The Exact Saturation Number for the Diamond

Abstract: What is the smallest size of a family of subsets of $[n]$ such that it does not contain an induced copy of $Q_2$ as a poset (known as the \textit{diamond}), but adding a new set creates such a copy? It is easy to see that a maximal chain has this property, and thus the answer is at most $n+1$. Despite the simplicity of the diamond structure, the lower bound stagnated at $\sqrt n$ for quite some time, until recently the authors obtained a linear lower bound. In this paper, we fully solve this question showing that such a family must have size at least $n+1$.

• The paper proves the main result that every induced-diamond-saturated family in the Boolean lattice has at least n+1 sets, matching the maximal-chain construction and establishing sat*(n,D₂)=n+1.
• The authors combine minimal and maximal decompositions, a peeling process, trace arguments, and disjointness lemmas to convert local diamond witnesses into a global linear lower bound.
• The result confirms a strong case of the bounded-versus-linear saturation dichotomy, while leaving the classification of all extremal families and exact values for higher Boolean lattices open.

Overview

This paper by Ivan and Jaffe determines the induced saturation number of the diamond poset $\mathcal{D}_2$ (the two-dimensional Boolean lattice) exactly: for every $n$, $\text{sat}^*(n,\mathcal{D}_2)=n+1$ (2604.06521). The upper bound is elementary—a maximal chain of subsets of $[n]$ contains no induced diamond, yet adding any missing set creates one. The substance of the paper is the matching lower bound, which closes a gap that had persisted at $\sqrt{n}$ for years before the authors' recent linear bound of $(n+1)/5$ [diamondlinear]. The result also confirms a strong instance of the prevailing dichotomy conjecture in poset saturation, which predicts that every finite poset has either bounded or linearly growing saturation number [keszegh2021induced].

Background and context

A family $\mathcal{F}\subseteq\mathcal{P}([n])$ is $\mathcal{P}$-saturated if it avoids an induced copy of $\mathcal{P}$ but any outside set completes one; $\text{sat}^*(n,\mathcal{P})$ is the minimum size of such a family. The notion was introduced by Ferrara et al. [ferrara2017saturation], and the general theory exhibits a coarse dichotomy: the saturation number is either bounded or at least $n$0 [freschi2023induced], while always polynomially bounded [polynomial]. Linearity had previously been established only for a short list of posets—the butterfly, the antichain, non-chain complete multipartite posets, the $n$1 poset, and the diamond—and among these the diamond resisted exact analysis longest.

The paper's structural contribution goes beyond the specific value. The authors note that even identifying which features make a poset's saturation number unbounded remains open; their earlier work showed that adding at most three "special" points to any poset yields a poset with linear saturation number [gluing], and that linear sums propagate unboundedness under certain conditions [sums].

Framework

The proof assumes $n$2 is diamond-saturated with $n$3 (otherwise a lemma of Ferrara et al. already gives $n$4). Under this assumption there exist a minimal and a maximal element that are incomparable. The authors then define two symmetric decompositions:

• Minimal side: $n$5 (minimal elements), $n$6 (sets that are minimal elements of diamonds whose other three members lie in $n$7), $n$8 (maximal elements of $n$9 not contained in any member of $\text{sat}^*(n,\mathcal{D}_2)=n+1$0), and $\text{sat}^*(n,\mathcal{D}_2)=n+1$1, the sets in $\text{sat}^*(n,\mathcal{D}_2)=n+1$2 serving as middle elements of diamonds generated by members of $\text{sat}^*(n,\mathcal{D}_2)=n+1$3.
• Maximal side: the dual objects $\text{sat}^*(n,\mathcal{D}_2)=n+1$4 obtained by passing to complements.

Two facts anchor the argument. First, $\text{sat}^*(n,\mathcal{D}_2)=n+1$5 is $\text{sat}^*(n,\mathcal{D}_2)=n+1$6-saturated, so every subset of $\text{sat}^*(n,\mathcal{D}_2)=n+1$7 either contains a member of $\text{sat}^*(n,\mathcal{D}_2)=n+1$8, is contained in one, or is contained in a member of $\text{sat}^*(n,\mathcal{D}_2)=n+1$9. Second, $[n]$0 and $[n]$1 are disjoint, as are $[n]$2 and $[n]$3. Supporting lemmas supply local structure: for each $[n]$4 with $[n]$5 minimal, some pair $[n]$6 lies in $[n]$7 above $[n]$8; each $[n]$9 forces at least $\sqrt{n}$0 members of $\sqrt{n}$1 of size at least $\sqrt{n}$2; and each $\sqrt{n}$3 admits, for each $\sqrt{n}$4, a member of $\sqrt{n}$5 inside $\sqrt{n}$6 containing $\sqrt{n}$7.

Lower bounding $\sqrt{n}$8

The central device is a peeling process on $\sqrt{n}$9. Writing $(n+1)/5$0 for the subfamily of $(n+1)/5$1 containing $(n+1)/5$2, the authors iteratively select an element $(n+1)/5$3 whose trace family $(n+1)/5$4 is inclusion-minimal among traces on the current support, remove it, and continue until exhaustion. This produces disjoint blocks $(n+1)/5$5 partitioning $(n+1)/5$6, where $(n+1)/5$7 is the set of ground elements appearing in no minimal set, together with $(n+1)/5$8 distinct members $(n+1)/5$9 satisfying $\mathcal{F}\subseteq\mathcal{P}([n])$0.

Under the assumption $\mathcal{F}\subseteq\mathcal{P}([n])$1 (where $\mathcal{F}\subseteq\mathcal{P}([n])$2), an inductive argument shows that each $\mathcal{F}\subseteq\mathcal{P}([n])$3 yields a distinct pair of sets in $\mathcal{F}\subseteq\mathcal{P}([n])$4: one meeting $\mathcal{F}\subseteq\mathcal{P}([n])$5 in exactly $\mathcal{F}\subseteq\mathcal{P}([n])$6 and another incomparable companion. The construction augments a maximal trace set $\mathcal{F}\subseteq\mathcal{P}([n])$7 with a carefully chosen filler $\mathcal{F}\subseteq\mathcal{P}([n])$8 from $\mathcal{F}\subseteq\mathcal{P}([n])$9—chosen via a covering-counting lemma guaranteeing avoidance of all minimal sets—so that the resulting set must be covered by some $\mathcal{P}$0, whose generating diamond then supplies the two desired middle elements. Distinctness across all $\mathcal{P}$1 follows from the intersection pattern with the blocks.

Combining the pieces gives $\mathcal{P}$2, and, when $\mathcal{P}$3, the stronger bound $\mathcal{P}$4 with all contributing $\mathcal{P}$5-sets containing $\mathcal{P}$6. By complement symmetry, analogous bounds hold for $\mathcal{P}$7.

Disjointness and the final contradiction

To combine the two sides additively, the paper establishes that when $\mathcal{P}$8, the families $\mathcal{P}$9 and $\mathcal{P}$0 are disjoint. Three propositions deliver this:

• $\mathcal{P}$1 when $\mathcal{P}$2: a set that is simultaneously minimal and maximal forces, for each of its elements $\mathcal{P}$3, three distinct members of $\mathcal{P}$4 differing from it exactly by $\mathcal{P}$5, giving $\mathcal{P}$6.
• $\mathcal{P}$7 (and dually): immediate, since membership would contradict minimality.
• $\mathcal{P}$8 when $\mathcal{P}$9: a double-counting argument using Lemma sizeB constructs $\text{sat}^*(n,\mathcal{P})$0 forced members of $\text{sat}^*(n,\mathcal{P})$1 around a hypothetical common element $\text{sat}^*(n,\mathcal{P})$2; since $\text{sat}^*(n,\mathcal{P})$3 and $\text{sat}^*(n,\mathcal{P})$4, at least two coincide, pinning down the overlap exactly as $\text{sat}^*(n,\mathcal{P})$5 of total size $\text{sat}^*(n,\mathcal{P})$6—but the top element $\text{sat}^*(n,\mathcal{P})$7 of the second diamond escapes this collection, forcing $\text{sat}^*(n,\mathcal{P})$8.

With disjointness in hand, the main theorem follows by contradiction from $\text{sat}^*(n,\mathcal{P})$9 in two cases. If $n$00, the additive bounds give $n$01, and a size-separation argument (via Lemma sizeA) shows $n$02, yielding $n$03. If $n$04 (say), the strengthened bound plus Lemma nicelemmame produces a forest of pairs $n$05 over $n$06; accounting for overlaps between this forest, $n$07, and $n$08 gives $n$09, and any maximal element $n$10 lies outside all three collections, pushing the count to $n$11—a contradiction. Hence $n$12.

Limitations and open questions

The proof is specific to the diamond; the authors state plainly that it reveals little about the structure of extremal diamond-saturated families. Beyond maximal chains, $n$13 together with all singletons, and $n$14 together with all co-singletons, no diamond-saturated family of size exactly $n$15 is known, and they conjecture these are the only ones—with the sharper conditional claim that any such family avoiding both $n$16 and $n$17 has size at least $n$18 for a universal constant $n$19. Whether a size-$n$20 example avoiding $n$21 and $n$22 exists is left open. For higher Boolean lattices, they exhibit a construction showing $n$23 and conjecture generally that $n$24 for $n$25; neither the $n$26 case nor the general form is settled here. The method also relies on the assumption $n$27 to obtain disjointness of the two sides, so the argument does not directly transfer to settings where that hypothesis fails.

Conclusion

The paper settles the induced saturation number of the diamond at exactly $n$28, upgrading the known linear lower bound to the precise optimum and providing the first exact value for a poset whose analysis had proven resistant. Technically, the contribution rests on a nested-trace decomposition of the minimal elements, a filling scheme that converts coverage by $n$29 into distinct witnesses in $n$30, and disjointness results that allow the minimal and maximal sides to be summed. The classification of extremal families and the extension to general hypercubes $n$31 remain open.