发表机构
Minnan Normal University; Guilin University of Technology（闽南师范大学; 桂林理工大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文构造了一个可数零维度量空间，其Hoare幂空间不是co-sober，从而否定了Xu及He和Zhao的相关猜想，并建立了Hoare幂空间co-sober性与Scott空间sober性的等价条件。
AI 中文摘要
Xu 询问co-sober空间的Hoare幂空间，特别是$T_2$-空间或度量空间的Hoare幂空间，是否必须是co-sober。我们给出否定答案。He和Zhao最近构造了一个可数$T_1$空间$X$，其开集格的Scott拓扑不是sober的。我们观察到他们的拓扑具有可数子基。经典的Ponomarev构造随后产生一个从可数零维可度量化空间$M$到$X$的开连续满射。直接和逆像使得Scott空间$\Sigma\mathcal{O}(X)$成为$\Sigma\mathcal{O}(M)$的收缩核，因此$\Sigma\mathcal{O}(M)$不是sober的。最后，我们证明对于每个$T_0$-空间$Y$，Hoare幂空间$\PH(Y)$是co-sober的当且仅当Scott空间$\Sigma\mathcal{O}(Y)$是sober的。因此，$\PH(M)$不是co-sober的。因此，同样的例子对He和Zhao关于可数Hausdorff空间的开集格的Scott sobriety问题给出了否定回答。
英文摘要
Xu asked whether the Hoare power space of a co-sober space, and in particular of a $T_2$-space or a metric space, must be co-sober. We give a negative answer. He and Zhao recently constructed a countable $T_{1}$ space $X$ whose lattice of open sets has a non-sober Scott topology. We observe that their topology has a countable subbase. The classical Ponomarev construction then yields an open continuous surjection from a countable zero-dimensional metrizable space $M$ onto $X$. Direct and inverse images then make the Scott space $Σ\mathcal{O}(X)$ a retract of $Σ\mathcal{O}(M)$, so $Σ\mathcal{O}(M)$ is not sober. Finally, we show that, for every $T_{0}$-space $Y$, the Hoare power space $\PH(Y)$ is co-sober if and only if the Scott space $Σ\mathcal{O}(Y)$ is sober. Consequently, $\PH(M)$ is not co-sober. Therefore, the same example answers negatively a question of He and Zhao about Scott sobriety of the lattices of open sets of countable Hausdorff spaces.
Comments6 pages
