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[4] (also see [10]), who proved that (2) (2α(G) − 1) · η(G) ≥ |G| . This result was improved by Kawarabayashi et al. [6] to (3) (2α(G) − 1) · η(G) ≥ |G| + ω(G) . Assuming α(G) ≥ 3, Kawarabayashi et al. [6] proved that (4) (4α(G) − 3) · η(G) ≥ 2|G|, which was further improved by Kawarabayashi and Song [7] to (5) (2α(G) − 2) · η(G) ≥ |G|. The following theorem is the main contribution of this note.

Theorem 1. Every graph G with η(G) ≥ 5 satisfies (2α(G) − 1)(2η(G) − 5) ≥ 2|G|− 5 .

Observe that Theorem 1 represents an improvement over (2), (4) and (5) whenever 2 2 |G| η(G) ≥ 5 and |G|≥ 5 η(G) . For example, Theorem 1 implies that α(G) > 7 for every |G| graph G with η(G) ≤ 6, whereas each of (2), (4) and (5) imply that α(G) > 12 .

2. Proof of Theorem 1

Theorem 1 employs the following lemma by Duchet and Meyniel [4]. The proof is included for completeness.

Lemma 1 ([4]). 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 two from D, and there is a neighbour w of v at distance one from D. Let D := D ∪{v, w} and S′ := S ∪{v}. Thus D′ is connected and contains an independent set S′ such that |D′| =2|S′|− 1. 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 G with η(G) ≤ t satisfies p·α(G) ≥ |G|. Then every graph G with η(G) ≥ t satisfies 2|G|− p 1 α(G) ≥ + . 4η(G)+2p − 4t 2 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 |D| = 2|S|− 1. |D|+1 Now α(G) ≥ |S| = 2 . Thus we are done if |D| +1 2|G|− p 1 (6) ≥ + . 2 4η(G)+2p − 4t 2 INDEPENDENT SETS IN GRAPHS WITH AN EXCLUDED CLIQUE MINOR 3

Now assume that (6) does not hold. That is, 2|G|− p |D|≤ . 2η(G)+ p − 2t Thus (2η(G)+ p − 2t − 2)|G| + p |G \ D| = |G| − |D|≥ . 2η(G)+ p − 2t Since D is dominating and connected, η(G \ D) ≤ η(G) − 1. Thus by induction, 2|G \ D|− p 1 α(G) ≥ α(G \ D) ≥ + 4η(G \ D)+2p − 4t 2 2(2η(G)+ p − 2t − 2)|G| +2p p 1 ≥ − + (2η(G)+ p − 2t)(4η(G) − 4+2p − 4t) 4η(G) − 4+2p − 4t 2 2|G|− p 1 = + . 4η(G)+2p − 4t 2 This completes the proof. 

Lemma 3. Suppose that Hadwiger’s Conjecture is true for some integer t. Then every graph G with η(G) ≥ t satisfies (2η(G) − t)(2α(G) − 1) ≥ 2|G|− t .

Proof. If Hadwiger’s Conjecture is true for t then t · α(G) ≥ |G| for every graph G with η(G) ≤ t. Thus Lemma 2 with p = t implies that every graph G with η(G) ≥ t satisfies 2|G|− t 1 α(G) ≥ + , 4η(G) − 2t 2 which implies the result. 

Theorem 1 follows from Lemma 3 with t = 5 since Hadwiger’s Conjecture holds for t = 5 [14].

3. Concluding Remarks The proof of Theorem 1 is substantially simpler than the proofs of (3)–(5), ignoring its dependence on the proof of Hadwiger’s Conjecture with t = 5, which in turn is based on the four-colour theorem. A bound that still improves upon (2), (4) and (5) but with a completely straightforward proof is obtained from Lemma 3 with s = 3: Every graph G with η(G) ≥ 3 satisfies (2η(G) − 3)(2α(G) − 1) ≥ 2|G|− 3. We finish with an open problem. The method of Duchet and Meyniel [4] was generalised by Reed and Seymour [12] to prove that the fractional chromatic number χf (G) ≤ 2η(G).

For sufficiently large η(G), is χf (G) ≤ 2η(G) − c for some constant c ≥ 1?

References

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Departament de Matematica` Aplicada II, Universitat Politecnica` de Catalunya, Barcelona, Spain E-mail address: [email protected]