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Explicit Witnesses at Every Gap of the Depth Filtration of βNβ\mathbb{N}

Carl Aza

Source record

Source: arXiv

Published: Sep 30, 2026

arXiv: 2609.38696

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Source abstract

Let Σ1=N∗Σ_{1} = \mathbb{N}^* and Σk+1=N∗+Σk‾Σ_{k+1} = \overline{\mathbb{N}^* + Σ_{k}} be the cumulative depth filtration of βNβ\mathbb{N}, the analogue for (N,+)(\mathbb{N},+) of a chain of closed ideals that Protasov and Protasova studied for discrete groups, where strict descent follows from a theorem of Lutsenko and Protasov. For every kk we give an explicit set whose closure meets ΣkΣ_{k} but not Σk+1Σ_{k+1}. Fix the doubly exponential sequence en=22ne_{n} = 2^{2^{n}}, partition it into kk subsequences E0,…,Ek−1E_{0}, \dots, E_{k-1} by the residue of the index modulo kk, and set Ak=E0+⋯+Ek−1A_{k} = E_{0} + \cdots + E_{k-1}. We prove that any sum q0+⋯+qk−1q_{0} + \cdots + q_{k-1} of free ultrafilters with Et∈qtE_{t} \in q_{t} lies in Σk∖Σk+1Σ_{k} \setminus Σ_{k+1}. The engine is a master lemma, proved by induction on jj: if a sum F1+⋯+FjF_{1} + \cdots + F_{j} of subsequences of {en}\{e_{n}\} with pairwise disjoint index sets belongs to a free ultrafilter ss, then s∉Σj+1s \notin Σ_{j+1}. The proof rests on a single rigidity of the doubly exponential sequence: a fixed difference forces the largest index in any shift-intersection, once it is large, to cancel within its own subsequence, which makes every shift-intersection descend by at least one level. The same witnesses lie in the gaps of the pure filtration.

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