FOM: Re: GCH.

Jan Mycielski jmyciel at euclid.Colorado.EDU
Thu Jun 22 13:54:03 EDT 2000


			AXIOMS WHICH IMPLY GCH

	The axioms which I have proposed on f.o.m. in Dec. 3, 1999 and 
May 8, 2000, were too strong. H. Woodin proved that their conjunction is 
inconsistent in ZFC.
	However, it is easy to correct them. The new axioms are simpler,
hopefully they are consistent in ZFC, and they imply the full GCH.
	Let P(X) denote the powerset of X.
	The new axioms are the following:

	A. For all cardinals alpha all OD subsets of P(alpha) which are of
cardinality larger than alpha have the cardinality 2^alpha.

	B. For all cardinals alpha 
		 |P(alpha) intersect OD| > alpha.

	The axiom A is suggested by the following well known theorems. 
All uncountable analytic sets of real numbers have perfect subsets
(Luzin). More generally, assuming appropriate large cardinal axioms, all
uncountable sets of reals which are in the class L[R] have perfect subsets
(Martin, Steel and Woodin).
	The axiom B is implied by the following more precise and (I
believe) more suggestive axiom:

	C. For all cardinals alpha
	|{Th((V_beta, epsilon, xi)_(xi < alpha)) : beta in Ord}| > alpha.

	It is easy to show that V = OD implies C. So we know that ZFC + C
is consistent (if ZFC is consistent). We hope that 
	ZFC + A + C + (any natural enough large cardinal axiom) 
is also consistent. We have also several other propositions related to A
which may have that property.

	THEOREM. ZFC + A + B implies GCH.

	Proof. By B we have an initial segment S of the canonical well
ordering of the set (P(alpha) intersect OD) such that |S| = alpha+. Of
course S is in OD. Hence, by A, |S| = 2^alpha. Thus alpha+ = 2^alpha. QED

						Jan Mycielski





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