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BRTCamb111810' - BOOLEAN RELATION THEORY by Harvey M...

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Unformatted text preview: BOOLEAN RELATION THEORY by Harvey M. Friedman Distinguished University Professor Mathematics, Philosophy, Computer Science Ohio State University [email protected] http://www.math.ohio-state.edu/ ~friedman/ University of Cambridge Cambridge, England November 10, 2010 Chapter 1 Introduction to BRT 1.1. General Formulation We begin with two Theorems at the heart of BRT = Boolean Relation Theory. THIN SET THEOREM. Let k ≥ 1 and f:N k ➞ N. There exists an infinite set A ⊆ N such that f[A k ] ≠ N. COMPLEMENTATION THEOREM. Let k ≥ 1 and f:N k ➞ N. Suppose that for all x ∈ N k , f(x) > max(x). There exists an infinite set A ⊆ N such that f[A k ] = N\A. Proof of TST: Let f:N k ➞ N. Color every x ∈ N k with f(x) if f(x) ∈ [0,ot(k)]; ot(k)+1 otherwise. By Ramsey’s theorem, let A ⊆ N be infinite, where the color of x ∈ A k depends only on the order type of x. Then the possible colors of x ∈ A k form a subset of [0,ot(k)] of cardinality ≤ ot(k), together with, possibly ot(k)+1. Hence f[A k ] does not even include [0,ot(k)]. QED TST IMPLIES INFINITE RAMSEY THEOREM? This is still wide open. We showed that TST is not provable in ACA . SHARP THIN SET THEOREM. Let k ≥ 1 and f:N k ➞ N. There exists an infinite set A ⊆ N such that f[A k ] does not include [0,ot(k)]. THEOREM. Sharp TST implies infinite Ramsey theorem. Proof: Assume TST. Let a 2-coloring of strictly increasing x ∈ N k be given, with colors 0,1. Color x ∈ N k by the order type of x if x is not strictly increasing (as an integer in [2,ot(k)]); the color of x otherwise. QED We proved that TST in dimension 2 is not provable in WKL . It is open whether TST in dimension 3 implies ACA . An old result of ours is that infinite Ramsey theorem is equivalent to ACA’ over RCA . ACA’ = ACA + ( ∀ n,x)(the n-th Turing jump of x exists). COMPLEMENTATION THEOREMS COMPLEMENTATION THEOREM. Let k ≥ 1 and f:N k ➞ N. Suppose that for all x ∈ N k , f(x) > max(x). There exists an infinite set A ⊆ N such that f[A k ] = N\A. In fact, A is unique. Build A by recursion. Suppose we have decided membership in A for all 0 ≤ i < n. Put n in A if and only if n ∉ f[A] up to now. These decisions are stable because f is strictly dominating. Every for simple f, the unique A may be complicated. For a study of this, it is better to use the upper image. Let f:N k ➞ Z. Write fA for f[A k ], and f < A for {f(x) > max(x): x ∈ A k }. COMPLEMENTATION THEOREM’. Let k ≥ 1 and f:N k ➞ Z. There exists an infinite set A ⊆ N such that f < A = N\A. A is unique. What is the structure of A if f is integral affine? Integral polynomial? BACK TO BOOLEAN RELATION THEORY We now use suggestive notation to restate the Thin Set Theorem and the Complementation Theorem....
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BRTCamb111810' - BOOLEAN RELATION THEORY by Harvey M...

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