2015
DOI: 10.1016/j.jcta.2015.06.004
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Boolean algebras and Lubell functions

Abstract: Let 2 [n] denote the power set of [n], where [n] = {1, 2, . . . , n}. A collection B ⊂ 2 [n] forms a d-dimensional Boolean algebra if there exist pairwise disjoint sets X 0 , X 1 , . . . , X d ⊆ [n], all non-empty with perhaps the exception of. Let b(n, d) be the maximum cardinality of a family F ⊂ 2 X that does not contain a d-dimensional Boolean algebra. Gunderson, Rödl, andIn this paper, we use the Lubell function as a new measurement for large families instead of cardinality. The Lubell value of a family o… Show more

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Cited by 14 publications
(16 citation statements)
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“…For a poset P , let L n (P ) be the maximum value lu n (F ) among families F ⊆ B n such that F is P -free. The following result has been used implicitly by many authors such as Axenovich and Walzer [2, Theorem 6] and Johnston, Lu and Milans [26,Theorem 3]. Although the result is straightforward, we provide the explicit statement in the case of 1-uniform Boolean Ramsey numbers along with a proof for completeness.…”
Section: -Uniform Boolean Ramsey Numbersmentioning
confidence: 91%
“…For a poset P , let L n (P ) be the maximum value lu n (F ) among families F ⊆ B n such that F is P -free. The following result has been used implicitly by many authors such as Axenovich and Walzer [2, Theorem 6] and Johnston, Lu and Milans [26,Theorem 3]. Although the result is straightforward, we provide the explicit statement in the case of 1-uniform Boolean Ramsey numbers along with a proof for completeness.…”
Section: -Uniform Boolean Ramsey Numbersmentioning
confidence: 91%
“…In general, Gunderson, Rödl and Sidorenko [16] showed that b(n, d) = O d (2 n n −1/2 d ), where the constant hidden in O d (.) is super exponential in d. This was improved by Johnston, Lu and Milans [18] to b(n, d) = O(2 n n −1/2 d ). Here, we present a short proof of the latter bound b(n, d) = O(2 n n −1/2 d ), and show a similar bound for a natural generalization of the problem in grids.…”
Section: Applicationsmentioning
confidence: 94%
“…Conjecture 1 is equivalent to the statement that for each poset P , the Turán function satisfies La * (n, P ) = O( n ⌊n/2⌋ ) = O(2 n / √ n). It is known (see [13]) that for each poset P , there exists α > 0 such that La * (n, P ) = O(2 n /n α ). In general, proving upper bounds on La * (n, P ) appears to be more challenging than proving upper bounds on La(n, P ).…”
Section: Conjecture 1 Every Poset Has a Finite Turán Thresholdmentioning
confidence: 99%