Linear Saturation for $\mathcal N$ via Butterflies

TL;DR

This paper proves that the induced saturation number for the 4-point poset $ ext{sat}^*(n, ext{N})$ grows linearly with n, advancing understanding of poset saturation in combinatorics.

Contribution

It establishes linear lower bounds for the induced saturation number of the poset $ ext{N}$, confirming linearity for this specific case.

Findings

Established that $ ext{sat}^*(n, ext{N}) ext{ is at least } rac{n+6}{4}$.

Previously known bounds were $2 oot n ext{ and } 2n$.

Proved a structural property of $ ext{N}$-saturated families involving antichains.

Abstract

Given a finite poset $\mathcal P$, how small can a family $\mathcal F$ of subsets of $[n]$ be such that $\mathcal F$ does not contain an induced copy of $\mathcal P$, but $\mathcal F\cup\{X\}$ contains such a copy for all $X\in\mathcal P([n])\setminus\mathcal F$? This is known as the induced saturation number of $\mathcal P$, denoted by $\text{sat}^*(n,\mathcal P)$. The main conjecture in the area is that the induced saturation number for any poset is either bounded, or linear. In this paper we establish linearity for the induced saturation number of the 4-point poset $\mathcal N$. Previously, it was known that $2\sqrt n\leq\text{sat}^*(n,\mathcal N)\leq 2n$. We show that $\text{sat}^*(n,\mathcal N)\geq\frac{n+6}{4}$. A crucial role in the proof is played by a structural feature of $\mathcal N$-saturated families, namely that if the family contains two antichains, one completely above the other, then it must also contain a `middle' point -- greater than one antichain and less than the other.