Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 19 / 1

A well-quasi-ordering is a preorder $(Q,\le )$ such that every infinite sequence $(x_n)$ has indices $i<j$ with $x_i\le x_j$. Reflexivity and transitivity are part of the definition; antisymmetry is unnecessary. An infinite sequence without such a pair is a bad sequence.

The perfect subsequence lemma strengthens this definition: every infinite sequence in a well-quasi-order has an infinite nondecreasing subsequence, meaningHere is a proof that does not assume the lemma implicitly. There must be an index $i$ with infinitely many later indices $j$ satisfying $x_i\le x_j$. Otherwise each index would have only finitely many such successors. Starting with any index, choose the next one beyond the union of those finite successor sets for the previously chosen indices. This constructs a bad sequence, contradicting the well-quasi-ordering property.

Choose $i_0$ with infinitely many successors above it, and restrict to those successors. Apply the same argument to that infinite subsequence, obtaining $i_1>i_0$ with infinitely many later successors above it within the restricted sequence. Continue. Every subsequent restriction lies inside all previous upper cones, so $x_{i_r}\le x_{i_s}$ for every $r<s$. This proves the lemma, including for a genuine quasi-order with distinct equivalent elements. Both alternatives below use it.

For finite rooted trees, use rooted-tree homeomorphic embedding: an injective vertex map preserves ancestors and lowest common ancestors, so edges may become paths and distinct branches remain distinct. The root can map to a vertex below the host root. More generally label vertices in a well-quasi-ordering $Q$ and require each source label to precede its image label. We prove this stronger labelled version of Kruskal's tree theorem, from which the unlabelled result follows by taking one label.

We first need the finite-word closure, and prove it rather than merely invoking Higman lemma. If finite words over $Q$ were not a well-quasi-ordering under order-preserving subsequence embedding with increased labels, choose a minimal bad sequence $w_0,w_1,\dots$, minimizing the length of $w_n$ among words admitting a bad continuation of the already chosen prefix. No $w_n$ is empty. Write $w_n=v_n{a}_n$ by removing its last letter. The perfect subsequence lemma gives $n_0<n_1<\cdots$ with $a_{n_r}\le a_{n_s}$ for $r<s$.

The sequence $w_0,\dots ,w_{n_0-1},v_{n_0},v_{n_1},\dots$ is still bad. An earlier $w_i$ embedding into a later $v_{n_r}$ would embed into $w_{n_r}$, contradicting the original sequence. An embedding of $v_{n_r}$ into $v_{n_s}$ extends by the last-letter comparison to an embedding of $w_{n_r}$ into $w_{n_s}$. Both possibilities are excluded. This contradicts minimality at position $n_0$, since $v_{n_0}$ is shorter. Thus the finite-word closure holds. The perfect subsequence lemma also proves finite product closure: first pass to a nondecreasing subsequence in one coordinate, then in the next, preserving all previous comparisons.

Suppose now that the labelled trees admit a bad sequence $T_0,T_1,\dots$. Choose a minimal bad sequence by vertex number. Let $\mathcal{U}$ be all the proper descendant-rooted subtrees of these trees. We claim that $\mathcal{U}$ is a well-quasi-ordering. If it had a bad sequence of subtrees $U_r$, its parent indices can be made strictly increasing, say $U_r$ comes from $T_{n_r}$. Indeed each finite initial collection of parent trees has only finitely many subtrees, and a bad sequence cannot contain infinitely many terms from a finite collection. Passing to a subsequence therefore makes $n_0<n_1<\cdots$.

Then $T_0,\dots ,T_{n_0-1},U_0,U_1,\dots$ is bad. A prefix tree embedding into $U_r$ would embed into its parent $T_{n_r}$, contrary to the original badness, and the tail is bad by construction. But $U_0$ has fewer vertices than $T_{n_0}$, contradicting minimality. This proves the claim.

Give the children of each root any fixed order. Its forest of child subtrees is a finite word in $\mathcal{U}$. The word theorem and finite product closure show that the pairs consisting of the root label and the child-subtree word are a well-quasi-ordering. Thus some $i<j$ have comparable root labels and an embedding of the child word of $T_i$ into that of $T_j$. Map the source root to the target root and use the obtained embeddings inside distinct target child subtrees. The paths from the target root into those distinct branches are disjoint except at the root, so this is a label-monotone tree embedding. It contradicts badness. Hence

For Friedman's finite form of Kruskal's theorem, fix $k\in \mathbb{N}$. There is $N(k)$ such that any $N(k)$ trees with $|T_i|\le k+i$ contain $i<j$ with $T_i⪯T_j$. Use indices starting at zero; shifting them only changes $k$. To prove this uniform finite bound, form the finite bad-sequence tree of finite bad prefixes obeying the size bounds. At position $i$ there are only finitely many rooted-tree isomorphism types with at most $k+i$ vertices, so every node has finitely many successors.

If no bound existed, this tree would have nodes of arbitrarily large depth. A node with arbitrarily long extensions has a child with arbitrarily long extensions, since it has only finitely many children. Choosing such a child repeatedly produces an infinite branch, as in the König infinity lemma. The branch is an infinite bad tree sequence, contradicting the theorem just proved. Therefore the required $N(k)$ exists. The same proof works with any fixed finite label alphabet and any specified size bound finite at each position.

The PDF's displayed inequality uses the relation it purports to define and therefore is not itself a complete definition. The absorption order consistent with that inequality and with the requested result isThis is the reverse of the usual natural order of an incline. We specify it explicitly: using instead $x+y=y$ would make the finite-generation conclusion false in general.

Idempotence gives $x+x=x$, proving reflexivity. If $x+y=x$ and $y+z=y$, then associativity givesso the relation is transitive. If also $y+x=y$, commutativity forces $x=y$. Thus it is even a partial order. In this order $x+y⪯x$ follows because $(x+y)+x=x+y$. Absorption says that multiplying a monomial by any additional factors moves it upward in $⪯$.

Let $g_1,\dots ,g_r$ generate the incline. Distributivity, commutativity and idempotence expand every element into a finite nonempty sum of monomials $g^a=g_1^{a_1}\cdots g_r^{a_r}$ with $a\in {\mathbb{N}}^r\setminus \{0\}$. A zero exponent means that factor is omitted; no multiplicative identity is assumed. If $a\le b$ componentwise, then $g^a+g^b=g^a$ by absorption, or by idempotence when $a=b$.

We prove the needed reverse-inclusion well-quasi-ordering of monomial ideals directly. Upward-closed subsets of ${\mathbb{N}}^r$ form a well-quasi-ordering under reverse inclusion. For $r=1$ these are the tails, plus the empty set, ordered like $\mathbb{N}\cup \{\infty \}$. For the induction, write an upset $U\subseteq {\mathbb{N}}^{r+1}$ as increasing slices $U_0\subseteq U_1\subseteq \cdots$ in ${\mathbb{N}}^r$. Their union is an upset with a finite minimal basis by Dickson lemma. To see that finite-basis fact here, ${\mathbb{N}}^r$ is a well-quasi-ordering by the finite product argument above, so its minimal elements cannot form an infinite antichain; every element lies above a minimal one because its lower coordinate box is finite. All basis elements of the union appear by some finite slice, so the slices eventually stabilize.

Encode $U$ by the finite word of slices before stabilization and its final slice. By the finite-word theorem proved in the other alternative, and finite product closure, these encodings are a well-quasi-ordering with slice comparisons given by reverse inclusion. For two comparable encodings, let the earlier pre-stabilization slice $U_k$ match the later slice $V_{f(k)}$, where $f(k)\ge k$. ThenBeyond the earlier stabilization index, its final slice contains the later final slice and hence every later slice. Therefore $U\supseteq V$. This proves the induction, including empty slices and the empty upset.

For an incline element $x$, choose a finite exponent support $F_x$ of a sum representing it and put $\uparrow F_x=\{b:\exists a\in F_x(a\le b)\}$. Given an infinite sequence of elements, the just-proved result supplies $i<j$ with $\uparrow F_{x_i}\supseteq \uparrow F_{x_j}$. Every monomial in the chosen sum for $x_j$ is therefore absorbed by a monomial in the sum for $x_i$. Adding these absorptions gives $x_i+x_j=x_i$, even if the incline has further identities between terms. HenceThe order direction matters. On the positive integers take $x+y=\min (x,y)$ and multiplication to be ordinary addition of integers. This is an incline generated by $1$. Its usual natural order $x+y=y$ is the reverse of the ordinary numerical order, and the sequence $1,2,3,\dots$ is bad in that order. Its reverse absorption order is the ordinary numerical order, consistent with the theorem above.