Independent sets in graphs with an excluded clique minor

Let G be a graph with n vertices, with independence number α , and with no K t +1 -minor for some t ≥ 5 . It is proved that (2 α − 1)(2 t − 5) ≥ 2 n − 5 . This improves upon the previous best bound whenever n ≥ 25 t 2 .


Introduction
In 1943, Hadwiger [7] made the following conjecture, which is widely considered to be one of the most important open problems in graph theory (i) ; see [19] for a survey.
Progress on the t = 6 case has been recently been obtained by Kawarabayashi and Toft [10] (without using the Four-Colour Theorem).The best known upper bound is χ(G) ≤ c • η(G) log η(G) for some constant c, independently due to Kostochka [11] and Thomason [17,18].

Proof of Theorem 1
Theorem 1 employs the following lemma by Duchet and Meyniel [5].The proof is included for completeness.

Lemma 1 ([5]
) Every connected graph G has a connected dominating set D and an independent set S ⊆ D such that |D| = 2|S| − 1.
Proof: Let D be a maximal connected set of vertices of G such that D contains an independent set S of G and |D| = 2|S| − 1.There is such a set since D := S := {v} satisfies these conditions for each vertex v.We claim that D is dominating.Otherwise, since G is connected, there is a vertex v at distance 2 from D, and there is a neighbour w of v at distance 1 from D. Let D ′ := D ∪ {v, w} and S ′ := S ∪ {v}.Thus D ′ is connected and contains an independent set S ′ such that Hence D is not maximal.This contradiction proves that D is dominating.✷ The next lemma is the key to the proof of Theorem 1.
Lemma 2 Suppose that for some integer t ≥ 1 and for some real number p ≥ t, every graph Proof: We proceed by induction on η(G) − t.If η(G) = t the result holds by assumption.Let G be a graph with η(G) > t.We can assume that G is connected.By Lemma 1, G has a connected dominating set D and an independent set S ⊆ D such that Now assume that (6) does not hold.That is,

1 4
|V (G)| for every planar graph G.It is interesting that the only known proof of this result depends on the Four-Colour Theorem.The best bound not using the Four-Colour Theorem is α