Let a a α α e then in m a κ and for all x 1 x n κ

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Let A = {aα: αE}. Then in M*, |A| = κ, and for all x1,...,xn< κ, f(x1,...xn) x1,...,xnf(x1,...,xn) A. Clearly min(A) f(A\{min(A)}). QED THEOREM 1.4.8. The Thin Set Theorem fails on (SAFCN(3,),NOPSUB()), (CSAFCN(3,),UNOPSUB()). If the continuum hypothesis holds then the Thin Set Theorem fails on (FCN(2,),cSUB()). Proof: Let f:3be given by f(x,y,z) = 1/(x-y) + 1/(x-z) if defined; 0 otherwise. Then f is semialgebraic. Let a,b , a < b. We claim that f[(a,b)] = . To see this, let u . Fix x (a,b). We can find y,z (a,b) such that 1/(x-y) and 1/(x-z) are any two reals with sufficiently large absolute values. Hence we can find y,z (a,b) such that f(x,y,z) = u. Let f:3be given by x(y-z). Then f is continuous and semialgebraic. Let A be an unbounded open subset of . We claim that fS = . To see this, let u . Let (a,b) A, where a < b. Choose z A such that |z| > |u/(b-a)|. Then |u/z| < b-a. Let x,y (a,b), where x-y = u/z. Then f(x,y,z) = u. The final claim is by Theorem 1.4.4. QED Note that the counterexamples above are in 3 dimensions. THEOREM 1.4.9. The Thin Set Theorem holds on (SAFCN(2,),UNOPSUB()). The Thin Set Theorem fails on (CSAFCN(2,),DEOPSUB()) and (RAFCN(2,),UNOPSUB()). Proof: Let E 2be semialgebraic. We say that A 2is small if and only if (x >> 0)(y >> x)((x,y) A). We claim that for any disjoint semialgebraic A,B 2, A is small or B is small.
307 To see this, let A,B 2be pairwise disjoint semiaglebraic sets, where A,B are not small. Then ¬(x >> 0)(y >> x)((x,y) A). ¬(x >> 0)(y >> x)((x,y) B). By the o-minimality of the field of real numbers, (x >> 0)¬(y >> x)((x,y) A). (x >> 0)¬(y >> x)((x,y) B). Again by o-minimality, (x >> 0)(y >> x)((x,y) A). (x >> 0)(y >> x)((x,y) B). This violates A B = . For A 2, let rev(A) = {(x,y): (y,x) A}. We now claim that for any three pairwise disjoint semialgebraic A,B,C 2, A and rev(A) is small; or B and rev(B) is small; or C and rev(C) is small. To see this, by the previous claim, among every pair of sets drawn from A,B,C, at least one is small. Hence at least two among A,B,C are small. By symmetry, assume A,B are small. Note that rev(A),rev(B) are disjoint and semialgebraic. Hence rev(A) is small or rev(B) is small. This establishes the claim. For the first claim of the Theorem, let f:2be semialgebraic. Consider f-1(0),f-1(1),... . These sets are semialgebraic and pairwise disjoint. By the above, we see that there exist infinitely many i such that f-1(i) and rev(f-1(i)) are small. Also note that for all i 0, f-1(i) contains all (x,x), x sufficiently large, or excludes all (x,x), x sufficiently large. Because of mutual disjointness, all but at most one f-1(i) has the property that it excludes all (x,x), x sufficiently large.

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