For a finite poset P and n≥ 1, define operatornamesat^(n, P) to be the minimum of | F| over all families F⊆ 2^[n] such that F contains no induced copy of P, but for every A∈ 2^[n]setminus F, the family F cupA contains an induced copy of P. The conjecture asserts that, for every finite poset P, either operatornamesat^(n, P)=O(1) as n→∞, or there exist constants c_ P>0 and n_0 such that operatornamesat^(n, P)≥ c_ Pn for all n≥ n_0.

Bounded-linear dichotomy for induced poset saturation numbers

For a finite poset $\mathcal P$ and $n\ge 1$ , define $\operatorname{sat}^{*}(n,\mathcal P)$ to be the minimum of $|\mathcal F|$ over all families $\mathcal F\subseteq 2^{[n]}$ such that $\mathcal F$ contains no induced copy of $\mathcal P$ , but for every $A\in 2^{[n]}\setminus\mathcal F$ , the family $\mathcal F\cup\{A\}$ contains an induced copy of $\mathcal P$ . The conjecture asserts that, for every finite poset $\mathcal P$ , either $\operatorname{sat}^{*}(n,\mathcal P)=O(1)$ as $n\to\infty$ , or there exist constants $c_{\mathcal P}>0$ and $n_{0}$ such that $\operatorname{sat}^{*}(n,\mathcal P)\ge c_{\mathcal P}n$ for all $n\ge n_{0}$ .

References

Primary source

arXiv

Additional references

The Bounded-Linear Dichotomy Holds for all 4 Point Posets — arXiv — Maria-Romina Ivan, Sean Jaffe

extremal-combinatorics poset-theory

Posted 1 day ago · edited 1 day ago

Progress summary

Refreshed 1 day ago

Claimed progress

An unrefereed preprint claims the smallest nontrivial cases are classified, but the conjecture for all finite posets remains open.

The problem asks whether every finite poset has induced saturation number either bounded independently of $n$ or at least linear in $n$ . The latest preprint claims this dichotomy for every four-point poset, including the previously unresolved case with an isolated point.

Known results

For every fixed finite poset $P$ , either $\operatorname{sat}^{*}(n,P)=O(1)$ or $\operatorname{sat}^{*}(n,P)\ge \min\{2\sqrt{n},n/2+1\}$ ; the conjectured lower bound $n+1$ remains open in general.
The $n+1$ lower bound is known for posets with legs and their duals.
For complete bipartite posets, $\operatorname{sat}^{*}(n,K_{s,t})=O(n)$ for fixed $s,t$ , refuting the earlier superlinear expectation already for $K_{2,2}$ .
A gluing result shows that adding at most three elements can produce a poset with saturation growth at most linear.

October 2026 four-point classification

Maria-Romina Ivan and Sean Jaffe report a complete four-point classification, including the isolated-point case. This is substantive progress toward the general dichotomy, but the preprint is unrefereed and no independent verification was found.

Current status (as of October 2026): The four-point case is claimed solved, while the bounded-versus-linear dichotomy for arbitrary finite posets remains open.

Sources

No solutions have been posted yet.