Re: Partitions of Reals

From: Robert Israel (israel_at_math.ubc.ca)
Date: 02/02/05 

Date: 1 Feb 2005 23:52:38 -0800

ululuca@tiscali.it wrote:
> Anyone knows anything about this problem, please

> There exists a non-trivial partition of the positive reals (so
without
> 0) into sets A and B (A, B disjoint, A U B = R+) such that A and B
are
> both closed under addition and multiplication.

It may be worth mentioning that, even if you remove the "closed under
multiplication" requirement, any such A and B would have to be Lebesgue
nonmeasurable. In fact, neither A nor B could contain a measurable set

of nonzero measure. This is because for any such set C, C+C contains
an interval (while on the other hand it's easy to show that A and B
must be dense).

A consequence is that any example will have to involve some form of
the Axiom of Choice: it's consistent with Zermelo-Frankel set theory
(without Choice) that every set is Lebesgue measurable.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada V6T 1Z2