  • Xiaoquan XuSchool of Mathematics and Statistics
    Minnan Normal University
    Zhangzhou, Fujian 363000, P.R. China
    e-mail
  • Xiaoyong XiSchool of Mathematics and Statistics
    Jiangsu Normal University
    Xuzhou, Jiangsu 221116, P.R. China
    e-mail
  • Dongsheng ZhaoMathematics and Mathematics Education
    National Institute of Education
    Nanyang Technological University
    1 Nanyang Walk
    Singapore 637616
    e-mail

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