Well Quasi Orders in a Categorical SettingWell-quasiorderings: From Theory to Applications

Dr M Benini

Università degli Studi dell’Insubriamarco.benini@uninsubria.it

September 25rd, 2015

Introduction

A well quasi order can be characterised in at least three different ways, whichare equivalent in classical logic with a sufficiently strong notion of set:■ Every proper descending chain is finite and any antichain is finite;■ Any infinite sequence contains an increasing pair;■ Any infinite sequence contains an infinite ascending chain.

Quasi orders can be identified with the small categories having at most onearrow between any pair of objects. In turn, quasi orders and monotone mapsform a category QOrd.Well quasi orders identify a full subcategory of QOrd.

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Quasi orders

In QOrd:■ the initial object is 0= ⟨;;;⟩;■ the terminal object is 1, the discrete category with one element;■ the product A ×B is ⟨A×B;

{((a′,b′),(a′′,b′′)) : a′ ≤A a′′ & b′ ≤B b′′}⟩;

■ the coproduct A +B is the disjoint union of A and B;■ the equaliser of f ,g : A →B is

{x ∈A : f (x)= g(x)

}ordered by ≤A;

■ the coequaliser of f ,g : A →B is B/≈, with ≈ the minimal equivalencerelation containing

{(f (x),g(x)) : x ∈A}

, ordered by the reflexive andtransitive closure of

{([x ]≈, [y ]≈) : x ≤B y

};

■ the exponential object BA is the set of monotone maps A→B, wheref ≤ g exactly when f (x)≤B g(x) for all x ∈A;

■ it is easy to find a counterexample showing that QOrd does not have asubobject classifier.

We denote quasi orders by A = ⟨A;≤A⟩.

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Quasi orders

Since the forgetful functor QOrd→Set has left and right adjoints, theshape of limits and colimits is forced. Also, orders do not add much: thesubcategory Ord of orders is reflective in QOrd.This situation is, essentially, the same in the categories we will be interestedin. In particular, the shape of limits and colimits, when they exist, is forcedeither by a forgetful functor to Set, or by an adjunction which is therestriction of the one in QOrd. Moreover, we will not speak about orderssince the subcategories of orders are reflective in the subcategories of quasiorders we are going to deal with.It is reasonable to ask to which extent the structure of quasi orders transfersto the full subcategories we are interested in, especially well quasi orders.Also, it is reasonable to ask for perfect counterexamples when the structuredoes not transfer.

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Well quasi orders and related categories

■ AQOrd is the full subcategory of QOrd whose objects are the quasiorders whose antichains are all finite;

■ WFQOrd is the full subcategory of QOrd whose objects are the quasiorders whose proper descending chains are all finite;

■ WQOrd is the full subcategory of QOrd whose objects are the quasiorders having both the properties above, so this is the category of wellquasi orders.

Being finite quasi orders, the initial (terminal) object of QOrd lies in allthese subcategories, thus they have initial (terminal) objects, too.Also, it is immediate to see that the construction of coproducts andequalisers in QOrd can be replicated in all the above subcategories.Products and coequalisers do not follow immediately.And exponentiation is not evident at all. . .

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Products

Products in the subcategories of interest, when they exist, must have thesame shape as in QOrd because it must be the right adjoint of therestriction of the diagonal functor.In WQOrd, finite products exist thanks to Dickson’s Lemma.The same holds in WFQOrd, by considering any proper descending chain inthe candidate product, and constructing out of it two descending chains inthe components, whose proper kernel must be finite, forcing the originalchain to be finite too.But, AQOrd does not have all the binary products: consider ⟨N;≤⟩ and the(total) lexicographic order O on the free monoid over 0< 1. Let xn = 0 . . .01be the word in O having n zeroes. Then {xn}n∈ω is an infinite descendingchain in O , so

{(n,xn)

}n∈ω defines an infinite antichain in the product.

This counterexample is perfect: pairing elements from an infinite ascendingchain with elements from an infinite descending chain is the only way toconstruct a counterexample in AQOrd.

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CoequalisersIt is immediate to see that the shape of coequalisers in AQOrd, WFQOrd,and WQOrd in the same as in QOrd because the forgetful functor to Sethas a right adjoint.Since every antichain in the coequaliser of f ,g : A →B in AQOrd inducesan antichain in B, it must be finite. So, AQOrd has coequalisers.But WFQOrd does not. In fact, consider ⟨N;=⟩, the set of naturals with thediscrete order, and the disjoint union of ω copies of a< b < c. Letf (n)= cn+1 and g(n)= an, for all n ∈N. Calculating the coequaliser:

c0 c1 c2

b0

OO

b1

OO

b2

OO

· · ·

a0

OO=

a1

OO=

a2

OO

Thus, {bi }i∈ω is an infinite proper descending chain in WFQOrd.As before, combining an infinite set of finite descending chains along aninfinite antichain is the only way to construct a counterexample in WFQOrd.

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Coequalisers

In fact, WQOrd has coequalisers. The proof is not immediate: first one hasto notice that any descending chain

{[ei ]≈

}i∈ω in the coequaliser object of

f ,g : A →B can be transformed into another descending chain such thateither g(xn)≥Q en+1 ≥Q f (xn+1) for each n ∈ω, or f (xn)≥Q en+1 ≥Q g(xn+1)for each n ∈ω. That is, there is sequence {xi }i in A which bounds theelements of {ei }i in B by means of f and g .Since {xi }i is infinite, there is a subsequence which forms an infiniteascending chain, and it has the form {xi }i>m for some m ∈ω. Thus, using thefact that f , g , and the candidate coequaliser map z 7→ [z ]≈ are monotone. . .

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Coequalisers

. . . one gets that

[g(xm)]≈��

[em+1]≈oo [f (xm+1)]≈��

oo

[g(xm+1)]≈��

=

[em+2]≈oo [f (xm+2)]≈��

oo

=...

��

......

��[g(xn−1)]≈

��

=

[en]≈oo [f (xn)]≈��

oo

=

[g(xn)]≈��

=

[en+1]≈oo [f (xn+1)]≈��

oo

=...

......

Then, for n>m+1, [en]≈ and [en+1]≈ are equivalent. So, any descendingchain of length ω is not proper.

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Exponentiation

AQOrd, WFQOrd, and WQOrd do not have exponentiation.In fact, AQOrd does not have products, so it cannot have exponentiation.And, in WQOrd (or WFQOrd), consider the set of natural numbers withthe standard ordering, and the order 2 given by 0< 1.Also, consider the family {fi : ⟨N;≤⟩→ 2}i∈N, defined by

fi (x)={1 if x ≥ i0 otherwise .

Whenever i < j , fi (x)> fj(x) for all x ∈N, that is fi > fj , since■ if x ≥ j > i , fi (x)= 1= fj(x);■ if x < i < j , fi (x)= 0= fj(x);■ if i ≤ x < j , fi (x)= 1> 0= fj(x).Thus f0 > f1 > f2 > ·· · is an infinite proper descending chain.

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Summary

0 1 × + ⊃ eq coeqQOrd X X X X X X X

AQOrd X X - X - X XWFQOrd X X X X - X -WQOrd X X X X - X X

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The Higman’s Lemma

Lemma 1 (Higman)⟨A∗;≤∗⟩ is a well quasi order if and only if ⟨A;≤A⟩ is so.One direction is obvious: when ⟨A∗;≤∗⟩ is a well quasi order, so is ⟨A;≤A⟩.Categorically, consider

A∗⊥,> = ⟨A∗t {⊥,>} ;≤∗t

{(⊥,x) ,(x ,>) : x ∈A∗t {⊥,>}

}⟩ ,

the well quasi order obtained by adding a global maximum and minimum.Let f : ⟨A∗;≤∗⟩→A∗

⊥,> be

f ([x1, . . . ,xn])=

[x1] if n= 1⊥ if n= 0> otherwise.

The equaliser of f and the canonical inclusion ⟨A∗;≤∗⟩→A∗⊥,> is isomorphic

to ⟨A;≤A⟩. Thus, ⟨A;≤A⟩ is a well quasi order.

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The Higman’s Lemma

The second step is to show that

Proposition 2⟨A∗;≤∗⟩ is a well founded quasi order if and only if ⟨A;≤A⟩ is a well foundedquasi order.The proof goes by proving that there a suitable coequaliser can beconstructed so that it is isomorphic to ⟨A;≤A⟩.It is relevant to remark the symmetry of the first two steps: the formerfollows from the construction of an equaliser, while the latter from theconstruction of a coequaliser. This is not by chance!The rest of the proof follows the standard track, noticing that infiniteantichains are, in fact, a special case of bad sequences.

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Exponentiation, again

It is not difficult to infer

Proposition 3If A is a well quasi order, the collection of finite and infinite sequences overA ordered by embedding has the finite antichain property.And, assuming the Axiom of Choice, it is evident that each function can bethought of as a sequence of values indexed by its domain, so

Corollary 4The exponential object A B in QOrd, with B any quasi order, has finiteantichains whenever A is a well quasi order.

Thus, the counterexample to the existence of exponential objects in WQOrdis prototypical: every counterexample must violate the finite properdescending chain property.

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The end

Questions?

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