# FOM: Re: Challenge on RVM

Karlis Podnieks podnieks at cclu.lv
Wed Jul 14 09:49:48 EDT 1999

```-----Original Message-----
From: Joe Shipman <shipman at savera.com>
To: fom at math.psu.edu <fom at 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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