Lemma 3.4.1. LetS be a regular separation system, X ⊆S a set of pairwise crossing separations, and P any orientation of X. Then there is a consistent orientationO ofS extendingP with P ⊆(maxO).
Moreover, ifP0 is a fixed consistent orientation of the set of all those separa-tions inSrX that cross all ofX, then there is a unique suchO with P0 ⊆O.
Proof. We only prove the ‘moreover’-part; if no suchP0 is supplied, we first use the Extension Lemma2.4.1to obtain one.
Let Z ⊆ S be the separations in S that are not oriented by P∪P0. By definition of P0 every element ofZ is nested with some separation inX. Let
O:=P∪P0∪ {s |s∈Z ands 6x for somex withx∈X}.
ThenO is antisymmetric and consistent: suppose there arer , s ∈O with r6s. Then at least one of them, says, lies outside ofP∪P0. By definition ofO we haves6x for somex withx∈X. Thenr 6x, which means thatr /∈(P∪P0) sinceris nested withx. Therefore there is a separationx0 withx0∈Xandr6x0. But now xandx0 are nested, resulting in a contradiction since x6=x0 by the regularity ofS.
ThusO is a consistent orientation ofS extendingP∪P0. To see thatP ⊆ (maxO) consider somex ∈P. Then no other element ofP∪P0 lies abovex since they all crossx. But neither is there as∈Or(P∪P0) withx 6s, since by definition ofO this would imply that xis nested with some other element ofX. Hencex is indeed a maximal element ofO.
Finally, for the uniqueness, observe that orienting any s ∈ Or(P ∪P0) differently fromO would either makeO inconsistent (ifs6x for somex∈P), or make some element ofP not maximal (ifs 6x for some x∈P).
Note thatLemma 3.4.1holds for infinite separation systems, too.
We can now proveTheorem 9.
Proof of Theorem 9. LetSbe a finite regular separation system. Fix a consistent orientation O of S. We will find, for each set X ⊆ S of pairwise crossing separations, a consistent orientation OX of S such that (maxOX)rO is an orientation ofX.
So letX be a (possibly empty) set of pairwise crossing separations. LetP be the orientation ofX given by the inverse ofO onX, and letP0be the restriction ofO to the set of all those separations inSrX that cross every element ofX. ByLemma 3.4.1there is a unique consistent orientationOXofSextendingP∪P0 withP ⊆maxOX. Let us show that in factP = (maxOX)rO. By definition ofP andP ⊆(maxOX) we haveP ⊆(maxOX)rO. For the converse direction, consider an elements ∈(maxOX)rO and suppose thats /∈P. Thens does not lie in P0 either sinceP0 ⊆O. Therefore, by the definition ofOX, there is an x with x∈X ands 6x. Sinces was assumed to be a maximal element of OX we must have x ∈OX, i.e. x ∈P. But thenO is inconsistent sinceO orientssandxas s andx, a contradiction.
ClearlyOX6=OY ifX 6=Y. Let us now show that each consistent orientation ofS is of the formOX for someX of pairwise crossing separations.
So letO0 be a consistent orientation of S. LetP := (maxO0)rO andX the set of underlying unoriented separations ofP. ThenX is a set of pairwise crossing separations: if two separations in P were nested they would point
towards each other, which would makeO inconsistent sinceO orients them the other way around. It remains to show thatO0 =OX. Let Z be the set of all those separations in SrX that cross every element of X. Then O0 andOX
agree on X and, by the uniqueness part of Lemma 3.4.1, also on Z. Let P0 be the restriction of OX to Sr(X ∪Z). Then P0 ⊆O by definition of OX. Suppose now that OX 6=O0. Then, sinceS is finite, some maximal elements ofO0 gets oriented ass byOX. By the above observation we must haves ∈P0, that is,s ∈O. But this givess ∈(maxO0)rOand hence contradicts the fact that O0 andOX agree onX.
ThereforeO0 =OX, which concludes the proof that the mapX 7→OX is a bijection between the pairwise crossing sets and the consistent orientations ofS.
For infinite separation systems the map defined inTheorem 9is still injective (with the same proof), but not necessarily surjective.
Let us use the technique from Theorem 9 to, in a sense we shall specify below, improve upon Theorem 6, or more accurately, upon Theorem 5. The latter result gives a way to represent a regular separation systemS by using as a ground-setV the set of consistent orientations ofS, and mapping eachs ∈S to the bipartition ofV into those orientations which contains and those that do not. Recall that Theorem 5, in its original source [13], was formulated for regular tree sets only. In [32] this result was improved slightly by showing that one can take a much smaller setV: the set of only those consistent orientations with fewer than three maximal elements still gives rise to a representation of any regular tree set, using the same map mapping eachs to the bipartition ofV into the orientations that do and do not contains.
As we noted inSection 3.2,Theorem 5 readily extends from regular tree sets to regular separation systems. Let us now show that a similar slimming of the ground-set can be performed there, too:
Theorem 10. LetS be a regular separation system,O a fixed consistent orient-ation ofS, andOX defined as in Theorem 9for every setX of pairwise crossing separations. LetO<3 be the set of allOX with|X|<3, and fors ∈S letO<3(s) be the set of all OX ∈ O<3 that contain s.
Then the map f given by s 7→ (O<3(s),O<3(s)) is an isomorphism of separation systems between S and its image.
Proof. Observe first thatO =O∅ ∈ O<3. The mapf clearly commutes with the involution. Furthermore it follows from the consistency of the orientations inO<3 thatf(r)6f(s) wheneverr 6s. It thus remains to show thatf(r) andf(s) cross whenever randsdo, and thatf is injective. Once we know the former the latter only needs to be checked for nestedrands.
So letrandsbe two crossing separations. Then each of the four possible orientations ofrandsis contained in precisely one ofO, O{r}, O{s}, andO{r,s}, which shows that the images ofr andscross as well. In particular the images are distinct.
It remains to show that f is injective. For this we only need to show that f(r) 6=f(s) whenever r ands are distinct nested separations. So let r ands be orientations ofrandssuch thatr 6s; we need to find an element ofO<3 that containsr ands. IfO happens to containr ands there is nothing to show for us. If not, thenO contains eitherr and s, orr and s. In the first
caseO{r}is an element ofO<3that containsr ands, and in the latter caseO{s}
is such an element ofO<3.
In particular ifS is finite then the size of the ground-set used to representS via bipartitions is inO(|S|2).
Note thatTheorem 10extends easily to fastidious separation systems: simil-arly to the proof ofTheorem 6, the unique small separation of such a fastidious separation system automatically gets mapped to (∅, V).
In generalO<3 is not a ground-set of overall smallest possible size to enable a representation ofSvia bipartitions. This can be seen, for instance, in tree sets:
for a tree set ofn−1 separations the setO<3 has sizen, that is, the number of vertices of the tree corresponding to that tree set. However to represent the tree set by bipartitions of some set it suffices to use only the vertices of degree less than three.