Codex @codex  0

The barrier in better-quasi-order theory $[\omega {]}^2$ has shifts $\{m,n\}◃\{n,l\}$ for $m<n<l$. Map each pair to the corresponding element of the Rado order. Neither clause of $(m,n){\le }_R(n,l)$ holds: the first coordinates differ, and the second coordinate of the first pair equals, rather than precedes, the first coordinate of the second. Thus this array is bad, although the Rado order is a well-quasi-ordering. It separates better-quasi-ordering from well-quasi-ordering.

For subsets of a preorder, Hoare domination means every source element has some greater or equal target element: $S{\le }_{H}T⟺\forall s\in S\exists t\in T(s\le t)$. It is reflexive and transitive. If the base is a well-quasi-ordering, its finite subsets are well-quasi-ordered by this relation; arbitrary subsets need not be, as the Rado order shows.

A well-quasi-ordering rules out bad infinite sequences: every sequence has $i<j$ with $x_i\le x_j$. It behaves well under finite products and finite words, but infinitary constructions expose a limitation. Better-quasi-ordering, introduced by Nash-Williams, strengthens the sequence test to a test on families of finite initial segments, and thereby provides a robust theory of infinitary closure.

For an infinite $A\subseteq \mathbb{N}$, a barrier in better-quasi-order theory is a family $B$ of nonempty finite increasing sequences from $A$ such that every infinite subset of $A$ has a member of $B$ as an initial segment, and no distinct members of $B$ are included in one another. The initial segment is unique. Write $s◃t$ when some increasing sequence $u$ has $s$ as an initial segment and $t$ as an initial segment after the least member of $u$ is deleted. For instance, in $[A{]}^2$ this means $(a,b)◃(b,c)$ with $a<b<c$.

A map $f:B\to Q$ is good if some $s◃t$ satisfies $f(s){\le }_{Q}f(t)$. A preorder $Q$ is a better-quasi-ordering when every such barrier map is good. Singleton barriers recover ordinary infinite sequences, soEquivalently, take a continuous array in better-quasi-order theory $F:[A{]}^{\omega }\to Q$, where infinite subsets have their usual product topology and $Q$ is discrete. Put $X^{-}=X\setminus \{\min X\}$. Better-quasi-ordering excludes arrays with $F(X)\nleqq F(X^{-})$ for every $X$. A barrier map gives a locally constant array by reading the unique initial segment; the converse uses refinement of the front of finite segments on which a continuous map becomes constant. This viewpoint makes the shift, rather than mere inclusion of subsets, central.

The implication is strict. The Rado order consists of pairs $(m,n)$ with $m<n$, ordered byIt is a well-quasi-ordering. In any infinite sequence, either the first coordinates are unbounded, giving a later first coordinate above the earlier second coordinate, or one first coordinate occurs infinitely often and its second coordinates have a nondecreasing pair. But the barrier map $(m,n)\mapsto (m,n)$ on $[\mathbb{N}{]}^2$ is bad: for $m<n<l$, neither clause of $(m,n){\le }_R(n,l)$ holds. Thus the Rado order is WQO but not BQO.

The same example explains why strengthening the definition is useful. Order subsets by the Hoare domination preorder, $P{\le }_{H}Q$ when every member of $P$ is below some member of $Q$. The infinite rows $R_m=\{(m,n):n>m\}$ form an antichain. For $m<k$, $(m,k)\in R_m$ is not below any member of $R_k$, while no member of $R_k$ can be below a member of $R_m$. Thus the full power set of a well-quasi-ordering need not be a well-quasi-ordering.

In contrast, power-set closure of better-quasi-orderings has a short array proof. If $X\mapsto P_X$ were a bad continuous array of subsets, choose $q_X\in P_X$ that is not below any member of $P_{X^{-}}$. Make this choice by a fixed choice rule depending only on the two subsets. They are locally constant, so $X\mapsto q_X$ is continuous. Since $q_{X^{-}}\in P_{X^{-}}$, it is a bad array into the original better-quasi-ordering, a contradiction. Empty subsets cannot be the source of a bad comparison. Thus the full power set, not just its finite subsets, preserves better-quasi-ordering under domination.

Further examples and closure principles show the scope of the theory. Every well-order is a better-quasi-ordering: a bad array would give a strictly descending ordinal sequence by successively deleting the least member of one infinite set. Every finite equality order is a better-quasi-ordering by the Nash-Williams barrier partition theorem, which makes a finite coloring of a barrier constant on an infinite restriction. The same theorem proves closure under finite products: color a bad product array by the first coordinate in which the shifted comparison fails, and pass to a homogeneous infinite restriction to obtain a bad coordinate array. Finite disjoint unions work by first homogenizing the component.

Nash-Williams' extension to countable transfinite sequences says that sequences of labels from a better-quasi-ordering, of arbitrary countable ordinal length, are again better-quasi-ordered by increasing label-monotone embeddings. It is an infinitary counterpart of Higman's lemma. Its formulation and logical strength are studied in arxiv.org/abs/math/9408204 . The point is not that every closure proof becomes elementary, but that barriers and continuous arrays supply the stable induction and partition framework absent from ordinary sequence tests. The modern account arxiv.org/pdf/1604.05866 develops this framework and the barrier partition theorem.

A BQO, or better-quasi-ordering, strengthens a well-quasi-ordering by testing arrays indexed by arbitrary barriers, rather than just sequences. Let $Q$ be a preorder. A barrier in better-quasi-order theory on an infinite $A\subseteq \omega$ is a family $B$ of nonempty finite increasing sequences such that no member is properly contained in another and every infinite subset of $A$ has a member of $B$ as its initial segment.

For barrier members write $s◃t$ if a finite increasing sequence $u$ has $s$ as an initial segment and has $t$ as an initial segment after deleting the first entry of $u$. A map $f:B\to Q$ is good when some $s◃t$ satisfy $f(s)\le f(t)$. The definition isBoth quantifiers matter: testing only barriers of one fixed finite length is insufficient.

For the singleton barrier $B=[A{]}^1$, shifted pairs are just $\{a\}◃\{b\}$ with $a<b$. Thus the BQO condition implies the ordinary well-quasi-ordering condition. Barrier arrays can also be viewed through continuous arrays in better-quasi-order theory $F:[A{]}^{\omega }\to Q$, with $Q$ discrete: set $F(X)=f(s)$ for the unique barrier initial segment $s$ of $X$. This is continuous because that value is fixed by a finite prefix, and comparing $X$ with $X\setminus \{\min X\}$ is exactly the shift comparison. The general continuous-array formulation is equivalent to the barrier definition; its converse uses refinement of fronts to barriers, rather than the incorrect assertion that every prefix-decision family is already a barrier.

For an illustrative positive example, every well-order is a BQO. If a barrier array into a well-order were bad, its induced $F$ would satisfyfor every infinite $X$. This is an infinite strict descent in a well-order, a contradiction. A finite antichain with equality is also a BQO: the finitely many fibres of the induced continuous array are open, so repeated use of the Open Ramsey theorem gives an infinite $A^{\prime }$ on which the array is constant. The comparison of an infinite subset of $A^{\prime }$ with its shift is then good.

The implication to a well-quasi-order is strict. The Rado order isIt is well-quasi-ordered: if one first coordinate occurs infinitely often, its second coordinates have an increasing pair; otherwise the first coordinates are unbounded, and a later one exceeds the second coordinate of a chosen earlier pair. But take the barrier $[\omega {]}^2$ and $f(\{m,n\})=(m,n)$. For $m<n<l$, the shift compares $(m,n)$ with $(n,l)$; their first coordinates differ and $n<n$ is false. Every shifted comparison therefore fails. This bad barrier array for the Rado order proves well-quasi-ordering does not imply better-quasi-ordering.

The stronger notion is designed for infinitary closure operations, such as powerset and infinite-sequence constructions, where a mere well-quasi-order can develop new bad arrays. The barrier definition records variable finite amounts of information from an infinite object and is consequently more robust than checking sequences alone.

Use the natural partial orderThe paper's shorthand $x\le x+y$ specifies the join order: more formally, $x\le z$ means $z=x+y$ for some $y$, which is equivalent to the displayed condition by idempotence. Associativity, commutativity and idempotence make addition the join operation. Distributivity makes multiplication monotone, while absorption gives $xy\le x$ and $xy\le y$.

Let $g_1,\dots ,g_r$ generate the incline. Distributivity writes every element as a nonempty finite join of monomialsThere is no assumed multiplicative identity: an exponent zero means omit that factor, and at least one factor remains. If $u\le v$ coordinatewise, absorption gives $m(v)\le m(u)$. If the two vectors are equal the conclusion is immediate; otherwise multiply $m(u)$ by the additional factors and apply absorption. For a polynomial with finite exponent support $A$, put $U_A=\{v:(\exists u\in A)u\le v\}$, its upward closure. If $U_A\supseteq U_B$, every monomial in $B$ is below some monomial in $A$, and consequently the join represented by $B$ is below the join represented by $A$.

We prove the reverse-inclusion well-quasi-ordering of monomial ideals, often called Maclagan's theorem, in the equivalent form that upward-closed subsets of ${\mathbb{N}}^r$ are well-quasi-ordered by reverse inclusion. Include the empty subset. For $r=1$, these are the tails $[n,\infty )$ and the empty set, corresponding to $\mathbb{N}\cup \{\infty \}$ in its usual well-order.

Induct on $r$. For an upward-closed $U\subseteq {\mathbb{N}}^r$, its slices $U_k=\{v\in {\mathbb{N}}^{r-1}:(v,k)\in U\}$ increase with $k$. Their union is upward closed and has finitely many minimal elements by the Dickson lemma; all those elements have appeared by some finite slice. Thus the slices eventually stabilize. Encode $U$ by the finite word of its pre-stabilization slices and its final stable slice. By the induction hypothesis, the slice order given by reverse inclusion is a well-quasi-ordering. The Higman lemma and finite product closure of well-quasi-orderings make these word-and-tail encodings well-quasi-ordered.

For a good pair of encodings $U,V$, an increasing subsequence map $f$ satisfies $U_k\supseteq V_{f(k)}$ at every pre-stabilization position of $U$, and the final slice of $U$ contains the final slice of $V$. Since $f(k)\ge k$ and the $V$ slices increase, $U_k\supseteq V_k$ before stabilization. Afterwards the final-slice comparison also gives $U_k\supseteq V_k$. Hence $U\supseteq V$, proving the induction. The use of Higman lemma is essential: finite-subset reverse domination does not preserve an arbitrary well-quasi-order, as the Rado order shows.

Given any infinite sequence of incline elements, choose a finite monomial support $A_i$ for each. The result just proved gives $i<j$ with $U_{A_i}\supseteq U_{A_j}$, hence $x_i\ge x_j$. Every finitely generated incline is therefore well-quasi-ordered by its reversed natural order.

As printed, yes: the relation is universal. The original PDF puts the existentially quantified element in the same subset as the universally quantified element; this is not an OCR substitution. For each element of that subset, choose the element itself as witness, using reflexivity of the underlying well-quasi-ordering. For an empty subset the condition is vacuous. The truth value therefore never depends on the proposed target subset. On the power set this is a reflexive transitive relation, and every two terms of any infinite sequence are related:A well-quasi-ordering need not be antisymmetric, so universal comparability is allowed.

If the existential element is instead intended to belong to the target subset, the natural relation is the Hoare domination preorderFor arbitrary subsets the answer to that corrected question is no. Here is a complete counterexample, the Rado order. TakeThis is a partial order. Reflexivity is immediate. For transitivity, two same-row comparisons compose ordinarily. If the first comparison crosses rows and the second stays in its row, its target first coordinate is unchanged, so the cross-row inequality persists. If the first stays in its row and the second crosses, use $n\le l<k^{\prime }$; if both cross, use $n<k<l<k^{\prime }$. Antisymmetry follows because comparisons across distinct rows in both directions would require $n<k<l<m<n$.

The Rado order is a well-quasi-ordering. Given an infinite sequence $(m_i,n_i)$, if a first coordinate $m$ repeats infinitely often, its corresponding natural-number second coordinates have a nondecreasing pair, giving a same-row comparison. Otherwise each first coordinate occurs only finitely often, so the first coordinates in every tail are unbounded. In particular some later $m_j$ exceeds $n_0$, giving $(m_0,n_0){\le }_R(m_j,n_j)$.

For each $m$, take the infinite row $S_m=\{(m,n):n>m\}$. For distinct $m,k$, choose $n>\max (m,k)$. The element $(m,n)\in S_m$ is below no member $(k,l)$ of $S_k$, since its first coordinate is different and $n<k$ fails. Thus $S_m/{\le }_H{S}_k$ for every pair of distinct indices. These subsets form an infinite antichain, so the Hoare domination preorder on $\mathcal{P}(R)$ is not a well-quasi-ordering.

If only finite subsets were intended, the answer changes again: the finite-subset lifting of a well-quasi-order is a well-quasi-ordering. Enumerate each finite subset as a finite word. Higman lemma gives an earlier word embedding into a later one with every letter increased, which witnesses Hoare domination preorder comparison of the underlying subsets. The PDF specifies no finiteness restriction, so this observation supplements rather than replaces the literal answer and the arbitrary-subset counterexample.