Alphabeta Math
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Ordering targets and excluding H1 from H2

PDF p. 3: Section 3, paragraph beginning Assume without loss

Statement

Let H1,H2 be non-isomorphic connected finite simple graphs, each containing an edge. Label them so that β(H1)≥β(H2) and, if equality holds, ∣H1∣≥∣H2∣. Then H2∈Forb⁡(H1).

Here ∣H∣ is the number of vertices of H. A graph is connected if every two vertices are joined by a path. A block is a maximal 2-connected subgraph, or the subgraph consisting of a bridge and its endpoints; 2-connected means having at least three vertices, being connected, and remaining connected after any one vertex is deleted. A bridge is an edge whose deletion increases the number of connected components. Since each target has an edge, β(H) is the largest number of vertices in a block of H. Finally, G∈Forb⁡(H) means that G contains no induced copy of H: there is no injective vertex map preserving and reflecting adjacency between distinct vertices.

Facts & Assumptions

[F1]

An induced embedding is an injective vertex map preserving and reflecting adjacency; in particular it is an ordinary embedding, an injective vertex map preserving edges. An induced copy exists exactly when such an embedding exists. Induced embeddings and induced copies of a graph Ordinary embeddings as injections preserving edges

[F2]

An ordinary embedding of one nonempty finite simple graph into another implies that the largest-block parameter of the first is at most that of the second. The parameter b in that result equals β under the block convention used here. Block containment and monotonicity under subgraphs Blocks and the parameter beta

[F3]

An injection between finite vertex sets is a bijection onto its image; the image has the cardinality of the domain and, as a subset of the target, has cardinality at most that of the target. The cardinality | A| of a finite set A subset of a finite set is finite, with | B| ≤ | A|, and equality holds if and only if B = A

[F4]

An induced embedding between finite simple graphs of equal vertex order is an isomorphism. Rigidity of mutual ordinary embeddings

[F5]

Membership in Forb⁡(H) is precisely the absence of an induced copy of H. Induced forbidden-subgraph class

Proof

Given: Non-isomorphic connected finite simple graphs H1,H2, each with an edge, ordered as in the statement.

Goal: Show that H2 contains no induced copy of H1.

1.1givenF1F2assume-contra

Suppose that H2 contains an induced copy of H1. Its induced embedding is an ordinary embedding, so block monotonicity gives β(H1)≤β(H2). The given reverse inequality therefore forces equality. In particular, if the first inequality in the ordering were strict, the supposed copy would already be impossible.

2.1givenstep 1.1F1F3

In the equality case, the ordering gives ∣H1∣≥∣H2∣. The induced embedding is injective, so finite cardinality gives ∣H1∣≤∣H2∣. Hence the vertex orders are equal.

3.1givenstep 2.1F4F5discharge-contradiction∎

The induced embedding between these equal-order finite graphs is an isomorphism, contradicting the assumption that H1 and H2 are non-isomorphic. Thus no induced copy exists, and H2∈Forb⁡(H1) by definition.

Depends on

Used by · 3

Dependency tree · 2 levels

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