This question is not category theory, but I am sure there are people on the net who know much more set theory than me. If you assume the generalized continuum hypothesis, it is evident that any power cardinal $\kappa$ has the property that $\lambda < \kappa$ implies $2^\lambda <= \kappa$. Otherwise, that property is obvious to me only for inaccessible cardinals. My question is whether there are a large class of cardinals for which that is true even without the GCH? Michael
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.
Thanks! Two different people misconstrued my question as asking if m < k is 2^m <= 2^k. You actually answered an even more exigent question since I asked only that m < k imply 2^m <= k, but your answer is even better, since the strict inequality is preserved. You can't make beth in tex, can you? Not without Hebrew fonts. Anyway, thanks again. Michael +++++++++++++++++++++++++++++++++++++++++
okay, then how about taking a regular (beth)_l with l limit. say f(0) = (beth)_0 f(a+1) = (beth)_{f(a)} f(l) = lim_{a < l} (beth)_a and take a fixpoint of f. again, this depends on Replacement and Choice, but not GCH or the existence of inaccessibles.
This is in reply to Mike Barr's question about cardinals kappa such that, for all lambda<kappa, the power set of lambda has cardinality <= kappa. (I assume that I've correctly reconstituted the question after Bitnet revised the TeX codes in it.) There are lots (i.e., a closed unbounded class) of such cardinals kappa. To get one (as large as you like), start with any cardinal, iterate exponentiation (with base 2) to produce a countable sequence of larger cardinals, and take the supremum of this sequence. This clearly has the desired property (in fact with < in place of <=). Furthermore, this property is preserved by suprema. So you get lots of such cardinals, without needing any assumptions (like inaccessibles) that go beyond ZFC. Andreas Blass
This is in reply to Mike Barr's revised question, requiring regularity, and Dan Leivant's reply. Unfortunately, regular beths with limit subscript are inaccessible, which Mike wanted to avoid. Even more unfortunately, I believe it is known to be consistent (relative to very large cardinals) that there are no cardinals of the sort Mike wants. Specifically, it is consistent that GCH fails everywhere, a result of Foreman and Woodin. In such a universe, the requirement "for all lambda < kappa, 2^lambda<=kappa " implies that kappa is a limit cardinal; if you also require kappa to be regular, then it is weakly inaccessible. I believe that the violations of GCH in the Foreman-Woodin model are mild enough so that such kappas must actually be strongly inaccessible, but I'm not certain about this. Incidentally, the fact that the Foreman-Woodin model requires large cardinals for its construction does not mean that this model has any inaccessibles. It does mean that there are cardinals that are inaccessible (and measurable, and much more) in certain submodels. Andreas Blass
participants (3)
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Andreas R. Blass -
barr＠triples.Math.McGill.CA -
Daniel.Leivant＠B.GP.CS.CMU.EDU