Talk given in the symposium "Well-quasiorderings: From Theory to Applications", DMV, Hamburg, 25 September 2015

1. Well Quasi Orders in a Categorical Setting
Well-quasiorderings: From Theory to Applications
Dr M Benini
Università degli Studi dell’Insubria
marco.benini@uninsubria.it
September 25rd, 2015

2. Introduction
A well quasi order can be characterised in at least three different ways, which
are 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 one
arrow between any pair of objects. In turn, quasi orders and monotone maps
form a category QOrd.
Well quasi orders identify a full subcategory of QOrd.
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3. 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 equivalence
relation containing (f (x),g(x)) : x ∈ A , ordered by the reflexive and
transitive closure of ([x]≈,[y]≈) : x ≤B y ;
the exponential object BA
is the set of monotone maps A → 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 does not have a
subobject classifier.
We denote quasi orders by A = 〈A;≤A〉.
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4. Quasi orders
Since the forgetful functor QOrd → Set has left and right adjoints, the
shape of limits and colimits is forced. Also, orders do not add much: the
subcategory Ord of orders is reflective in QOrd.
This situation is, essentially, the same in the categories we will be interested
in. In particular, the shape of limits and colimits, when they exist, is forced
either by a forgetful functor to Set, or by an adjunction which is the
restriction of the one in QOrd. Moreover, we will not speak about orders
since the subcategories of orders are reflective in the subcategories of quasi
orders we are going to deal with.
It is reasonable to ask to which extent the structure of quasi orders transfers
to the full subcategories we are interested in, especially well quasi orders.
Also, it is reasonable to ask for perfect counterexamples when the structure
does not transfer.
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5. Well quasi orders and related categories
AQOrd is the full subcategory of QOrd whose objects are the quasi
orders whose antichains are all finite;
WFQOrd is the full subcategory of QOrd whose objects are the quasi
orders whose proper descending chains are all finite;
WQOrd is the full subcategory of QOrd whose objects are the quasi
orders having both the properties above, so this is the category of well
quasi orders.
Being finite quasi orders, the initial (terminal) object of QOrd lies in all
these subcategories, thus they have initial (terminal) objects, too.
Also, it is immediate to see that the construction of coproducts and
equalisers 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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6. Products
Products in the subcategories of interest, when they exist, must have the
same shape as in QOrd because it must be the right adjoint of the
restriction 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 in
the candidate product, and constructing out of it two descending chains in
the components, whose proper kernel must be finite, forcing the original
chain 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...01
be the word in O having n zeroes. Then {xn}n∈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 an infinite ascending
chain with elements from an infinite descending chain is the only way to
construct a counterexample in AQOrd.
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7. Coequalisers
It 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 Set
has a right adjoint.
Since every antichain in the coequaliser of f ,g : A → B in AQOrd induces
an antichain in B, it must be finite. So, AQOrd has coequalisers.
But WFQOrd does not. In fact, consider 〈N;=〉, the set of naturals with the
discrete order, and the disjoint union of ω copies 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 descending chains along an
infinite antichain is the only way to construct a counterexample in WFQOrd.
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8. Coequalisers
In fact, WQOrd has coequalisers. The proof is not immediate: first one has
to 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 that
either 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 the
elements of {ei }i in B by means of f and g.
Since {xi }i is infinite, there is a subsequence which forms an infinite
ascending chain, and it has the form {xi }i>m for some m ∈ ω. Thus, using the
fact that f , g, and the candidate coequaliser map z → [z]≈ are monotone. . .
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9. 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 descending
chain of length ω is not proper.
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10. 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 with
the 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 ≥ i
0 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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11. Summary
0 1 × + ⊃ eq coeq
QOrd
AQOrd - -
WFQOrd - -
WQOrd -
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12. 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∗
{⊥, };≤∗ (⊥,x),(x, ) : x ∈ A∗
{⊥, } 〉 ,
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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13. 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 founded
quasi order.
The proof goes by proving that there a suitable coequaliser can be
constructed so that it is isomorphic to 〈A;≤A〉.
It is relevant to remark the symmetry of the first two steps: the former
follows from the construction of an equaliser, while the latter from the
construction of a coequaliser. This is not by chance!
The rest of the proof follows the standard track, noticing that infinite
antichains are, in fact, a special case of bad sequences.
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14. Exponentiation, again
It is not difficult to infer
Proposition 3
If A is a well quasi order, the collection of finite and infinite sequences over
A ordered by embedding has the finite antichain property.
And, assuming the Axiom of Choice, it is evident that each function can be
thought of as a sequence of values indexed by its domain, so
Corollary 4
The exponential object A B
in QOrd, with B any quasi order, has finite
antichains whenever A is a well quasi order.
Thus, the counterexample to the existence of exponential objects in WQOrd
is prototypical: every counterexample must violate the finite proper
descending chain property.
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15. The end
Questions?
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