On Wed, 12 Jan 2011 20:11:50 -0500 (EST), ksoileau <firstname.lastname@example.org> wrote:
> >We say that a set X is EVEN if there exist disjoint A and B such that >A union B equals X and A and B have the same cardinality, otherwise >we >say that X is ODD. Are all infinite sets even? prove or disprove, >assuming the Axiom of Choice. The countably infinite case is trivial, >so assume X is uncountable.
Use Zorn's Lemma: Consider the set of all pairs (A,f) where A is a subset of X and f is a 1-1 map from A into X\A. Define the order, <, by (A,f) < (B,g) iff A is a subset of B and f is the restriction of g to A. This is a partial order and I have convinced myself that it satisfies the hypotheses of Zorn's Lemma (for any chain of such pairs consider the union of the sets in each these pairs), so there is a maximal element: (M,h).
The range of h is contained in X\M, and if there are 2 elements x,y of X\M not in h(M), then add x to M to get M' and extend h by mapping x to y to get h'. Then (M,h) < (M',h'), contradicting the maximality of M.
Therefore X\M differs from h(M) by at most one element, and so have the same cardinality. Since M and h(M) also have the same cardinality, it follows that M and X\M have the same cardinality.
Dan To reply by email, change LookInSig to luecking