Dickson's Lemma, Higman's Theorem and Beyond: a survey of some basic results in order theory

TL;DR

This survey reviews fundamental results in order theory, focusing on proofs that certain orders lack descending chains and antichains, highlighting key theorems like Dickson's Lemma and Higman's Theorem.

Contribution

It compiles and explains essential proofs of order-theoretic properties, emphasizing their significance and interrelations in the field.

Findings

Orders without descending chains or antichains are characterized.

Key theorems like Dickson's Lemma and Higman's Theorem are discussed.

Foundational results in order theory are systematically presented.

Abstract

We provide proofs for the fact that certain orders have no descending chains and no antichains.

Equations

$$S_j:=\{y_j\mid (y_i{)}_{i\in {\mathbb{N}}_0}\text{ is a bad sequence from A with }(y_0,\dots ,y_{j-1})=(a_0,\dots ,a_{j-1})\},$$

$$c(\{i,j\})\,(k):=\left\{\begin{array}[]{rl}1&\text{ if }\bm{a}^{(i)}_{k}\leq\bm{a}^{(j)}_{k},\\ 2&\text{ if }\bm{a}^{(i)}_{k}>\bm{a}^{(j)}_{k}.\end{array}\right.$$

$$\Phi_{F}(\bm{a}):=\left\{\begin{array}[]{rl}\min\{c\in\mathbb{N}_{0}|(\bm{a},c)\in F\}&\text{ if there exists $c^{\prime}\in\mathbb{N}$ such that $(\bm{a},c^{\prime})\in F$ },\\ \infty&\text{ otherwise.}\end{array}\right.$$

$${\Phi }_{F_j}({\mathbf{a}}^{(i,j)})<{\Phi }_{F_i}({\mathbf{a}}^{(i,j)}).$$

$$c(\{i,j,k\})\,(l):=\left\{\begin{array}[]{rl}1&\text{if }\bm{a}^{(i,j)}_{l}\leq\bm{a}^{(j,k)}_{l},\\ 2&\text{if }\bm{a}^{(i,j)}_{l}>\bm{a}^{(j,k)}_{l}.\end{array}\right.$$

$${\mathbf{a}}_l^{(t_0,t_1)}>{\mathbf{a}}_l^{(t_1,t_2)}>{\mathbf{a}}_l^{(t_2,t_3)}>\cdots .$$

$${\Phi }_{F_{t_r}}({\mathbf{a}}^{(t_r,t_{r+1})})>{\Phi }_{F_{t_{r+1}}}({\mathbf{a}}^{(t_r,t_{r+1})}).$$

$${\Phi }_{F_{t_{r+1}}}({\mathbf{a}}^{(t_r,t_{r+1})})\ge {\Phi }_{F_{t_{r+1}}}({\mathbf{a}}^{(t_{r+1},t_{r+2})}).$$

$$a^{-1}X:=\{y\in A^{\ast }\mid ay\in X\}.$$

$$\mathrm{Som}(X):=\{a\in A\mid \exists u\in A^{\ast }:au\text{ is a minimal element of }X\}.$$

$$\uparrow f[X]\subseteq \uparrow f[Y]\Rightarrow X\subseteq Y.$$

$$f(x_1,\dots ,x_m):=\underset{x_1}{\underbrace{11\dots 1}}\underset{x_2}{\underbrace{22\dots 2}}\dots \underset{x_m}{\underbrace{mm\dots m}}=1^{x_1}{2}^{x_2}\dots m^{x_m}$$

Full text

Supported by the Austrian Science Fund (FWF): P29931.

1. Introduction

We investigate some finiteness conditions of a partially ordered set $({A},{\leq})$. As usual, we write $a\geq b$ for $b\leq a$, and $a<b$ (or $b>a$) for $a\leq b$ and $a\neq b$. Furthermore, $a\perp b$ stands for ($a\not\leq b$ and $b\not\leq a$); in this case, we call $a$ and $b$ uncomparable. A sequence $(a_{i})_{i\in\mathbb{N}_{0}}$ from $A$ is a descending chain if $a_{i}>a_{i+1}$ for all $i\in\mathbb{N}_{0}$, it is an antichain if for all $i,j\in\mathbb{N}_{0}$, $a_{i}\leq a_{j}$ implies $i=j$, and it is an ascending chain if for all $i\in\mathbb{N}_{0}$, $a_{i}<a_{i+1}$. A subset $U$ of $A$ is called upward closed if $u\in U$, $a\in A$, and $u\leq a$ imply $a\in U$. For a subset $B$ of $A$, we define the upward closed set generated by $B$ by ${\uparrow}B:=\{a\in A\mid\exists b\in B:b\leq a\}$. By $U({A},{\leq})$ or simply $U(A)$ we denote the set of upward closed subsets of $A$. This set can be ordered by set inclusion $\subseteq$.

One frequently uses the fact that certain partially ordered sets have no descending chain and no antichain; such orders are called well partial orders. [Lav76, AH07] provide powerful techniques to establish that a given order is a well partial order. In this note, we restrict our attention to some particular ordered sets: The first set that we consider is the set $\mathbb{N}_{0}^{m}$ of vectors of natural numbers of some fixed length $m$, which we order by $(a_{1},\ldots,a_{m})\leq(b_{1},\ldots,b_{m})$ if $a_{i}\leq b_{i}$ for all $i\in\{1,\ldots,m\}$, and we provide proofs of the following well known facts:

Theorem 1.1**.**

Let $m\in\mathbb{N}$.

1. (1)

$(\mathbb{N}_{0}^{m},\leq)$* has no descending chain and no antichain [Dic13, Lemma A].* 2. (2)

$(U(\mathbb{N}_{0}^{m},\leq),\subseteq)$* has no ascending chain and no antichain [Mac01], [AP04, Corollary 1.8].*

It is easy to see that $(\mathbb{N}_{0}^{m},\leq)$ has no descending chain. In 1913, Dickson proved that $(\mathbb{N}_{0}^{m},\leq)$ has no antichain, a fact that lies at the basis of the theory of Gröbner bases [BL83, Buc70]. This fact can be stated differently. We call an ideal of the polynomial ring $\mathbb{Q}[x_{1},\ldots,x_{N}]$ monomial if it is generated by monomials. Then Dickson’s Lemma states that every monomial ideal is finitely generated (which of course also follows from Hilbert’s Basis Theorem). A somewhat surprising fact is that the set of monomial ideals of $\mathbb{Q}[x_{1},\ldots,x_{n}]$ has no antichain, which has been proved in [Mac01], but probably much before in an order theoretic setting: in this setting, the result states that the set of upward closed subsets of $(\mathbb{N}_{0}^{m},\leq)$ has no antichain. A direct proof is given at the end of Section 4. Another proof using an ordering of words over a finite alphabet that goes back to G. Higman [Hig52] is given in Section 6. This is the second type of order relations we study: For a finite alphabet $A$, we say that a word $u\in A^{*}=\bigcup_{n\in\mathbb{N}_{0}}A^{n}$ embeds into a word $v$ if $u$ can be obtained from $v$ by cancelling some letters and in this case write $u\leq_{e}v$. For example $u=aabbca$ embeds into $v=\underline{a}b\underline{a}\underline{b}a\underline{b}\underline{c}\underline{a}c$. It follows from [Hig52] that for a finite set $A$, $(A^{*},\leq_{e})$ has no antichain. Also, the upward closed subsets of $(A^{*},\leq_{e})$ have no antichain [NW64]. Formally, we define when $x\leq_{e}y$ holds by recursion on the length of $x$. First, the empty word $\emptyset$ satisfies $\emptyset\leq y$ for all $y\in A^{*}$. If $x=au$ with $a\in A$ and $u\in A^{*}$, then $x\leq_{e}y$ if there are words $v,w\in A^{*}$ such that $y=vaw$ and $u\leq_{e}w$. Then we have:

Theorem 1.2**.**

Let $A$ be a finite set.

1. (1)

$(A^{*},\leq_{e})$* has no descending chain and no antichain [Hig52].* 2. (2)

$(U(A^{*},\leq_{e}),\subseteq)$* has no ascending chain and no antichain.*

We will give a proof of this theorem in Section 5. We also investigate the following ordering of words used in [AMM14]. Let $A$ be a finite set, and let $B:=(A\times\{0\})\cup(A\times\{1\})$. We define a mapping $\varphi:A^{*}\to B^{*}$ by $\varphi(a_{1},\ldots,a_{n}):=(b_{1},\ldots,b_{n})$ with $b_{i}:=(a_{i},0)$ if $a_{i}\not\in\{a_{1},\ldots,a_{i-1}\}$ and $b_{i}:=(a_{i},1)$ if $a_{i}\in\{a_{1},\ldots,a_{i-1}\}$. For $u=(a_{1},\ldots,a_{n})$, we use $S(u)$ to denote the set of letters that occur in $u$, formally $S(u):=\{a_{i}\mid i\in\{1,\ldots,n\}\}$. For $u,v\in A^{*}$, we say that $u\leq_{E}v$ if $\varphi(u)\leq_{e}\varphi(v)$ and $S(u)=S(v)$.

Theorem 1.3**.**

Let $A$ be a finite set.

1. (1)

$(A^{*},\leq_{E})$* has no descending chain and no antichain [AMM14, Lemma 3.2].* 2. (2)

$(U(A^{*},\leq_{E}),\subseteq)$* has no ascending chain and no antichain.*

The first listed author has used these results on several occasions: in [Aic10], the absence of antichains in $(A^{*},\leq_{e})$ is used to establish that every constantive Mal’cev clone on a finite set is finitely related. In [AM13], it is proved that the set of admissible higher commutator operations on a finite congruence lattice is well partially ordered by a natural ordering. The variant of Higman’s embedding ordering considered in Theorem 1.3 was used in [AMM14] to generalize the results from [Aic10], resulting in a proof that every finite algebra with few subpowers is finitely related, and [AM16] uses the same ordering to prove that every subvariety of the variety generated by such an algebra is finitely generated. Recently, McDevitt [McD18] gave a construction of a large class of word orderings generalizing $\leq_{e}$ and $\leq_{E}$, and he established a sufficient criterion [McD18, Proposition 47] for such orders to be well partial orders, thereby generalizing Theorems 1.2(1) and 1.3(1).

An open question in clone theory is whether there is a finite set with an antichain of clones containing a Mal’cev operation. One motivation for proving the absence of antichains is that this absence often allows testing certain properties of a structure by considering whether it contains finitely many forbidden substructures [RS04]. The aim of this note is to establish the order theoretic results that are listed above and used in [Aic10, AM13, AMM14, AM16] in a rather direct way. In particular, we will not resort to the theory of better quasi orderings [Lav76]. The note is self-contained, in particular we introduce the well known and very useful concept of minimal bad sequences due to [NW63], although this is done in a similar way at several other places (cf. [AH07]). However, we will not give a proof of the following theorem due to F.P. Ramsey [Ram29] (cf. [Neš95]): Denoting the $2$-element subsets of $\mathbb{N}_{0}$ by \big{(}\begin{smallmatrix}\mathbb{N}_{0}\\ 2\end{smallmatrix}\big{)}, Ramsey’s Theorem states that for every finite set $T$ and for every c:\big{(}\begin{smallmatrix}\mathbb{N}_{0}\\ 2\end{smallmatrix}\big{)}\to T, there is an infinite subset $Y$ of $\mathbb{N}_{0}$ such that the restriction of $c$ to \big{(}\begin{smallmatrix}Y\\ 2\end{smallmatrix}\big{)} is constant.

Most of the results and proofs in this note are well known and can be found, e.g., in [AH07]. However, we believe that the the two proofs of Theorem 1.1(2) and the proof of Theorem 1.2(2) are new. The derivation of Theorem 1.3 from Theorem 1.2 using quasi-embeddings (cf. [AH07]) was suggested to the author by N. Ruškuc in 2015 (cf. [Ruš15]). Considering [Lav76, NW64, AH07], the reader will easily find out that the theory of well quasi orders and better quasi orders is much deeper than scratched in the present note. The aim of the present note is to give easily accessible proofs for some basic and very useful results in this theory. The first listed author has faced the need for such proofs when teaching the basics of Gröbner Basis Theory, e.g., the existence of universal Gröbner bases (cf. [Aic09]) or, in universal algebra [BS81], the order theoretical foundations of the finite relatedness of finite algebras with few subpowers [AMM14].

2. Basics of order theory

It can easily be shown, using, however, some form of the axiom of choice, that a partially ordered set $({A},{\leq})$ has no descending chain if and only if every finite subset $X$ of $A$ contains a minimal element. A sequence $(b_{i})_{i\in\mathbb{N}_{0}}$ is a subsequence of $(a_{i})_{i\in\mathbb{N}_{0}}$ if there is a mapping $t:\mathbb{N}_{0}\to\mathbb{N}_{0}$ that satisfies $i<j\Rightarrow t(i)<t(j)$ for all $i,j\in\mathbb{N}_{0}$ and for all $i\in\mathbb{N}_{0}$, we have $b_{i}=a_{t(i)}$. In this case, $(b_{i})_{i\in\mathbb{N}_{0}}=(a_{t(i)})_{i\in\mathbb{N}_{0}}$.

Lemma 2.1**.**

Let ${\mathbb{A}}=({A},{\leq})$ be a partially ordered set, and let $S=(a_{i})_{i\in\mathbb{N}_{0}}$ be a sequence from $A$. Then $S$ has a subsequence $T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$ such that one of the following conditions holds:

1. (1)

$T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$* is constant, which means that for all $i,j\in\mathbb{N}_{0}$, $a_{t(i)}=a_{t(j)}$.* 2. (2)

$T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$* is a descending chain, which means that for all $i,j\in\mathbb{N}_{0}$, $i<j\Rightarrow a_{t(i)}>a_{t(j)}$.* 3. (3)

$T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$* is an ascending chain, which means that for all $i,j\in\mathbb{N}_{0}$, $i<j\Rightarrow a_{t(i)}<a_{t(j)}$.* 4. (4)

$T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$* is an antichain, which means that for all $i,j\in\mathbb{N}_{0}$, $i<j\Rightarrow a_{t(i)}\perp a_{t(j)}$.*

Proof: We define a coloring $c$ of the $2$-element subsets of $\mathbb{N}_{0}$. Let $i,j\in\mathbb{N}_{0}$ with $i<j$. We set $c(\{i,j\}):=1$ if $a_{i}=a_{j}$, $c(\{i,j\}):=2$ if $a_{i}<a_{j}$, $c(\{i,j\}):=3$ if $a_{i}>a_{j}$, and $c(\{i,j\}):=4$ if $a_{i}\perp a_{j}$. By Ramsey’s Theorem, $\mathbb{N}$ contains an infinite subset $Y$ such that $c$ is constant on \big{(}\begin{smallmatrix}Y\\ 2\end{smallmatrix}\big{)}. We let $t:\mathbb{N}_{0}\to Y$ be an injective increasing function from $\mathbb{N}_{0}$ into $Y$. Then $T:=(a_{t(i)})_{i\in\mathbb{N}_{0}}$ is the required subsequence. ∎

3. Bad sequences

Definition 3.1**.**

Let $({A},{\leq})$ be a partially ordered set. A sequence $(a_{i})_{i\in\mathbb{N}}$ from $A$ is good if there are $i,j\in\mathbb{N}_{0}$ such that $i<j$ and $a_{i}\leq a_{j}$, and it is bad if it is not good.

Lemma 3.2**.**

Let ${\mathbb{A}}=({A},{\leq})$ be an ordered set. The following are equivalent:

1. (1)

Every sequence from $A$ is good. 2. (2)

${\mathbb{A}}$* has no descending chain and no antichain.*

Proof: Assume that every sequence from $A$ is good. Then ${\mathbb{A}}$ has no descending chain and no antichain, since such chains are all bad. Now assume that ${\mathbb{A}}$ has no descending chain and no antichain, and let $(a_{i})_{i\in\mathbb{N}_{0}}$ be a sequence from $A$. Then by Lemma 2.1, $(a_{i})_{i\in\mathbb{N}_{0}}$ has a subsequence $T=(a_{t(i)})_{i\in\mathbb{N}_{0}}$ that is constant, a descending chain, an ascending chain, or an antichain. Descending chains and antichains are excluded by the assumptions, thus $T$ is either constant or ascending. In both cases, $a_{t(1)}\leq a_{t(2)}$, and hence $T$ is good. ∎

We call a sequence $(a_{i})_{i\in\mathbb{N}_{0}}$ a minimal bad sequence if it is a bad sequence, and for every $i\in\mathbb{N}_{0}$ and for every $b<a_{i}$, every sequence starting with $(a_{0},a_{1},\ldots,a_{i-1},b)$ is good.

Lemma 3.3**.**

Let ${\mathbb{A}}=({A},{\leq})$ be an ordered set. If ${\mathbb{A}}$ has no descending chain and it has an infinite antichain, then it has a minimal bad sequence.

Proof: We assume that $(A,\leq)$ has an antichain. This antichain is a bad sequence. Inductively, we can define a minimal bad sequence. We define $a_{0}$ to be a minimal element with respect to $\leq$ of S_{0}:=\{y_{0}\mid(y_{i})_{i\in\mathbb{N}_{0}}\text{ is a bad sequence from A}\}. For $j\in\mathbb{N}$, having defined $(a_{0},\ldots,a_{j-1})$, we let

[TABLE]

and we choose $a_{j}$ to be a minimal element of $S_{j}$ with respect to $\leq$. ∎

Lemma 3.4**.**

Let $({A},{\leq})$ be a partially ordered set. The following are equivalent:

1. (1)

$({A},{\leq})$* has no descending chain and no antichain.* 2. (2)

$({U(A,\leq)},{\subseteq})$* has no ascending chain.*

Proof: Let $(a_{i})_{i\in\mathbb{N}}$ be a descending chain or an antichain from $A$, and let $B_{i}:={\uparrow}\{a_{1},\ldots,a_{i}\}$ for $i\in\mathbb{N}_{0}$. Then $(B_{i})_{i\in\mathbb{N}}$ is an ascending chain. For the other implication, let $(C_{i})_{i\in\mathbb{N}_{0}}$ be an ascending chain, and choose $c_{i}\in C_{i+1}\setminus C_{i}$. We show that $(c_{i})_{i\in\mathbb{N}_{0}}$ is bad. Suppose $i<j$ and $c_{i}\leq c_{j}$. Since $C_{i+1}$ is upward closed, we then have $c_{j}\in C_{i+1}$, and thus $c_{j}\in C_{j}$. This contradicts the choice of $c_{j}$ in $C_{j+1}\setminus C_{j}$. Hence $({A},{\leq})$ has a bad sequence, and thus by Lemma 3.2, it must have a descending chain or an antichain. ∎

4. Dicksons’s ordering

In this section we provide a direct proof for Theorem 1.1 by using Ramsey’s theorem. In Section 6 we show that these results can also be obtained as a consequence of Theorem 1.2 by using quasi-embeddings.

Proof of Theorem 1.1(1)..

For $m\in\mathbb{N}$ and $k\in\{1,\ldots,m\}$ we denote the $m$-tuple $(a_{1},\dots,a_{m})\in\mathbb{N}_{0}^{m}$ by $\bm{a}$ and the $k$ th component of $\bm{a}$ by $\bm{a}_{k}$.

It can be easily seen that $(\mathbb{N}_{0}^{m},\leq)$ has no descending chain.

We assume that $(\mathbb{N}_{0}^{m},\leq)$ has an antichain $(\bm{a}^{(i)})_{i\in\mathbb{N}_{0}}$. Now we color the $2$-element subsets of $\mathbb{N}_{0}$ with the elements of ${\{1,2\}}^{\{1,\dots,m\}}$ as colors. For $i<j$, we set

[TABLE]

By Ramsey’s Theorem, we find a subsequence $(\bm{a}^{t(i)})_{i\in\mathbb{N}_{0}}$ and a color $C\in\{1,2\}^{\{1,\dots,m\}}$ such that for all $i,j\in\mathbb{N}_{0}$ with $i<j$, we have $c(\{t(i),t(j)\})=C$.

Assume there exists $k\in{\{1,\dots,m\}}$ such that $C(k)=\,\,2$. Then $\bm{a}^{(t_{0})}_{k}>\bm{a}^{(t_{1})}_{k}>\bm{a}^{(t_{2})}_{k}>\cdots$ in contradiction to the fact that $(\mathbb{N}_{0},\leq)$ has no descending chain.

Hence $C(k)=1$ for all $k$ and therefore $\bm{a}^{(t_{0})}\leq\bm{a}^{(t_{1})}\leq\bm{a}^{(t_{2})}\leq\cdots$, contradicting our assumption of $(\bm{a}^{(i)})_{i\in\mathbb{N}_{0}}$ being an antichain. ∎

Proof of Theorem 1.1(2)..

By Theorem 1.1(1), $(\mathbb{N}_{0}^{m},\leq)$ has no descending chain and no antichain. Hence Lemma 3.4 yields that $(U(\mathbb{N}_{0}^{m},\leq),\subseteq)$ has no ascending chain. It remains to show that $(U(\mathbb{N}_{0}^{m},\leq),\subseteq)$ has no antichain. If $m=1$, the set of upward closed subsets of $\mathbb{N}_{0}$ is totally ordered, and hence there cannot be an antichain. In the case $m\geq 2$, we encode each upward closed subset of $\mathbb{N}_{0}^{m}$ by a function from $\mathbb{N}_{0}^{m-1}$ to $\mathbb{N}_{0}$. For each upward closed subset $F$ of $\mathbb{N}_{0}^{m}$ we define a function $\Phi_{F}:\mathbb{N}_{0}^{m-1}\to\mathbb{N}_{0}\cup\{\infty\}$ by

[TABLE]

for $\bm{a}\in\mathbb{N}_{0}^{m-1}$. First we show that for all $\bm{a},\bm{b}\in\mathbb{N}_{0}^{m-1}$ such that $\bm{a}\leq\bm{b}$, also $\Phi_{F}(\bm{a})\geq\Phi_{F}(\bm{b})$ holds. Therefore, let $c:=\Phi_{F}(\bm{a})$. We assume $c\neq\infty$. We have $(\bm{a},c)\in F$. Since $F$ is an upward closed set, also $(\bm{b},c)\in F$, and hence $\Phi_{F}(\bm{b})\leq c=\Phi_{F}(\bm{a})$. Moreover, for the upward closed subsets $F,G$ of $\mathbb{N}_{0}^{m}$ the inclusion $F\subseteq G$ holds if and only if $\Phi_{F}(\bm{a})\geq\Phi_{G}(\bm{a})$ for all $\bm{a}\in\mathbb{N}_{0}^{m-1}$.

Let $(F_{i})_{i\in\mathbb{N}_{0}}$ be an antichain in $(U(\mathbb{N}_{0}^{m},\leq),\subseteq)$. Thus for $i,j\in\mathbb{N}$ with $i<j$, $F_{j}\not\subseteq F_{i}$ holds. Accordingly, there exists $\bm{a}^{(i,j)}\in\mathbb{N}_{0}^{m-1}$ such that

[TABLE]

Now we color the $3$-element subsets of $\mathbb{N}_{0}$ with the elements of ${\{1,2\}}^{\{1,\dots,m-1\}}$ as colors. For $l\in\{1,\ldots,m-1\}$, we denote the $l$ th component of $\bm{a}^{(i,j)}$ by $\bm{a}^{(i,j)}_{l}$. For $i<j<k$ we define the coloring of $\{i,j,k\}$ in the following way:

[TABLE]

By Ramsey’s Theorem, we find an infinite subset $T$ of $\mathbb{N}_{0}$, $T=\{t_{1},t_{2},\dots\}$, $t_{1}<t_{2}<\dots$, and a color $C\in\{1,2\}^{\{1,\dots,m-1\}}$ such that for all $i,j,k\in\mathbb{N}_{0}$ with $i<j<k$, we have $c(\{t_{i},t_{j},t_{k}\})=C$. We now show that $C(l)=\,\,1$ for all $l\in\{1,\ldots,m-1\}$.

In contradiction to that, we assume that there exists $l$ such that $C(l)\,=\,\,2$. Then

[TABLE]

holds. Hence we have constructed a descending chain of natural numbers, which is impossible.

Thus for all $r\in\mathbb{N}_{0}$ the inequality $\bm{a}^{(t_{r},t_{r+1})}\leq\bm{a}^{(t_{r+1},t_{r+2})}$ holds. Now let $r\in\mathbb{N}_{0}$. Because of the choice of $\bm{a}^{(t_{r},t_{r+1})}$, we have

[TABLE]

Since $\bm{a}^{(t_{r},t_{r+1})}\leq\bm{a}^{(t_{r+1},t_{r+2})}$, we also have

[TABLE]

Hence the sequence $(\Phi_{F_{t_{i}}}(\bm{a}^{(t_{i},t_{i+1})}))_{i\in\mathbb{N}_{0}}$ is a descending chain in $\mathbb{N}_{0}\cup\{\infty\}$, which is impossible.

Consequently there cannot exist an antichain of upward closed subsets of $\mathbb{N}_{0}^{m}$. ∎

5. Higman’s ordering

Of the results listed in Section 1, we now derive Theorem 1.2.

Proof of Theorem 1.2(1)..

Since $u<_{e}v$ implies that the length of $u$ is strictly smaller than the length of $v$, $(A^{*},\leq_{e})$ has no descending chain. We assume that $(A^{*},\leq_{e})$ has an antichain. Then by Lemma 3.3, we find a minimal bad sequence $U=(u_{i})_{i\in\mathbb{N}_{0}}$ be in $A^{*}$ Such a sequence cannot contain the empty word, since $u_{i}=\emptyset$ implies $u_{i}\leq_{e}u_{i+1}$, and therefore $U$ is good. Hence we can write $u_{i}=a_{i}v_{i}$ with $a_{i}\in A$ and $v_{i}\in A^{*}$. Since $A$ is finite, there is a subsequence $(a_{t(i)})_{i\in\mathbb{N}_{0}}$ and $b\in A$ such that $a_{t(i)}=b$ for all $i\in\mathbb{N}_{0}$. The sequence $(u_{1},u_{2},\ldots,u_{t(0)-1},v_{t(0)},v_{t(1)},\ldots)$ is smaller, hence good. Therefore, we find $i,j\in\mathbb{N}_{0}$ such that either $i<j<t(0)$ and $u_{i}\leq_{e}u_{j}$, or $i<t(0)$ and $u_{i}\leq_{e}v_{t(j)}$, or $i<j$ and $v_{t(i)}\leq_{e}v_{t(j)}$. In the case $i<j<t(0)$ and $u_{i}\leq_{e}u_{j}$, $U$ is good, contradicting the assumptions. If $i<t(0)$ and $u_{i}\leq_{e}v_{t(j)}$, then $u_{i}\leq_{e}v_{t(j)}\leq_{e}a_{j}v_{t(j)}=u_{t(j)}$, and thus $U$ is good, contradicting the assumptions. If $i<j$ and $v_{t(i)}\leq_{e}v_{t(j)}$, then $u_{t(i)}=a_{(t(i))}v_{t(i)}=bv_{t(i)}\leq_{e}bv_{t(j)}=a_{(t(j))}v_{t(j)}=u_{t(j)}$, and thus $U$ is good, again contradicting the assumptions. ∎

Before proving Theorem 1.2(2), we need some preparation. For $X\subseteq A^{*}$ and $a\in A$, we define $a^{-1}X$ by

[TABLE]

The set of starting letters of minimal elements of $X$ is defined by

[TABLE]

Lemma 5.1**.**

Let $X,Y$ be a upward closed subsets of $({A^{*}},{\leq_{e}})$, and let $a\in X$. Then we have:

1. (1)

$X\subseteq a^{-1}X$. 2. (2)

If $a\in\operatorname{Som}(X)$, then $X\subset a^{-1}X$. 3. (3)

If $X\neq A^{*}$ and for all $b\in\operatorname{Som}(X)$, $b^{-1}X\subseteq b^{-1}Y$. Then $X\subseteq Y$.

Proof.

For item (1), let $x\in X$. Then $x\leq_{e}ax$, and therefore, since $X$ is upward closed, $ax\in X$. Thus $x\in a^{-1}X$.

For item (2), we choose $y\in A^{*}$ such that $ay$ is a minimal element of $X$ with respect to $\leq_{e}$. Then $y\in a^{-1}X$. Since $y<_{e}ay$ and $ay$ is minimal in $X$, we have $y\not\in X$. Thus the inclusion $X\subset a^{-1}X$ is indeed proper.

For item (3), we fix $x\in X$. Then there is a minimal element $y$ of $X$ with respect to $\leq_{e}$ such that $y\leq_{e}x$. In the case that $y$ is the empty word, $X=A^{*}$, which is excluded by the assumptions. If $y$ is not empty, there is $b\in A$ and $z\in A^{*}$ such that $y=bz$. Then $b\in\operatorname{Som}(X)$ and $z\in b^{-1}X$. Therefore $z\in b^{-1}Y$, and therefore $bz\in Y$. Thus $y\in Y$, and since $Y$ is upward closed, $x\in Y$. ∎

Proof of Theorem 1.2(2)..

By Theorem 1.2(1), $(A^{*},\leq)$ has no descending chain and no antichain. Hence Lemma 3.4 yields that $(U(A^{*},\leq_{e}),\subseteq)$ has no ascending chain. We call a sequence $(X_{i})_{i\in\mathbb{N}_{0}}$ co-good if there are $i,j\in\mathbb{N}$ with $i<j$ and $X_{i}\supseteq X_{j}$, and co-bad otherwise. We assume that $U(A^{*})$ has an antichain. This antichain is a co-bad sequence. Inductively, we can define a maximal co-bad sequence. We define $X_{0}$ to be a maximal element with respect to $\subseteq$ of S_{0}:=\{Y_{0}\mid(Y_{i})_{i\in\mathbb{N}_{0}}\text{ is a co-bad sequence of U(A^{})}\}. This maximal element exists because $(U(A^{*},\leq_{e}),\subseteq)$ has no ascending chain. For $j\in\mathbb{N}$, having defined $(X_{0},\ldots,X_{j-1})$, we let S_{j}:=\{Y_{j}\mid(Y_{i})_{i\in\mathbb{N}_{0}}\text{ is a co-bad sequence of U(A^{}) with }(Y_{0},\ldots,Y_{j-1})=(X_{0},\ldots,X_{j-1})\}, and we choose $X_{j}$ to be a maximal element of $S_{j}$ with respect to $\subseteq$.

Since the sequence $(\operatorname{Som}(X_{i}))_{i\in\mathbb{N}_{0}}$ can take at most $2^{|A|}$ values, there is $B\subseteq A$ and a subsequence $(X_{t(i)})_{i\in\mathbb{N}_{0}}$ such that $\operatorname{Som}(X_{t(i)})=B$ for all $i\in\mathbb{N}_{0}$. Now we color the two element subsets of $\mathbb{N}_{0}$ with the elements of $\{1,2\}^{B}$ as colors. For $i<j$ and $b\in B$, we set $c(\{i,j\})\,(b):=1$ if $b^{-1}X_{i}\not\supseteq b^{-1}X_{j}$, $c(\{i,j\})\,(b):=2$ if $b^{-1}X_{i}\supseteq b^{-1}X_{j}$. We restrict our coloring to the two element subsets of $t[\mathbb{N}_{0}]$. By Ramsey’s Theorem, we find a subsequence $(X_{t(r(i))})_{i\in\mathbb{N}_{0}}=:(X_{s(i)})_{i\in\mathbb{N}_{0}}$ and a color $C\in\{1,2\}^{B}$ such that for all $i,j\in\mathbb{N}_{0}$ with $i<j$, we have $c(\{s(i),s(j)\})=C$. Furthermore, $\operatorname{Som}(X_{s(i)})=B$ for all $i\in\mathbb{N}_{0}$.

Let us first consider the case that we have $b\in B$ such that $C(b)=1$. To this end, we consider the sequence $Y:=(X_{1},X_{2},\ldots,X_{s(0)-1},b^{-1}X_{s(0)},b^{-1}X_{s(1)},\ldots)$. Since $b\in\operatorname{Som}(X_{s(0)})$, Lemma 5.1(2) yields $X_{s(0)}\subset b^{-1}X_{s(0)}$. By the maximality of $(X_{i})_{i\in\mathbb{N}_{0}}$, the sequence $Y$ is co-good. Therefore, we find $i,j\in\mathbb{N}_{0}$ such that either $i<j<s(0)$ and $X_{i}\supseteq X_{j}$, or $i<s(0)$ and $X_{i}\supseteq b^{-1}X_{s(j)}$, or $i<j$ and $b^{-1}X_{s(i)}\supseteq b^{-1}X_{s(j)}$. The case $i<j<s(0)$ and $X_{i}\supseteq X_{j}$ cannot occur because the sequence $(X_{i})_{i\in\mathbb{N}_{0}}$ is co-bad. In the case $i<s(0)$ and $X_{i}\supseteq b^{-1}X_{s(j)}$, Lemma 5.1 (1) yields $X_{i}\supseteq b^{-1}X_{s(j)}\supseteq X_{s(j)}$. Again this cannot occur because $(X_{i})_{i\in\mathbb{N}_{0}}$ is co-bad. In the case $i<j$ and $b^{-1}X_{s(i)}\supseteq b^{-1}X_{s(j)}$, we obtain $c(\{s(i),s(j)\})\,(b)=2$, contradicting the assumption $C(b)=1$.

Hence we have $C(b)=2$ for all $b\in B$. Therefore, for all $b\in\operatorname{Som}(X_{s(1)})$, we have $b^{-1}(X_{s(0)})\supseteq b^{-1}(X_{s(1)})$. Since $(X_{i})_{i\in\mathbb{N}_{0}}$ is co-bad, we have $X_{s(1)}\not\supseteq X_{s(2)}$. From this, we conclude $X_{s(1)}\neq A^{*}$. Now Lemma 5.1 (3) yields $X_{s(0)}\supseteq X_{s(1)}$, contradicting the fact that $(X_{i})_{i\in\mathbb{N}_{0}}$ is co-bad. ∎

6. Other well partially ordered sets

We will derive the other results stated in Section 1 by using quasi-embeddings:

Definition 6.1**.**

[AH07] Let ${\mathbb{A}}=(A,\leq_{A})$ and ${\mathbb{B}}=(B,\leq_{B})$ be partially ordered sets. A mapping $f:A\to B$ is a quasi-embedding from ${\mathbb{A}}$ into ${\mathbb{B}}$ if for all $a_{1},a_{2}\in A:f(a_{1})\leq_{B}f(a_{2})\Rightarrow a_{1}\leq_{A}a_{2}$.

Lemma 6.2**.**

[AH07]** Let $({A},{\leq_{A}})$ and $({B},{\leq_{B}})$ be partially ordered sets. Let $f:A\to B$ be a quasi-embedding from $({A},{\leq_{A}})$ into $({B},{\leq_{B}})$. Then we have:

1. (1)

If $({A},{\leq_{A}})$ has an antichain, then $({B},{\leq_{B}})$ has an antichain. 2. (2)

If $({A},{\leq_{A}})$ has a descending chain, then $({B},{\leq_{B}})$ has an antichain or a descending chain.

Proof: For (1), we let $(a_{i})_{i\in\mathbb{N}_{0}}$ be an antichain. We define $b_{i}:=f(a_{i})$ and we claim that $(b_{i})_{i\in\mathbb{N}}$ is an antichain. To this end, let $j,k\in\mathbb{N}_{0}$ with $b_{j}\leq b_{k}$. Then $f(a_{j})\leq f(a_{k})$, and therefore $a_{j}\leq a_{k}$. Since $(a_{i})_{i\in\mathbb{N}_{0}}$ is an antichain, $j=k$, completing the proof that $(b_{i})_{i\in\mathbb{N}_{0}}$ is an antichain.

For (2), let $(a_{i})_{i\in\mathbb{N}_{0}}$ be a descending chain. According to Lemma 2.1, it is sufficient to show that that the sequence $(b_{i})_{i\in\mathbb{N}_{0}}:=(f(a_{i}))_{i\in\mathbb{N}_{0}}$ has no subsequence that is constant or an ascending chain. Seeking a contradiction, we let $(b_{t(i)})_{i\in\mathbb{N}_{0}}$ be a subsequence with $b_{t(1)}\leq b_{t(2)}\leq\cdots$. Since $f$ is a quasi-embedding, $a_{t(1)}\leq_{A}a_{t(2)}$, which contradicts the fact that $(a_{i})_{i\in\mathbb{N}_{0}}$ is descending. ∎

Lemma 6.3**.**

Let $({A},{\leq_{A}})$ and $({B},{\leq_{B}})$ be partially ordered sets. Let $f:A\to B$ be a quasi-embedding from $({A},{\leq_{A}})$ into $({B},{\leq_{B}})$. Let $\phi:U(A,\leq_{A})\to U(B,\leq_{B})$ be defined by $\phi(X):={\uparrow}f[X]=\{b\in B\mid\exists x\in X:b\geq_{B}f(x)\}$. Then $\phi$ is a quasi-embedding from $(U(A,\leq_{A}),\subseteq)$ into $(U(B,\leq_{B}),\subseteq)$.

Proof.

We have to show that for all $X,Y\in U(A)$, we have

[TABLE]

To this end, we fix $X,Y\in U(A)$ and assume ${\uparrow}f[X]\subseteq{\uparrow}f[Y]$. We let $x\in X$. Then $f(x)\in f[X]$. Hence $f(x)\in{\uparrow}f[X]$, and therefore $f(x)\in{\uparrow}f[Y]$. Therefore, there exists $y\in Y$ such that $f(x)\geq_{B}f(y)$. By the assumptions on $f$, $x\geq_{A}y$. Since $Y$ is upward closed, we obtain $x\in Y$. This completes the proof of (6.1). ∎

Corollary 6.4**.**

Let $(A,\leq_{A})$ and $(B,\leq_{B})$ be partially ordered sets, and let $f$ be a quasi-embedding from $(A,\leq_{A})$ into $(B,\leq_{B})$. If $(B,\leq_{B})$ has no descending chain and no antichain and $(U(B,\leq_{B}),\subseteq)$ has no antichain, then $(A,\leq_{A})$ has no descending chain and no antichain, and $(U(A,\leq_{A}),\subseteq)$ has no ascending chain and no antichain.

Proof.

From Lemma 6.2(1), we obtain that $({A},{\leq_{A}})$ has no antichain. From Lemma 6.2(2), we obtain that $({A},{\leq_{A}})$ has no descending chain. Lemma 3.4 now yields that $(U(A,\leq_{A}),\subseteq)$ has no ascending chain. By Lemma 6.3, there exists a quasi-embedding from $(U(A,\leq_{A}),\subseteq)$ into $(U(B,\leq_{B}),\subseteq)$. Applying Lemma 6.2(1) to this quasi-embedding, we obtain that $(U(A,\leq_{A}),\subseteq)$ has no antichain. ∎

Proof of Theorem 1.1..

We first construct a quasi-embedding from ${\mathbb{A}}:=(\mathbb{N}_{0}^{m},\leq)$ into ${\mathbb{B}}=(\{1,\ldots,m\}^{*},\leq_{e})$. We define $f:A\to B$ by

[TABLE]

for all $(x_{1},\ldots,x_{m})\in\mathbb{N}_{o}^{m}$. Now if $1^{x_{1}}2^{x_{2}}\ldots m^{x_{m}}\leq_{e}1^{y_{1}}2^{y_{2}}\ldots m^{y_{m}}$, then $(x_{1},\ldots,x_{m})\leq(y_{1},\ldots,y_{m})$. Thus $f$ is a quasi-embedding. By Corollary 6.4 and Theorem 1.2, ${\mathbb{A}}$ has no antichain and no descending chain and $(U({\mathbb{A}}),\subseteq)$ has no ascending chain and no antichain. ∎

Proof of Theorem 1.3..

We define a mapping $f:A^{*}\to B^{*}\times\mathcal{P}(A)$ by $f(u):=(\varphi(u),S(u))$. We order the set $B^{*}\times\mathcal{P}(A)$ by $(u,S)\leq(v,T)$ if $u\leq_{e}v$ and $S=T$. Then $f$ is a quasi-embedding from $(A^{*},\leq_{E})$ into $B^{*}\times\mathcal{P}(A)$. Since $(B^{*},\leq_{e})$ has no antichain and $\mathcal{P}(A)$ is finite, $B^{*}\times\mathcal{P}(A)$ has no antichain. Hence Lemma 6.2(1) implies that $(A^{*},\leq_{E})$ has no antichain. Since $u<_{E}v$ implies that the length of $u$ is strictly smaller than the length of $v$, $(A^{*},\leq_{E})$ has no descending chain. Thus, using Lemma 3.4, we get that $(U(A^{*},\leq_{E}),\subseteq)$ has no ascending chain.

It remains to show that $(U(A^{*},\leq_{E}),\subseteq)$ has no antichain. Using Corollary 6.4, it is sufficient to show that $C:=(U(B^{*}\times\mathcal{P}(A),\leq),\subseteq)$ has no antichain. Seeking a contradiction, we let $(M_{i})_{i\in\mathbb{N}_{0}}$ be an antichain in $C$, and for $S\in\mathcal{P}(A)$, let $X(M_{i},S):=\{w\in B^{*}\mid(w,S)\in M_{i}\}$. Then $(\langle X(M_{i},S)\mid S\in\mathcal{P}(A)\rangle)_{i\in\mathbb{N}_{0}}$ is a sequence in $(U(B^{*},\leq_{e}))^{\mathcal{P(A)}}$. Since $\mathcal{P}(A)$ is finite, and since $U(B^{*},\leq_{e})$ has no ascending chain and no antichain by Theorem 1.2(2), a modification of the argument in the proof of Theorem 1.1(1) can be used to show that $(U(B^{*},\leq_{e}))^{\mathcal{P(A)}}$ has no ascending chain and no antichain. Hence by the dual of Lemma 3.2, there are $i,j\in\mathbb{N}$ with $i<j$ such that $X(M_{i},S)\supseteq X(M_{j},S)$ for all $S\in\mathcal{P}(A)$. Therefore $M_{i}\supseteq M_{j}$, contradicting the fact that $(M_{i})_{i\in\mathbb{N}_{0}}$ is an antichain. ∎

7. Acknowledgements

The authors would like to thank J. Farley and N. Ruškuc for numerous discussions on the theory of ordered sets. Part of this research was done while the first listed author was visiting P. Aglianò at the University of Siena, Italy.