AI 中文总结
本文通过构造嵌入与实例，证明完备布尔代数的Scott空间未必是co-sober的，否定了相关公开问题，同时验证了对应空间的非可定义性与良滤性。
AI 中文摘要
本文首先证明，对于完备布尔代数$L$，$L$的补图是一个KC-空间（作为子空间），且该图的每个非单点紧不可约子空间都会在平方代数$L\times L$的Scott空间中生成一个非主k-不可约紧饱和集。随后我们证明，每个紧序列US-空间都可以嵌入到某个合适的完备布尔代数的补图中；该嵌入由有限尾约束偏序集及其正则开完备化构造而成。将此构造应用于van Douwen的紧Fréchet反Hausdorff US-空间，得到一个完备布尔代数$B$，其Scott空间$ΣB$不是余可定义的（co-sober），从而否定地回答了关于完备布尔代数Scott空间的一个问题。该Scott空间$ΣB$同时也是非可定义的（non-sober），并且作为完备格的Scott空间，它是良滤过的（well-filtered）。
英文摘要
In this paper, we first prove that for a complete Boolean algebra $L$, the complement graph of $L$ is a \emph{KC}-space (as a subspace) and each non-singleton compact irreducible subspace of that graph generates a non-principal $k$-irreducible compact saturated set in the Scott space of the square algebra $L\times L$. We then show that every compact sequential \emph{US}-space embeds into the complement graph of a suitable complete Boolean algebra. The embedding is built from a finite tail-constraint poset and its regular-open completion. Applying the construction to van Douwen's compact Fréchet anti-Hausdorff \emph{US}-space gives a complete Boolean algebra $B$ whose Scott space $Σ~\!\!B$ is not co-sober, thereby answering negatively a question on Scott spaces of complete Boolean algebras. The same Scott space $Σ~\!\!B$ is non-sober and, as a Scott space of a complete lattice, is well-filtered.
Comments12 pages
