单位正方形上的形式正方形将FS-域与RB-域区分开来
Formal squares over the unit square separate FS-domains from RB-domains
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中文总结 AI 辅助
该研究解决了RB-域是否为FS-域的逆问题，通过构造特定域证明其为FS-域而非RB-域，采用定量方法，独立于他人工作且方法不同。
中文摘要 AI 辅助
每个RB-域都是FS-域，而其逆是否成立是一个长期存在的开放问题。我们证明：平面内中心位于单位正方形$[0,1]^2$中的闭轴对齐正方形构成的域，附加整个平面后按逆包含关系排序，该域是FS-域但不是RB-域。证明是定量的：有限网格论证首先表明，在每个有限板$[0,1]^2\times[0,m]$上，对每个近似恒等式和任意$\varepsilon>0$，近似恒等式中的一个元素的半径超额（即输出半径减去输入半径）一致至多为$\varepsilon$；相比之下，对每个收缩和任意$m>0$，在板$[0,1]^2\times[0,m]$的某个点处，半径超额至少为$m/(4m+1)$。该解决方案独立于Chen、Kou和Lyu近期的工作获得，且采用了不同的方法。
英文摘要
Every RB-domain is an FS-domain. Whether the converse holds was a long-standing open problem. We prove that the domain of closed axis-parallel squares in the plane whose centres lie in the unit square $[0,1]^2$, with the whole plane adjoined and ordered by reverse inclusion, is an FS-domain but not an RB-domain. The proof is quantitative. A finite-grid argument first shows that, on every finite slab $[0,1]^2\times[0,m]$, for every approximate identity and every $\varepsilon>0$, one member of the approximate identity has radius excess (i.e., the output radius minus the input radius) uniformly at most $\varepsilon$. By contrast, for every deflation and every $m>0$, the radius excess is at least $m/(4m+1)$ at some point of the slab $[0,1]^2\times[0,m]$. This solution was obtained independently of the recent work of Chen, Kou, and Lyu and uses a different method.
