15 Apr
1991
15 Apr
'91
3:48 p.m.
Let (beth)_0 = (aleph)_0, (beth)_(a+1) = 2^(beth)_a, (beth)_l = lim { (beth)_a | a < l } for l a limit ordinal Then for any cardinal k = (beth)_l (l limit) m < k ==> 2^m < k Note that by Replacement there are as many such cardinals k as there are (limit) ordinals, whereas ZFC does not prove the existence of even one inaccessible cardinal.