A note on the size of free families
Abstract
The poset consists of four distinct sets such that , , and where is not necessarily a subset of . A family , considered as a subposet of the dimensional Boolean lattice , is free if it does not contain as a subposet. Let be the size of a largest free family in . Katona and Tarján proved that , where and is the size of a singleerrorcorrecting code with constant weight . In this note, we prove for even and , , which improves the bound on in the second order term for some values of and should be an improvement for an infinite family of values of , depending on the behavior of the function .
Keywords: forbidden subposets, errorcorrecting codes
1 Introduction
The dimensional Boolean lattice, , denotes the partially ordered set (poset) , where and, for every finite set , denotes the set of subsets of . For posets, and , we say is a (weak) subposet of if there exists an injection that preserves the partial ordering. That is, whenever in , we have in . If is a subposet of such that contains no subposet , we say is free.
free posets (or free families) have been extensively studied, beginning with Sperner’s theorem in 1928. Sperner [7] proved that the size of the largest antichain in is . Erdős [2] generalized this result to chains. Katona and Tarján [6] addressed the problem of free families and got an asymptotic result. Griggs and Katona [5] addressed free families, obtaining Theorem 1 below. See Griggs and Li [4] for a survey of the progress on free families. Let denote the size of the largest free family in .
The main result of this note is Theorem 4, in which, for some values of , we improve the bounds on in the secondorder term. The poset consists of four distinct sets such that , , and . However, is not necessarily a subset of . See Figure 1. The earliest extremal result on free families is Theorem 1.
Theorem 1 (Griggs and Katona [5]).
The construction for the lower bound of Theorem 1 comes directly from a previous result of Katona and Tarján [6] from 1983 on free families. The poset consists of three elements such that and . It is clear that because any free family is also free.
To establish the lower bound, Katona and Tarján used a constantweight code construction due to Graham and Sloane [3] from 1980. In the proof of Theorem 4, we obtain a lower bound that appears to be larger than the current known bound. However, whether it is an improvement depends on the behavior of some functions wellknown in coding theory. In order to discuss our results we need some brief coding theory background.
1.1 Coding Theory Background
Let denote the size of the largest family of vectors of length such that each vector has exactly ones and the Hamming distance between any pair of distinct vectors is at least . This is the same as the size of the largest family of subsets of such that each subset has size exactly and the symmetric difference of any pair of distinct sets is at least .
The quantity is important in the field of errorcorrecting codes. In fact, computes the size of a singleerrorcorrecting code with constant weight . Henceforth, we will use “SEC code” as shorthand for “singleerrorcorrecting code.”
The first nontrivial value of for is . Graham and Sloane [3] give a lower bound construction for .
Theorem 2 (Graham and Sloane [3]).
1.2 Main Result
Katona and Tarján [6] estimated the following lower bound for free families.
Theorem 3.
Let . Then,
The following theorem is our main result of the note.
Theorem 4.
Let be even and let . Then,
(1) 
Remark 5.
This is potentially an improvement when is even. We note that the same 3level construction works for odd and . This gives a family of size nontrivially in three layers. However, since in the odd case, this does not provide an improvement to the known bounds.
We believe that, for , the quantity is strictly unimodal as a function of as long as . This strict unimodality has been established [1] for and known bounds suggest that it is the case for larger values of as well. If unimodality holds, then would achieve its maximum uniquely at or . Therefore, we expect (1) to also be a strict improvement over Theorem 3 in the case where is even. However, to our knowledge, the unimodality of has never been established and seems to be a highly nontrivial problem.
Proof of Theorem 4.
Given , let be a constant weight SEC code of size . Define and . Claim 6 gives some important properties of .
Claim 6.

Both and are SEC codes with constant weight and , respectively.

If and , .
Proof.
(i). Let . Then since and their symmetric difference must be at least in order for to be a EC code. Thus, is a SEC code. By a similar argument, is a SEC code.
In order to finish the proof, we just need to show that the family is free.
To that end, suppose there is a subposet with elements where , and (see Figure 1). Where is the element ?
We know that because it has to have elements below it and the elements of are all minimal in . We know that because that would force and, being subsets of would require , a contradiction to being a SEC code. Therefore, .
Now, where is ? We know that because . We know because that would force and thus would force , this is a contradiction to the fact that is a SEC code. Therefore, .
In order for the copy of to exist, , which implies and so . Recall, however, that and are distinct members of and so have symmetric difference at least 4, a contradiction. ∎
Acknowledgements
We would like to extend our thanks to Kirsten Hogenson and SungYell Song for providing helpful conversations.
References
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