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Jun 22, 2022 at 8:13 history edited CommunityBot
replaced http://math.uga.edu/~pete with http://alpha.math.uga.edu/~pete
Jan 25, 2010 at 13:24 history edited Pete L. Clark CC BY-SA 2.5
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Jan 25, 2010 at 13:23 comment added Joel David Hamkins About your AC remark and cardinal arithmetic: If ZF, you can prove that kappa^2=kappa for any infinite (well-ordered) cardinality, so yes, kappa\times\aleph_0 = kappa for such kappa. If you mean "cardinal" in the not-AC sense, however, then if kappa is infinite but Dedekind finite, it cannot have \kappa\times\aleph_0 = \kappa, since this implies a countably infinite subset. But to get the proper class of models, as you suggest, one needs to use only the well-ordered cardinals.
Jan 25, 2010 at 13:14 history answered Pete L. Clark CC BY-SA 2.5