A complete Heyting algebra whose Scott space is non-sober
A complete Heyting algebra whose Scott space is non-sober
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DOI：
10.4064/fm704-4-2020
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发表时间：
2019-03
影响因子：
0.6
通讯作者：
Xiaoquan Xu;Xiaoyong Xi;Dongsheng Zhao
中科院分区：
文献类型：
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作者：
Xiaoquan Xu;Xiaoyong Xi;Dongsheng Zhao
证明了：（1）对任意完备格L，L的Scott空间的所有非空饱和紧子集的集合D（L）是完备Heyting代数（包含序相反）;（2）若完备格L的Scott空间是非sober空间，则D（L）的Scott空间是非sober空间.利用这些结果和Isbell关于非sober完备格的例子，我们推出了存在一个完备Heyting代数的Scott空间是非sober的，从而肯定地回答了Jung提出的一个问题.我们还将证明一个$T_0$空间是良滤的当且仅当它的上空间（集合$\mathcal{D}（X）$的所有非空饱和紧子集$X$配备上Vietoris拓扑）是良滤的，这回答了另一个公开问题。
We prove that (1) for any complete lattice $L$, the set $\mathcal{D}(L)$ of all nonempty 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 the 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 give a positive answer to a problem posed by Jung. We will also prove that a $T_0$ space is well-filtered iff its upper space (the set $\mathcal{D}(X)$ of all nonempty saturated compact subsets of $X$ equipped with the upper Vietoris topology) is well-filtered, which answers another open problem.