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Sacks forcing collapses \({\mathfrak c}\) to \({\mathfrak b}\). (English) Zbl 0797.03053
The author improves some result of A. Roslanowski and S. Shelah and answers a question from their paper. The main result is that a Sacks algebra is nowhere \(({\mathfrak b},{\mathfrak c},{\mathfrak c})\)-distributive, which implies that Sacks forcing collapses \(\mathfrak c\) to \(\mathfrak b\).

MSC:
03E40 Other aspects of forcing and Boolean-valued models
03C25 Model-theoretic forcing
03E25 Axiom of choice and related propositions
06A07 Combinatorics of partially ordered sets
06E05 Structure theory of Boolean algebras
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