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Disjoint labels
Let be a finite family of objects. Give each object a nonempty label set of size at most , and suppose each label occurs in at most objects. Then there is a subfamily with pairwise disjoint label sets and .
Indeed, take a maximal such subfamily. Every unchosen object meets a label of a chosen object. For each chosen object, the union of its at most labels occurs in at most objects. Counting these covering families proves the bound, even when they overlap.
More generally, if objects have label sets of size at most and every set of at most forbidden labels can be avoided by some object of each of prescribed types, one can choose an object of each type with pairwise disjoint labels. Choose types in order; the union of previously chosen labels has size at most . Empty labels and cause no problem. The simple avoidance estimate behind this argument is that a set of labels excludes at most objects when each label has incidence at most .
Row fans
Suppose each object has a row and a nonempty label set of size at most , with . Suppose there are at most rows, each vertex occurs in for at most objects, and within any fixed row each label occurs in at most objects. Let , , and let be forbidden. If
there are fans, each consisting of objects of the same row with disjoint label sets. Their whole sets, consisting of the row and all their labels, avoid and are pairwise disjoint between fans.
To choose the next fan, discard every object meeting or a previously selected fan. The total forbidden set has size at most , so (1) leaves more than objects. Some row has more than of them. Within that row, greedily select objects. The labels already used exclude at most candidates at any stage, so selection continues. The labels are nonempty, so the selected objects are distinct. The new whole set has at most vertices. This proves the induction and all disjointness assertions.
Separating ordered roles
If each object specifies an ordered pair of distinct vertices , independent fair two-coloring puts in class zero and in class one with probability . Some coloring retains at least objects. For prescribed ordered pairs per object, apply this argument successively to the remaining family, using a separate coloring for each pair. At least objects remain, and each prescribed first role is separated from every prescribed second role for its coloring, even across different remaining objects. This last cross-object conclusion is why the colorings are retained, rather than merely testing distinctness within each object.
Weighted common neighborhoods
Let be finite sets, let , and give a nonnegative weight . Suppose , , and whenever . For any , if
there are distinct whose common neighborhood has weight greater than .
For a set of size , there are ordered -tuples. The union bound over pairs of equal coordinates gives at most noninjective tuples. Thus at least injective tuples lie in . Count the weighted incidences of with such tuples in two orders. The total is at least , while at most tuples are possible. If every common weight were at most , this would contradict (2).
The weighted common-neighborhood statement holds with real nonnegative weights as well as integer weights. The application uses integer path multiplicities.
Source and scope
Complete elementary proofs of the instances used from
FiniteLabelPacking, FiniteDisjointSubfamily, OrientedRolePartition,
FiniteRolePartitions, FiniteRowFans, and FiniteWeightedCommon, pinned
Lean lines 6066–6154, 6603–6667, 7084–7173, 7838–7952, and 8815–8918.
The row-fan statement above includes nonempty labels, as in its application
to positive path tails; the formal interface also permits empty labels.
The tuple collision estimate is the elementary
KSTUpper.noninjective_count input, not an appeal to the full
Kővári–Sós–Turán theorem. See the
exposition, p. 5, for the outline
these selections supply.
Used by. Suffix fans; Heavy common neighborhoods; Light-path count.
Bears on. #571.