On $k$-antichains in the unit $n$-cube

TL;DR

This paper investigates the size of special subsets called $k$-antichains in the unit $n$-cube, establishing an upper bound on their Hausdorff measure and conjecturing the existence of extremal examples, verified in two dimensions.

Contribution

It introduces the concept of $k$-antichains in the unit $n$-cube and provides an upper bound on their Hausdorff measure, advancing understanding of their geometric properties.

Findings

Hausdorff measure of $k$-antichains is at most $kn$

Bound is asymptotically sharp

Conjecture verified for $n=2$

Abstract

A \emph{chain} in the unit $n$-cube is a set $C\subset [0,1]^n$ such that for every $\mathbf{x}=(x_1,\ldots,x_n)$ and $\mathbf{y}=(y_1,\ldots,y_n)$ in $C$ we either have $x_i\le y_i$ for all $i\in [n]$, or $x_i\ge y_i$ for all $i\in [n]$. We consider subsets, $A$, of the unit $n$-cube $[0,1]^n$ that satisfy \[ \text{card}(A \cap C) \le k, \, \text{ for all chains } \, C \subset [0,1]^n \, , \] where $k$ is a fixed positive integer. We refer to such a set $A$ as a $k$-antichain. We show that the $(n-1)$-dimensional Hausdorff measure of a $k$-antichain in $[0,1]^n$ is at most $kn$ and that the bound is asymptotically sharp. Moreover, we conjecture that there exist $k$-antichains in $[0,1]^n$ whose $(n-1)$-dimensional Hausdorff measure equals $kn$ and we verify the validity of this conjecture when $n=2$.

Equations

$$\text{card}(A\cap C)\le k,\text{ for all chains }C\subset [0,1{]}^n,$$

$${\mathcal{H}}^0(A)\le \sum _{i=1}^k\left(\genfrac{}{}{0ex}{}{n}{\lfloor \frac{n-k}{2}\rfloor +i}\right).$$

$$A=\bigcup _{i=1}^k\left\{\mathbf{x}=(x_1,\dots ,x_n)\in \{0,1{\}}^n:\sum _{i=1}^n{x}_i=\lfloor \frac{n-k}{2}\rfloor +i\right\}.$$

$${\mathcal{H}}^{n-1}(A)\le n.$$

$$A_p=\left\{\mathbf{x}\in [0,1{]}^n:\Vert \mathbf{x}{\Vert }_p=1\right\},$$

$${\mathcal{H}}^{n-1}(A)\le kn.$$

$$B=\{\mathbf{x}\in A:\text{ there is no }\mathbf{y}\in A\setminus \{x\}\text{ satisfying }\mathbf{y}\le \mathbf{x}\}.$$

$$W(g,h):=\{(x,y)\in [0,1{]}^2:g(x)\le y\le h(x)\}.$$

$$R[\mathbf{x},\mathbf{y}]:=\{(z_1,z_2)\in {\mathbb{R}}^2:z_1\in [x_1,y_1]\text{ and }{z}_2\in [y_2,x_2]\}$$

$$f(x):=\frac{1}{2}(g(x)+h(x)),\text{ for }x\in [0,1].$$

$$0<g^{-1}(f(x_n))<x_n\text{ and }0<f^{-1}(h(x_n))<x_n.$$

$$x:=\underset{n\to \infty }{\lim }{x}_n\text{ and }y:=\underset{n\to \infty }{\lim }{y}_n.$$

$$0<x<x_n,\text{ for every }n\in \mathbb{N}.$$

$$\underset{n\to \infty }{\lim }(x_n,f(x_n))=(0,f(0))=(0,1),$$

$$\underset{n\to \infty }{\lim }(y_n,f(y_n))=(1,f(1))=(1,0).$$

$$\sum _{n=1}^{\infty }(x_n-x_{n+1})=\frac{1}{2}\text{ and }\sum _{n=1}^{\infty }(y_{n+1}-y_n)=\frac{1}{2}$$

$$\sum _{n=1}^{\infty }(f(x_{n+1})-f(x_n))=1-f\left(1/2\right)\text{ and }\sum _{n=1}^{\infty }(f(y_n)-f(y_{n+1}))=f\left(1/2\right)$$

$$\sum_{n=1}^{\infty}\bigg{(}(x_{n}-x_{n+1})+(y_{n+1}-y_{n})+(f(x_{n+1})-f(x_{n}))+(f(y_{n})-f(y_{n+1}))\bigg{)}=2\,.$$

$$Gr(D)$$

$$\displaystyle\subset\bigcup_{n=1}^{\infty}\bigg{(}R[(x_{n+1},f(x_{n+1})),(x_{n},f(x_{n}))]\cup R[(y_{n},f(y_{n})),(y_{n+1},f(y_{n+1}))]\bigg{)}$$

$$\subset W(g,h)$$

$${\mathcal{H}}^1(Gr(D))=\sum _{n=1}^{\infty }({\mathcal{H}}^1(Gr(d_{x,n}))+{\mathcal{H}}^1(Gr(d_{y,n})))=2$$

$$f_1(x)>f_2(x)>\cdots >f_{2k}(x),\text{ for every }x\in (0,1).$$

$${\mathcal{H}}^1(A)=\sum _{i=1}^k{\mathcal{H}}^1(Gr(D_i)=2k,$$

Full text

[TABLE]

where $k$ is a fixed positive integer. We refer to such a set $A$ as a $k$-antichain. We show that the $(n-1)$-dimensional Hausdorff measure of a $k$-antichain in $[0,1]^{n}$ is at most $kn$ and that the bound is asymptotically sharp. Moreover, we conjecture that there exist $k$-antichains in $[0,1]^{n}$ whose $(n-1)$-dimensional Hausdorff measure equals $kn$ and we verify the validity of this conjecture when $n=2$.

Keywords and phrases: $k$-antichains, Hausdorff measure, singular function

Mathematics Subject Classification (2010): 05D05; 28A78; 05C35; 26A30

1 Prologue, related work and main results

Let $[n]$ denote the set of positive integers $\{1,\ldots,n\}$, and $2^{[n]}$ denote the collection of all subsets of $[n]$. Given two points $\mathbf{x}=(x_{1},\ldots,x_{n})$ and $\mathbf{y}=(y_{1},\ldots,y_{n})$ in $\mathbb{R}^{n}$, we write $\mathbf{x}\leq\mathbf{y}$ if $x_{i}\leq y_{i}$, for all $i\in[n]$. Given a subset $S\subset\mathbb{R}^{n}$, we say that a set $C\subset S$ is a chain in $S$ if for all $\mathbf{x},\mathbf{y}\in C$ it either holds $\mathbf{x}\leq\mathbf{y}$ or $\mathbf{y}\leq\mathbf{x}$. Given a non-negative real number $s$, we denote by $\mathcal{H}^{s}(\cdot)$ the $s$-dimensional Hausdorff outer measure (see [9, p. 81 and p. 1–2]). Notice that $\mathcal{H}^{0}(\cdot)$ is counting measure. Finally, given a positive integer $k$ and a set $S\subset\mathbb{R}^{n}$, a $k$-antichain in $S$ is a set $A\subset S$ such that $\mathcal{H}^{0}(A\cap C)\leq k$, for all chains $C\subset S$. An $1$-antichain is simply referred to as an antichain.

This work is motivated by a particular result from extremal set theory. Extremal set theory (see [1, 5]) is a rapidly growing branch of combinatorics which is concerned with the problem of obtaining sharp estimates on the size of a collection $\mathcal{F}\subset 2^{[n]}$, subject to constraints that are expressed in terms of union, intersection or inclusion. A particular line of research is driven by the idea that several results from extremal combinatorics have continuous counterparts. This is an idea that goes back to the 70’s (see [17]) and, since its conception, has resulted in reporting several analogues of results from extremal combinatorics both in a “measure-theoretic context” (see, for example, [3, 4, 6, 7, 12, 13, 14, 16]) as well as in a “vector space context” (see, for example, [2, 11, 15]) In this note we report yet another measure-theoretic analogue of a result from extremal combinatorics.

Before being more precise, let us remark that one can associate a binary vector of length $n$ to every $F\subset[n]$: simply put $1$ in the $i$-th coordinate if $i\in F$, and [math] otherwise. Notice that this correspondence is bijective, and one may choose to not distinguish between subsets of $[n]$ and elements of $\{0,1\}^{n}$. In other words, any statement regarding collections $\mathcal{F}\subset 2^{[n]}$ can be turned to a statement regarding subsets $F\subset\{0,1\}^{n}$, and vice versa.

Perhaps the most fundamental result in extremal set theory is due to Sperner [20]. It provides a sharp upper bound on the cardinality of an antichain in $\{0,1\}^{n}$. Sperner’s theorem is a well-known and celebrated result that has been generalised in a plethora of ways (see [5] for a textbook devoted to the topic). A particular extension of Sperner’s theorem is due to Paul Erdős, and reads as follows.

Theorem 1.1** (Erdős [8]).**

Fix a positive integer $k\in[n]$. If $A$ is a $k$-antichain in $\{0,1\}^{n}$, then

[TABLE]

Notice that the bound provided by Theorem 1.1 is sharp and is attained by the set

[TABLE]

In other words, Erdős’ result provides a sharp upper bound on the size of a $k$-antichain in the binary $n$-cube $\{0,1\}^{n}$. In this article we investigate a continuous analogue of Theorem 1.1. There are several ways to consider Theorem 1.1 in a continuous setting (see [16] for an alternative direction), but the main idea is to examine what happens when one replaces the binary $n$-cube $\{0,1\}^{n}$ with the unit $n$-cube $[0,1]^{n}$ in Theorem 1.1. What is the maximum “size” of a $k$-antichain in the unit $n$-cube $[0,1]^{n}$? Since we are dealing with subsets of $[0,1]^{n}$ and we have to choose an adequate notion of “size”. A first choice could be the $n$-dimensional Lebesgue measure, denoted $\mathcal{L}^{n}(\cdot)$. However, it is not difficult to see, using Lebesgue’s density theorem, that the $\mathcal{L}^{n}$-measure of a $k$-antichain equals zero. Given this fact, it is therefore natural to ask for sharp upper bounds on the Hausdorff dimension and the corresponding Hausdorff measure of a $k$-antichain in the unit $n$-cube. In the case of antichains this has been considered in [7], where the following continuous analogue of Sperner’s theorem has been reported.

Theorem 1.2** (Engel et al. [7]).**

If $A$ is an antichain in $[0,1]^{n}$, then

[TABLE]

In particular, the Hausdorff dimension of an antichain is at most $n-1$. Let us remark that the bound provided by Theorem 1.2 is asymptotically sharp. Indeed, as is observed in [7], this can be seen by considering the boundary of $\ell^{p}$-unit balls, i.e., by considering the sets

[TABLE]

as $p\to\infty$. Notice that $A_{p}$ is an antichain in $[0,1]^{n}$, but $A_{\infty}$ is not. Moreover, notice that $\mathcal{H}^{n-1}(A_{\infty})=n$. Now, it is not difficult to see that the $p$-ball $B_{p}=\{\mathbf{x}\in\mathbb{R}^{n}:\|\mathbf{x}\|_{p}\leq 1\}$ converges, with respect to the Hausdorff distance, to the $\infty$-ball $B_{\infty}=\{\mathbf{x}\in\mathbb{R}^{n}:\|\mathbf{x}\|_{\infty}\leq 1\}$. Furthermore, it is known (see [19, p. 219]) that whenever a sequence of convex bodies $B_{i}$ converges, with respect to the Hausdorff distance, to a convex body $B$, then it follows that $\mathcal{H}^{n-1}(\partial B_{i})$ converges to $\mathcal{H}^{n-1}(B)$. Hence $\mathcal{H}^{n-1}(A_{p})$ tends to $n$, as $p\to\infty$, and therefore one can find an antichain in $[0,1]^{n}$ whose $\mathcal{H}^{n-1}$-measure is arbitrarily close to $n$. There remains the question of whether there exists an antichain whose $\mathcal{H}^{n-1}$-measure is equal to $n$. The following conjecture has been put forward in [7].

Conjecture 1.3** (Engel et al. [7]).**

There exists an antichain in $[0,1]^{n}$ such that $\mathcal{H}^{n-1}(A)=n$.

When $n=1$ this conjecture is clearly true, and when $n=2$ it is observed in [7] that the validity of Conjecture 1.3 is an immediate consequence of the following, well-known, result. Recall that a singular function $f:[a,b]\to[c,d]$ is a strictly decreasing function whose derivative equals zero almost everywhere.

Theorem 1.4** (Folklore).**

Let $f:[a,b]\to[c,d]$ be a singular function and let $G_{f}=\{(x,f(x)):x\in[a,b]\}$ be its graph. Then $\mathcal{H}^{1}(G_{f})=(b-a)+(d-c)$.

We refer the reader to [18, p. 101] for details regarding the existence of singular functions, and to [10, p. 810] for a sketch of a proof of Theorem 1.4. Since the graph of a singular function $f:[0,1]\to[0,1]$ is clearly an antichain in $[0,1]^{2}$, it follows that Conjecture 1.3 holds true when $n=2$.

In this note we focus on $k$-antichains in $[0,1]^{n}$, for $k>1$. Using Theorem 1.2, we obtain the following upper bound on the maximum “size” of a $k$-antichain in the unit $n$-cube.

Theorem 1.5**.**

Fix a positive integer $k\geq 1$. If $A$ is a $k$-antichain in $[0,1]^{n}$, then

[TABLE]

Using a similar argument as the one used in the remarks after Theorem 1.2, it can be shown that the upper bound provided by Theorem 1.5 is asymptotically sharp, and it is therefore natural to ask whether there exist $k$-antichains in $[0,1]^{n}$ whose $\mathcal{H}^{n-1}$-measure is equal to $kn$. We conjecture that the answer is in the affirmative, for all $n\geq 2$, and in this note we verify the validity of this conjecture for $n=2$.

Theorem 1.6**.**

There exists a $k$-antichain in $[0,1]^{2}$ such that $\mathcal{H}^{1}(A)=2k$.

2 Proofs

Proof of Theorem 1.5.

It is enough to show that there exist $k$ sets $A_{1},\ldots,A_{k}\subset[0,1]^{n}$ such that $A=\cup_{i=1}^{k}A_{i}$ and each $A_{i}$ is an antichain. Theorem 1.5 then follows from Theorem 1.2. We prove the required result by induction on $k$. The case $k=1$ is clear. Assuming that the result holds true for $k-1>1$, we prove it for $k$. Let $B$ be the set consisting of all minimal elements of $A$. That is, let

[TABLE]

Clearly, $B$ is an antichain and it is enough to show that $A\setminus B$ is a $(k-1)$-antichain in $[0,1]^{n}$; the result then follows from the induction hypothesis. Assume, towards a contradiction, that $A\setminus B$ is not a $(k-1)$-antichain. This implies that there exists a chain $C\subset[0,1]^{n}$ such that $\mathcal{H}^{0}((A\setminus B)\cap C)\geq k$. Let $\mathbf{y}\in(A\setminus B)\cap C$ be a minimal element, i.e, $\mathbf{y}$ is such that there does not exist $\mathbf{z}$, which is distinct from $\mathbf{y}$, satisfying $\mathbf{z}\in(A\setminus B)\cap C$ and $\mathbf{z}\leq\mathbf{y}$. Notice that the existence of $\mathbf{y}$ follows from the fact that, since $A$ is a $k$-antichain, $(A\setminus B)\cap C$ is a finite set. Since $\mathbf{y}\notin B$ it follows that there exists $\mathbf{x}\in A$ such that $\mathbf{x}\neq\mathbf{y}$ and $\mathbf{x}\leq\mathbf{y}$. Now set $D:=\{\mathbf{x}\}\cup(A\setminus B)\cap C$ and notice that $D$ is a chain that satisfies $\mathcal{H}^{0}(D\cap A)\geq k+1$, contrariwise to the fact that $A$ is a $k$-antichain. The result follows. ∎

We proceed with the proof of Theorem 1.6. This requires some additional piece of notation. Given two functions $g,h:[0,1]\to[0,1]$, let

[TABLE]

Given a function $g:[0,1]\to[0,1]$, let $Gr(g)=\{(x,y)\in[0,1]^{2}:y=g(x)\}$ be its graph. If $A\subset[0,1]^{2}$, we denote its interior by $\operatorname{int}(A)$. Finally, given two points $\mathbf{x}=(x_{1},x_{2}),\mathbf{y}=(y_{1},y_{2})\in\mathbb{R}^{2}$ with $x_{1}<y_{1}$ and $x_{2}>y_{2}$, let

[TABLE]

be the rectangle “determined” by the points $\mathbf{x},\mathbf{y}$. The proof of Theorem 1.6 relies upon the following.

Lemma 2.1**.**

Let $g,h:[0,1]\to[0,1]$ be strictly decreasing and continuous bijections such that $g(x)<h(x)$, for all $x\in(0,1)$. Then there exists a strictly decreasing function $D:(0,1)\to(0,1)$ such that

• (a)

$g(x)\leq D(x)\leq h(x)$* for every $x\in(0,1)$,*

• (b)

$\mathcal{H}^{1}(Gr(D))=2$.

Proof.

Consider the function $f:[0,1]\to[0,1]$ defined by

[TABLE]

Clearly, $f$ is a strictly decreasing, continuous, bijection and $g(x)<f(x)<h(x)$ holds true for every $x\in(0,1)$. We will show that we can inductively construct sequences $\{x_{n}\}_{n}$ and $\{y_{n}\}_{n}$ that satisfy the following five conditions:

• (i)

$\frac{1}{2}=x_{1}>x_{2}>\dots>0$ and $\frac{1}{2}=y_{1}<y_{2}<\dots<1$,

• (ii)

$R[(x_{n+1},f(x_{n+1})),(x_{n},f(x_{n}))]\subset W(g,h)$, $n\in\mathbb{N}$,

• (iii)

$R[(y_{n},f(y_{n})),(y_{n+1},f(y_{n+1}))]\subset W(g,h)$, $n\in\mathbb{N}$,

• (iv)

$R[(x_{n+1},f(x_{n+1})),(x_{n},f(x_{n}))]\not\subset\operatorname{int}(W(g,h))$, $n\in\mathbb{N}$,

• (v)

$R[(y_{n},f(y_{n})),(y_{n+1},f(y_{n+1}))]\not\subset\operatorname{int}(W(g,h))$, $n\in\mathbb{N}$.

We first show how to construct the sequence $\{x_{n}\}_{n}$. Begin by setting $x_{1}=\frac{1}{2}$. Now, assuming we have already constructed $x_{1},\dots,x_{n}$ satisfying (i), (ii) and (iv), we show how to construct $x_{n+1}$. By (i) we have $1>x_{n}>0$. Since $g,f,h$ are strictly decreasing functions and $g(x)<f(x)<h(x)$ holds true, for every $x\in(0,1)$, it follows that

[TABLE]

Now set $x_{n+1}:=\max\{g^{-1}(f(x_{n})),f^{-1}(h(x_{n}))\}$. Clearly, it holds $0<x_{n+1}<x_{n}$ as well as $R[(x_{n+1},f(x_{n+1})),(x_{n},f(x_{n}))]\subset W(g,h)$ and $R[(x_{n+1},f(x_{n+1})),(x_{n},f(x_{n}))]\not\subset\operatorname{int}(W(g,h))$. So $x_{1},\dots,x_{n+1}$ satisfy (i), (ii) and (iv). Thus we finished the construction of the sequence $\{x_{n}\}$. The sequence $\{y_{n}\}_{n}$ can be constructed similarly; we leave the details to the reader.

Since the sequences $\{x_{n}\}$ and $\{y_{n}\}$ are monotone and bounded, there exists the limits

[TABLE]

We now show that $x=0$. Assume, towards a contradiction, that $x\neq 0$. Clearly, it holds

[TABLE]

Since $(x,f(x))\in\operatorname{int}(W(g,h))$, there exists $\delta>0$ such that for every $y,z\in(x-\delta,x+\delta)$ satisfying $y<z$ we have $R[(y,f(y)),(z,f(z))]\subset\operatorname{int}(W(g,h))$. By (1) it follows that there exists $n\in\mathbb{N}$ such that $x_{n-1}\in(x-\delta,x+\delta)$. Then (iv) implies that $x_{n}\leq x-\delta<x$ which contradicts (2). Hence it holds $x=0$. In a similar way, it can be shown that $y=1$.

Since $f$ is continuous we have

[TABLE]

[TABLE]

Since $x_{1}=y_{1}$ it follows that

[TABLE]

as well as

[TABLE]

and therefore we conclude

[TABLE]

Now Theorem 1.4 implies that for every $n\in\mathbb{N}$ there exist strictly decreasing functions $d_{x,n},d_{y,n}$ that satisfy the following four conditions:

• (A)

$d_{x,n}:[x_{n+1},x_{n}]\to[f(x_{n}),f(x_{n+1})]$,

• (B)

$d_{y,n}:[y_{n},y_{n+1}]\to[f(y_{n+1}),f(y_{n})]$,

• (C)

$\mathcal{H}^{1}(Gr(d_{x,n}))=(x_{n}-x_{n+1})+(f(x_{n+1})-f(x_{n}))$,

• (D)

$\mathcal{H}^{1}(Gr(d_{y,n}))=(y_{n+1}-y_{n})+(f(y_{n})-f(y_{n+1}))$.

Gluing those functions together, we obtain desired function $D:(0,1)\to(0,1)$. Indeed, by (A), (B), (ii) and (iii) we have

[TABLE]

and so $D$ satisfies (a). Using (3), (C) and (D) we conclude that

[TABLE]

and therefore $D$ also satisfies (b). The lemma follows. ∎

We are now ready to prove Theorem 1.6.

Proof of Theorem 1.6.

Clearly, there exist continuous and strictly decreasing bijections $f_{i}:[0,1]\to[0,1]$, $i\in[2k]$, such that

[TABLE]

By Lemma 2.1 we can find for every $i\in[k]$ strictly decreasing functions $D_{i}:(0,1)\to(0,1)$ such that

• ($\alpha$)

$f_{2i}(x)\leq D_{i}(x)\leq f_{2i-1}(x)$ for every $x\in(0,1)$,

• ($\beta$)

$\mathcal{H}^{1}(Gr(D_{i}))=2$.

Now consider the set $A:=\bigcup_{i=1}^{k}Gr(D_{i})$. Since $D_{i}$ is a strictly decreasing function, it follows that $Gr(D_{i})$ is an antichain for every $i\in[k]$, and therefore $A$ is $k$-antichain. Since $Gr(D_{i})\subset(0,1)^{2}$ for every $i\in[k]$, we have $A\subset(0,1)^{2}$. By ($\alpha$) and (4) we have $Gr(D_{i})\cap Gr(D_{j})=\emptyset$ for every $i,j\in[k]$, $i\neq j$. Thus, by ($\beta$) we have

[TABLE]

as desired. ∎

3 Concluding remarks

As mentioned in the introduction, there are several ways to consider Theorem 1.1 in a continuous setting, and an alternative direction has been considered in [16]. It is shown in [16] that given $s\in[0,1]$ and $\beta\geq 0$ there exists a set $A\subset[0,1]^{n}$ that satisfies $\dim_{H}(A)=n-1+s$ and $\mathcal{H}^{s}(A\cap C)\leq\beta$, for all chains $C\subset[0,1]^{n}$. Here, $\dim_{H}(\cdot)$ denotes Hausdorff dimension (see [9, p. 86]). Given this result, the following problem arises naturally.

Problem 3.1** (Mitsis et al. [16]).**

Fix $s\in[0,1]$ and $\beta\geq 0$. Let $A\subset[0,1]^{n}$ be a measurable set such that $\dim_{H}(A)=n-1+s$ and $\mathcal{H}^{s}(A\cap C)\leq\beta$, for all chains $C\subset[0,1]^{n}$. What is a sharp upper bound on $\mathcal{H}^{n-1+s}(A)$?

The case $s=0,\beta=1$ has been considered in [7]. The case $s=0,\beta\in\mathbb{N}$ has been the content of the present article. The case $s=1,\beta\in(0,n]$ has been considered in [16]. The problem remains open for all other values of the parameters $s,\beta$.