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*To*: "FOM" <fom@math.psu.edu>*Subject*: FOM: Re: Challenge on RVM*From*: "Karlis Podnieks" <podnieks@cclu.lv>*Date*: Wed, 14 Jul 1999 16:49:48 +0300*Sender*: owner-fom@math.psu.edu

-----Original Message----- From: Joe Shipman <shipman@savera.com> To: fom@math.psu.edu <fom@math.psu.edu> Date: 1999 July 14 12:11 Subject: FOM: Challenge on RVM ... Why do set theorists regard CH as so profoundly undecided when it is settled by a reasonable axiom? ... KP> Perhaps, there are alternative ways of "deciding" CH on an intuitive basis. For example, Freiling's Axiom of Symmetry could serve as an "obvious" argument in favour of "non CH". See http://www.jmas.co.jp/FAQs/sci-math-faq/continuum. A fragment follows: ""The following is from Bill Allen on Freiling's Axiom of Symmetry. This is a good one to run your intuitions by. Let A be the set of functions mapping Real Numbers into countable sets of Real Numbers. Given a function f in A, and some arbitrary real numbers x and y, we see that x is in f(y) with probability 0, i.e. x is not in f(y) with probability 1. Similarly, y is not in f(x) with probability 1. Let AX be the axiom which states ``for every f in A, there exist x and y such that x is not in f(y) and y is not in f(x)" The intuitive justification for AX is that we can find the x and y by choosing them at random. In ZFC, AX = not CH. proof: If CH holds, then well-order R as r_0, r_1, .... , r_x, ... with x < aleph_1. Define f(r_x) as r_y : y >= x } . Then f is a function which witnesses the falsity of AX. If CH fails, then let f be some member of A. Let Y be a subset of R of cardinality aleph_1. Then Y is a proper subset. Let X be the union of all the sets f(y) with y in Y, together with Y. Then, as X is an aleph_1 union of countable sets, together with a single aleph_1 size set Y, the cardinality of X is also aleph_1, so X is not all of R. Let a be in R X, so that a is not in f(y) for any y in Y. Since f(a) is countable, there has to be some b in Y such that b is not in f(a). Thus we have shown that there must exist a and b such that a is not in f(b) and b is not in f(a). So AX holds. Freiling's proof, does not invoke large cardinals or intense infinitary combinatorics to make the point that CH implies counter-intuitive propositions. Freiling has also pointed out that the natural extension of AX is AXL (notation mine), where AXL is AX with the notion of countable replaced by Lebesgue Measure zero. Freiling has established some interesting Fubini-type theorems using AXL."" Karlis Podnieks http://www.ltn.lv/~podnieks/ University of Latvia, Institute of Mathematics and Computer Science

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