Alphabeta Math
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Theorem 1.1: both vertex-Ramsey distinctions

PDF p. 1: Abstract; Theorem 1.1; PDF p. 2: Section 3, ordinary part; PDF p. 3: Section 3, induced part; PDF p. 4: Section 3, conclusion

Statement

Let H1,H2 be non-isomorphic finite simple graphs, each with at least one vertex, and let k be a positive integer. Then Rk(H1)≠Rk(H2) and Rindk(H1)≠Rindk(H2).

Graphs are isomorphic if a bijection between their vertex sets preserves and reflects adjacency. For a graph H, Rk(H) is the collection of nonempty finite simple graphs G for which every function V(G)→{1,…,k} has a monochromatic ordinary copy of H: a subgraph isomorphic to H whose vertices all have one colour. The induced class Rindk(H) uses the same condition with induced copies, which also preserve nonadjacency. Equality of classes means equality of membership for every such graph G.

Facts & Assumptions

[F1]

For non-isomorphic nonempty finite simple graphs and any positive integer k, their ordinary k-colour vertex-Ramsey classes are distinct. Ordinary part of Theorem 1.1

[F2]

For non-isomorphic nonempty finite simple graphs and any positive integer k, their induced k-colour vertex-Ramsey classes are distinct. Induced part of Theorem 1.1

Proof

Given: Non-isomorphic finite simple graphs H1,H2, each with at least one vertex, and a positive integer k.

Goal: Prove both stated inequalities of vertex-Ramsey classes.

1.1givenF1

The given graphs and colour count satisfy the hypotheses of F1. Therefore Rk(H1)≠Rk(H2).

2.1givenF2step 1.1∎

The same hypotheses satisfy F2. Therefore Rindk(H1)≠Rindk(H2). Combining the two inequalities proves the statement.

Source notes

The article citation attributes the combined theorem. The argument above composes the separate ordinary and induced results.

Depends on

Dependency tree · 2 levels

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Sources