Nearly Projective Boolean Algebras

Heindorf, Lutz; Shapiro, Leonid B.

doi:10.1007/bfb0094103

Cited by 28 publications

(45 citation statements)

References 0 publications

Supporting

Mentioning

Contrasting

Order By: Relevance

“…Some of the conditions in the next proposition were proved by Koppelberg [10]; see also Balcar, Jech and Zapletal [3] or Heindorf and Shapiro [8]. For the sake of completeness we give its proof.…”

Section: Lemma 11mentioning

confidence: 80%

See 1 more Smart Citation

Boolean algebras admitting a countable minimally acting group

Błaszczyk¹,

Kucharski²,

Turek³

2014

Open Mathematics

View full text Add to dashboard Cite

The aim of this paper is to show that every infinite Boolean algebra which admits a countable minimally acting group contains a dense projective subalgebra. MSC:54B35, 54A05, 52A01, 54D30 Regular and relatively complete subalgebrasAll Boolean algebras considered here are assumed to be infinite. Boolean algebraic notions, excluding symbols for Boolean operations, follow the Koppelberg's monograph [9]. In particular, if (B ∧ ∨ − 0 1) is a Boolean algebra, then B + = B \ {0} denotes the set of all non-zero elements of B. A set A ⊆ B is a subalgebra of the Boolean algebra B, A ≤ B for short, if 1 0 ∈ A and A is closed under Boolean operations or, equivalently, − ∈ A for all ∈ A. We shall write A ∼ = B whenever A and B are isomorphic Boolean algebras. A non-empty set X ⊆ B + is called a partition of B whenever ∧ = 0 for distinct ∈ X and B X = 1 *

show abstract

Section: Lemma 11mentioning

confidence: 80%

“…An algebraic proof of Haydon's Theorem can be found in Koppelberg [11] and also in Heindorf and Shapiro [8].…”

Section: Theorem 14 (Haydon's Theorem)mentioning

confidence: 99%

Boolean algebras admitting a countable minimally acting group

Błaszczyk¹,

Kucharski²,

Turek³

2014

Open Mathematics

View full text Add to dashboard Cite

show abstract

“…Fact (6) By our hypothesis the ultrafilters D n are not nowhere dense and so by Lemma for every n < ω we can choose a function f n : ω → ω> 2 such that (∀B ∈ D n )(∃u ∈ ω> 2)(∀ν ∈ ω> 2)(∃k ∈ B)(u ν ¢ f n (k)).…”

Section: Remarksmentioning

confidence: 99%

“…In "forcing language" condition (a) says that there exists a nontrivial σ-centered forcing not adding Cohen reals A subalgebra B of a Boolean algebra A is called regular whenever for every X ⊆ B, sup B X = 1 implies sup A X = 1; see e.g. Heindorf and Shapiro [6]. Clearly, every dense subalgebra is regular.…”

mentioning

confidence: 99%

Regular subalgebras of complete Boolean algebras

Błaszczyk

Shelah

2001

J. symb. log.

View full text Add to dashboard Cite

Abstract. It is proved that the following conditions are equivalent:(a) there exists a complete, atomless, σ-centered Boolean algebra, which does not contain any regular, atomless, countable subalgebra, (b) there exists a nowhere dense ultrafilter on ω. Therefore the existence of such algebras is undecidable in ZFC. In "forcing language" condition (a) says that there exists a nontrivial σ-centered forcing not adding Cohen reals A subalgebra B of a Boolean algebra A is called regular whenever for every X ⊆ B, sup B X = 1 implies sup A X = 1; see e.g. Heindorf and Shapiro [6]. Clearly, every dense subalgebra is regular. Although every complete Boolean algebra contains a free Boolean algebra of the same size (see the Balcar-Franek Theorem; [2]), not always such an embedding is regular. For instance, if B is a measure algebra, then it contains a free subalgebra of the same cardinality as B, but B cannot contain any infinite free Boolean algebra as a regular subalgebra. Indeed, measure algebras are weakly σ-distributive but free Boolean algebras are not, and a regular subalgebra of a weakly σ-distributive one is again σ-distributive. Thus B does not contain any free Boolean algebra. On the other hand, measure algebras are not σ-centered. So, a natural question arises whether there exists a σ-centered, complete, atomless Boolean algebra B without regular free subalgebras. Since countable atomless Boolean algebras are free and every free Boolean algebra contains a countable regular free subalgebra, it is enough to ask whether B contains a countable regular subalgebra. In the paper we prove that such an algebra exists iff there exists a nowhere dense ultrafilter.

show abstract

“…For a set A ⊆ H(χ) for χ large enough, we write dcl H(χ),∈,< [A] for the Skolem closure (Skolem hull) of A in the structure H(χ), ∈, < , where < is a well-ordering of H(χ) (for details, see [16], 400-402, or [15], 165-170).…”

Section: Theoremmentioning

confidence: 99%

Almost disjoint pure subgroups of the Baer-Specker group

Kolman¹,

Shelah²

1999

Abelian Groups and Modules

View full text Add to dashboard Cite

Abstract. We prove in ZFC that the Baer-Specker group Z ω has 2 ℵ 1 nonfree pure subgroups of cardinality ℵ 1 which are almost disjoint: there is no non-free subgroup embeddable in any pair.In this short paper we prove the following result. Theorem 1. There exists a family G = {G α : α < 2 ℵ1 } of non-isomorphic nonfree pure subgroups of the Baer-Specker group Z ω such that:Recall that the Baer-Specker group Z ω is the abelian group of functions from the natural numbers into the integers (see [1] and [18]). It contains the canonical pure free subgroup Z ω = ⊕ n<ω Z. The group Z ω is not κ-free for any cardinal κ > ℵ 1 , but it is ℵ 1 -free, so the groups G α in Theorem 1 are almost free.Theorem 1 answers a question of the first author, and has its place in the line of recent research dealing with the lattice structure of the pure subgroups of Z ω (see , who showed that under various set-theoretic hypotheses, there exist families of maximal possible size of almost free abelian groups which are pairwise almost disjoint. Following [11], we say that two groups A and B are almost disjoint if whenever H is embeddable as a subgroup in both A and B, then H is free. Clearly if A and B are non-free and almost disjoint, then they are non-isomorphic in a very strong way. On the other hand, the intersection of two almost disjoint groups of size ℵ 1 need not necessarily be countable, so grouptheoretic almost disjointness differs from its set-theoretic homonym. Theorem 1 establishes in ZFC that the Baer-Specker group contains large families of almost disjoint almost free non-free pure uncountable subgroups.We thank the referee for constructive comments.

show abstract

Nearly Projective Boolean Algebras

Abstract: To our parentsAfter the main text was ready, Sakae Fuchino kindly wrote an appendix on set-theoretic methods in the field, which also includes some of his recent independence results.

Cited by 28 publications

References 0 publications

Boolean algebras admitting a countable minimally acting group

Boolean algebras admitting a countable minimally acting group

Regular subalgebras of complete Boolean algebras

Almost disjoint pure subgroups of the Baer-Specker group

Contact Info

Product

Resources

About