From: uranus!ikastan@uunet.uu.net (Ilias Kastanas) Subject: Re: trick property Date: 9 Apr 2000 22:29:04 GMT Newsgroups: sci.math.research Summary: [missing] In article <8ca0gl$4nt$1@news.si.fct.unl.pt>, manuel wrote: @ Consider the monotonous infinite sequences u_(n) of distinct natural @numbers. @ @ Can you find a property P such that: @I) for all u_(n) there exist a subsequence that satisfy P. @ii) for all u_(n) there exist a subsequence that doesn´t satisfy P. @ @Example: if you define property P to be " u_(n +1) - u_(n) >=2" it is easy @to find a sequence a_n such that all her subsequences satisfy P, @contraditing ii). Fred Galvin has already answered this; let me streamline a bit. Consider a wellordering < of the u 's and define P(u) <=> "for some subsequence u' of u, u' < u". [If v and all its subsequences satisfied P, then v > v' > v'' > ...; if ~P, then v(0), , v(2), , v(4), ... > v(0), v(1), v(2), , v(4), ... > v(0), v(1), v(2), v(3), v(4), ... > ... Hence P is not Ramsey.] The Ramsey property does obtain if one adds various definability assumptions; maybe Fred was too modest to mention that a standard result in this vein is the Galvin-Prikry theorem (Borel sets are Ramsey). Ilias ============================================================================== From: Fred Galvin Subject: Re: trick property Date: Sun, 9 Apr 2000 22:22:06 -0500 Newsgroups: sci.math.research On 9 Apr 2000, Ilias Kastanas wrote: > Fred Galvin has already answered this; let me streamline a > bit. Consider a wellordering < of the u 's and define P(u) <=> > "for some subsequence u' of u, u' < u". You mean u' < u. No doubt that is the most elegant example of a non-Ramsey set. Here is another one: By Problem 5348 (Amer. Math. Monthly 72 (1965), 1136; Amer. Math. Monthly 74 (1967), 730-731), there is a function f:N^N --> N such that, for each x in N^N, the equation x_n = f(x_{n+1}, x_{n+2}, x_{n+3}, ...) holds for all but finitely many n. Given such a function f, let P be the property u_1 = f(u_2, u_3, u_4, ...). :-)