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Independence is monotone under induced embeddings

Statement

Let J and G be finite simple graphs. If J has an induced embedding into G, then α(J)≤α(G). In particular, α(G)<α(J) excludes an induced copy of J in G.

Facts & Assumptions

[F1]

An induced embedding is an injection on vertices that preserves adjacency and nonadjacency between distinct vertices; an induced copy arises from such an embedding. See Induced embeddings and induced copies of a graph.

[F2]

An independent set has no adjacent distinct vertices. For a finite simple graph, its independence number is the maximum cardinality of an independent vertex set, and this maximum is attained. See Cliques, independent sets, clique number and independence number.

[F3]

A bijection from a finite set to another set makes their cardinalities equal. See The cardinality | A| of a finite set.

[F4]

An injection satisfies f(x)=f(y)⇒x=y; a surjection takes every value in its codomain; a bijection has both properties. The image f(S) consists of the values f(s) for s∈S. See Injection, surjection, bijection.

Proof

Given: Finite simple graphs J and G and an induced embedding f:V(J)→V(G).

Goal: Show α(J)≤α(G) and the stated exclusion of an induced copy.

1.1F1F2F4choose

Choose an independent set S⊆V(J) with ∣S∣=α(J). If two distinct vertices of f(S) were adjacent, their distinct preimages in S would be adjacent by the defining equivalence for an induced embedding, contrary to independence. Thus f(S) is independent in G.

2.1F1F2F3F4step 1.1∎

The restriction f:S→f(S) is injective because f is injective, and it is surjective by the definition of f(S). Hence it is a bijection, so ∣f(S)∣=∣S∣=α(J). By maximality of α(G), α(J)≤α(G). Consequently, if α(G)<α(J), no induced embedding, and hence no induced copy of J in G, can exist.

Depends on

Used by · 1

Dependency tree · 2 levels

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