A COMPACT SPACE IS NOT ALWAYS SI-COMPACT

被引:2
|
作者
He, Zhengmao [1 ]
Wang, Kaiyun [1 ]
机构
[1] Shaanxi Normal Univ, Sch Math & Stat, Xian, Peoples R China
关键词
Scott topology; compact saturated set; quasicontinuous domain; T-lattice; SI-compactness;
D O I
10.1216/rmj.2022.52.2041
中图分类号
O1 [数学];
学科分类号
0701 ; 070101 ;
摘要
We consider the poset Q(P) of all nonempty compact saturated subsets of the Scott space of a poset P, equipped with the reverse inclusion order. We also introduce the notion of T-lattices, that is, a complete lattice L is called a T-lattice if for any x is an element of L\ {1L}, up arrow x \ {x} is an element of Q(L). We prove that if P and Q are quasicontinuous domains or T-lattices, P is order isomorphic to Q if and only if (Q(P), superset of) is order isomorphic to (Q(Q), superset of). We also prove that for a compact space X, if the Scott space of the open set lattice O(X) is sober, then X is SI-compact. Using this result and the Isbell example of a non-sober complete lattice, we present a compact sober space which is not SI-compact. This gives a positive answer to an open problem posed by Zhao and Ho.
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页码:2041 / 2051
页数:11
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