On an extremal problem for poset dimension

Let $f(n)$ be the largest integer such that every poset on $n$ elements has a $2$-dimensional subposet on $f(n)$ elements. What is the asymptotics of $f(n)$? It is easy to see that $f(n)\geqslant n^{1/2}$. We improve the best known upper bound and show $f(n)=\mathcal{O}(n^{2/3})$. For higher dimensions, we show $f_d(n)=\mathcal{O}\left(n^\frac{d}{d+1}\right)$, where $f_d(n)$ is the largest integer such that every poset on $n$ elements has a $d$-dimensional subposet on $f_d(n)$ elements.


Introduction
Every partially ordered set on n elements has a chain or an antichain of size at least n 1/2 , this is an immediate consequence of Dilworth's Theorem or its easier dual counterpart. Chains and antichains are very special instances of 2-dimensional posets. Surprisingly, the following simple problem is open: Let f (n) be the largest integer such that every poset on n elements has a 2-dimensional subposet on f (n) elements. What is the asymptotics of f (n)?
Although this sounds like a natural extremal-type question for posets, it was posed only in 2010, by François Dorais [1]. Clearly, n 1/2 f (n) n. Reiniger and Yeager [5] proved a sublinear upper bound, that is f (n) = O(n 0.8295 ). Their construction is a lexicographic power of standard examples.
The main idea behind our contribution was a belief that a (k × k)-grid is asymptotically the largest 2-dimensional subposet of the (k × k × k)-cube. This led us to the following theorem: Recall that the dimension dim(P ) of a poset P is the least integer d such that elements of P can be embedded into R d in such a way that x < y in P if and only if the point of x is below the point of y with respect to the product order on R d . Equivalently, the dimension of P is the least d such that there are d linear extensions of P whose intersection is P . By convention, whenever we say a poset is d-dimensional, we mean its dimension is at most d.
Reiniger and Yeager [5] also studied the guaranteed size of the largest d-dimensional subposet of poset on n elements. Let f d (n) be the largest integer such that every poset on n elements has a d-dimensional subposet on f d (n) elements. They proved, in particular, By the n d -grid we mean the poset on the ground set Moreover, it is easy to see that the n d+1 -grid contains as a subposet the n d -grid -simply fix one coordinate to an arbitrary value. We prove that this is asymptotically the largest d-dimensional subposet of the n d+1 -grid. For d 7, this observation improves on the best known upper bound for the asymptotics of f d (n). .
In order to show this we apply a multidimensional version of the theorem by Marcus and Tardos [3] saying that the number of 1-entries in an n × n (0, 1)-matrix that avoids a fixed permutation matrix P is O(n). The multidimensional version was proved by Klazar and Marcus [2], and then independently by Methuku and Pálvölgyi [4], who applied it to another extremal problem related to subposets, i.e. they proved that for every poset P the size of any family of subsets of [n] that does not contain P as a subposet is at most O n n/2 .

Dimension two
If we ignore a multiplicative constant, Theorem 1 becomes a special case of Theorem 2. Still, we provide a short and simple proof of Theorem 1, as we believe it might provide a better insight to the core of the problem.
Proof of Theorem 1. First we argue for values of n such that n = r 3 for some r ∈ N. Then at the end of the proof we address the general case.
Let C r be the poset with the ground set [r] 3 , where (x 1 , y 1 , z 1 ) Cr (x 2 , y 2 , z 2 ) if (z 1 z 2 ) and (y 1 < y 2 or (y 1 = y 2 and x 1 = x 2 )), see Figure 1.  Figure 1. A subposet of C 3 composed of all elements with z-coordinate equal to i (on the left) and the poset C 3 itself (on the right). Consider any subposet S of C r such that |S| 4r 2 . We will prove that dim(S) > 2 by showing that S contains as a subposet the poset 1 of dimension 3 presented in Figure 2.
Let S 1 be a poset obtained from S by removing every element (x, y, z) such that S contains no element (x, y, z ) with z < z. Note that |S 1 | 3r 2 , as for every pair (x, y) ∈ [r] 2 at most one element is removed. Now by the pigeonhole principle, we get that S 1 contains a subposet S 2 on at least 3r elements such that all elements of S 2 have the same z-coordinate.
Let A be any point in S 2 with the minimal y-coordinate and let S 3 be the subposet of S 2 obtained by removing all points with the same y-coordinate as A. As there can be at most r points with the same y-coordinate, |S 3 | 2r. By the pigeonhole principle for r − 1 containers, S 3 contains three points with the same y-coordinate, say B 1 = (x 1 , y, z), B 2 = (x 2 , y, z), B 3 = (x 3 , y, z). Thanks to the removal rule that led to the creation of S 1 , the poset S contains points C 1 = (x 1 , y, z 1 ), C 2 = (x 2 , y, z 2 ), C 3 = (x 3 , y, z 3 ) for some z 1 , z 2 , z 3 < z.
One can easily verify that the subposet {A, B 1 , B 2 , B 3 , C 1 , C 2 , C 3 } of S is the poset in Figure 2. Since it has dimension 3, we have dim(S) > 2, which concludes the proof for n being a perfect cube. Now, fix any n ∈ N, and let r = 3 √ n . Note that f is non-decreasing, thus With a more tedious analysis, which involves one more forbidden subposet and removal of both lowest and highest z-coordinate points in each (x, y)-column, we can prove a slightly stronger upper bound, i.e. f (n) = 3n 2/3 + o n 2/3 . However, we do not know how to improve on the asymptotics of f .

Higher dimensions
In this section we prove Theorem 2. In order to do this we apply a multidimensional version of the theorem by Marcus and Tardos [3], proved by Klazar and Marcus [2]. First, we recall their result. The original terminology can be simplified because our argument does not use arbitrary sized matrices and we can focus only on multidimensional analogues of square matrices.
We call a subset of [n] d a d-dimensional (0, 1)-matrix.
For two d-dimensional such that x is the x i -th element of L i for each i ∈ {1, 2, . . . , d + 1}. Note that the natural product order of elements of P is isomorphic to S d+1 . Now, take any d-dimensional subposet of the n d+1 -grid and denote by A the set of its elements. In particular, the subposet does not contain S d+1 as a subposet. Note that it implies that A avoids P , thus by Theorem 3 the size of the subposet is O(n d ).
Note that the proof above does not exploit any specific properties of the standard example, apart from its dimension. In particular, it implies that every (d + 1)-dimensional poset P can be found in every subposet of the n d+1 -grid of size Ω(n d ), with the constant hidden in the asymptotic notation depending on the choice of P .