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Titlebook: Cardinal Functions on Boolean Algebras; J. Donald Monk Book 1990 Springer Basel AG 1990 algebra.Boolean algebra.cardinal function.function

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樓主: bradycardia
31#
發(fā)表于 2025-3-27 00:38:55 | 只看該作者
32#
發(fā)表于 2025-3-27 01:25:38 | 只看該作者
Hereditary Density,We begin again with some equivalent definitions, which are similar to the case of hereditary Lindel?f degree. Recall from page 42 the definition of left-separated sequence.
33#
發(fā)表于 2025-3-27 08:21:17 | 只看該作者
34#
發(fā)表于 2025-3-27 12:21:45 | 只看該作者
35#
發(fā)表于 2025-3-27 14:52:00 | 只看該作者
36#
發(fā)表于 2025-3-27 19:24:37 | 只看該作者
37#
發(fā)表于 2025-3-28 00:10:25 | 只看該作者
38#
發(fā)表于 2025-3-28 04:35:31 | 只看該作者
https://doi.org/10.1007/0-387-31609-4t 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
39#
發(fā)表于 2025-3-28 09:52:30 | 只看該作者
https://doi.org/10.1007/978-981-99-7879-3h 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
40#
發(fā)表于 2025-3-28 12:49:15 | 只看該作者
Feeding, Foraging, and Predation, an ultrafilter on .]. Clearly then, by topological duality, .(.) = sup(.A, .). For a weak product we have.. To show this, it suffices to show that . = |.| for the “new” ultrafilter .. This ultrafilter is defined as follows. For each subset . of ., let .. be the element of Π... such that ... = 1 if
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