well-quasi-ordering

In mathematics, specifically order theory, a well-quasi-ordering or wqo on a set $X$ is a quasi-ordering of $X$ for which every infinite sequence of elements $x_0, x_1, x_2, \ldots$ from $X$ contains an increasing pair $x_i \leq x_j$ with $i < j.$

Motivation

Well-founded induction can be used on any set with a well-founded relation, thus one is interested in when a quasi-order is well-founded. (Here, by abuse of terminology, a quasiorder $\le$ is said to be well-founded if the corresponding strict order $x\le y\land y\nleq x$ is a well-founded relation.) However the class of well-founded quasiorders is not closed under certain operations—that is, when a quasi-order is used to obtain a new quasi-order on a set of structures derived from our original set, this quasiorder is found to be not well-founded. By placing stronger restrictions on the original well-founded quasiordering one can hope to ensure that our derived quasiorderings are still well-founded.

An example of this is the power set operation. Given a quasiordering $\le$ for a set $X$ one can define a quasiorder $\le^$ on $X$'s power set $P(X)$ by setting $A \le^ B$ if and only if for each element of $A$ one can find some element of $B$ that is larger than it with respect to $\le$. One can show that this quasiordering on $P(X)$ needn't be well-founded, but if one takes the original quasi-ordering to be a well-quasi-ordering, then it is.

Formal definition

A well-quasi-ordering on a set $X$ is a quasi-ordering (i.e., a reflexive, transitive binary relation) such that any infinite sequence of elements $x_0, x_1, x_2, \ldots$ from $X$ contains an increasing pair $x_i \le x_j$ with $i< j$. The set $X$ is said to be well-quasi-ordered, or shortly wqo.

A well partial order, or a wpo, is a wqo that is a proper ordering relation, i.e., it is antisymmetric.

Among other ways of defining wqo's, one is to say that they are quasi-orderings which do not contain infinite ''strictly decreasing'' sequences (of the form $x_0> x_1> x_2> \cdots$) nor infinite sequences of ''pairwise incomparable'' elements. Hence a quasi-order (''X'', ≤) is wqo if and only if (''X'', <) is well-founded and has no infinite antichains.

Ordinal type

Let $X$ be well partially ordered. A (necessarily finite) sequence $(x_1, x_2, \ldots, x_n)$ of elements of $X$ that contains no pair $x_i \le x_j$ with $i< j$ is usually called a ''bad sequence''. The ''tree of bad sequences'' $T_X$ is the tree that contains a vertex for each bad sequence, and an edge joining each nonempty bad sequence $(x_1, \ldots, x_, x_n)$ to its parent $(x_1, \ldots, x_)$. The root of $T_X$ corresponds to the empty sequence. Since $X$ contains no infinite bad sequence, the tree $T_X$ contains no infinite path starting at the root. Therefore, each vertex $v$ of $T_X$ has an ordinal height $o(v)$, which is defined by transfinite induction as $o(v) = \lim_ (o(w)+1)$. The ''ordinal type'' of $X$, denoted $o(X)$, is the ordinal height of the root of $T_X$.

A ''linearization'' of $X$ is an extension of the partial order into a total order. It is easy to verify that $o(X)$ is an upper bound on the ordinal type of every linearization of $X$. De Jongh and Parikh proved that in fact there always exists a linearization of $X$ that achieves the maximal ordinal type $o(X)$.

Examples

* $(\N, \le)$, the set of natural numbers with standard ordering, is a well partial order (in fact, a well-order). However, $(\Z, \le)$, the set of positive and negative integers, is not a well-quasi-order, because it is not well-founded (see Pic.1).
* $(\N, , )$, the set of natural numbers ordered by divisibility, is not a well-quasi-order: the prime numbers are an infinite antichain (see Pic.2).
* $(\N^k, \le)$, the set of vectors of $k$ natural numbers (where $k$ is finite) with component-wise ordering, is a well partial order ( Dickson's lemma; see Pic.3). More generally, if $(X, \le)$ is well-quasi-order, then $(X^k,\le^k)$ is also a well-quasi-order for all $k$.
* Let $X$ be an arbitrary finite set with at least two elements. The set $X^*$ of words over $X$ ordered lexicographically (as in a dictionary) is not a well-quasi-order because it contains the infinite decreasing sequence $b, ab, aab, aaab, \ldots$. Similarly, $X^*$ ordered by the prefix relation is not a well-quasi-order, because the previous sequence is an infinite antichain of this partial order. However, $X^*$ ordered by the subsequence relation is a well partial order. (If $X$ has only one element, these three partial orders are identical.)
* More generally, $(X^*,\le)$, the set of finite $X$-sequences ordered by embedding is a well-quasi-order if and only if $(X, \le)$ is a well-quasi-order ( Higman's lemma). Recall that one embeds a sequence $u$ into a sequence $v$ by finding a subsequence of $v$ that has the same length as $u$ and that dominates it term by term. When $(X,=)$ is an unordered set, $u\le v$ if and only if $u$ is a subsequence of $v$.
* $(X^\omega,\le)$, the set of infinite sequences over a well-quasi-order $(X, \le)$, ordered by embedding, is not a well-quasi-order in general. That is, Higman's lemma does not carry over to infinite sequences. Better-quasi-orderings have been introduced to generalize Higman's lemma to sequences of arbitrary lengths.
* Embedding between finite trees with nodes labeled by elements of a wqo $(X, \le)$ is a wqo ( Kruskal's tree theorem).
* Embedding between infinite trees with nodes labeled by elements of a wqo $(X, \le)$ is a wqo ( Nash-Williams' theorem).
* Embedding between countable scattered linear order types is a well-quasi-order ( Laver's theorem).
* Embedding between countable boolean algebras is a well-quasi-order. This follows from Laver's theorem and a theorem of Ketonen.
* Finite graphs ordered by a notion of embedding called " graph minor" is a well-quasi-order ( Robertson–Seymour theorem).
* Graphs of finite tree-depth ordered by the induced subgraph relation form a well-quasi-order, as do the cographs ordered by induced subgraphs.

Constructing new wpo's from given ones

Let $X_1$ and $X_2$ be two disjoint wpo sets. Let $Y=X_1\cup X_2$, and define a partial order on $Y$ by letting $y_1\le_Y y_2$ if and only if $y_1,y_2\in X_i$ for the same $i\in\$ and $y_1 \le_ y_2$. Then $Y$ is wpo, and $o(Y) = o(X_1) \oplus o(X_2)$, where $\oplus$ denotes natural sum of ordinals.

Given wpo sets $X_1$ and $X_2$, define a partial order on the Cartesian product $Y=X_1\times X_2$, by letting $(a_1,a_2)\le_Y (b_1,b_2)$ if and only if $a_1\le_ b_1$ and $a_2\le_ b_2$. Then $Y$ is wpo (this is a generalization of Dickson's lemma), and $o(Y) = o(X_1)\otimes o(X_2)$, where $\otimes$ denotes natural product of ordinals.

Given a wpo set $X$, let $X^*$ be the set of finite sequences of elements of $X$, partially ordered by the subsequence relation. Meaning, let $(x_1,\ldots,x_n)\le_ (y_1,\ldots,y_m)$ if and only if there exist indices 1\le i_1<\cdots such that $x_j \le_X y_$ for each $1\le j\le n$. By Higman's lemma, $X^*$ is wpo. The ordinal type of $X^*$ is Republished in: $o(X^*)=\begin\omega^,&o(X) \text;\\ \omega^,&o(X)=\varepsilon_\alpha+n \text\alpha\textn;\\ \omega^,&\text.\end$

Given a wpo set $X$, let $T(X)$ be the set of all finite rooted trees whose vertices are labeled by elements of $X$. Partially order $T(X)$ by the tree embedding relation. By Kruskal's tree theorem, $T(X)$ is wpo. This result is nontrivial even for the case $, X, =1$ (which corresponds to unlabeled trees), in which case $o(T(X))$ equals the small Veblen ordinal. In general, for $o(X)$ countable, we have the upper bound $o(T(X))\le\vartheta(\Omega^\omega o(X))$ in terms of the $\vartheta$ ordinal collapsing function. (The small Veblen ordinal equals $\vartheta(\Omega^\omega)$ in this ordinal notation.)

Wqo's versus well partial orders

In practice, the wqo's one manipulates are quite often not orderings (see examples above), and the theory is technically smoother if we do not require antisymmetry, so it is built with wqo's as the basic notion. On the other hand, according to Milner 1985, ''no real gain in generality is obtained by considering quasi-orders rather than partial orders... it is simply more convenient to do so.''

Observe that a wpo is a wqo, and that a wqo gives rise to a wpo between equivalence classes induced by the kernel of the wqo. For example, if we order $\Z$ by divisibility, we end up with $n\equiv m$ if and only if $n=\pm m$, so that $(\Z,, )\approx(\N,, )$.

Infinite increasing subsequences

If $(X, \le)$ is wqo then every infinite sequence $x_0, x_1, x_2, \ldots,$ contains an infinite increasing subsequence $x_ \le x_\le x_ \le \cdots$ (with $n_0< n_1< n_2< \cdots$). Such a subsequence is sometimes called perfect.
This can be proved by a Ramsey argument: given some sequence $(x_i)_i$, consider the set $I$ of indexes $i$ such that $x_i$ has no larger or equal $x_j$ to its right, i.e., with i. If $I$ is infinite, then the $I$-extracted subsequence contradicts the assumption that $X$ is wqo. So $I$ is finite, and any $x_n$ with $n$ larger than any index in $I$ can be used as the starting point of an infinite increasing subsequence.

The existence of such infinite increasing subsequences is sometimes taken as a definition for well-quasi-ordering, leading to an equivalent notion.

Properties of wqos

* Given a quasiordering $(X,\le)$ the quasiordering $(P(X), \le^+)$ defined by $A \le^+ B \iff \forall a \in A, \exists b \in B, a \le b$ is well-founded if and only if $(X,\le)$ is a wqo.
* A quasiordering is a wqo if and only if the corresponding partial order (obtained by quotienting by $x \sim y \iff x\le y \land y \le x$) has no infinite descending sequences or antichains. (This can be proved using a Ramsey argument as above.)
* Given a well-quasi-ordering $(X,\le)$, any sequence of upward-closed subsets $S_0 \subseteq S_1 \subseteq \cdots \subseteq X$ eventually stabilises (meaning there exists $n \in \N$ such that $S_n = S_ = \cdots$; a subset $S \subseteq X$ is called ''upward- closed'' if $\forall x,y \in X, x \le y \wedge x \in S \Rightarrow y \in S$): assuming the contrary $\forall i \in \N, \exists j \in \N, j > i, \exists x \in S_j \setminus S_i$, a contradiction is reached by extracting an infinite non-ascending subsequence.
* Given a well-quasi-ordering $(X,\le)$, any subset $S$ of $X$ has a finite number of minimal elements with respect to $\le$, for otherwise the minimal elements of $S$ would constitute an infinite antichain.

See also

*
*
*

Notes

Here ''x'' < ''y'' means: $x\le y$ and $x \neq y.$

References

*
*
*
*
*
*

{{Order theory

Order theory
Wellfoundedness