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a4notes - Math 486 Homework 4 Notes Paul Shafer...

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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 4-colorable if every finite subgraph is 4-colorable. Proof via K¨onig’s lemma: Let G = ( { a 0 , a 1 , . . . } , E ) be a countable graph. Let G n denote the subgraph induced by { a 0 , . . . , a n - 1 } . Define a tree T 4 (= finite strings over { 0 , 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 4-coloring 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 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 4-colorable by assumption. Let c : { a 0 , . . . , a n - 1 } → { 0 , 1 , 2 , 3 } be a 4-coloring 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 } , where | σ n | = n . Define the coloring c : { a 0 , a 1 , . . . } → { 0 , 1 , 2 , 3 } by c ( a i ) = σ n ( i ) for any (or all) n > i , and note that c is
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