AUGER
发表于 2025-3-28 14:52:13
Notes on the Infinity Laplace Equation subalgebra or homomorphic image of ., then |Ult.| ≤ |Ult.|. For weak products we have .. The situation for full products is more complicated: ., where . = Σ. d... This follows from the following two facts: .where “→” means “is isomorphically embeddable in”, and “U” means “disjoint union”. Next, cle
诙谐
发表于 2025-3-28 18:48:08
Overview: 978-3-7643-2495-7978-3-0348-6381-0
我怕被刺穿
发表于 2025-3-28 23:07:02
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慢跑
发表于 2025-3-29 06:24:36
Notes on the Infinity Laplace Equation subalgebra or homomorphic image of ., then |Ult.| ≤ |Ult.|. For weak products we have .. The situation for full products is more complicated: ., where . = Σ. d... This follows from the following two facts: .where “→” means “is isomorphically embeddable in”, and “U” means “disjoint union”. Next, clearly |Ult ⊕. = Π. |Ult..|.
说明
发表于 2025-3-29 09:49:30
Introduction,functions are the cardinality of the algebra A, and sup{|X| : . is a family of pairwise disjoint elements of .}. We have selected 21 such functions as the most important ones, and several others are mentioned as we go along. Let us mention right away an ambiguity in these notes: .. For each function
Anthropoid
发表于 2025-3-29 13:27:54
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虚构的东西
发表于 2025-3-29 15:53:30
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CREEK
发表于 2025-3-29 20:01:22
Irredundance, true for . a homomorphic image of .. Concerning the derived operations, we note just the obvious facts that Irr.. = Irr., Irr.. = ω, Irr.-A = ω, and . Irr.. = Irr.. Obviously any chain is irredundant; so Length . ≤ Irr.. The difference can be large, e.g. in a free BA. By Theorem 4.25 of Part I of t
invade
发表于 2025-3-30 03:46:55
Cardinality,t its behaviour under ultraproducts are the same as the well-known and difficult problems concerning the cardinality of ultraproducts in general. Card. is a non-obvious function. Clearly Card.. ≤ 2. for every infinite BA, and Card.. = ω for many BAs, e.g. for free BAs and interval algebras. But Card
FILTH
发表于 2025-3-30 07:22:22
Tightness,h that 0 < y < .; hence there is an ultrafilter .. such that . ∈ .. but .. ≠ .. Let . = {.. : . ∈ .}. Thus . ⊆ ⋃ .. Suppose that . is a finite subset of . such that . ⊆ .. But it is a very elementary exercise to show that no ultrafilter is included in a finite union of other, different, ultrafilters