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Unformatted text preview: 1 3.2. Some Useful Lemmas. DEFINITION 3.2.1. The standard pairing function on N is the function P:N 2 → N due (essentially) to Cantor: P(n,m) = (n 2 +m 2 +2nm+n+3m)/2 ≥ n,m. It is well known that P is a bijection, and also that for all n ≥ 0, [0,n(n+1)/2) ⊆ P[[0,n) 2 ]. In addition, P is strictly increasing in each argument. Let T:N 2 → N be such that T(2n,2m) = P(n,m), T(2n,2m+1) = T(2n+1,2m) = T(2n+1,2m+1) = 2n+2m+2. Then for all n ≥ 0, [0,n(n+1)/2) ⊆ T[([0,2n) ∩ 2N) 2 ]. Hence for all n ≥ 8, every element of [0,n 2 /8) is realized as a value of T at even pairs from [0,n). It is clear that T(2n,2m) ≥ (n 2 +2n)/2,(m 2 +2m)/2 ≥ 2n,2m. Hence for n,m ≥ 2, T(n,m) ≥ n,m. LEMMA 3.2.1. There exists 3ary f ∈ ELG ∩ SD such that the following holds. Let A ⊆ N be nonempty, where fA ∩ 2N ⊆ A. Then fA is cofinite. We can also require that for all n ≥ 0, f(n,n,n) ∈ 2N. Proof: We define f ∈ ELG ∩ SD as follows. Let p,q ∈ [2 n ,2 n+1 ), n ≥ 0. Define f(2 n ,p,q) = min(2 n+1 +T(p2 n ,q 2 n ),2 n+2 ). Note that for n ≥ 8, as p,q vary over the even elements of [2 n ,2 n+1 ), every value in [2 n+1 ,2 n+2 ) is realized. Also note that for all n ≥ 0, f(2 n ,2 n ,2 n ) = 2 n+1 . For all n > 0, define f(n,n,n) to be the least 2 k ≥ 2n; f(0,0,0) = 2. For all n < m < r, define f(r,n,n) = 2r+1, f(r,n,m) = 2r+2, f(r,n,r) = 2r+3, f(r,m,n) = 2r+4, f(r,r,n) = 2r+5. For all triples a,b,c, if f(a,b,c) has not yet been defined, define f(a,b,c) = 2a,b,c+1. It is obvious that f ∈ SD. To see that f ∈ ELG, we need only examine the definition of f(2 n ,p,q), p,q ∈ [2 n ,2 n+1 ), where n is sufficiently large. If p,q ∈ [2 n ,2 n +2 n1 ), then obviously f(2 n ,p,q) ≥ 2 n+1 ≥ 42 n ,p,q/3. If p,q ∉ [2 n ,2 n +2 n 1 ), then f(2 n ,p,q) ≥ 2 n+1 +T(p2 n ,q2 n ) ≥ 2 n+1 + 2 n1 ≥ 5p/4,5q/4. Also,f(2 n ,p,q) ≤ 2 n+2 ≤ 2p,2q. Therefore f ∈ ELG. 2 Let A ⊆ N be nonempty, where fA ∩ 2N ⊆ A. Let f(min(A),min(A),min(A)) = 2 k ≥ 2. Then 2 k ∈ fA ∩ 2N. Therefore 2 k ∈ A. Suppose j ≥ k and 2 j ∈ A. Then f(2 j ,2 j ,2 j ) = 2 j+1 ∈ fA. We have thus established by induction that for all j ≥ k, 2 j ∈ A....
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This note was uploaded on 08/05/2011 for the course MATH 366 taught by Professor Joshua during the Fall '08 term at Ohio State.
 Fall '08
 JOSHUA
 Math

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