A grid generalisation of the Kruskal-Katona theorem

TL;DR

This paper extends the Kruskal-Katona theorem to a grid setting, identifying extremal subsets that minimize the $d$-shadow for sets within multi-valued coordinate spaces.

Contribution

It provides an exact characterization of extremal sets minimizing the $d$-shadow in grid-structured spaces, generalizing classical combinatorial results.

Findings

Sets $oxed{[t]_r^n}$ are extremal for fixed $t$.

Sets with at least $r$ zeros are extremal in the unrestricted case.

The results generalize the Kruskal-Katona theorem to multi-valued grids.

Abstract

For a set $A\subseteq\left[k\right]^{n}=\left\{ 0,\dots,k-1\right\} ^{n}$, we define the $d$-shadow of $A$ to be the set of points obtained by flipping to zero one of the non-zero coordinates of some point in $A$. Let $\left[k\right]_{r}^{n}$ be the set of those points in $\left[k\right]^{n}$ with exactly $r$ non-zero coordinates. Given the size of $A$, how should we choose $A\subseteq\left[k\right]_{r}^{n}$ so as to minimise the $d$-shadow? Note that the case $k=2$ is answered by the Kruskal-Katona theorem. Our aim in this paper is to give an exact answer to this question. In particular, we show that the sets $\left[t\right]_{r}^{n}$ are extremal for every $t$. We also give an exact answer to the 'unrestricted' question when we just have $A\subseteq\left[k\right]^{n}$, showing for example that the set of points with at least $r$ zeroes is extremal for every $r$.

Equations

$$\Gamma \left(\left(a_1,\dots a_n\right)\right)=\left\{\left(a_1-1,a_2\dots ,a_n\right),\left(a_1,a_2-1,\dots ,a_n\right)\dots ,\left(a_1,a_2\dots ,a_n-1\right)\right\}\cap F,$$

$$t_{s}y=y_1\dots y_{s-1}t{y}_s\dots y_{n-1},$$

$$A_{s,t}=\left\{y\in {\left[k\right]}^{n-1}:t_{s}y\in A\right\}.$$

$$d\left(C_s\left(A\right)\right)=d\left(\bigcup _{t=0}^{k-1}{t}_s{B}_{s,t}\right)=\left(\left(\bigcup _{t=1}^{k-1}{0}_s{B}_{s,t}\right)\cup 0_{s}d\left(B_{s,0}\right)\right)\cup \left(\bigcup _{t=1}^{k-1}{t}_{s}d\left(B_{s,t}\right)\right),$$

$$d\left(A\right)=d\left(\bigcup _{t=0}^{k-1}{t}_s{A}_{s,t}\right)=\left(\left(\bigcup _{t=1}^{k-1}{0}_s{A}_{s,t}\right)\cup 0_{s}d\left(A_{s,0}\right)\right)\cup \left(\bigcup _{t=1}^{k-1}{t}_{s}d\left(A_{s,t}\right)\right).$$

$$\left(\bigcup _{t=1}^{k-1}{0}_s{B}_{s,t}\right)\cup 0_{s}d\left(B_{s,0}\right),1_{s}d\left(B_{s,1}\right),\dots ,{\left(k-1\right)}_{s}d\left(B_{s,k-1}\right)$$

$$\left|d\left(\bigcup _{t=0}^{k-1}{t}_s{B}_{s,t}\right)\right|=\left|\left(\bigcup _{t=1}^{k-1}{0}_s{B}_{s,t}\right)\cup 0_{s}d\left(B_{s,0}\right)\right|+\sum _{t=1}^{k-1}\left|t_{s}d\left(B_{s,t}\right)\right|,$$

$$\left|d\left(\bigcup _{t=0}^{k-1}{t}_s{A}_{s,t}\right)\right|=\left|\left(\bigcup _{t=1}^{k-1}{0}_s{A}_{s,t}\right)\cup 0_{s}d\left(A_{s,0}\right)\right|+\sum _{t=1}^{k-1}\left|t_{s}d\left(A_{s,t}\right)\right|.$$

$$\left|t_{s}d\left(A_{s,t}\right)\right|=\left|d\left(A_{s,t}\right)\right|\ge \left|d\left(B_{s,t}\right)\right|=\left|t_{s}d\left(B_{s,t}\right)\right|.$$

$$\left|\left(\bigcup _{t=1}^{k-1}{0}_s{B}_{s,t}\right)\cup 0_{s}d\left(B_{s,0}\right)\right|=\max \left(\left|B_{s,1}\right|,\dots ,\left|B_{s,k-1}\right|,\left|d\left(B_{s,0}\right)\right|\right).$$

$$\left|\left(\bigcup _{t=1}^{k-1}{0}_s{A}_{s,t}\right)\cup 0_{s}d\left(A_{s,0}\right)\right|\ge \max \left(\left|A_{s,1}\right|,\dots ,\left|A_{s,k-1}\right|,\left|d\left(A_{s,0}\right)\right|\right)$$

$$\left|\bigcup _{t=1}^{k-1}{0}_s{A}_{s,t}\cup 0_{s}d\left(A_{s,0}\right)\right|\ge \left|\bigcup _{t=1}^{k-1}{0}_s{B}_{s,t}\cup 0_{s}d\left(B_{s,0}\right)\right|.$$

$$\left|d\left(A\right)\right|\ge \left|d\left(C_s\left(A\right)\right)\right|,$$

$$X=A\setminus A_0=\left\{x_1{x}_2\in A:x_1\ne 0\text{ and }{x}_2\ne 0\right\}.$$

$$d\left(X\right)=\left\{0x_1,\dots ,0x_r,y_{1}0,\dots ,y_{s}0\right\}.$$

$$X\subseteq \left\{y_j{x}_i:1\le j\le s,1\le i\le r\right\},$$

$$\left|d\left(A\right)\right|\ge 1+\lceil \sqrt{4\left(\left|A\right|-2k+1\right)}\rceil .$$

$$\left|d\left(C\right)\right|\le 1+\left(r+1\right)+r=1+\lceil \sqrt{4\left(\left|A\right|-2k+1\right)}\rceil ,$$

$$\left|d\left(C\right)\right|\le 1+2\left(r+1\right)=1+\lceil \sqrt{4\left(\left|A\right|-2k+1\right)}\rceil .$$

$$x\le y,y\in A,x_i=y_i\text{ for some }i\Rightarrow x\in A.$$

$$B_s=\left\{x\in {\left[k\right]}^n:\left|R_0\left(x\right)\right|=s\right\},$$

$$\left|B_{\ge r+1}\cup \left({\left[s\right]}^n\cap B_r\right)\right|<\left|A\right|\le \left|B_{\ge r+1}\cup \left({\left[s+1\right]}^n\cap B_r\right)\right|,$$

$$B_{\ge r+1}\cup \left({\left[s\right]}^n\cap B_r\right)\subseteq A\subseteq B_{\ge r}\cup \left(\left[\left({\left[s+1\right]}^n\cap B_r\right)\right]\right).$$

$$z_{t}=\begin{cases}\begin{array}[]{cc}v_{t}&\text{if }t\neq j\\ x_{j}&\text{if }t=j\end{array}.\end{cases}$$

$$z_{t}=\begin{cases}\begin{array}[]{cc}v_{t}&\text{if }t\neq p\\ 1&\text{if }t=p\end{array}.\end{cases}$$

$$C=B_{\ge r+1}\cup \left({\left[s\right]}^n\cap B_r\right)\cup \left(\bigcup _{S\in \mathcal{A},S{<}_{bin}{T}_1}{C}_S\right)\cup \left\{y:y\le x,y\in T_1\right\}.$$

$$X=B_{\ge 1}\cup \left({\left[s\right]}^n\cap B_0\right)\cup \bigcup _{S{\le }_{bin}{T}_1}{C}_S.$$

$$\left|d\left(B\right)\right|=\left|{\left[nk\right]}_{\le r-1}^n\right|+\left|d\left(A\right)\right|$$

$$\left|d\left(X\right)\right|=\left|{\left[nk\right]}_{\le r-1}^n\right|+\left|d\left(C\right)\right|,$$

$$d\left(A\right)\cap {\left[nk\right]}_{\le r-1}^n=d\left(C\right)\cap {\left[nk\right]}_{\le r-1}^n=\varnothing .$$

Full text

Our aim in this paper is to give an exact answer to this question. In particular, we show that the sets $\left[t\right]_{r}^{n}$ are extremal for every $t$. We also give an exact answer to the ’unrestricted’ question when we just have $A\subseteq\left[k\right]^{n}$, showing for example that the set of points with at least $r$ zeroes is extremal for every $r$.

1 Introduction

Let $\left\{0,1\right\}^{n}$ be the set of all sequences $x=x_{1}\dots x_{n}$ of length $n$ with $x_{i}\in\left\{0,1\right\}$ for all $i$. For $A\subseteq\left\{0,1\right\}^{n}$, define the lower shadow of $A$ to be the set of points obtained from any of its point by changing one coordinate of any point $x\in A$ from $1$ to [math], and denote it by $\partial^{-}A$. Define the rank of a point $x$ to be $w\left(x\right)=\left|\left\{i:x_{i}=1\right\}\right|$, and define $\left\{0,1\right\}_{r}^{n}=\left\{x\in\left\{0,1\right\}^{n}:w\left(x\right)=r\right\}$. Note that the lower shadow operator decreases the rank of a point by one.

One can similarly define the *upper shadow *of $A$ to be the set of points obtained from any of its point by changing one coordinate of any point $x\in A$ from [math] to $1$, and denote the upper shadow by $\partial^{+}A$. Again, it is clear that the upper shadow operator increases the rank of a point by one.

For a given $r$, it is natural to ask how should one choose $A\subseteq\left\{0,1\right\}_{r}^{n}$ of a given size in order to minimise the size of the lower shadow. This question was answered by Kruskal [6] and Katona [4]. Define the colexicographic order $\leq_{c}$ on $\left\{0,1\right\}_{r}^{n}$ as follows. For points $x,y\in\left\{0,1\right\}_{r}^{n}$, let $X=\left\{i:x_{i}=1\right\}$ and $Y=\left\{i:y_{i}=1\right\}$, and we set $x\leq_{c}y$ if $X=Y$ or $\max\left(X\Delta Y\right)\in Y$. The Kruskal-Katona theorem states that for a set $A\subseteq\left\{0,1\right\}_{r}^{n}$ of a given size, the lower shadow of $A$ is minimised when $A$ is chosen to be an initial segment of colexicographic order.

Define the lexicographic order $\leq_{l}$ on $\left\{0,1\right\}_{r}^{n}$ by setting $x\leq_{l}y$ if $X=Y$ or $\min\left(X\Delta Y\right)\in X$, where $X$ and $Y$ are defined as above. For $x\in\left\{0,1\right\}^{n}$, define $x^{c}\in\left\{0,1\right\}^{n}$ to be the point obtained by taking $\left(x^{c}\right)_{i}=1-x_{i}$ for all $i$. For a given $A\subseteq\left\{0,1\right\}_{r}^{n}$, define $\overline{A}$ by setting $\overline{A}=\left\{x^{c}:x\in A\right\}$. Note that we have $\left|\overline{A}\right|=\left|A\right|$ and $\overline{A}\subseteq\left\{0,1\right\}_{n-r}^{n}$. It is also easy to verify that $\partial^{+}A=\overline{\partial^{-}\overline{A}}$, and hence it follows that $\left|\partial^{+}A\right|=\left|\partial^{-}\overline{A}\right|$. Thus the question on minimising the size of the upper shadow in $\left\{0,1\right\}_{r}^{n}$ can be transformed into a question on minimising the size of the lower shadow in $\left\{0,1\right\}_{n-r}^{n}$. Hence the Kruskal-Katona theorem implies that the initial segments of the lexicographic order minimise the upper shadow.

There are many natural generalisations of the lower and upper shadow for points in $\left[k\right]^{n}$, and one such generalisation can be obtained in the following way. Let $\left[k\right]=\left\{0,\dots,k-1\right\}$ and $\left[k\right]^{n}=\left\{0,\dots,k-1\right\}^{n}$. Define the rank of a point $x\in\left[k\right]^{n}$ to be $w\left(x\right)=\left|\left\{i:x_{i}\geq 1\right\}\right|$. For $0\leq r\leq n$ set $\left[k\right]_{r}^{n}=\left\{x\in\left[k\right]^{n}:w\left(x\right)=r\right\}$.

Define the $d$-shadow of a point $x\in\left[k\right]^{n}$ to be the set of those points obtained from $x$ by flipping one of the coordinates of $x$ which is in $\left\{1,\dots,k-1\right\}$ to [math]. We denote the $d$ shadow of a point $x$ by $d\left(\left\{x\right\}\right)$. For $A\subseteq\left[k\right]^{n}$ we define the $d$-shadow of $A$ by setting $d\left(A\right)=\bigcup_{x\in A}d\left(\left\{x\right\}\right)$. For example, $d\left(\left\{012\right\}\right)=\left\{002,010\right\}$ and $d\left(\left\{000\right\}\right)=\emptyset$. Note that the rank of a point in $d\left(\left\{x\right\}\right)$ is one lower than the rank of $x$. It is clear that $d$ agrees with $\partial^{-}$ when $k=2$, so this operator indeed generalises the lower shadow.

There is a superficial resemblance to a result of Clements, as we now describe. Define the $d^{+}$*-shadow *of a point $x\in\left[k\right]^{n}$ by setting $d^{+}\left(\left\{x\right\}\right)$ to be the set of those points obtained from $x$ by changing one of the coordinates of $x$ which equals [math] to any number in $\left\{1,\dots,k-1\right\}$, and we set $d\left(A\right)=\bigcup_{x\in A}d\left(\left\{x\right\}\right)$. Again it is clear that the rank of a point in $d^{+}\left(\left\{x\right\}\right)$ is one larger than the rank of $x$.

Clements [2] found an order in $\left[k\right]_{r}^{n}$ whose initial segments have minimal $d^{+}$-shadow. Recall that the ordinary lower and upper shadow can be related to each other by using the fact that $\partial^{+}A=\overline{\partial^{-}\overline{A}}$. However, for $k\geq 3$, it is clear that there is no similar natural relation between $d^{+}$- and $d$-shadows. That is, given the Clements’ result for $d^{+}$-shadow, there seems to be no way to deduce results related to $d$-shadow.

There is also a superficial resemblance to the Clements-Lindström Theorem [3]. Let $k_{1},\dots,k_{n}$ be integers such that $1\leq k_{1}\leq\dots\leq k_{n}$, and let $F$ be the set of all integer sequences $\left(a_{1},\dots,a_{n}\right)$ with $0\leq a_{i}\leq k_{i}$ for all $i$. Define the shadow operator $\Gamma$ by setting

[TABLE]

and $\Gamma\left(A\right)=\bigcup_{a\in A}\Gamma\left(a\right)$ for $A\subseteq F$. Let $F_{r}$ be the set of those sequences $\left(a_{1},\dots,a_{n}\right)\in F$ with $\sum_{i=1}^{n}a_{i}=r$. Generalise the lexicographic order by writing $\left(a_{1},\dots,a_{n}\right)<_{l}\left(b_{1},\dots,b_{n}\right)$ if there exists $i$ such that $a_{j}=b_{j}$ for all $j<i$ and $a_{i}<b_{i}$. The Clements-Lindström theorem states the initial segments of the lexicographic order minimise the size of the $\Gamma$-shadow on $F_{r}$.

The aim of this paper is to find an order on $\left[k\right]_{r}^{n}$ whose initial segments have minimal $d$-shadow. In fact, we do this by first solving the unrestricted version, i.e. we find an order on $\left[k\right]^{n}$ whose initial segments have minimal $d$-shadow. Once we have proved the result for $\left[k\right]^{n}$, the result for $\left[k\right]_{r}^{n}$ follows easily.

We start by defining the order, whose initial segments minimise the $d$-shadow for the unrestricted version. For each $i$ define $R_{i}\left(x\right)=\left\{j:x_{j}=i\right\}$. For fixed $k$, define an order $\leq$ on $\left[k\right]^{n}$ by setting $x\leq y$ if $x=y$ or one of the following conditions holds

$\left|R_{0}\left(x\right)\right|>\left|R_{0}\left(y\right)\right|$ 2. 2.

$\left|R_{0}\left(x\right)\right|=\left|R_{0}\left(y\right)\right|$ and for the largest $i$ satisfying $R_{i}\left(x\right)\neq R_{i}\left(y\right)$ we have $\max\left(R_{i}\left(x\right)\Delta R_{i}\left(y\right)\right)\in R_{i}\left(y\right)$.

As usual, we say that $x<y$ if $x\leq y$ and $x\neq y$. Now we are ready to state the unrestricted version of our theorem.

Theorem 1**.**

Let $A$ be a subset of $\left[k\right]^{n}$, and let $B$ be an initial segment of the $\leq$-order in $\left[k\right]^{n}$ with $\left|B\right|=\left|A\right|$. Then $\left|d\left(A\right)\right|\geq\left|d\left(B\right)\right|$.

The proof of Theorem 1 is an inductive proof. As the proof of Theorem 1 is rather long, we split it into four subsections. In the first subsection, we introduce certain codimension-1 compression operators, and we prove that they cannot increase the size of the shadow. In particular, because of the compression operators, it suffices to prove Theorem 1 for those sets $A$ that are stable under the compression operators.

Compression operators have been much utilised in other papers as well, see e.g. [1, 5]. For example, there are very straightforward proofs using compression operators for Harper’s vertex-isoperimetric inequality on the hypercube and for the edge-isoperimetric inequality on the hypercube. In these examples it is very straightforward to prove the desired isoperimetric inequality for the sets that are stable under the compression operators. However, proving the inequality for the sets that are stable under the compression operators is highly nontrivial in our main theorem. In particular, the main part of the proof consists of dealing with the sets that are stable under the compression operators and proving that Theorem 1 holds for such sets.

Even though $n=1$ is the base case in the proof, it turns out to be convenient to consider the case $n=2$ individually as well. Hence we consider the case $n=2$ in the second subsection. This special case turns out to be reasonably straightforward when restricted to the sets that are stable under the compression operators.

We start the proof of the general case $n\geq 3$ by making some observations on the structure of the sets that are stable under the compression operators. As an example, we prove that it suffices to restrict our attention to the sets $A$ for which there exists $r$ such that $B_{\geq r}\subseteq A\subseteq B_{\geq r+1}$ and which are also stable under the compression operators. Hence we may write $A$ as $A=B_{\geq r}\cup D$ for some $D\subseteq B_{r+1}$, and the fact that $A$ is stable under the compression operators also restricts the structure of $D$.

Let $A$ be a set of the form $A=B_{\geq r}\cup D$ for some $D\subseteq B_{r+1}$. Note that we have $\left|d\left(A\right)\right|=\left|B_{\geq r+1}\right|+\left|d\left(D\right)\right|$, and hence we can focus on analysing $d\left(D\right)$. The fourth subsection is dedicated to analysing $d\left(D\right)$ by using previous and some new structural observations. At this stage of the proof, we need to split into smaller subcases depending on the size and the structure of $A$.

Denote the restriction of the $\leq$-order on $\mathbb{N}_{r}^{n}=\left\{x=x_{1}\dots x_{n}:w\left(x\right)=r\right\}$ also by $\leq$, where we define $\mathbb{N}=\left\{0,1,\dots\right\}$. This is well-defined, as it is easy to check that $\left[m\right]_{r}^{n}$ is an initial segment of the restriction of $\leq$ on $\left[k\right]_{r}^{n}$ for all $k\geq m$, and the orders coincide on $\left[m\right]_{r}^{n}$. Now we can state our main theorem.

Theorem 2**.**

Let $A$ be a subset of $\left[k\right]_{r}^{n}$, and let $B$ be an initial segment of the $\leq$-order in $\left[k\right]_{r}^{n}$ with $\left|B\right|=\left|A\right|$. Then $\left|d\left(A\right)\right|\geq\left|d\left(B\right)\right|$.

We end this section by introducing some notation that we use throughout the paper. We write $\left[n\right]^{\left(r\right)}=\left\{A\subseteq\left[n\right]:\left|A\right|=r\right\}$ and $\left[n\right]^{\left(\leq r\right)}=\left\{A\subseteq\left[n\right]:\left|A\right|\leq r\right\}$. For convenience, we often write $B_{r}=\left[k\right]_{n-r}^{n}=\left\{x\in\left[k\right]^{n}:\left|R_{0}\left(x\right)\right|=r\right\}$ and $B_{\geq r}=\bigcup_{i=r}^{n}B_{i}$ for the set of sequences with exactly $r$ zeroes and at least $r$ zeroes respectively. Note that $B_{r}$ depends on the ground set $\left[k\right]$, but since the value of $k$ is often clear the dependence is not highlighted. For $x\in\left[k\right]^{n}$, we write $d\left(x\right)$ instead of $d\left(\left\{x\right\}\right)$.

For $a_{i}\in\left[k\right]$ and for positive integers $t_{i}\in\mathbb{N}$ we define $\left(t_{1}\cdot a_{1}\right)\left(t_{2}\cdot a_{2}\right)\dots\left(t_{r}\cdot a_{r}\right)$ to represent the point $a_{1}\dots a_{1}a_{2}\dots a_{2}\dots a_{r}\dots a_{r}$, which has $t_{1}$ $a_{1}$’s immediately followed by $t_{2}$ $a_{2}$’s, and so on. For example, we have $\left(3\cdot 0\right)\left(2\cdot 4\right)56=0004456$.

For $y=y_{1}\dots y_{n-1}\in\left[k\right]^{n-1}$, $s\in\left[n\right]$ and $t\in\left[k\right]$ define

[TABLE]

and for $Y\subseteq\left[k\right]^{n-1}$ we write $t_{s}Y=\left\{t_{s}y\,:\,y\in Y\right\}$. Note that we have $t_{s}x\leq t_{s}y$ if and only if $x\leq y$.

Finally define the binary order $\leq_{bin}$ on ${\cal P}\left(\mathbb{N}\right)$ by setting $X\leq_{b}Y$ if $X=Y$ or $\max\left(X\Delta Y\right)\in Y$. We write $X<_{bin}Y$ if $X\leq_{bin}Y$ and $X\neq Y$. Note that the second condition in the definition of the $\leq$-order is equivalent to saying that for the largest $i$ for which $R_{i}\left(x\right)\neq R_{i}\left(y\right)$ we have $R_{i}\left(x\right)<_{bin}R_{i}\left(y\right)$.

2 Proof of Theorem 1

For convenience, we say that a set $A\subseteq\left[k\right]^{n}$ is *extremal *if the size of $d\left(A\right)$ is minimal among all subsets of $\left[k\right]^{n}$ of the same size. The proof of Theorem 1 is an inductive proof. Note that the case $n=1$ is trivial. Throughout the Section 2 we assume that Theorem 1 holds for $\left[k\right]^{n-1}$, and our aim is to prove it for $\left[k\right]^{n}$.

2.1 The compression operators

For $A\subseteq\left[k\right]^{n}$, $t\in\left[k\right]$ and $s\in\left[n\right]$ define

[TABLE]

Let $B_{s,t}\subseteq\left[k\right]^{n-1}$ be an initial segment of the $\leq$-order with $\left|B_{s,t}\right|=\left|A_{s,t}\right|$, and set $C_{s,t}=t_{s}B_{s,t}$. Define the $C_{s}$*-compression *of $A$ by setting $C_{s}\left(A\right)=\bigcup_{t=0}^{k-1}C_{s,t}$. We start by proving that $C_{s}$-compressions cannot increase the size of the $d$-shadow.

Claim 1*.*

For all $A\subseteq\left[k\right]^{n}$ and $s\in\left[n\right]$ we have $\left|C_{s}\left(A\right)\right|=\left|A\right|$ and $\left|d\left(A\right)\right|\geq\left|d\left(C_{s}\left(A\right)\right)\right|$.

Proof.

For a given $s$, note that the sets $\left(C_{s,t}:t\in\left[k\right]\right)$ are pairwise disjoint, as every $x\in C_{s,t}$ satisfies $x_{s}=t$. Since $\left|C_{s,t}\right|=\left|A_{s,t}\right|$ for all $t\in\left[k\right]$, it follows that $\left|C_{s}\left(A\right)\right|=\left|A\right|$.

Note that we have

[TABLE]

and similarly we have

[TABLE]

Observe that the $k$ sets

[TABLE]

are pairwise disjoint sets as their $s^{th}$ coordinates disagree pairwise. Hence we have

[TABLE]

and similarly we have

[TABLE]

Since Theorem 1 holds for $\left[k\right]^{n-1}$, it follows that

[TABLE]

Note that the $d$-shadow of an initial segment is also an initial segment, and initial segments are nested. Hence we have

[TABLE]

Combining the trivial estimate

[TABLE]

with (5), (6) and the fact that $\left|A_{s,i}\right|=\left|B_{s,i}\right|$ for all $0\leq i\leq k-1$, it follows that

[TABLE]

Thus pairing up the terms in (3) and (4) in the natural way and applying (5) and (8) gives that

[TABLE]

which completes the proof. ∎

We say that $T\subseteq\left[k\right]^{n}$ is compressed if $C_{s}\left(T\right)=T$ holds for all $s\in\left\{1,\dots,n\right\}$. We now make the standard observation that it suffices to prove Theorem 1 for compressed sets.

Claim 2*.*

Let $A$ be a subset of $\left[k\right]^{n}$. Then there exists a compressed set $B\subseteq\left[k\right]^{n}$ with $\left|B\right|=\left|A\right|$ and $\left|d\left(A\right)\right|\geq\left|d\left(B\right)\right|$.

Proof.

Consider a sequence $\left(A_{m}\right)$ with $A_{0}=A$ obtained as follows. Given $A_{m}$, if there exists $s\in\left[n\right]$ for which $C_{s}\left(A_{m}\right)\neq A_{m}$ we set $A_{m+1}=C_{s}\left(A_{m}\right)$. Otherwise we set $A_{m+1}=A_{m}$.

Let $K_{i}$ be the $i^{th}$ set in $\left[k\right]^{n}$ with respect to the $\leq$-order. Define $f\left(A\right)=\sum_{i=1}^{k^{n}}i\mathbb{I}\left\{K_{i}\in A\right\}$, where $\mathbb{I}\left\{K_{i}\in A\right\}$ denotes the indicator function of the event $K_{i}\in A$. By the construction of the compression operator $C_{s}$, it is easy to verify that we have $f\left(C_{s}\left(A\right)\right)\leq f\left(A\right)$ for all $s\in\left\{1,\dots,n\right\}$, and for a given $s$ the equality holds if and only if we have $C_{s}\left(A\right)=A$. Since $f\left(A\right)$ is always a non-negative integer, it follows that the sequence $f\left(A_{m}\right)$ is eventually constant. Thus there exists $r$ for which we have $f\left(C_{s}\left(A_{r}\right)\right)=f\left(A_{r}\right)$ for all $s$, and hence we have $C_{s}\left(A_{r}\right)=A_{r}$ for all $s$. Therefore $A_{r}$ is compressed.

By Claim 1, if follows that we have $\left|d\left(A_{i}\right)\right|\geq\left|d\left(A_{i+1}\right)\right|$ for all $0\leq i\leq r-1$. Hence it follows that $\left|d\left(A\right)\right|\geq\left|d\left(A_{r}\right)\right|$, which completes the proof. ∎

From now on, let $A\subseteq\left[k\right]^{n}$ denote a compressed set and let $C$ denote the initial segment of $\left[k\right]^{n}$ of the same size. By Claim 2, it suffices to prove that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$.

2.2 The special case $n=2$.

Before moving on to the general case, we prove Theorem 1 when $n=2$. This turns out to be convenient as $n=2$ is too small in one part of the general argument.

Claim 3*.*

When $n=2$, Theorem 1 holds for a compressed set $A$.

Proof.

The claim is trivial if $\left|A\right|=1$. Let $C$ denote the initial segment of $\left[k\right]^{n}$ of size $\left|A\right|$. If $2\leq\left|A\right|\leq 2k-1$, then $C$ is a subset of $B_{\geq 1}$ and hence it follows that $d\left(C\right)=\left\{00\right\}$. Thus it is evident that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$.

Now suppose that $\left|A\right|\geq 2k$. We write $A$ as $A=A_{0}\cup X$, where $A_{0}=A\cap B_{\geq 1}$ and

[TABLE]

Since $\left|B_{\geq 1}\right|=2k-1$, it follows that $\left|X\right|\geq\left|A\right|-2k+1$, and in particular $X$ is non-empty.

Let $x_{1},\dots,x_{r}\in\left\{1,\dots,k-1\right\}$ and $y_{1},\dots,y_{s}\in\left\{1,\dots,k-1\right\}$ be chosen so that

[TABLE]

Then we certainly have

[TABLE]

which implies that $\left|X\right|\leq rs$. Since for non-negative integers $r$ and $s$ we have $r+s\geq\left\lceil\sqrt{4rs}\right\rceil$, it follows that $\left|d\left(X\right)\right|\geq\left\lceil\sqrt{4\left|X\right|}\right\rceil$.

Since $A$ is compressed and $\left|A\right|>1$, it follows that $A$ contains a point $x_{1}x_{2}\neq 00$ with $x_{1}=0$ or $x_{2}=0$. In particular, we must have $00\in d\left(A\right)$. Hence it follows that $d\left(A\right)=\left\{00\right\}\cup d\left(X\right)$. As $\left|X\right|\geq\left|A\right|-2k+1$, we have

[TABLE]

If $r^{2}+2k\leq\left|A\right|\leq r^{2}+r+2k-1$ for some $1\leq r\leq k-2$, it is easy to verify that $C$ satisfies $C\subseteq B_{\geq 1}\cup\left\{x_{1}x_{2}:1\leq x_{1}\leq r+1,\,1\leq x_{2}\leq r\right\}$. In particular, we have

[TABLE]

and hence it follows that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$. If $r^{2}+r+2k\leq\left|A\right|\leq\left(r+1\right)^{2}+2k-1$ for some $1\leq r\leq k-2$, it is easy to verify that $C$ satisfies $C\subseteq B_{\geq 1}\cup\left\{x_{1}x_{2}:1\leq x_{1},x_{2}\leq r+1\right\}$. Again, we have

[TABLE]

Thus it follows that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$, which completes the proof. ∎

2.3 General observations

From now on we suppose that $n\geq 3$, and our aim is to prove Theorem 1 for a compressed set $A\subseteq\left[k\right]^{n}$. As in the previous section, the proof is trivial if $\left|A\right|=1$ or $2\leq\left|A\right|\leq n\left(k-1\right)+1$, as in these cases the $d$-shadow of the initial segment has size [math] or $1$ respectively. Thus we may assume that $\left|A\right|\geq n\left(k-1\right)+2$.

We say that $B\subseteq\left[k\right]^{n}$ is a down-set if for any $y\in B$ and for a point $x\in\left[k\right]^{n}$ satisfying $x_{j}\leq y_{j}$ for all $j$ we have $x\in B$. In this subsection we make some observations on the structure of compressed set $A$. We start by proving that $A$ is a down-set, and in fact this follows from a slightly stronger statement. Secondly, recall that for an initial segment $C$ there exists $r$ satisfying $B_{\geq r}\subseteq C\subseteq B_{\geq r+1}$. For compressed sets the same conclusion does not necessarily hold, but as a second structural claim we prove that for compressed sets there exists $r$ for which $d\left(B_{\geq r}\right)\subseteq d\left(A\right)\subseteq d\left(B_{\geq r+1}\right)$. Since we only consider the size of the $d$-shadow of $A$, in a sense this is ’equally good’ for our purposes as having $B_{\geq r}\subseteq A\subseteq B_{\geq r+1}$. Note that the second observation allows us to focus only on $d\left(A\cap B_{r+1}\right)$.

Before proving that $A$ and $d\left(A\right)$ are down-sets, we start by a straightforward observation that turns out to be very useful. Given points $x,y$ with $x\leq y$ and $y\in A$, suppose that there exists $i$ for which $x_{i}=y_{i}$, and for convenience set $t=x_{i}$. Since $A$ is compressed, it follows that $A_{i,t}$ is an initial segment. Let $x^{\prime}$ and $y^{\prime}$ be the points obtained from $x$ and $y$ by removing the $i^{th}$ coordinate, and note that $y^{\prime}\in A_{i,t}$. Since $x\leq y$ and $x_{i}=y_{i}$, it follows that $x^{\prime}\leq y^{\prime}$. Since $A$ is compressed and $y^{\prime}\in A_{i,t}$, it follows that $x^{\prime}\in A_{i,t}$ and hence we also have $x\in A$. In summary, we have proved that

[TABLE]

This turns out to be a very useful fact that we will use throughout the rest of the proof. As a simple consequence, we now obtain that $A$ and $d\left(A\right)$ are down-sets.

Claim 4*.*

Let $A\subseteq\left[k\right]^{n}$ be a compressed set. Then $A$ is a down-set, and furthermore $d\left(A\right)$ is also a down-set.

Proof.

Let $y\in A$ and let $x\in\left[k\right]^{n}$ be a point satisfying $x_{i}\leq y_{i}$ for all $i$. Let $z\in\left[k\right]^{n}$ be obtained by taking $z_{1}=x_{1}$ and $z_{s}=y_{s}$ for all $s\geq 2$, and note that we certainly have $x\leq z\leq y$. Since $y_{2}=z_{2}$ and $x_{1}=z_{1}$, two applications of (10) implies that $z\in A$ and $x\in A$, which completes the proof of the first part.

Let $y\in d\left(A\right)$ and let $x\in\left[k\right]^{n}$ for which $x_{i}\leq y_{i}$ for all $i$. Choose $v\in A$ satisfying $y\in d\left(v\right)$, and let $a$ be the unique index for which $v_{a}\neq 0$ but $y_{a}=0$. Let $u$ be the point obtained by setting $u_{a}=v_{a}$ and $u_{j}=x_{j}$ for all $j\neq a$. Then for $j\neq a$ we have $u_{j}=x_{j}\leq y_{j}=v_{j}$. Since we also have $u_{a}=v_{a}$, the first part implies that $u\in A$. Since $x_{a}\leq y_{a}=0$, we must have $x_{a}=0$. Hence we have $x\in d\left(u\right)\subseteq d\left(A\right)$, which completes the proof. ∎

Before moving on to the second structural fact, we introduce some notation that will be used throughout the rest of the paper. For $x=x_{1}\dots x_{n}\in\mathbb{N}^{n}$, set $m\left(x\right)=\max\left(x_{1},\dots,x_{n}\right)$ and recall that for all $i\in\mathbb{N}$ we defined $R_{i}\left(x\right)=\left\{j:x_{j}=i\right\}$. Define $c\left(x\right)=\left(m\left(x\right),R_{m\left(x\right)}\left(x\right),\left|R_{0}\left(x\right)\right|\right)$. That is, the first coordinate of $c\left(x\right)$ is $\max\left(x_{1},\dots,x_{n}\right)$, the second coordinate is the set of all positions where this maximum is attained and the last coordinate is the number of $x_{i}$’s that equal 0. Define the component of $x$ by $C_{x}=\left\{y\in\left[k\right]^{n}:c\left(y\right)=c\left(x\right)\right\}$.

Note that we have $x\in\left[k\right]^{n}$ if and only if $C_{x}\subseteq\left[k\right]^{n}$, as every $y\in C_{x}$ has the same maximum value of a coordinate. Moreover, since every $y\in C_{x}$ also shares the same positions of the maximum coordinates and same number of coordinates that equal 0, it follows that for any other class $C_{z}$ either all the points of $C_{x}$ occur before the points of $C_{z}$ with respect to the $\leq$-order, or all the points of $C_{x}$ occur after the points of $C_{z}$. Hence we can order the classes inside $\left[k\right]^{n}$ as $C_{1},\dots,C_{m}$ such that for any $i\neq j$, $x\in C_{i}$ and $y\in C_{j}$ we have $x\leq y$ if and only if $i<j$.

Note that for all $x,y\in C_{i}$ there exists $r$ such that $x_{r}=y_{r}$. Indeed, any $r\in R_{m\left(x\right)}\left(x\right)=R_{m\left(y\right)}\left(y\right)$ works. Hence (10) implies that for any $i$ we have $A\cap C_{i}=\emptyset$ or there exists $y_{i}\in C_{i}$ such that $A\cap C_{i}=\left\{x\in C_{i}\,:\,x\leq y_{i}\right\}$.

For fixed $s$ and $t$, the classes of the form $\left(s,\,A,\,t\right)$ for $A\in\left[n\right]^{\left(\leq n-t\right)}$ occur consecutively with respect to the $\leq$-order. Furthermore, from the definition of the $\leq$-order it is easy to verify that these classes occur in the order induced by the binary order on $\left[n\right]^{\left(\leq n-t\right)}$. In particular, if $A_{i}$ and $A_{i+1}$ are two consecutive sets in binary order with $\left|A_{i}\right|\leq n-t$ and $\left|A_{i+1}\right|\leq n-t$, then $\left(s,\,A_{i},t\right)$ immediately precedes $\left(s,\,A_{i+1},\,t\right)$ in the order of classes.

Recall that $B_{s}$ is defined to be the set of those points with exactly $s$ coordinates that equal 0, i.e.

[TABLE]

and $B_{\geq s}=\bigcup_{i=s}^{n}B_{i}$. Note that if $X$ is an initial segment of the $\leq$-order, there exists $r$ such that $B_{\geq r+1}\subseteq X\subset B_{\geq r}$. Our next aim is to prove that for any compressed set $A$ there exists $r$ so that $d\left(B_{r}\right)\subseteq d\left(A\right)\subseteq d\left(B_{r+1}\right)$.

Claim 5*.*

Let $A$ be a compressed set, and let $0\leq p\leq n$ be the minimal index for which $A\cap B_{p}\neq\emptyset$. Then we have $d\left(B_{\geq p+1}\right)\subseteq d\left(A\right)\subseteq d\left(B_{p}\right)$. In particular, if $r$ is chosen such that $\left|B_{\geq r+1}\right|<\left|A\right|\leq\left|B_{\geq r}\right|$, it suffices to prove Theorem 1 for compressed sets $A$ that satisfy $B_{\geq r+1}\subseteq A\subseteq B_{\geq r}$.

Proof.

Let $p$ be the minimal index for which $A\cap B_{p}\neq\emptyset$, and let $u$ be the minimal point under the $\leq$-order in $B_{p}$. Since $A$ is a down-set and $u\in B_{p}$, it follows that every coordinate of $u$ must be either [math] or $1$. Hence there exists a set $X$ of size $n-p$ so that $u_{i}=\mathbb{I}\left\{i\in X\right\}$.

Let $x\in B_{\geq p+2}=d\left(B_{\geq p+1}\right)$. If $R_{0}\left(x\right)\cap X\neq\emptyset$, choose $i\in R_{0}\left(x\right)\cap X$ and consider the point $y$ obtained by taking $y_{j}=x_{j}$ for $j\neq i$ and $y_{i}=1$. Since $i\in X$, it follows that $u_{i}=1=y_{i}$. On the other hand, since $\left|R_{0}\left(y\right)\right|=p+2-1=p+1>p=\left|R_{0}\left(u\right)\right|$ it follows that $y\leq u$. Hence (10) implies that $y\in A$, and thus we have $x\in d\left(A\right)$.

If $R_{0}\left(x\right)\cap X=\emptyset$, choose any $i\in R_{0}\left(x\right)$ and again consider the point $y$ obtained by taking $y_{j}=x_{j}$ for $j\neq i$ and $y_{i}=1$. Let $j\in R_{0}\left(x\right)\setminus\left\{i\right\}$, and note that such $j$ exists as $\left|R_{0}\left(x\right)\right|\geq p+2\geq 2$. As $j\not\in X$, we have $u_{j}=0=x_{j}=y_{j}$. Similarly as in the first case, we have $\left|R_{0}\left(y\right)\right|>\left|R_{0}\left(u\right)\right|$ and thus it follows that $y\leq u$. Hence (10) implies that $y\in A$, and therefore we have $x\in d\left(A\right)$, which completes the proof of the first part.

For the second part, suppose that $A^{\prime}$ is a compressed set satisfying $\left|B_{\geq r+1}\right|<\left|A^{\prime}\right|\leq\left|B_{\geq r}\right|$ for some $r$. Let $C$ denote the initial segment of the $\leq$-order with $\left|C\right|=\left|A^{\prime}\right|$, and assume that Theorem 1 holds for any compressed set $A\subseteq\left[k\right]^{n}$ satisfying $\left|A\right|=\left|A^{\prime}\right|$ and $B_{\geq r+1}\subseteq A\subseteq B_{\geq r}$. Let $p$ denote the least integer for which $A^{\prime}\cap B_{p}\neq\emptyset$. Since $\left|A^{\prime}\right|>\left|B_{\geq r+1}\right|$, it follows that $p\leq r$.

If $p<r$, the first part implies that $d\left(B_{\geq r}\right)\subseteq d\left(B_{\geq p+1}\right)\subseteq d\left(A^{\prime}\right)$, and since $C\subseteq B_{\geq r}$ it certainly follows that $\left|d\left(A^{\prime}\right)\right|\geq\left|d\left(C\right)\right|$. If $p=r$, the choice of $p$ implies that $A^{\prime}\subseteq B_{\geq r}$, and the first part implies that $d\left(B_{\geq r+1}\right)\subseteq d\left(A^{\prime}\right)$. Let $D^{\prime}=A^{\prime}\cap B_{r}$, and note that the last two observations imply that $d\left(A^{\prime}\right)=d\left(B_{\geq r+1}\right)\cup d\left(D^{\prime}\right)$. Let $C^{\prime}=C\cap B_{r}$, and note that since $\left|A^{\prime}\right|=\left|C\right|$ and $B_{\geq r+1}\subseteq C$, it follows that $\left|C^{\prime}\right|\leq\left|D^{\prime}\right|$. Let $D$ be the set of $\left|C^{\prime}\right|$ smallest points of $D^{\prime}$ under the $\leq$-order, and let $A=B_{\geq r+1}\cup D$. It is easy to see that $\left|A\right|=\left|C\right|=\left|A^{\prime}\right|$ and that $A$ is compressed, and by the construction of $A$ it follows that $B_{\geq r+1}\subseteq A\subseteq B_{\geq r}$. Furthermore, since $D\subseteq D^{\prime}$ and $d\left(B_{\geq r+1}\right)\subseteq d\left(A^{\prime}\right)$, it follows that $d\left(A\right)\subseteq d\left(A^{\prime}\right)$. In particular, we must have $\left|d\left(A^{\prime}\right)\right|\geq\left|d\left(A\right)\right|$, which completes the proof. ∎

2.4 Proof of Theorem 1 when $n\geq 3$.

Let $A$ be a compressed set, let $r$ be chosen such that $\left|B_{\geq r+1}\right|<\left|A\right|\leq\left|B_{\geq r}\right|$ and let $C$ be the the initial segment of the $\leq$-order with $\left|C\right|=\left|A\right|$. Recall that by Claim 5 we may assume that we have $B_{\geq r+1}\subseteq A\subseteq B_{\geq r}$. From now on we set $D=A\cap B_{r}$. Since $\left|d\left(A\right)\right|=\left|B_{\geq r+2}\right|+\left|d\left(D\right)\right|$ and $\left|d\left(C\right)\right|=\left|B_{\geq r+2}\right|+\left|d\left(C\cap B_{r}\right)\right|$, it suffices to focus on comparing $d\left(D\right)$ with $d\left(C\cap B_{r}\right)$.

We split the proof into two cases depending on whether $\left|A\right|\leq\left|B_{\geq 1}\right|$ or $\left|A\right|>\left|B_{\geq 1}\right|$.

Case 1. $\left|A\right|\leq\left|B_{\geq 1}\right|.$

Since $\left|A\right|\leq\left|B_{\geq 1}\right|$, it follows that $A=B_{\geq r+1}\cup D$ for some $D\subseteq B_{r}$ and $r\geq 1$. Recall that Theorem 1 is trivial when $\left|A\right|\leq n\left(k-1\right)+1$, and hence we may assume that $r\leq n-2$.

Note that each of the sets $B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)$ is an initial segment of the $\leq$-order, and recall that $\left[s\right]^{n}=\left\{0,\dots,s-1\right\}^{n}$. We start by comparing the sets $d\left(D\right)$ and $d\left(\left[s\right]^{n}\cap B_{r}\right)$ for an appropriately chosen $s$.

Claim 6*.*

Let $s=\max\left\{m\left(x\right):x\in D\right\}=\max\left\{x_{i}:x_{1}\dots x_{n}\in D\right\}$. Then we have $d\left(\left[s\right]^{n}\cap B_{r}\right)\subseteq d\left(A\right)$. In particular, if $r$ and $s$ are chosen so that

[TABLE]

then it suffices to prove Theorem 1 for compressed sets $A$ that satisfy

[TABLE]

Proof.

When $s=1$ the claim is equivalent to the condition $B_{\geq r+1}\subseteq d\left(A\right)$ which is certainly true. Now suppose that $s\geq 2$, and let $x$ be the least point under the $\leq$-order in $D$ with $m\left(x\right)=s$. Since $A$ is a down-set, the minimality of $x$ implies that there exists a unique $i$ satisfying $x_{i}=s$.

Let $v\in d\left(\left[s\right]^{n}\cap B_{r}\right)$. Since $\left|R_{0}\left(v\right)\right|=r+1\geq 2$, it follows that there exists $j\neq i$ with $v_{j}=0$. If $x_{j}\neq 0$, consider the point $z$ obtained by setting

[TABLE]

Otherwise, pick any $p\neq j$ with $v_{p}=0$, and consider $z$ obtained by setting

[TABLE]

Since $v_{t}\in\left[s\right]$ for all $t$ and $j$ is chosen so that $x_{j}\neq s$, it follows that in either case we have $z\in\left[s\right]^{n}$. In both cases we certainly have $v\in d\left(z\right)$ by the construction of $z$. Since $v\in d\left(\left[s\right]^{n}\cap B_{r}\right)=\left[s\right]^{n}\cap B_{r+1}$, it follows that $\left|R_{0}\left(x\right)\right|=\left|R_{0}\left(z\right)\right|$. Since $m\left(x\right)>m\left(z\right)$, we have $z\leq x$.

In the first case we have $x_{j}=z_{j}$, and hence (10) implies that $z\in A$. In the second case we have $z_{j}=v_{j}=x_{j}=0$, so again (10) implies that $z\in A$. Hence it follows that $v\in d\left(A\right)$, which completes the proof of the first part.

For the second part, suppose that $A^{\prime}$ is a compressed set satisfying $\left|B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right|<\left|A^{\prime}\right|\leq\left|B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)\right|$ for some $r$ and $s$. Recall that by Claim 5 we may assume that $B_{\geq r+1}\subseteq A^{\prime}\subseteq B_{\geq r}$. Let $C$ be the initial segment of the $\leq$-order with $\left|C\right|=\left|A^{\prime}\right|$, and note that $C\subseteq B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)$.

Let $t=\max\left(m\left(x\right):x\in A^{\prime}\cap B_{r}\right)$. Since $\left|A^{\prime}\right|>\left|B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right|$, we must have $t\geq s$ (recall that $\left[s\right]^{n}=\left\{0,\dots,s-1\right\}^{n}$). If $t\geq s+1$, the first part implies that $d\left(B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)\right)\subseteq d\left(A^{\prime}\right)$. In particular, it follows that $d\left(C\right)\subseteq d\left(A^{\prime}\right)$ and thus $\left|d\left(A^{\prime}\right)\right|\geq\left|d\left(C\right)\right|$, as required.

If $t=s$, it follows that $A^{\prime}\subseteq B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)$. Let $X=\left\{x\in A\cap B_{r}:m\left(x\right)=s\right\}$, and note that $X\neq\emptyset$. Let $Y$ be the set of $k$ least points of $X$ under the $\leq$-order, where $k=\left|A^{\prime}\right|-\left|B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right|$, and consider $A$ defined by $A=B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\cup Y$. It is clear that $A$ is compressed, and we have $\left|A\right|=\left|A^{\prime}\right|=\left|C\right|$.

By the construction of $A$ it follows that $B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\subseteq A\subseteq B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)$. By Claim 5 and the first part, we have that $d\left(B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right)\subseteq d\left(A^{\prime}\right)$. In particular, it follows that $d\left(B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right)\cup d\left(X\right)\subseteq d\left(A^{\prime}\right)$. Since $d\left(Y\right)\subseteq d\left(X\right)$, we must have $d\left(A\right)=d\left(B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\right)\cup d\left(Y\right)\subseteq d\left(A^{\prime}\right)$. Hence it follows that $\left|d\left(A^{\prime}\right)\right|\geq\left|d\left(A\right)\right|$, which completes the proof. ∎

Let $s=\max\left(m\left(x\right):x\in D\right)$. From now on we suppose that $A$ is a compressed set that satisfies $B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\subset A\subseteq B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)$.

We split our proof into two subcases based on whether $s=\max\left(m\left(x\right):x\in D\right)$ satisfies $s=1$ or $s\geq 2$. When $s=1$, Theorem 1 is actually equivalent to the Kruskal-Katona Theorem, and hence the case $s\geq 2$ is the main part of the proof.

Case 1.1. $s=1.$

Since $s=1$, it follows that $D\subseteq\left\{0,1\right\}^{n}\cap B_{r}$. However, $\left\{0,1\right\}^{n}\cap B_{r}$ is just the set of those points in $\left\{0,1\right\}^{n}$ with exactly $r$ coordinates that equal [math], i.e. the set $\left\{0,1\right\}_{n-r}^{n}$. It is easy to check that the colexicographic order and the $\leq$-order coincide on $\left\{0,1\right\}_{n-r}^{n}$. Thus Theorem 1 holds when $s=1$, as the Kruskal-Katona theorem states that the initial segments of the colexicographic order minimise the lower shadow on $\left\{0,1\right\}_{n-r}^{n}$.

Case 1.2. $s\geq 2.$

Since $s\geq 2$, it follows that for any point $x$ changing a value of a non-zero coordinate to 1 cannot increase the set $R_{s}\left(x\right)$. As a consequence of Claim 6, from now on we only need to focus on those points in $D$ that contain $s$ as a coordinate.

Let ${\cal A}=\left[n\right]^{\left(\leq n-r\right)}\setminus\left\{\emptyset\right\}$, and for $X\in{\cal A}$ denote the class $C_{i}$ corresponding to $\left(s,X,r\right)$ by $C_{X}$. Note that ${\cal A}$ characterizes all the classes containing points with $s$ as a maximal coordinate and $r$ zeroes. As noted before, the order of classes under the $\leq$-order is the order induced on ${\cal A}$ by the binary order.

Let $T\in{\cal A}$ be the largest set under the binary order with $C_{T}\cap D\neq\emptyset$. We start by proving that for any $S\in{\cal A}$ with $S<_{bin}T$ we have $d\left(C_{S}\right)\subseteq d\left(D\right)$. If $S\cap T\neq\emptyset$, the claim follows easily as $C_{S}\subseteq D$ holds by (10). We start by proving the claim when $T$ contains only one element, as in this case such an easy argument does not exist and the stronger conclusion $C_{s}\subseteq D$ may not always hold. In Claim 8 we deduce that for any $S\in{\cal A}$ with $S<_{bin}T$ we have $d\left(C_{S}\right)\subseteq d\left(D\right)$ regardless of the size of $T$.

Claim 7*.*

Let $X\in{\cal A}$ be a set satisfying $C_{X}\cap D\neq\emptyset$ and $\left|X\right|=1$. Then for any $S\in{\cal A}$ with $S<_{bin}X$ we have $d\left(C_{S}\right)\subseteq d\left(D\right)$.

Proof.

Let $i$ be chosen so that $X=\left\{i\right\}$, and define the particular point $a=s_{i}\left(\left(\left(n-r-1\right)\cdot 1\right)\left(r\cdot 0\right)\right)$. Note that $a$ is the least point in $C_{X}$, and hence (10) implies that $a\in D$.

Let $x\in d\left(C_{S}\right)$. Since $r\geq 1$, it follows that $\left|R_{0}\left(x\right)\right|=r+1\geq 2$. Hence there exists distinct elements $l$ and $m$ with $x_{l}=x_{m}=0$. In particular, we may assume that $m\neq i$.

Let $y$ be the point obtained by taking $y_{j}=x_{j}$ for $j\neq m$ and $y_{m}=1$, and $z$ be the point obtained by taking $z_{j}=x_{j}$ for $j\neq l$ and $z_{l}=1$. Note that we have $R_{s}\left(y\right)\subseteq S$ and $R_{s}\left(z\right)\subseteq S$, and recall that $R_{s}\left(a\right)=X$. Since $S<_{bin}X$, these conditions imply that $y\leq a$ and $z\leq a$. Note that by the construction of $y$ and $z$ we have $y_{m}=1$ and $z_{m}=x_{m}=0$. Since $m\neq i$, it follows that $a_{m}\in\left\{0,1\right\}$ and hence we either have $a_{m}=y_{m}$ or $a_{m}=z_{m}$. Thus (10) implies that we have $y\in D$ or $z\in D$, and hence in either case it follows that $x\in d\left(D\right)$, which completes the proof. ∎

Claim 8*.*

Let $T\in{\cal A}$ be the largest element under the binary order with $C_{T}\cap D\neq\emptyset$. Then for any $S\in{\cal A}$ with $S<_{bin}T$ we have $d\left(C_{S}\right)\subseteq d\left(D\right)$.

Proof.

If $\left|T\right|=1$, the result follows immediately from Claim 7. Hence we may assume that $\left|T\right|>1$, and define $T_{1}=\left\{\max T\right\}$. Note that $T_{1}\cap T\neq\emptyset$ and $T_{1}<_{bin}T$. Hence (10) implies that $C_{T_{1}}\subseteq D$.

Let $S\in{\cal A}$ for which $S<_{bin}T$. If $\max S=\max T$, then (10) implies that $C_{S}\subseteq D$, and in particular it follows that $d\left(C_{S}\right)\subseteq d\left(D\right)$. If $\max S<\max T$, it follows that $S<_{bin}T_{1}$. By applying Claim 7 for the set $T_{1}$ which contains only one point, it follows that $d\left(C_{S}\right)\subseteq d\left(D\right)$, which completes the proof. ∎

Now we are ready to finish the proof of Case 1.2. Let $C$ be the initial segment of the $\leq$-order with $\left|C\right|=\left|A\right|$. Recall that $r$ and $s$ are chosen so that $B_{\geq r+1}\cup\left(\left[s\right]^{n}\cap B_{r}\right)\subseteq A\subseteq B_{\geq r+1}\cup\left(\left[s+1\right]^{n}\cap B_{r}\right)$. Since $C$ is an initial segment, it follows that there exists $T_{1}\in{\cal A}$ and $x\in T_{1}$ so that

[TABLE]

As $\left|A\right|=\left|C\right|$, it follows that $T_{1}\leq_{bin}T$. If $T_{1}<T$, we have $d\left(C_{S}\right)\subseteq d\left(A\right)$ for all $S$ with $S\in{\cal A}$ and $S\leq_{bin}T_{1}$ by Claim 8. In particular, it follows that $d\left(C\right)\subseteq d\left(A\right)$, and therefore $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$. If $T_{1}=T$, then we have $A\cap C_{T}=\left\{y:y\leq z,y\in C_{T}\right\}$ for some $z\in C_{T}$ with $x\leq z$ as $C$ is an initial segment. Hence we have $d\left(\left\{y:y\leq x,y\in T_{1}\right\}\right)\subseteq d\left(A\right)$, and it follows that $d\left(C\right)\subseteq d\left(A\right)$. This completes the proof of Case 1.2.

Case 2. $\left|A\right|>\left|B_{\geq 1}\right|.$

Recall that $A$ is of the form $A=B_{\geq 1}\cup D$ for some $D\subseteq B_{0}=\left\{1,\dots,k-1\right\}^{n}$. Note that Theorem 1 holds when $\left|A\right|=\left|B_{\geq 1}\right|+1$, as $\left|d\left(x\right)\right|=n$ for any $x\in B_{0}$. From now on we assume that $\left|A\right|>\left|B_{\geq 1}\right|+1$. Hence it follows that $\left|D\right|>1$, and thus $s=\max\left(m\left(x\right):x\in D\right)$ satisfies $s\geq 2$.

Ideally, we would like to use an approach that is similar to the one used in Case 1. However, it turns out that the statements of Claims 6 and 8 need to be slightly modified. We start with a preliminary result which states that (10) also holds for points inside $d\left(D\right)$. Then we move on to prove an appropriate version of Claim 6. When $T\neq\left\{1\right\}$ it follows that we again have $d\left(\left\{1,\dots,s-1\right\}^{n}\right)\subseteq d\left(A\right)$, but when $T=\left\{1\right\}$ one specific point might be missing from the shadow.

Claim 9*.*

Let $y\in d\left(D\right)$, and let $x$ be a point satisfying $\left|R_{0}\left(x\right)\right|=1$ and $x\leq y$. Furthermore, suppose that there exists $i$ for which $x_{i}=y_{i}$. Then we have $x\in d\left(D\right)$.

Proof.

We may certainly assume that $x<y$. Since $y\in d\left(D\right)$, there exists $v\in D$ with $y\in d\left(v\right)$. Thus there exists $b$ satisfying $v_{j}=y_{j}$ for $j\neq b$ and $y_{b}=0$, and since $A$ is a down-set, we may assume that $v_{b}=1$. Note that $b$ is the unique index for which $y_{b}=0$ as $\left|R_{0}\left(y\right)\right|=1$.

Since $\left|R_{0}\left(x\right)\right|=1$, there exists a unique index $a$ with $x_{a}=0$. Let $u$ be the sequence obtained by taking $u_{j}=x_{j}$ for $j\neq a$ and $u_{a}=1$. Note that we have $x\in d\left(u\right)$.

Our aim is to prove that $u\in D$. Let $t$ be the largest index for which $R_{t}\left(x\right)\neq R_{t}\left(y\right)$, and for such $t$ we have $\max\left(R_{t}\left(x\right)\Delta R_{t}\left(y\right)\right)\in R_{t}\left(y\right)$. Observe that we must have $t\geq 1$, as for any point $z$ the sets $R_{t}\left(z\right)$ are disjoint and their union equals $\left\{1,\dots,n\right\}$.

First note that by the construction of the points $u$ and $v$ it follows that $R_{s}\left(u\right)=R_{s}\left(x\right)$ and $R_{s}\left(v\right)=R_{s}\left(y\right)$ for $s\geq 2$. In particular, it follows that R_{s}\left(u\right)=R_{s}$$\left(v\right) for $s>t$.

We start by proving that $t=1$ implies $u=v$. Indeed, by the previous observation we have $R_{s}\left(u\right)=R_{s}\left(v\right)$ for all $s>1$, and we also have $R_{0}\left(u\right)=R_{0}\left(v\right)=\emptyset$. Since for any point $c$ the sets $R_{s}\left(c\right)$ are disjoint and their union equals $\left\{1,\dots,n\right\}$, we must also have $R_{1}\left(u\right)=R_{1}\left(v\right)$. In particular, it follows that $u=v$.

If $t\geq 2$, we have $R_{t}\left(u\right)=R_{t}\left(x\right)$ and $R_{t}\left(v\right)=R_{t}\left(y\right)$. Hence $\max\left(R_{t}\left(u\right)\Delta R_{t}\left(v\right)\right)\in R_{t}\left(v\right)$, and thus it follows that $u<v$. Recall that there exists $i$ satisfying $x_{i}=y_{i}$. If $i\not\in\left\{a,b\right\}$, it follows that $u_{i}=x_{i}$ and $v_{i}=y_{i}$ by the construction of the points $u$ and $v$. In particular, we have $u_{i}=v_{i}$, and thus (10) implies that $u\in D$.

Now suppose that $i\in\left\{a,b\right\}$. Since $x_{a}=0$ and $y_{b}=0$, we must have $x_{i}=y_{i}=0$. However, recall that both $R_{0}\left(x\right)$ and $R_{0}\left(y\right)$ contain only one point. Hence we must have $a=b=i$, and thus by the construction of $u$ and $v$ it follows that $u_{i}=v_{i}=1$. As before, (10) implies that $u\in D$, which completes the proof as $x\in d\left(u\right)$. ∎

Claim 10*.*

If $T\neq\left\{1\right\}$, we have $d\left(\left\{1,\dots,s-1\right\}^{n}\right)\subseteq d\left(A\right)$. 2. 2.

If $T=\left\{1\right\}$, we have $d\left(\left\{1,\dots,s-1\right\}^{n}\right)\setminus\left\{0\left(\left(n-1\right)\cdot\left(s-1\right)\right)\right\}\subseteq d\left(A\right)$.

Proof.

We start with the case $T\neq\left\{1\right\}$, and recall that $s\geq 2$. Let $j=\max\left(T\right)$, which by assumption is at least 2. Define the particular point $a=s_{j}\left(\left(n-1\right)\cdot 1\right)$ which is the least point in $C_{\left\{j\right\}}$. Let $z$ be a point in $C_{T}\cap D$. Since $a\leq z$ and $a_{j}=z_{j}$, (10) implies that $a\in D$. Let $b=s_{j-1}\left(\left(n-1\right)\cdot 1\right)$, and note that $b$ is well-defined as $j>1$. Since $2\left(n-1\right)>n$, there exists $i$ for which $a_{i}=b_{i}=1$ by the pigeonhole principle. Thus (10) implies that $b\in D$.

Let $x\in d\left(\left\{1,\dots,s-1\right\}\right)$, and let $i$ be the unique index satisfying $x_{i}=0$. Let $y$ be obtained by taking $y_{r}=x_{r}$ for $r\neq i$ and $y_{i}=1$. Since $s\geq 2$, it follows that $y\in\left\{1,\dots,s-1\right\}^{n}$, so we have $y\leq a$ and $y\leq b$. If $i\neq j$ we have $a_{i}=y_{i}=1$, and if $i=j$ we have $b_{i}=y_{i}=1$. Since $a,b\in D$, we must have $y\in D$ in either case by (10). This completes the proof of the first part.

Now suppose that $T=\left\{1\right\}$. For convenience define the points $w=0\left(\left(n-1\right)\cdot\left(s-1\right)\right)$ and $a=s\left(\left(n-1\right)\cdot 1\right)$. Note that $a$ is the least point in $C_{T}$, and thus we have $a\in D$. For every $2\leq i\leq n$ define the points $b_{i}=1_{i}\left(\left(n-1\right)\cdot\left(s-1\right)\right)$. Then $\left(b_{i}\right)_{i}=1=a_{i}$ and $b_{i}\leq a$ for all $i$, so (10) implies that $b_{i}\in d\left(D\right)$ for all $i$. For $2\leq i\leq n$ define $c_{i}=0_{i}\left(\left(n-1\right)\cdot\left(s-1\right)\right)$, and note that $c_{i}\in d\left(b_{i}\right)$ for all $i$. Hence it follows that $c_{i}\in d\left(D\right)$ for all $2\leq i\leq n$.

Let $x$ be a point with $x\in d\left(\left\{1,\dots,s-1\right\}^{n}\right)$ and $x\neq w$. If $x_{i}=0$ for some $i\geq 2$, then we have $x_{j}\leq\left(c_{i}\right)_{j}$ for every $j$. Since $d\left(A\right)$ is a down-set by Claim 4, it follows that $x\in d\left(D\right)$.

Now suppose that $x$ is a point with $x_{1}=0$ and $x\neq w$. Since $x\neq w$ it follows that $R_{s-1}\left(x\right)$ is a proper subset of $\left\{2,\dots,n\right\}$. Note that $R_{s-1}\left(c_{2}\right)=\left\{1,3,\dots,n\right\}$ is larger than any proper subset of $\left\{2,\dots,n\right\}$ under the binary order, and hence we must have $x\leq c_{2}$. Let $k\in\left\{2,\dots,n\right\}\setminus R_{s-1}\left(x\right)$, and consider $v$ obtained by taking $v_{j}=\left(c_{2}\right)_{j}$ for $j\neq k$ and $v_{k}=x_{k}$. Then we have $x\leq v\leq c_{2}$, and hence applying Claim 9 twice implies that $v\in d\left(D\right)$ and $x\in d\left(D\right)$. Thus we have $d\left(\left\{1,\dots,s-1\right\}^{n}\right)\setminus\left\{w\right\}\subseteq d\left(D\right)$, which completes the proof. ∎

Define the particular points $w=0\left(\left(n-1\right)\cdot\left(s-1\right)\right)$ and $a_{i}=i\left(\left(n-1\right)\cdot\left(s-1\right)\right)$ for $1\leq i\leq s-1$. It is easy to verify that $w\in d\left(x\right)$ for $x\in\left[s\right]^{n}\cap B_{0}$ if and only if $x=a_{i}$ for some $i$. In order to deal with the case $T=\left\{1\right\}$, we prove a result that adding suitably many consecutive points to an initial segment must increase the size of the $d$-shadow. This is done in Claim 12, but first we need a preliminary result. For $x\in\left[k\right]^{n}$, define the $x^{+}$ to be the successor of $x$, i.e. $x^{+}$ is the least point under the $\leq$-order for which $x<x^{+}$.

Claim 11*.*

Let $x\in D$, and suppose that we also have $y=x^{+}\in D$. Then the following claims are true.

There exists an unique $i$ satisfying $y_{i}>x_{i}$. 2. 2.

$y_{j}<x_{j}$ implies that $y_{j}=1$. 3. 3.

If $y_{j}\neq 1$ for all $j$, it follows that $y_{j}>x_{i}$ for all $j$, where $i$ is the unique index satisfying $y_{i}>x_{i}$.

Proof.

Let $t$ be the largest index satisfying $R_{t}\left(x\right)\neq R_{t}\left(y\right)$. Since $x<y$ we must have $R_{t}\left(x\right)<_{bin}R_{t}\left(y\right)$. Let $i=\max\left(R_{t}\left(x\right)\Delta R_{t}\left(y\right)\right)$, and note that $i\in R_{t}\left(y\right)$ implies that $y_{i}>x_{i}$.

Consider the point $z$ obtained by taking $z_{j}=\min\left(x_{j},y_{j}\right)$ for all $j\neq i$ and $z_{i}=y_{i}$. By the construction of $z$ we have $z_{j}\leq y_{j}$ for all $j$, and hence we have $z\leq y$. For all $s>t$ we have $R_{s}\left(x\right)=R_{s}\left(y\right)$, and hence by the construction of $z$ it also follows that $R_{s}\left(x\right)=R_{s}\left(z\right)$ for all $s>t$. Similarly, it is easy to see that $R_{t}\left(x\right)\cap R_{t}\left(y\right)\subseteq R_{t}\left(z\right)$. Since we also have $i\in R_{t}\left(z\right)$, it follows that $\max\left(R_{t}\left(x\right)\Delta R_{t}\left(z\right)\right)=i\in R_{t}\left(z\right)$, which implies that $x<z$. Combining these two observations we obtain that $x<z\leq y$, and since $y=x^{+}$ it follows that $y=z$. Thus for any $j\neq i$ we have $y_{j}=\min\left(x_{j},y_{j}\right)\leq x_{j}$, which completes the proof of the first part.

Let $X=\left\{j:y_{j}<x_{j}\right\}$, and let $u$ be the sequence obtained by taking $u_{j}=1$ for all $j\in X$ and $u_{j}=y_{j}$ for $j\not\in X$. Since $y\in D$, it follows that $u_{j}\leq y_{j}$ for all $j$, and hence we have $u\leq y$. Let $t$ and $i$ be defined as in the previous part. Since $R_{s}\left(x\right)=R_{s}\left(y\right)$ for all $s>t$, it follows that $R_{s}\left(x\right)=R_{s}\left(u\right)$ for all $s>t$. Furthermore, it is easy to see that we have $\left\{i\right\}\cup\left(R_{t}\left(x\right)\cap R_{t}\left(y\right)\right)\subseteq R_{t}\left(u\right)$. Hence it follows that $\max\left(R_{t}\left(u\right)\Delta R_{t}\left(x\right)\right)=i$, and therefore we have $x<u$. Combining these two observations we obtain that $x<u\leq y$, and since $y=x^{+}$ it follows that $y=u$. Hence $y_{j}<x_{j}$ implies that $y_{j}=1$, which proves the second part.

Suppose that we have $y_{j}\neq 1$ for all $j$. Combining the first and the second part, it follows that there exists unique $i$ so that $y_{j}=x_{j}$ for all $j\neq i$ and $y_{i}>x_{i}$. Since $y$ is the successor of $x$, it evidently follows that $y_{i}=x_{i}+1$. Our aim is to prove that $x_{i}$ is strictly smaller than any other $x_{j}$. First assume that there exists $k$ so that $x_{k}<x_{i}$, and consider $v$ obtained by taking $v_{j}=x_{j}$ for $j\neq k$ and $v_{k}=x_{k}+1$. The we certainly have $x<v$. For convenience, set $a=x_{i}$ and $b=x_{k}$. By the construction of $v$ it follows that $R_{s}\left(x\right)=R_{s}\left(y\right)=R_{s}\left(v\right)$ for all $s\geq a+2$. However, since $b<a$ it follows that $R_{a+1}\left(y\right)=R_{a+1}\left(x\right)\cup\left\{i\right\}$, while $R_{a+1}\left(v\right)=R_{a+1}\left(x\right)$. Hence we also have $v<y$ which contradicts the fact that $y$ is the successor of $x$.

Again let $a=x_{i}$, and note that if $a=1$ the claim follows evidently. Now suppose that for all $j$ we have $x_{j}\geq a$ but $\left|R_{a}\left(x\right)\right|\geq 2$. Let $v$ be the point obtained by taking $v_{j}=x_{j}$ for all $j\not\in R_{a}\left(x\right)$, $v_{i}=a+1$ and $v_{j}=1$ for all $j\in R_{a}\left(x\right)\setminus\left\{i\right\}$. Note that for all $s\geq a+2$ we have $R_{s}\left(x\right)=R_{s}\left(v\right)=R_{s}\left(y\right)$. However, we also have $R_{a+1}\left(y\right)=R_{a+1}\left(v\right)=R_{a+1}\left(x\right)\cup\left\{i\right\}$, $R_{a}\left(y\right)=R_{a}\left(x\right)\setminus\left\{i\right\}$ and $R_{a}\left(v\right)=\emptyset$ since $a>1$. Since $R_{a}\left(x\right)$ contains at least two elements, it follows that $x<v<y$, which contradicts the fact that $y=x^{+}$.

Hence for all $j\neq i$ we have $x_{j}>x_{i}$. Since $y_{j}\neq 1$ for all $j$, the second part implies that $y_{j}\geq x_{j}$ for all $j$. Hence for all $j\neq i$ we have $y_{j}\geq x_{j}>x_{i}$, and by the first part we have $y_{i}>x_{i}.$ This completes the proof of the third part. ∎

Claim 12*.*

Let $x_{1},\dots,x_{L-1}\in\left\{1,\dots,L-1\right\}^{n}$ be consecutive points under the $\leq$-order. Let $X$ and $Y$ be initial segments of the $\leq$-order on $\left[k\right]^{n}$ defined by $X=\left\{y:y\leq x_{L-1}\right\}$ and $Y=\left\{y:y\leq x_{1}\right\}$. Then $\left|d\left(X\right)\right|=\left|d\left(Y\right)\right|$ if and only if $x_{i}=\left(i\right)\left(\left(n-1\right)\cdot\left(L-1\right)\right)$ for all $1\leq i\leq L-1$.

Proof.

Let $x\in\left\{1,\dots,L-1\right\}^{n}$, and consider the initial segment $Z$ defined by $Z=\left\{y:y<x\right\}$. We start by proving that $\left|d\left(Z\cup\left\{x\right\}\right)\right|=\left|d\left(Z\right)\right|$ if and only if $R_{1}\left(x\right)=\emptyset$. Suppose that there exists $z\in d\left(Z\cup\left\{x\right\}\right)\setminus d\left(Z\right)$, and let $j$ be the unique index for which $z_{j}=0$. Note that such $j$ is unique as $x\in B_{0}$. Let $a_{1},\dots,a_{L-1}$ be the points obtained by taking $\left(a_{i}\right)_{k}=z_{k}$ for $k\neq j$, and $\left(a_{i}\right)_{j}=i$. It is clear that $a_{1}<\dots<a_{L-1}$, and we have $z\in d\left(u\right)$ for $u\in\left\{1,\dots,L-1\right\}^{n}$ if and only if $u\in\left\{a_{1},\dots,a_{L-1}\right\}$. In particular, it follows that $z\in d\left(Z\cup\left\{x\right\}\right)\setminus d\left(Z\right)$ if and only if $Z\cap\left\{a_{1},\dots,a_{L-1}\right\}=\emptyset$ and $x\in\left\{a_{1},\dots,a_{L-1}\right\}$. Since $Z$ is an initial segment, we must have $x=a_{1}$. Thus $\left|d\left(Z\cup\left\{x\right\}\right)\right|=\left|d\left(Z\right)\right|$ implies that $R_{1}\left(x\right)=\emptyset$, and if $x=a_{1}$ then $z\in d\left(Z\cup\left\{x\right\}\right)\setminus d\left(Z\right)$ which proves the converse as well.

Let $X$ and $Y$ satisfy the conditions of the claim, and suppose that we have $\left|d\left(X\right)\right|=\left|d\left(Y\right)\right|$. Then the previous observation implies that we have $R_{1}\left(x_{i}\right)=\emptyset$ for all $2\leq i\leq L-1$.

Let $m\left(i\right)$ denote the value of the smallest coordinate of the point $x_{i}$. Since $R_{1}\left(x_{i}\right)=\emptyset$ for $2\leq i\leq L-1$, the third part of Claim 11 implies that $\left(x_{i+1}\right)_{j}>m\left(i\right)$ for all $j$. In particular, we have $m\left(i+1\right)>m\left(i\right)$, and thus $m$ is strictly increasing. Since $R_{1}\left(x_{2}\right)=\emptyset$ it follows that $m\left(2\right)\geq 2$, and hence we must have $m\left(i\right)\geq i$ for all $i$. In particular, it follows that $m\left(L-1\right)\geq L-1$, and thus it follows that $x_{L-1}=\left(n\cdot\left(L-1\right)\right)$. Since $x_{i}$’s are consecutive points, it is easy to verify that $x_{i}=\left(i\right)\left(\left(n-1\right)\cdot\left(L-1\right)\right)$ for all $1\leq i\leq L-1$. ∎

The following claim allows us to restrict to proving Theorem 1 for compressed sets $A$ satisfying $B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq A\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$ for some $s$.

Claim 13*.*

Let $A$ be a compressed set with $\left|A\right|\geq\left|B_{\geq 1}\right|+2$ and let $s\geq 2$ be chosen so that $\left|B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right|<\left|A\right|\leq\left|B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)\right|$. Then there exists a compressed set $B$ with $\left|A\right|=\left|B\right|$ satisfying $B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq B\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$ for which we have $\left|d\left(A\right)\right|\geq\left|d\left(B\right)\right|$.

Proof.

Let $q=\max\left(m\left(x\right):x\in D\right)$, and note that the condition on the size of $A$ implies that $q\geq s$. Let $T$ be the largest element under the binary order so that the class $C_{T}$ has a non-empty intersection with $A$.

First suppose that $T=\left\{1\right\}$, and let $C$ be the initial segment of the $\leq$-order with $\left|C\right|=\left|A\right|$. Define the particular points $w=0\left(\left(n-1\right)\cdot\left(q-1\right)\right)$ and $w_{i}=i\left(\left(n-1\right)\cdot\left(q-1\right)\right)$ for $1\leq i\leq q-1$. Thus Claim 10 implies that $d\left(\left[s\right]^{n}\right)\setminus\left\{w\right\}\subseteq d\left(A\right)$.

We start by verifying that $C\cap C_{T}\subseteq A\cap C_{T}$. If $q>s$, the inclusion is trivial as we have $C\cap C_{T}=\emptyset$. If $q=s$, since $T$ is the largest element under the binary order so that $C_{T}\cap A\neq\emptyset$, it follows that $A\setminus\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)=A\cap C_{T}$, and similarly $C\setminus\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)=C\cap C_{T}$. Since $C$ is an initial segment with $\left|C\right|=\left|A\right|$, we must have $\left|A\cap C_{T}\right|\geq\left|C\cap C_{T}\right|$. As $A\cap C_{T}$ is of the form $\left\{y\in C_{T}:y\leq x\right\}$ for some $x\in C_{T}$, the inclusion follows.

First consider the case when we also have $w\in d\left(A\right)$. Then $d\left(\left[s\right]^{n}\right)\subseteq d\left(A\right)$, and since $d\left(C\cap C_{T}\right)\subseteq d\left(A\right)$, it follows that $d\left(C\right)\subseteq d\left(A\right)$. Thus we may take $B=C$.

Now suppose that $w\not\in d\left(A\right)$. We first deal with the easy case $q>s$. Note that we have $d\left(C\right)\subseteq d\left(B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)\right)$. Since $q\geq s+1$, Claim 10 implies that $d\left(\left[s+1\right]^{n}\cap B_{0}\right)\setminus\left\{w\right\}\subseteq d\left(A\right)$, and hence it follows that $d\left(C\right)\setminus d\left(A\right)\subseteq\left\{w\right\}$. Since $q\geq s+1$, $A$ contains a point $u\in\left(\left[q\right]^{n}\setminus\left[s\right]^{n}\right)\cap B_{0}$. As $n\geq 2$ and $R_{0}\left(u\right)=\emptyset$, $d\left(u\right)$ contains a point whose coordinate is at least $s$. In particular, it follows that $d\left(u\right)\not\subseteq d\left(C\right)$ and hence we must have $\left|d\left(A\right)\setminus d\left(C\right)\right|\geq 1$. Combining this with $\left|d\left(C\right)\setminus d\left(A\right)\right|\leq 1$, it follows that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$, and hence we may take $B=C$.

Now suppose that we have $w\not\in d\left(A\right)$ and $q=s$. By using the observation $C\cap C_{T}\subseteq A\cap C_{T}$ and Claim 10, it follows that $d\left(C\right)\setminus d\left(A\right)=\left\{w\right\}$. Since $w\not\in d\left(A\right)$ and $q=s$, it follows that $\left\{w_{1},\dots,w_{s-1}\right\}\subseteq C\setminus A$, and in particular we have $\left|\left(C\setminus A\right)\cap\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)\right|\geq s-1$. Since $A\setminus\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)=A\cap C_{T}$ and $C\setminus\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)=C\cap C_{T}$, it follows that $\left|A\cap C_{T}\right|\geq\left|C\cap C_{T}\right|+s-1$.

Let $x$ be the largest point of $A$ and $y$ be the largest point of $C$ under the $\leq$-order. Define $X=\left\{z:z\leq x\right\}$, and note that $C=\left\{z:z\leq y\right\}$ as $C$ is an initial segment. Note that we have $x,y\in C_{T}$, and hence we have $\left|X\right|\geq\left|C\right|+\left(s-1\right)$. Thus Claim 12 with $L=s$ implies that $\left|d\left(X\right)\right|>\left|d\left(C\right)\right|$, unless $x=\left(2\right)\left(\left(n-1\right)\cdot\left(s\right)\right)$ and $y=\left(n\cdot s\right)$. However, these points are not in $C_{T}$ as $T=\left\{1\right\}$, and thus there exists $u\in d\left(X\right)\setminus d\left(C\right)$. Thus we have $u\in d\left(\left\{z:x<z\leq y\right\}\right)\subseteq d\left(C_{T}\cap A\right)\subseteq d\left(A\right)$, and hence it follows that $\left|d\left(A\right)\setminus d\left(C\right)\right|\geq 1$. Since $\left|d\left(C\right)\setminus d\left(A\right)\right|=1$, it follows that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$, as required.

Finally consider the case $T\neq\left\{1\right\}$. If $q>s$, Claim 10 implies that $d\left(\left[s+1\right]^{n}\cap B_{0}\right)\subseteq d\left(A\right)$, and thus we can take $B=C$. If $q=s$, let $X=A\setminus\left(B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right)$ and $k=\left|A\right|-\left|B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\right|$. Let $Y$ be the set of $k$ least points in $X$ under the $\leq$-order, and let $B=B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\cup Y$. We certainly have $\left|A\right|=\left|B\right|$, and since $q=s$ it follows that $B\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$. Furthermore, as $A$ is compressed it follows that $B$ is compressed as well. Since $T\neq\left\{1\right\}$, Claim 10 implies that $d\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq d\left(A\right)$. Since $Y\subseteq X$, it follows that $d\left(B\right)\subseteq d\left(A\right)$, and thus $B$ satisfies the conditions, which completes the proof. ∎

From now on we assume that $A$ is a compressed set for which there exists $s\geq 2$ so that $B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq A\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$. As before, we set $T$ to be the largest set under the binary order for which $C_{T}\cap A\neq\emptyset$. Our next aim is to prove an appropriate version of Claim 8.

Claim 14*.*

If $\left|T\right|\neq 1$, then for all $S$ with $S<_{bin}T$ we have $C_{S}\subseteq A$. 2. 2.

Let $T=\left\{j\right\}$ and $U=\left\{1,\dots,j-1\right\}$ where $j\neq 1$. Then for all $S$ with $S<_{bin}U$ we have $d\left(C_{S}\right)\subseteq d\left(A\right)$. Define the particular point $w_{j}=0_{j}\left(\left(\left(j-1\right)\cdot s\right)\left(\left(n-j\right)\cdot\left(s-1\right)\right)\right)$. Then we also have $d\left(C_{U}\right)\setminus\left\{w_{j}\right\}\subseteq d\left(A\right)$.

Proof.

We start by proving the first part. Let $j=\max\left(T\right)$ and set $U=\left\{1,\dots,j-1\right\}$. Since $\left|T\right|\neq 1$ it follows that $U\cap T\neq\emptyset$, and hence (10) implies that $C_{U}\subseteq A$. Given any $S\in{\cal A}$ with $S<_{bin}T$, note that we must either have $S\subseteq U$ or $j\in S$. If $S\subseteq U$, we also have $S\leq_{bin}U$ and since $C_{U}\subseteq A$, (10) implies that $C_{S}\subseteq A$ holds as well. If $j\in S$, then we certainly have $S\cap T\neq\emptyset$. Since $S<_{bin}T$, (10) implies that $C_{S}\subseteq A$. In particular, for all $S<_{bin}T$ we have $C_{S}\subseteq A$, which completes the proof of the first part.

Now suppose that $T=\left\{j\right\}$ for some $j$ with $j\not\in\left\{1,n\right\}$. The assumption $j<n$ is used in the proof when we are considering the $\left(j+1\right)^{th}$ coordinate of a sequence, and such applications are not highlighted in the proof. Again, let $U=\left\{1,\dots,j-1\right\}$. Define the particular points $a=s_{j}\left(\left(n-1\right)\cdot 1\right)$ and $b=\left(\left(j-1\right)\cdot s\right)\left(\left(n-j+1\right)\cdot 1\right)$. Note that $a$ is the least point under the $\leq$-order in $C_{T}$ and $b$ is a point in $C_{U}$. Since $C_{T}\cap A\neq\emptyset$, (10) implies that we also have $a\in A$. Since $U<_{bin}T$ and $a_{j+1}=b_{j+1}=1$, (10) implies that $b\in A$ and thus we have $C_{U}\cap A\neq\emptyset$. Let $S$ be a set satisfying $S<_{bin}T$ and $S\neq U$. Then we must have $S\subset U$, and since $C_{U}\cap A\neq\emptyset$, (10) implies that $C_{S}\subseteq A$. Thus it only suffices to consider $d\left(C_{U}\right)$.

Let $x\in d\left(C_{U}\right)\setminus\left\{w_{j}\right\}$ and let $i$ be the unique index satisfying $x_{i}=0$. If $i\neq j$, define $y$ by taking $y_{t}=x_{t}$ for all $t\neq i$ and $y_{i}=1$. Since $j\neq i$, it follows that $y_{i}=a_{i}=1$. We also have $R_{s}\left(y\right)\subseteq U$, and hence $y\leq a$. Since $a\in A$, (10) implies that we must also have $y\in A$, and hence we have $x\in d\left(A\right)$.

Now suppose that $i=j$. Since $x\in d\left(C_{U}\right)\setminus\left\{w_{j}\right\}$ and $i=j$, it follows that $x_{t}=s$ for all $t\leq j-1$, $x_{j}=0$, $1\leq x_{t}\leq s-1$ for all $t\geq j+1$, and there exists $k\geq j+1$ satisfying $x_{k}\leq s-2$. Let $u$ be the point obtained by setting $u_{t}=x_{t}$ for $t\neq j$ and $u_{j}=1$, and let $v$ be obtained by taking $v_{t}=x_{t}$ for all $t\not\in\left\{j,k\right\}$, $v_{j}=s-1$ and $v_{k}=1$. It is easy to see that $x\in d\left(u\right)$. Since $R_{s}\left(u\right)=R_{s}\left(v\right)$ and $R_{s-1}\left(v\right)=R_{s-1}\left(u\right)\cup\left\{j\right\}$, it follows that $u<v$. Since $k\neq i$, it follows that $v_{k}=a_{k}=1$, and hence (10) implies that $v\in A$. Since $u_{1}=v_{1}$, (10) implies that $u\in A$ and hence it follows that $x\in d\left(A\right)$, as required.

Finally suppose that $T=\left\{n\right\}$. Let $U=\left\{1,\dots,n-1\right\}$ and $V=\left\{2,\dots,n-1\right\}$. Note that for any $S<_{bin}T$ we have either $S=U$ or $S\leq_{bin}V$. Define the particular point $a$ by $a=\left(\left(n-1\right)\cdot 1\right)s$. Since $a$ is the least point in $C_{T}$, it follows that $a\in A$.

Given $S<_{bin}T$ and a point $x\in d\left(C_{S}\right)$, let $i$ be the unique index for which $x_{i}=0$. Note that we may certainly assume that we have $x\not\in d\left(C_{S^{\prime}}\right)$ for any proper subset $S^{\prime}$ of $S$, with the convention $C_{\emptyset}=\left[s\right]^{n}\cap B_{0}$. It is easy to see that this assumption is equivalent to assuming that $i\not\in S$.

If $S=U$, the only possible point is $x=\left(\left(n-1\right)\cdot s\right)0$ which is precisely the point $w_{n}$. Now suppose that $S=V$, and note that we must have $i\in\left\{1,n\right\}$. Let $u$ and $v$ be the points $u=\left(1\right)\left(\left(n-2\right)\cdot s\right)\left(s-1\right)$ and $v=\left(s-1\right)\left(\left(n-2\right)\cdot s\right)\left(1\right)$. Since $a_{1}=u_{1}=1$ and $u<a$, (10) implies that $u\in A$. Note that $R_{s}\left(u\right)=R_{s}\left(v\right)$ and $R_{s-1}\left(v\right)<_{bin}R_{s-1}\left(u\right)$, and hence it follows that $v<u$. Since $n\geq 3$, we have $u_{2}=v_{2}=s$, and hence (10) implies that $v\in A$.

For all $1\leq j\leq s-2$ and $1\leq k\leq s-2$ define the points $w_{j,k}=\left(j\right)\left(\left(n-2\right)\cdot s\right)\left(k\right)$. Since $R_{s-1}\left(w_{j,k}\right)=\emptyset$, it follows that $w_{j,k}<u$. As $\left(w_{j,k}\right)_{2}=u_{2}=s$, (10) implies that $w_{j,k}\in A$ for all $1\leq j,k\leq s-2$.

Since $0\in\left\{x_{1},x_{n}\right\}$, it follows that either $x=\left(0\right)\left(\left(n-2\right)\cdot s\right)\left(c\right)$ or $x=\left(d\right)\left(\left(n-2\right)\cdot s\right)\left(0\right)$ for some $1\leq c,d\leq s-1$. If $c=s-1$ or $d=s-1$, we have $x\in d\left(u\right)$ or $x\in d\left(v\right)$ respectively. If $c\leq s-2$ or $d\leq s-2$, it is easy to see that $x\in d\left(w_{1,c}\right)$ or $x\in d\left(w_{d,1}\right)$ respectively. In particular, it follows that $x\in d\left(A\right)$. This completes the proof when $S=V$.

In order to deal with the sets $S$ with $S<_{bin}V$, we first assume that $n\geq 4$. As noticed in the previous case, we have $C_{V}\cap A\neq\emptyset$. If $S\cap V\neq\emptyset$, (10) implies that we also have $C_{S}\subseteq A$ and hence the conclusion certainly follows. The only non-empty set $S$ satisfying $S<_{bin}V$ for which $S\cap V=\emptyset$ is $S=\left\{1\right\}$. However, as $n\geq 4$ it follows that $\left\{1\right\}<_{bin}\left\{1,2\right\}<_{bin}V$, and since $\left\{1\right\}\cap\left\{1,2\right\}\neq\emptyset$ and $\left\{1,2\right\}\cap V\neq\emptyset$ we also have $C_{\left\{1\right\}}\subseteq D$. This completes the proof when $n\geq 4$.

Now suppose that $n=3$. Then $V=\left\{2\right\}$, and thus the only non-empty set $S$ satisfying $S<_{bin}V$ is $S=\left\{1\right\}$. Let $x\in d\left(C_{\left\{1\right\}}\right)$ and let $i$ be the unique index satisfying $x_{i}=0$. Recall that we may assume that $i\in\left\{2,3\right\}$. Let $y$ be the point obtained by setting $y_{j}=x_{j}$ for $j\neq i$ and $y_{i}=1$. Our aim is to show that we have $y\in D$.

If $i=2$ and $y_{3}=s-1$, we have $y_{3}=u_{3}$. Since $y\leq u$, (10) implies that $y\in D$. If $i=2$ and $y_{3}=c$ for some $1\leq c\leq s-2$, then $y_{3}=c=\left(w_{1,c}\right)_{3}$, and since $y\leq w_{1,c}$ it follows that $y\in D$. Finally, if $i=3$, we have $y_{3}=v_{3}=1$, and since $y<v$ it follows that $y\in D$. Thus in all three cases we have $y\in D$, which completes the proof. ∎

Now we have all the necessary tools to finish the proof of Theorem

1. Let $A$ be a set satisfying $B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq A\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$, and let $C$ be the initial segment of $\leq$ with $\left|A\right|=\left|C\right|$. As $A$ and $C$ are of the same size, we must also have $B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\subseteq C\subseteq B_{\geq 1}\cup\left(\left[s+1\right]^{n}\cap B_{0}\right)$. Let $T_{1}$ be the largest class under the binary order satisfying $C_{T^{\prime}}\cap C\neq\emptyset$, and since $C$ is an initial segment we certainly have $T_{1}\leq_{bin}T$.

First suppose that $T_{1}<_{bin}T$, and set

[TABLE]

Note that we cannot have $T=\left\{1\right\}$ as $\left\{1\right\}$ is the least point under the binary order in ${\cal A}$.

If $\left|T\right|\geq 2$, for any $S\in{\cal A}$ with $S\leq_{bin}T_{1}$ we have $S<_{bin}T$, and hence Claim 14 implies that $C_{S}\subseteq A$. Hence it follows that $X\subseteq A$, and since $C\subseteq X$ we must have $C\subseteq A$. Since $A$ and $C$ have the same size, it follows that $A=C$. However, this contradicts the assumption $T_{1}<_{bin}T$, so this case cannot occur when $\left|T\right|\geq 2$.

If $T=\left\{j\right\}$ for some $j\neq 1$, define the particular point $w_{j}=0_{j}\left(\left(\left(j-1\right)\cdot s\right)\left(\left(n-j\right)\cdot\left(s-1\right)\right)\right)$. Then Claim 14 implies that for all $S<_{bin}\left\{1,\dots,j-1\right\}$ we have $d\left(C_{S}\right)\subseteq d\left(A\right)$, and for $U=\left\{1,\dots,j-1\right\}$ we have $d\left(C_{U}\right)\setminus\left\{w_{j}\right\}\subseteq d\left(A\right)$. Since $T_{1}<_{bin}T$, it follows that $T_{1}\leq_{bin}U$, and hence $d\left(X\right)\setminus\left\{w_{j}\right\}\subseteq d\left(A\right)$. In particular, we have $\left|d\left(X\right)\setminus d\left(A\right)\right|\leq 1$.

Pick any point $v\in C_{T}$, and consider $u\in d\left(v\right)$ obtained by flipping the first coordinate to [math]. Since $j\neq 1$, it follows that $u_{j}=s$. In particular, it follows that $u\not\in d\left(X\right)$ as $T_{1}\leq_{bin}U$. Hence $u\in d\left(A\right)\setminus d\left(X\right)$, and thus it follows that $\left|d\left(A\right)\setminus d\left(X\right)\right|\geq 1$. Hence we have $\left|d\left(A\right)\right|\geq\left|d\left(X\right)\right|$, and since $C\subseteq X$ this completes the proof in this case.

Hence we are only left with the case when $T_{1}=T$. It turns out to be convenient to split into two subcases based on the size of $T$. When $T=\left\{1\right\}$, there are no non-empty sets $S$ satisfying $S<_{bin}T$. Hence we can deal with the cases $T=\left\{1\right\}$ and $\left|T\right|\geq 2$ at the same time.

Case 2.1. $\left|T\right|\geq 2$ or $T=\left\{1\right\}.$

Let $C$ be an initial segment with $\left|C\right|=\left|A\right|$, and let $s$, $T$ and $T_{1}$ be chosen as before. Recall that we have $T_{1}=T$. We prove that under these assumptions it follows that $A=C$.

Let $X$ be defined by $X=B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\cup\left(\bigcup_{S<_{bin}T}C_{S}\right)$. Claim 14 implies that for all $S<_{bin}T$ we have $C_{S}\subseteq A$, and hence it follows that $X\subseteq A$. On the other hand, by the choice of $T$ we have $A\setminus X=A\cap C_{T}$ and $C\setminus X=C\cap C_{T}$. Since $\left|A\right|=\left|C\right|$ and both of these sets contain $X$ as a subset, it follows that $\left|A\cap C_{T}\right|=\left|C\cap C_{T}\right|$. Since both sets $A\cap C_{T}$ and $C\cap C_{T}$ are of the form $\left\{y\in C_{T}:y\leq x\right\}$ for some $x\in C_{T}$, it follows that $A\cap C_{T}=C\cap C_{T}$. Hence we must have $A=C$, as required.

Case 2.2. $T=\left\{j\right\}$ where $2\leq j\leq n.$

As before, let $C$ be the initial segment with $\left|C\right|=\left|A\right|$, let $s$, $T$ and $T_{1}$ be chosen as before, and recall that we have $T_{1}=T=\left\{j\right\}$ for some $2\leq j\leq n$. Define the particular points $w=0_{j}\left(\left(\left(j-1\right)\cdot s\right)\left(\left(n-j\right)\cdot\left(s-1\right)\right)\right)$ and $w_{i}=i_{j}\left(\left(\left(j-1\right)\cdot s\right)\left(\left(n-j\right)\cdot\left(s-1\right)\right)\right)$ for $1\leq i\leq s-1$. Note that if $w\in d\left(x\right)$ for some $x\in C_{S}$ with $S\leq_{bin}T$, then we must have $x=w_{i}$ for some $i$. As before, let $X=B_{\geq 1}\cup\left(\left[s\right]^{n}\cap B_{0}\right)\cup\bigcup_{S<_{bin}T}C_{S}$.

First consider the easy case when we have $w\in d\left(A\right)$. Then Claim 10 implies that $d\left(C_{S}\right)\subseteq d\left(A\right)$ for all $S$ with $S<_{bin}T$. Since $X\subseteq C$ and we have $C\setminus X\subseteq C_{T}$ and $A\setminus X\subseteq C_{T}$, it follows that $\left|A\cap C_{T}\right|\geq\left|C\cap C_{T}\right|$. By using the same argument as in Case 2.1, it follows that $C\cap C_{T}\subseteq A\cap C_{T}$ and thus it follows that $d\left(C\right)\subseteq d\left(A\right)$, as required.

Now suppose that $w\not\in d\left(A\right)$, and thus it follows that $\left\{w_{1},\dots,w_{s-1}\right\}\cap A\neq\emptyset$. Since we have $\left\{w_{1},\dots,w_{s-1}\right\}\subseteq X\subseteq C$, it follows that $\left|C\setminus X\right|\geq\left|A\setminus X\right|+s-1$, and hence we have $\left|A\cap C_{T}\right|\geq\left|C\cap C_{T}\right|+s-1$. Let $u$ and $v$ be chosen so that $A\cap C_{T}=\left\{z\in C_{T}:z\leq u\right\}$ and $C\cap C_{T}=\left\{z\in C_{T}:z\leq v\right\}$. Let $U$ be the initial segment $U=\left\{z:z\leq u\right\}$, and note that $C=\left\{z:z\leq v\right\}$. Since $\left|C\setminus X\right|\geq\left|A\setminus X\right|+s-1$, it follows that $\left|U\right|\geq\left|C\right|+s-1$.

Since $n\geq 3$, it follows that $\left(1\right)\left(\left(n-1\right)\cdot s\right)\not\in C_{T}$, and hence must have $v\neq\left(1\right)\left(\left(n-1\right)\cdot s\right)$. Since $\left|U\right|\geq\left|C\right|+s-1$, Claim 12 with $L=s+1$ implies that there exists $a\in d\left(U\right)\setminus d\left(C\right)$. Hence we must have $a\in d\left(A\cap C_{T}\right)$, and thus it follows that $\left|d\left(A\right)\setminus d\left(C\right)\right|\geq 1$. Together with $\left|d\left(C\right)\setminus d\left(A\right)\right|=1$, it follows that $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$. This completes the proof of Theorem 1. $\hfill\square$

3 Minimal $d$-shadow for a given rank

Recall that we defined $\left[k\right]_{r}^{n}=\left\{x\in\left[k\right]^{n}:w\left(x\right)=r\right\}$ to be the set of those sequences with exactly $r$ non-zero coordinates, and consider the restriction of the $\leq$-order on $\left[k\right]_{r}^{n}$. Since $\left|R_{0}\left(x\right)\right|=n-r$ is constant for all $x\in\left[k\right]_{r}^{n}$, it follows that for distinct $x$ and $y$ we have $x\leq y$ if and only if $\max\left(R_{j}\left(x\right)\Delta R_{j}\left(y\right)\right)\in R_{j}\left(y\right)$, where $j$ is the largest index for which $R_{j}\left(x\right)\neq R_{j}\left(y\right)$.

Note that for all $m\leq k$, $\left[m\right]_{r}^{n}$ is an initial segment of the $\leq$-order in $\left[k\right]_{r}^{n}$, and the restrictions of the $\leq$-order on $\left[m\right]_{r}^{n}$ and on $\left[k\right]_{r}^{n}$ coincide on $\left[m\right]_{r}^{n}$. Thus these orders extend naturally to an order on $\mathbb{N}_{r}^{n}=\left\{x=x_{1}\dots x_{n}:x_{i}\in\mathbb{N},w\left(x\right)=r\right\}$, which we will also denote by $\leq$.

The notion of $d$-shadow is still sensible for subsets of $\mathbb{N}_{r}^{n}$ as well. If $A\subseteq\mathbb{N}_{r}^{n}$, we certainly have $d\left(A\right)\subseteq\mathbb{N}_{r-1}^{n}$. We now use Theorem 1 to deduce that among the subsets of $\mathbb{N}_{r}^{n}$ of a given finite size, the initial segment of the $\leq$-order minimises the size of the $d$-shadow.

Theorem 2**.**

Let $A$ be a finite subset of $\mathbb{N}_{r}^{n}$ and let $C$ be the initial segment of the $\leq$-order on $\mathbb{N}_{r}^{n}$ with $\left|A\right|=\left|C\right|$. Then we have $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$.

Proof.

Let $k$ be the size of $A$. Since $A$ contains points of length $n$ and exactly $r$ coordinates that equal zero, by reordering coordinates if necessary we may assume that $A\subseteq\left[nk\right]_{r}^{n}$. Let $B=\left[nk\right]_{\leq r-1}^{n}\cup A$, and recall that we have $\left[nk\right]_{\leq r-1}^{n}=\left\{x\in\left[nk\right]^{n}:\left|R_{0}\left(x\right)\right|\geq n-r+1\right\}$. Let $X$ be the initial segment on $\left[nk\right]^{n}$ with $\left|X\right|=\left|B\right|$. Then $X=\left[nk\right]_{\leq r-1}^{n}\cup C$, where $C$ is the initial segment of the $\leq$-order on $\left[nk\right]_{r}^{n}$ with $\left|C\right|=\left|A\right|$. Thus $C$ is also the initial segment of $\leq$ on $\mathbb{N}_{r}^{n}$ with $\left|C\right|=\left|A\right|$.

Note that we have

[TABLE]

and

[TABLE]

as

[TABLE]

Since $\left|B\right|=\left|X\right|$ and $X$ is an initial segment of the $\leq$-order on $\left[nk\right]^{n}$, Theorem 1 implies that $\left|d\left(B\right)\right|\geq\left|d\left(X\right)\right|$. Thus we have $\left|d\left(A\right)\right|\geq\left|d\left(C\right)\right|$, which completes the proof. ∎