论文标题
并非每个可数的完整晶格都清醒
Not every countable complete lattice is sober
论文作者
论文摘要
对斯科特空间清醒的研究在域理论中具有相对悠久的历史。劳森（Lawson）和霍夫曼（Hoffmann）独立地证明了每个连续定向的完整寄居者（通常称为域）的Scott空间清醒。约翰斯通（Johnstone）建造了第一个导演的完整poset，其斯科特（Scott）空间是非塞子。不久之后，伊斯贝尔（Isbell）提供了一个完整的晶格，并在非索伯·斯科特（Scott）空间中提供了完整的晶格。基于Isbell的示例，Xu，XI和Zhao表明，甚至还有一个完整的Heyting代数，其Scott空间是非偷窃的。然后，Achim Jung询问每个可数的完整晶格是否都有清醒的Scott空间。 令$σp$为Poset $ P $的Scott空间。在本文中，我们首先证明了产品空间的拓扑$σp\ timesσq$与产品poset $ p \ times q $相吻合，如果设置$ id（p）$（p）$（p）$和$ id（q）$ posets $ p $ p $和$ q $的所有非繁琐的理想都是可数的。基于此结果，我们推断出一个定向的完整Poset $ P $具有清醒的Scott空间，如果$ id（p）$是可数的，并且空间$σp$连贯且过滤良好。因此，带有$ id（l）$ Countable的完整晶格$ l $具有清醒的Scott空间。通过使用获得的结果，我们构建了一个可数的完整晶格，其Scott空间是非偷窃的，因此对Jung的问题给出了负面答案。
The study of the sobriety of Scott spaces has got an relative long history in domain theory. Lawson and Hoffmann independently proved that the Scott space of every continuous directed complete poset (usually called domain) is sober. Johnstone constructed the first directed complete poset whose Scott space is non-sober. Not long after, Isbell gave a complete lattice with non-sober Scott space. Based on Isbell's example, Xu, Xi and Zhao showed that there is even a complete Heyting algebra whose Scott space is non-sober. Achim Jung then asked whether every countable complete lattice has a sober Scott space. Let $ΣP$ be the Scott space of poset $P$. In this paper, we first prove that the topology of the product space $ΣP\times ΣQ$ coincides with the Scott topology on the product poset $P\times Q$ if the set $Id(P)$ and $Id(Q)$ of all non-trivial ideals of posets $P$ and $Q$ are both countable. Based on this result, we deduce that a directed complete poset $P$ has a sober Scott space, if $Id(P)$ is countable and the space $ΣP$ is coherent and well-filtered. Thus a complete lattice $L$ with $Id(L)$ countable has a sober Scott space. Making use the obtained results, we then construct a countable complete lattice whose Scott space is non-sober and thus give a negative answer to Jung's problem.
