The canonical splinter theorem

We have seen inSection 4.1.3thatTheorem 18is already strong enough to imply most ofTheorem 20, but crucially does not guarantee the canonicity asserted in (iii). In this section we wish to prove a version ofTheorem 18 using a stronger set of assumptions from which we can deduceTheorem 20in full: we want to find, for a family A = (Ai | i ∈ I) of subsets of some universe U, a nested set N =N(A) meeting all the A_i that iscanonical, i.e., which only depends on invariants of A. More formally, we want to findN =N(A) in such a way that ifA⁰= (A⁰_i|i∈I) is another family of subsets of some other universeU⁰ that also meets the assumptions of our theorem, andϕ is an isomorphism of separation systems betweenS

i∈IA_iand S

i∈IA⁰_i withϕ(A_i) =A⁰_i for alli∈I, we ask that N(A⁰) = ϕ(N(A)). In particular, the nested set found by our theorem should not depend on the universe into which the familyAis embedded.

The assumptions ofTheorem 18are not sufficient to guarantee the existence of such a canonical set. Consider the example where we have just two separations,s andt, which are crossing and letA= (A1) = ({s, t}). Note thatA splinters, but there may be an automorphism that swaps the two separations, and so the choice of any single one of them would be non-canonical. Since the separations cross we cannot use both of them for our nested set either.

For obtaining a canonical nested set, one crucial ingredient will be the notion of extremal elements of a set of separations, which was already used in [17].

Given a set A⊆U of (unoriented) separations, an element a∈A isextremal inA, or anextremal elementofA, ifahas some orientationa that is a maximal element ofA. (Recall thatA is the set of orientations of separations inA.) The set of extremal elements of a set of separations is an invariant of separation systems in the following sense: if E is the set of extremal elements of some setA⊆S of separations, andϕis an isomorphism betweenS and some other separation system, thenϕ(E) is precisely the set of extremal separations ofϕ(A).

Moreover, the extremal separations of a set A⊆U are nested with each other under relatively weak assumptions: for instance, it suffices that for any two separations inAat least two of their corner separations also lie inA.

Let us formally state a set of assumptions under which we can prove a canonical version ofTheorem 18. Given two separationsrandsand two of their corner separationsc₁ andc₂, we say thatc₁ andc₂ arefrom different sides of r if, for orientations of c₁,r, andswithc₁= (r∧s), there is an orientation c₂ of c₂ such that either c₂= (r∧s) orc₂= (r∧s). Note thatc₁ andc₂ being from different sides ofrdoes not imply thatc₁and c₂ are distinct separations;

consider for instance the edge case thatr=s=c1=c2.

LetA= (Ai|i∈I) be a finite collection of non-empty finite subsets of U and let4be any partial order onI. We writei≺j if and only ifi4j andi6=j. We say thatAsplinters hierarchicallyif for allai∈Ai andaj ∈Aj the following two conditions hold:

(1) Ifi≺j, either some corner separation ofai andaj lies inAj, or two corner separations ofai andaj from different sides ofai lie inAi.

(2) If neither i ≺ j nor j ≺ i, there are k ∈ {i, j} and corner separations c1 and c2 of ai andaj from different sides ofak such thatc1 ∈Ak and c2∈Ai∪Aj.

a_i

a_j

a_i

a_j

a_i

a_j

a_i

a_j

Figure 4.2: The possible configurations in (2) in the definition of splinter hierarchically, up to symmetry.

In particular if 4 is the trivial partial order on I in which all i 6= j are incomparable then A splinters hierarchically if and only if (2) holds for all a_i ∈A_i anda_j ∈A_j; this special case which ignores the partial order onI is perhaps the cleanest form of an assumption that suffices for a canonical nested set meeting all A_i inA. The reason we need to allow a partial order4on I and the slightly weaker condition in (1) for comparable elements ofI is that otherwise we would not be able to deduce Theorem 20in full from our main theorem of this section due to a quirk in the way that robustness is defined for profiles in [17] (seeSection 4.1.6).

Our first lemma enables us to find a canonical nested set insideS

i∈IA_i for a collection of setsA_iwhose indexing set is an antichain:

Lemma 4.1.6. Let (Ai |i ∈I) be a collection of subsets of U that splinters hierarchically. If K⊆I is an antichain in4, then the set of extremal elements of S

k∈KAk is nested.

Proof. Suppose that K ⊆ I is an antichain and that for some i, j ∈K there are ai ∈ Ai and aj ∈ Aj such that ai and aj are extremal in S

k∈KAk but cross. Leta_ianda_jbe the orientations ofai andaj witnessing their extremality.

Sinceai andaj cross, there are three ways of orientingai andaj such that the supremum of this orientation is strictly larger thana_iora_j. Hence none of these corner separations can lie inAi∪Aj, since that would contradict the maximality of a_ior a_jin S

k∈KA_k. On the other hand, since neither i≺j norj ≺i, by condition (2) and the assumption that a_i anda_j cross there are at least two orientations ofa_ianda_j whose corresponding supremum lies inA_i∪A_j, causing a contradiction to the extremality ofa_i anda_j.

We are now able to prove a canonical splinter theorem by repeatedly applying Lemma 4.1.6 to the collection of theAi of4-minimal index that have not yet been met by the nested set constructed so far:

Theorem 19 (Canonical splinter theorem). Let U be a universe of separations and letA= (Ai|i∈I)be a collection of subsets ofU that splinters hierarchically with respect to a partial order4on I. Then there exists a nested setN =N(A) meeting everyAi inA.

Moreover,N(A)is canonical: ifϕis an isomorphism of separation systems betweenS

i∈IA_iand a subset of some universeU⁰ such that the familyϕ(A) :=

(ϕ(A_i) | i ∈ I) splinters hierarchically with respect to 4, then N(ϕ(A)) = ϕ(N(A)).

Proof. We proceed by induction on|I|. If|I|= 1 we can choose asN the set of extremal elements ofA_i, which is nested byLemma 4.1.6and clearly canonical.

So suppose that|I|>1 and that the claim holds for all smaller index sets.

LetK be the set of minimal elements ofI with respect to4. By Lemma 4.1.6 the setE =E(A) of extremal elements ofS

k∈KAk is nested. Let J ⊆I be the set of indices of all those Aj that do not meet E, and for j ∈ J let A⁰_j be the set of all elements of Aj that are nested with E. We claim that the collectionA⁰= (A⁰_j|j ∈J) splinters hierarchically with respect to4onJ. This follows from Lemma 2.1.1as soon as we show that eachA⁰_j is non-empty.

To see that eachA⁰_j is non-empty, forj ∈J letaj be an element ofAj that crosses as few elements of E as possible. We wish to show thataj is nested withE and thus aj ∈A⁰_j. So suppose thataj crosses some separation in E, that is, somea_i ∈A_i∩E withi∈IrJ. Sincei is a minimal element ofIwe have eitheri4j or thati andj are incomparable. We shall treat these cases separately.

Consider first the case thati≺j. By condition(1)of splintering hierarchically, either some corner separation ofa_i anda_j lies inA_j, or two corner separations of a_i anda_j from different sides ofa_i lie in A_i. The first of these possibilities contradicts the choice of aj, since that corner separation in Aj would cross fewer elements of E byLemma 2.1.1. On the other hand, the latter of these

possibilities contradicts the choice of a_i as an extremal element of S

k∈KA_k. Thus the casei4j is impossible.

Let us now consider the case that i and j are incomparable. Again, by the choice of aj, none of the corner separations ofai andaj can lie inAj by Lemma 2.1.1. Therefore condition(2) of splintering hierarchically yields the existence of a corner separation of ai and aj in Ai for each side of ai; this, however, contradicts the extremality ofai inS

k∈KAk as before.

Therefore each of the sets A⁰_j with j ∈ J is non-empty, and hence the collectionA⁰= (A⁰_j|j ∈J) splinters hierarchically with respect to4. Since|J|<

|I|we may apply the induction hypothesis to this collection to obtain a canonical nested setN⁰=N(A⁰) meeting allA⁰_j. NowN =N⁰∪E is a nested subset ofU which meets everyAifori∈I. It remains to show thatN is canonical.

To see thatN is canonical let ϕbe an isomorphism of separation systems between S

i∈IA_i and a subset of some universe U⁰ such that ϕ(A) splinters hierarchically with respect to4inU⁰. Then ϕ(E) =E(ϕ(A)), i.e., the set of extremal elements ofS

i∈Iϕ(A_i) is exactlyϕ(E). Therefore ϕ(E) meetsϕ(A_i) if and only if E meets A_i. Consequently the restriction of ϕ toS

j∈JA⁰_j is an isomorphism of separation systems between S

j∈JA⁰_j and its image in U⁰ with the property that ϕ(A⁰) splinters hierarchically with respect to 4on J. Moreover, forj∈J, the imageϕ(A⁰_j) ofA⁰_j is exactly the set of those separations in ϕ(A_j) that are nested withϕ(E).

Thus we can apply the induction hypothesis to find that N(ϕ(A⁰)) = ϕ(N(A⁰)). Together with the above observation that ϕ(E(A)) = E(ϕ(A)) this gives

ϕ(N(A)) =ϕ(E(A))∪ϕ(N(A⁰)) =E(ϕ(A))∪N(ϕ(A⁰)) =N(ϕ(A)), concluding the proof.