Some results on Specker’s problem

A note on singular cardinals without the axiom of choice

“…Э. Шпеккер поставил вопрос [8], непротиворечиво ли ∀ λ ℵ0 Cov(λ, ℵ0, 2 λ ); известны только частичные ответы (см. [1]). Мы изучаем, насколько сингулярными могут быть кардиналы без AC.…”

Long Borel hierarchies

“…168]) that assuming the axiom of choice for countable families, the length of the Borel hierarchy is ω 1 . To see that it has height at least ω 1 , he shows that for all α such that 1 ≤ α < ω 1 , Σ 0 α = Π 0 α , by constructing universal sets in each Borel class.…”

“…Komjáth asks if the Borel hierarchy can have length greater than ω 2 . This would require a model where both ω 1 and ω 2 have cofinality ω. In Gitik 1980 [5] a model of ZF is produced (assuming the consistency of ZFC plus unboundedly many strongly compact cardinals) in which every ℵ α has cofinality ω. Schindler [10] shows that the consistency strength of two successor singular cardinals is at least a Woodin cardinal (approximately).…”

“…Apter and Gitik [1] give a partial solution to a (presumably still open) problem of Specker: Is it possible to have a model of ZF in which P(ℵ α ) ∈ G α+1 for every ordinal α?…”