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Lemma 2.1: repetition characterizes connectivity

PDF p. 2: Lemma 2.1 and proof

Statement

Let k≥1 be an integer and let H be a finite simple graph with at least one vertex. Then H has the Ramsey repetition property at k if and only if H is connected.

Connected means that every two vertices of H are joined by a path. The repetition property means that, for every finite simple nonempty graph A and every integer q≥1, A∈Rindk(H) if and only if qA∈Rindk(H). Here qA is the disjoint union of q copies of A, and membership in Rindk(H) means that every colouring of the graph's vertices with colours from {1,…,k} contains a monochromatic induced copy of H. An induced copy preserves both adjacency and nonadjacency, and monochromatic means that all its vertices have the same colour.

Facts & Assumptions

[F1]

For a connected finite simple nonempty graph H and positive integer k, H has the Ramsey repetition property at k. Connected targets have repetition

[F2]

The connected components of a finite graph partition its vertex set. They are nonempty connected subgraphs; if the nonempty graph is disconnected, it has at least two components. Connected graphs and connected components defined by the existence of vertex paths The connected components of a graph partition its vertex set and are its maximal connected subgraphs

[F3]

If a finite simple graph H has t≥2 connected components and k is positive, there exist a finite simple graph A and a positive integer q such that A∉Rindk(H) but qA∈Rindk(H). Specifically, the witness lemma takes q=k(t−1)+1 and supplies an eligible A. Disjoint-copy witness against repetition

[F4]

The repetition property requires the two membership predicates for A and qA to agree for every permitted A and positive q. Ramsey repetition property at fixed k

Proof

Given: A positive integer k and a finite simple nonempty graph H.

Goal: Show that H has the repetition property at k exactly when H is connected.

1.1givenF1

If H is connected, F1 applies to the given H and k, so H has the repetition property at k.

2.1givenF2F3F4step 1.1∎

If H is disconnected, F2 gives t≥2 connected components. By F3, there are a permitted graph A and positive integer q for which qA belongs to Rindk(H) but A does not. This contradicts the universal equivalence required by F4, so H does not have the repetition property. Thus the repetition property implies connectivity by contraposition; together with step 1.1, this proves the equivalence.

Depends on

Used by · 1

Dependency tree · 2 levels

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Sources