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Unformatted text preview: Math 486 Homework 4 Notes Paul Shafer [email protected] February 26, 2008 Problem I.6.8 We give two proofs that a countable graph is 4colorable if every finite subgraph is 4colorable. Proof via K¨onig’s lemma: Let G = ( { a ,a 1 ,... } ,E ) be a countable graph. Let G n denote the subgraph induced by { a ,...,a n 1 } . Define a tree T ⊆ 4 <ω (= finite strings over { , 1 , 2 , 3 } ) as follows: σ ∈ T ⇔ ∀ i,j <  σ  (( a i ,a j ) ∈ E → σ ( i ) 6 = σ ( j )) . That is, σ is in T if and only if σ represents a 4coloring of G  σ  . To prove that T is a tree, we need to show that if σ ∈ T and τ ⊆ σ then τ ∈ T . This is clear because if ∀ i,j <  σ  (( a i ,a j ) ∈ E → σ ( i ) 6 = σ ( j )) holds for σ then it holds for any substring of σ . T is finitely branching because it a subtree of 4 <ω which is finitely branching. We prove that T is infinite by showing that for each n there is a σ in T of length n . G n is a finite subgraph of G and so is 4colorable by assumption. Let c : { a ,...,a n 1 } → { , 1 , 2 , 3 } be a 4coloring of G n . Let σ be the corresponding string. That is, define σ of length n by σ ( i ) = c ( a i ) for i = 0 ,...,n 1. Then σ ∈ T . T is an infinite, finitely branching tree and so has an infinite path P = { σ n  n ∈ ω...
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This homework help was uploaded on 02/23/2008 for the course MATH 4860 taught by Professor Dorais during the Spring '08 term at Cornell University (Engineering School).
 Spring '08
 DORAIS
 Math, Logic

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