variations on the higman's lemma

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Variations on the Higman’s Lemma JAIST Logic Seminar Dr M Benini Università degli Studi dell’Insubria [email protected] August 8 th , 2016

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Variations on the Higman’s LemmaJAIST Logic Seminar

Dr M Benini

Università degli Studi dell’[email protected]

August 8th, 2016

Well quasi orders

Quasi orders can be identified with the small categories having atmost one arrow between each pair of objects. In turn, quasiorders and monotone maps form a category QOrd.A well quasi order can be characterised in at least three differentways, which are equivalent in classical logic with a sufficientlystrong notion of set: Every proper descending chain is finite and every antichain isfinite;

Every infinite sequence contains an increasing pair; Every infinite sequence contains an infinite ascending chain.

Well quasi orders identify a full subcategory of QOrd.

( 2 of 30 )

Quasi orders

In QOrd: 0= ⟨;;;⟩; 1 is the discrete category with one element; A ×B = ⟨A×B;

((a′,b′),(a′′,b′′)) : a′ ≤A a′′ & b′ ≤B b′′

⟩; A +B is the disjoint union of A and B; the equaliser of f ,g : A →B is ⟨x ∈A : f (x)= g(x)

;≤A⟩;

the coequaliser of f ,g : A →B is B/≈, with ≈ the minimalequivalence relation containing

(f (x),g(x)) : x ∈A

, orderedby the reflexive and transitive closure of

([x ]≈, [y ]≈) : x ≤B y

;

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

it is easy to find a counterexample showing that QOrd doesnot have a subobject classifier.

( 3 of 30 )

Well quasi orders and related categories

AQOrd is the full subcategory of QOrd whose objects are thequasi orders with finite antichains;

WFQOrd is the full subcategory of QOrd whose objects arethe well-founded quasi orders, that is, those having finiteproper descending chains;

WQO is the full subcategory of QOrd whose objects are thequasi orders having both the properties above, so this is thecategory of well quasi orders.

( 4 of 30 )

Well quasi orders and related categories

Being finite quasi orders, the initial and terminal objects ofQOrd lie in all these subcategories, thus they have initial andterminal objects, too.Also, it is immediate to see that the construction of coproductsand equalisers in QOrd can be replicated in all thesesubcategories.On the contrary, products and coequalisers do not followimmediately. And exponentiation is not evident at all. . .

( 5 of 30 )

Products

Products in the subcategories of interest, when they exist, musthave the same shape as in QOrd because they arise from theright adjoint of the restriction of the diagonal functor.In WQO, finite products exist thanks to Dickson’s Lemma.The same holds in WFQOrd, by considering each properdescending chain in the candidate product, and constructing outof it two descending chains in the components, whose properkernel must be finite, forcing the original chain to be finite too.

( 6 of 30 )

Products

But, AQOrd does not have all the binary products: consider⟨N;≤⟩ and the (total) lexicographic order O on the free monoidover 0< 1. Let xn = 0 . . .01 be the word in O having n zeroes.Then xnn∈ω is an infinite descending chain in O , so

(n,xn)

n∈ω

defines an infinite antichain in the product.This counterexample is perfect: pairing elements from aninfinite ascending chain with elements from an infinite descendingchain is the only way to construct a counterexample in AQOrd.

( 7 of 30 )

Coequalisers

It is immediate to see that the shape of coequalisers in AQOrd,WFQOrd, and WQO is the same as in QOrd because theforgetful functor to Set has a right adjoint.Since every antichain in the coequaliser of f ,g : A →B inAQOrd induces an antichain in B, it must be finite. So,AQOrd has coequalisers.

( 8 of 30 )

CoequalisersBut WFQOrd does not. In fact, consider ⟨N;=⟩, the set ofnaturals with the discrete order, and the disjoint union of ωcopies of a< b < c . Let f (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 proper descendingchains along an infinite antichain is the only way to construct acounterexample in WFQOrd.( 9 of 30 )

Coequalisers

In fact, WQO has coequalisers. The proof is not immediate: firstone has to notice that each descending chain

[ei ]≈

i∈ω in the

coequaliser object of f ,g : A →B can be embedded into anotherdescending chain such that either g(xn)≥Q en+1 ≥Q f (xn+1) foreach 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 the elements of ei i inB by means of f and g .Since xi i is infinite, there is a subsequence which forms aninfinite ascending chain, and it has the form xi i>m for somem ∈ω. Thus, using the fact that f , g , and the candidatecoequaliser map z 7→ [z ]≈ are monotone. . .

( 10 of 30 )

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, everydescending chain of length ω is not proper.( 11 of 30 )

ExponentiationAQOrd, WFQOrd, and WQO do not have exponentiation.In fact, AQOrd does not have products, so it cannot haveexponentiation.And, in WQO (or WFQOrd), consider the set of naturalnumbers with the standard ordering, the order 2 given by 0< 1,and the family fi : ⟨N;≤⟩→ 2i∈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.( 12 of 30 )

Summary

0 1 × + ⊃ eq coeqQOrd X X X X X X X

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

WQO X X X X - X X

( 13 of 30 )

The Higman’s Lemma

Let A∗ be the set of finite sequences over the alphabet A, and let≤∗ be the ordering given by embedding, that is

[a1, . . . ,an]≤∗ [b1, . . . ,bm]

if and only if there is an injective and monotone mape : 1, . . . ,n → 1, . . . ,m such that ai ≤A be(i) for all 1≤ i ≤ n.

Lemma 1 (Higman)⟨A∗;≤∗⟩ is a well quasi order if and only if ⟨A;≤A⟩ is so.

( 14 of 30 )

The Higman’s Lemma

One direction is obvious: when ⟨A∗;≤∗⟩ is a well quasi order, sois ⟨A;≤A⟩.Categorically, consider

A∗⊥,> = ⟨A∗t ⊥,> ;≤∗t

(⊥,x) ,(x ,>) : x ∈A∗t ⊥,>

⟩ ,

the well quasi order obtained by adding a global maximum andminimum. 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.( 15 of 30 )

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 awell founded quasi order.The proof goes by proving that a suitable coequaliser can beconstructed such that it is isomorphic to ⟨A∗;≤∗⟩.We add one isolated element ∗ to ⟨A;≤A⟩, and then we considerthe collection of sequences of length n over this augmented quasiorder. By ordering them pointwise, we get another quasi order⟨An∗;≤×⟩. And, when ⟨A;≤A⟩ is well-founded, ⟨An∗;≤×⟩ iswell-founded, too.

( 16 of 30 )

The Higman’s Lemma

Thus, ⟨P;≤P⟩ =⊔i∈ω⟨Ai∗;≤×⟩ is a well-founded quasi order, since

each proper descending chain lies in a single component of thedisjoint union, so it is finite.Consider f ([x1, . . . ,xn])= [y1, . . . ,ym] where [y1, . . . ,ym] is themaximal subsequence of [x1, . . . ,xn] containing no ∗’s. It is clearthat f is monotone.Calculating the coequaliser of f and the identity of ⟨P;≤P⟩ inQOrd, we see that it is well-founded.

( 17 of 30 )

The Higman’s Lemma

For the third step, showing that if ⟨A;≤A⟩ is a well-quasi order,then ⟨A∗;≤∗⟩ has finite antichains, we may follow the standardproof by Nash-Williams.In fact, a bad sequence is a ‘relaxed form’ of infinite antichain.The third step is not satisfactory: we replicate the kernel of theNash-Williams’ proof to get the result, which suffices to provethe whole Higman’s Lemma, not just the property on antichains.

( 18 of 30 )

Proper descending chains

Let A be a quasi order. Define

D(A )= xi i∈I : xi i∈I is a proper descending chain in A

,

and xi i∈I ≤D(A )yi

i∈J if and only if there is an injective and

monotone map η : I → J from the ordinal I to the ordinal J suchthat, for each i ∈ I , xi ≤A yη(i).

Fact 3The structure D(A )= ⟨D(A );≤D(A )⟩ is a quasi order.

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Proper descending chains

Proposition 4If A is a well founded quasi order, so is D(A ).

Proof.Let S =

x1ii∈I1 ≥D(A ) · · · ≥D(A )

xn

ii∈In ≥D(A ) · · · be a

descending chain in D(A ). Then, fixed the embedding maps η,each element x1

i , i ∈ I1, is the starting point of a descendingchain in A , by definition of ≤D(A ). But, by hypothesis, eachdescending chain in A is either proper and finite, with length mi ,or it has a finite prefix of length mi followed by a possibly infinitetail of equivalent elements. Thus, since I1 is finite, being

x1

ii∈I1

a proper descending chain in A , there is m=max mi : i ∈ I1.Then, necessarily,

xn

ii∈In is equivalent to

xk

ii∈Ik , for every n,k

greater than m.( 20 of 30 )

Proper descending chains

Corollary 5A is a well founded quasi order, if and only if D(A ) is.

Proof.One direction is the previous proposition. The other direction isimmediate, considering the embedding A →D(A ), A 7→ A.

We notice that the property that proper descending chains in A

are finite, has been used only twice: (i) to establish that I1 isfinite; and (ii) to get the mi ’s.Thus, if we substitute D(A ) with the collection of finitesequences over A in the same proof, we get the Higman’sLemma on well founded quasi orders, for free.

( 21 of 30 )

Proper descending chains

Proposition 6If A is a well quasi order, so is D(A ).

Proof. (i)It suffices to show that D(A ) has the finite antichain property.So, let S =

x i

j

j∈Ji

i∈I

be an antichain in D(A ), and suppose,with no loss of generality, that Ji ≤ Jk whenever i ≤ k .

Then, for each i ,k ∈ I , i 6= k ,x i

j

j∈Ji

∥xk

j

j∈Jk

, that is, for alli ,k ∈ I , i 6= k , for every ηi ,k : Ji → Jk injective and monotone, forsome j∗ ∈ Ji , x i

j∗ 6≤ xkηi ,k(j∗), and for every ηk ,i : Jk → Ji injective

and monotone, for some j∗∗ ∈ Jk , xkj∗∗ 6≤ x i

ηk ,i (j∗∗). ,→

( 22 of 30 )

Proper descending chains

,→ Proof. (ii)Fix any sequence

ηi ,i+1

i∈I , eventually excluding the last element

if I is a limit ordinal. Then, there is ξ mapping each i ∈ I to Jisuch that x i

ξ(i) 6≤ x i+1ηi ,i+1(ξ(i)), obtained by choosing the j∗ above.

Consider the sequence C defined as(x0ξ(0),x0

ξ(0)

)∪(

x2(i+1)ξ(2(i+1)),x2(i+1)

η2i+1,2(i+1)(ξ(2i+1))

)i∈I

∪(x2i+1η2i ,2i+1(ξ(2i)),x2i+1

ξ(2i+1)

)i∈I

.

,→

( 23 of 30 )

Proper descending chains

,→ Proof. (iii)By induction on the initial prefix of I up to ω: x0

ξ(0) 6≤ x1η0,1(ξ(0)), so

(x0ξ(0),x0

ξ(0)

)6≤

(x1η0,1(ξ(0)),x1

ξ(1)

);

if i = 2k +1, x2k+1ξ(2k+1) 6≤ x2(k+1)

η2k+1,2(k+1)(ξ(2k+1)), so(x2k+1η2k ,2k+1(ξ(2k)),x2k+1

ξ(2k+1)

)6≤

(x2(k+1)ξ(2(k+1)),x2(k+1)

η2k+1,2(k+1)(ξ(2k+1))

),

that is, (x iηi−1,i (ξ(i−1)),x i

ξ(i)

)6≤

(x i+1ξ(i+1),x i+1

ηi ,i+1(ξ(i))

);

,→( 24 of 30 )

Proper descending chains

,→ Proof. (iv)

if i = 2k , k ≥ 1, then x2kξ(2k) 6≤ x2k+1

η2k ,2k+1(ξ(2k)), so(x2kξ(2k),x2k

η2k−1,2k(ξ(2k−1))

)6≤

(x2k+1η2k ,2k+1(ξ(2k)),x2k+1

ξ(2k+1)

),

that is, (x iξ(i),x i

ηi−1,i (ξ(i−1))

)6≤

(x i+1ηi ,i+1(ξ(i)),x i+1

ξ(i+1)

).

So, any two consecutive elements αi , αi+1 in C are such thatαi 6≤αi+1, for any i ∈ I ∩ω. ,→

( 25 of 30 )

Proper descending chains

,→ Proof. (v)Thus, the sequence Cω obtained restricting C to I ∩ω, is a badsequence in A ×A . But A is a well quasi order, so by Dickson’sLemma, A ×A is a well quasi order, too, forcing any badsequence to be finite. Then, necessarily, Cω is finite, so I∩ω<ω,i.e., I is finite. Thus, the antichain S is finite, being I itslength.

Corollary 7A is a well quasi order if and only if D(A ) is.

( 26 of 30 )

Higman’s Lemma, again

The same result can be derived from Higman’s Lemma bynoticing that the set D(A ) induces a full subcategory in the wellquasi order of finite sequences over A , which must be a wellquasi order, because it is the restriction of a well quasi order.However, the proof of the previous proposition does not dependon the fact that the elements are proper descending chains: infact, it just requires the elements to be in the well quasi order A .

Corollary 8If A is a well quasi order, the collection of sequences over A

ordered by embedding has the finite antichain property.And, of course, we get the Higman’s Lemma for free.

( 27 of 30 )

Exponentiation, finally

Let Ω(A ) be the collection of arbitrary sequences over A .Viewing each sequence as a function from an ordinal to theindexed element, we get,

Proposition 9The exponential object A I in QOrd, with I any ordinal, hasfinite antichains whenever A is a well quasi order.

Proof.The exponential object A I is isomorphic to a sub quasi order ofΩ(A ). Since Ω(A ) has the finite antichain property, the sameholds for A I .

( 28 of 30 )

Exponentiation, finally

Assuming the Axiom of Choice,

Corollary 10The exponential object A B in QOrd, with B any quasi order,has finite antichains whenever A is a well quasi order.Hence, the counterexample to the existence of exponentialobjects in WQO is, in fact, maximal, that is, any counterexamplewill violate the finite descending chain property, whereas nocounterexample could be found which contradicts the finiteantichain property, as far as we operate within a sufficiently richset theory.

( 29 of 30 )

The end

Questions?

( 30 of 30 )