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Pi01120905

# Pi01120905 - 1 P01 INCOMPLETENESS finite set equations by...

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1 P 0 1 INCOMPLETENESS: finite set equations by Harvey M. Friedman December 4, 2005 ‘Beautiful’ is a word used by mathematicians with a semi rigorous meaning. We give an ‘arguably beautiful’ explicitly P 0 1 sentence independent of ZFC. See Proposition A from section 1. 1. P 0 1 INDEPENDENT STATEMENT. We use [1,n] for the discrete interval {1,...,n}. Let A [1,n] k . We write A’ = [1,n] k \A. This treats [1,n] k as the ambient space. Let R [1,n] 3k [1,n] k . We define R<A> = {y [1,n] k : ( \$ x A 3 )(R(x,y))}. We say that R is strictly dominating if and only if for all x,y [1,n] k , if R(x,y) then max(x) < max(y). We start with a basic finite fixed point theorem. THEOREM 1.1. For all k,n 1 and strictly dominating R [1,n] 3k [1,n] k , there exists A [1,n] k such that R<A’> = A. Furthermore, A [1,n] k is unique. We can obviously take complements, obtaining what we call the ‘complementation theorem’ for R<A>. THEOREM 1.2. For all k,n 1 and strictly dominating R [1,n] 3k [1,n] k , there exists A [1,n] k such that R<A> = A’. Furthermore, A [1,n] k is unique. Here is a trivial modification of Theorem 1.1 with the same proof. We call this the ‘complementation theorem’ for R<A\{t} k >. THEOREM 1.3. For all k,n,t 1 and strictly dominating R [1,n] 3k [1,n] k , there exists A [1,n] k such that R<A\{t} k > = A’. Furthermore, A [1,n] k is unique.

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2 We now incorporate the free set concept. Let R [1,n] 3k [1,n] k . We say that A is R free if and only if A and R<A> are disjoint. The reader will recall the following familiar ‘maximal free set’ theorem. THEOREM 1.4. For all k,n 1, every R [1,n] 3k [1,n] k has a maximal free set. Note that the equation in Theorems 1.2, R<A> = A’, asserts that A is a very strong kind of free set. THEOREM 1.5. For all k,n 1, every strictly dominating R [1,n] 3k [1,n] k has a free set A such that R<A> = A’.
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Pi01120905 - 1 P01 INCOMPLETENESS finite set equations by...

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