We show that it is consistent with ZFC (relative to large cardinals) that every infinite Boolean algebra $B$ has an irredundant subset $A$ such that $2^{|A|} = 2^{|B|}$. This implies in particular that $B$ has $2^{|B|}$ subalgebras. We also discuss some more general problems about subalgebras and free subsets of an algebra. The result on the number of subalgebras in a Boolean algebra solves a question of Monk from [6]. The paper is intended to be accessible as far as possible to a general audience, in particular we have confined the more technical material to a "black box" at the end. The proof involves a variation on Foreman and Woodin's model in which GCH fails everywhere.
@article{1183744818,
author = {Cummings, James and Shelah, Saharon},
title = {A Model in which Every Boolean Algebra has many Subalgebras},
journal = {J. Symbolic Logic},
volume = {60},
number = {1},
year = {1995},
pages = { 992-1004},
language = {en},
url = {http://dml.mathdoc.fr/item/1183744818}
}
Cummings, James; Shelah, Saharon. A Model in which Every Boolean Algebra has many Subalgebras. J. Symbolic Logic, Tome 60 (1995) no. 1, pp. 992-1004. http://gdmltest.u-ga.fr/item/1183744818/