Related papers: Complete sigma-centered Boolean algebra not adding…
There exists a complete atomless Boolean algebra that has no proper atomless complete subalgebra.
In this paper we prove three theorems about the theory of Borel sets in models of ZF without any form of the axiom of choice. We prove that if B is a G-delta-sigma set, then either B is countable or B contains a perfect subset. Second, we…
How many endomorphisms does a Boolean algebra have? Can we find Boolean algebras with as few endomorphisms as possible? Of course from any ultrafilter of the Boolean algebra we can define an endomorphism, and we can combine finitely many…
It is unprovable that every complete subalgebra of a countably closed complete Boolean algebra is countably closed.
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…
The existence of a countably compact group without non-trivial convergent sequences in ZFC alone is a major open problem in topological group theory. We give a ZFC example of a Boolean topological group G without non-trivial convergent…
We consider homogeneity properties of Boolean algebras that have nonprincipal ultrafilters which are countably generated.It is shown that a Boolean algebra B is homogeneous if it is the union of countably generated nonprincipal ultrafilters…
In [Sh:89] we, answering a question of Monk, have explicated the notion of ``a Boolean algebra with no endomorphisms except the ones induced by ultrafilters on it'' (see section 2 here) and proved the existence of one with character density…
The goal of this paper is twofold. In addition to the results stated in the next paragraph, we present some classical results on absoluteness relevant to functional analysis that are well known to logicians but not nearly as well advertised…
We show that the big Ramsey degree of the Boolean algebra with 3 atoms within the countable atomless Boolean algebra is infinite.
We investigate classes of Boolean algebras related to the notion of forcing that adds Cohen reals. A >>Cohen algebra<< is a Boolean algebra that is dense in the completion of a free Boolean algebra. We introduce and study generalizations of…
We give two equivalent definitions of sigma algebras that are atomless conditionally to a smaller sigma algebra.
It is proved that the derivation algebra of a centerless perfect Lie algebra of arbitrary dimension over any field of arbitrary characteristic is complete and that the holomorph of a centerless perfect Lie algebra is complete if and only if…
A Souslin algebra is a complete Boolean algebra whose main features are ruled by a tight combination of an antichain condition with an infinite distributive law. The present article divides into two parts. In the first part a representation…
We give a necessary and sufficient condition for an atomless Boolean algebra to be countably generated, and use it to give new proofs of some some know facts due to Gaifman-Hales and Solovay and also due to Jech, Kunen and Magidor. We also…
We show that there are semi-Cohen Boolean algebras which cannot be completely embedded into Cohen Boolean algebras. Using the ideas from this proof, we give a simpler argument for a theorem of S. Koppelberg and S. Shelah, stating that there…
We prove that there is a lattice embedded from every countable distributive lattice into the Boolean algebra of computable subsets of $\mathbb{N}$. Along the way, we discuss all relevant results about lattices, Boolean algebras and…
It is consistent that every weakly distributive complete ccc Boolean algebra carries a strictly positive Maharam submeasure.
Complete Boolean algebras proved to be an important tool in topology and set theory. Two of the most prominent examples are B(kappa), the algebra of Borel sets modulo measure zero ideal in the generalized Cantor space {0,1}^kappa equipped…
We give an example of a regular and complete subalgebra of a Cohen algebra which is not Cohen.