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Related papers: On diamond-free subposets of the Boolean lattice

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In the theory of partially-ordered sets, the two-dimensional Boolean lattice is known as the diamond. In this paper, we show that, if $\mathcal{F}$ is a family in the $n$-dimensional Boolean lattice that has no diamond as a subposet, then…

Combinatorics · Mathematics 2015-03-13 Lucas Kramer , Ryan R. Martin

Let $La(n,P)$ be the maximum size of a family of subsets of $[n]=\{1,2,...,n\}$ not containing $P$ as a (weak) subposet. The diamond poset, denoted $B_{2}$, is defined on four elements $x,y,z,w$ with the relations $x<y,z$ and $y,z<w$.…

Combinatorics · Mathematics 2017-11-27 Dániel Grósz , Abhishek Methuku , Casey Tompkins

For a family $\mathcal{F}$ of subsets of [n]=\{1, 2, ..., n} ordered by inclusion, and a partially ordered set P, we say that $\mathcal{F}$ is P-free if it does not contain a subposet isomorphic to P. Let $ex(n, P)$ be the largest size of a…

Combinatorics · Mathematics 2016-05-24 Maria Axenovich , Jacob Manske , Ryan R. Martin

For a family of subsets of $[n]:={1,2,...,n}$, the Lubell function is defined as $\hb_n(\F):=\sum_{F\in\F}\binom{n}{|F|}^{-1}$. In \cite{GriLiLu}, Griggs, Lu, and the author conjectured that if a family $\F$ of subset of $[n]$ does not…

Combinatorics · Mathematics 2012-06-20 Wei-Tian Li

Given a finite poset P, we consider the largest size La(n,P) of a family of subsets of $[n]:=\{1,...,n\}$ that contains no subposet P. This problem has been studied intensively in recent years, and it is conjectured that $\pi(P):=…

Combinatorics · Mathematics 2011-09-07 Jerrold R. Griggs , Wei-Tian Li , Linyuan Lu

The $\mathcal{N}$ poset consists of four distinct sets $W,X,Y,Z$ such that $W\subset X$, $Y\subset X$, and $Y\subset Z$ where $W$ is not necessarily a subset of $Z$. A family $\mathcal{F}$ as a subposet of the $n$-dimensional Boolean…

Combinatorics · Mathematics 2017-04-18 Ryan R. Martin , Shanise Walker

We prove that the largest $Q_2$-free family of subsets of $[n]$ which contains sets of at most three different sizes has at most $(3 + 2\sqrt {3})N/3 + o(N) \approx 2.1547N + o(N)$ members, where $N = {n \choose {\lfloor n/2 \rfloor}}$.…

Combinatorics · Mathematics 2011-08-23 Jacob Manske , Jian Shen

Let $2^{[n]}$ denote the power set of $[n]:=\{1,2,..., n\}$. A collection $\B\subset 2^{[n]}$ forms a $d$-dimensional {\em Boolean algebra} if there exist pairwise disjoint sets $X_0, X_1,..., X_d \subseteq [n]$, all non-empty with perhaps…

Combinatorics · Mathematics 2013-07-15 Travis Johnston , Linyuan Lu , Kevin G. Milans

For every integer $n$ with $n \geq 6$, we prove that the Boolean dimension of a poset consisting of all the subsets of $\{1,\dots,n\}$ equipped with the inclusion relation is strictly less than $n$.

Combinatorics · Mathematics 2025-03-13 Marcin Briański , Jędrzej Hodor , Hoang La , Piotr Micek , Katzper Michno

For a given fixed poset $\mathcal P$ we say that a family of subsets of $[n]$ is $\mathcal P$-saturated if it does not contain an induced copy of $\mathcal P$, but whenever we add to it a new set, an induced copy of $\mathcal P$ is formed.…

Combinatorics · Mathematics 2025-04-01 Maria-Romina Ivan

Given a finite poset $P$, we consider the largest size $\lanp$ of a family $\F$ of subsets of $[n]:=\{1,...,n\}$ that contains no subposet $P$. This continues the study of the asymptotic growth of $\lanp$; it has been conjectured that for…

Combinatorics · Mathematics 2015-03-23 Jerrold R. Griggs , Wei-Tian Li

In this paper we show that for any poset $P$ that is not an antichain, the number of induced $P$-free families in the Boolean lattice $2^{[n]}$ is at most $ 2^{O(\mathrm{La}^*(n,P))}$, where $\mathrm{La}^*(n,P)$ denotes the the largest size…

Combinatorics · Mathematics 2026-03-25 Tao Jiang , Sean Longbrake , Liana Yepremyan

In this paper we introduce a particular lattice of subgroups called a "cyclic-diamond" and show that every finite non-cyclic group contains a cyclic-diamond as a sublattice of its lattice of subgroups. Turning to the infinite case, we show…

Group Theory · Mathematics 2023-07-14 Matt Alexander

Let $F$ be a family of subsets of $\{1,\ldots,n\}$. We say that $F$ is $P$-free if the inclusion order on $F$ does not contain $P$ as an induced subposet. The \emph{Tur\'an function} of $P$, denoted $\pi^*(n,P)$, is the maximum size of a…

Combinatorics · Mathematics 2014-08-05 Linyuan Lu , Kevin G. Milans

For a fixed poset $\mathcal P$ we say that a family $\mathcal F\subseteq\mathcal P([n])$ is $\mathcal P$-saturated if it does not contain an induced copy of $\mathcal P$, but whenever we add a new set to $\mathcal F$, we form an induced…

Combinatorics · Mathematics 2026-03-10 Maria-Romina Ivan , Sean Jaffe

Given a finite poset $\mathcal{P}$, a family $\mathcal{F}$ of elements in the Boolean lattice is induced-$\mathcal{P}$-saturated if $\mathcal{F}$ contains no copy of $\mathcal{P}$ as an induced subposet but every proper superset of…

Combinatorics · Mathematics 2019-08-06 Ryan R. Martin , Heather C. Smith , Shanise Walker

We prove upper and lower bounds on the local dimension of any pair of layers of the Boolean lattice, and show that the local dimension of the first and middle layers of the $n$-dimensional Boolean lattice is asymptotically $\frac{n}{\log_2…

Combinatorics · Mathematics 2020-05-05 David Lewis

For two posets $P$ and $Q$, we say $Q$ is $P$-free if there does not exist any order-preserving injection from $P$ to $Q$. The speical case for $Q$ being the Boolean lattice $B_n$ is well-studied, and the optiamal value is denoted as…

Combinatorics · Mathematics 2016-05-03 Jun-Yi Guo , Fei-Huang Chang , Hong-Bin Chen , Wei-Tian Li

In the Boolean lattice, Sperner's, Erd\H{o}s's, Kleitman's and Samotij's theorems state that families that do not contain many chains must have a very specific layered structure. We show that if instead of $\mathbb{Z}_2^n$ we work in…

Combinatorics · Mathematics 2018-10-03 Jason Long , Adam Zsolt Wagner

The Boolean lattice $2^{[n]}$ is the family of all subsets of $[n]=\{1,\dots,n\}$ ordered by inclusion, and a chain is a family of pairwise comparable elements of $2^{[n]}$. Let $s=2^{n}/\binom{n}{\lfloor n/2\rfloor}$, which is the average…

Combinatorics · Mathematics 2019-11-22 Benny Sudakov , Istvan Tomon , Adam Zsolt Wagner
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