Is ZF + CH + not-CC consistent?
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From: Herman Jurjus on 17 Nov 2009 14:17 Subject says it (CC stands for 'countable choice'). More generally, the question is: 'how much' of AxC is implied by CH? (As AC is equivalent to the trichotomy of cardinals, and CH imply trichotomy for sets not larger than P(N), someone might expect that ZF+CH perhaps implies some weak forms of choice.) -- Cheers, Herman Jurjus
From: David Hartley on 17 Nov 2009 16:50 In message <hdusrq$udh$1(a)news.eternal-september.org>, Herman Jurjus <hjmotz(a)hetnet.nl> writes >Subject says it (CC stands for 'countable choice'). > >More generally, the question is: 'how much' of AxC is implied by CH? > >(As AC is equivalent to the trichotomy of cardinals, and CH imply >trichotomy for sets not larger than P(N), someone might expect that >ZF+CH perhaps implies some weak forms of choice.) First, how do you state CH in ZF without AC? Is it: there is no cardinal k such that aleph_0 < k < c, there is no uncountable cardinal less than c, c = aleph_1, or every uncountable cardinal is greater than or equal to c At first glance, none of the first three rules out infinite, Dedekind-finite sets, so you may not have trichotomy for sets not larger than P(N). (All but the first obviously give it for sets smaller than or comparable to P(N).) -- David Hartley
From: Tim Little on 18 Nov 2009 02:23 On 2009-11-17, Herman Jurjus <hjmotz(a)hetnet.nl> wrote: > CH imply trichotomy for sets not larger than P(N) I don't see how CH rules out any set being incomparable in cardinality with P(N). The hypothesis merely asserts that no set is both smaller than P(N) and larger than N. - Tim
From: Herman Jurjus on 18 Nov 2009 02:53 David Hartley wrote: > In message <hdusrq$udh$1(a)news.eternal-september.org>, Herman Jurjus > <hjmotz(a)hetnet.nl> writes >> Subject says it (CC stands for 'countable choice'). >> >> More generally, the question is: 'how much' of AxC is implied by CH? >> >> (As AC is equivalent to the trichotomy of cardinals, and CH imply >> trichotomy for sets not larger than P(N), someone might expect that >> ZF+CH perhaps implies some weak forms of choice.) > > First, how do you state CH in ZF without AC? Is it: > there is no cardinal k such that aleph_0 < k < c, > there is no uncountable cardinal less than c, > c = aleph_1, > or every uncountable cardinal is greater than or equal to c > > > At first glance, none of the first three rules out infinite, > Dedekind-finite sets, so you may not have trichotomy for sets not larger > than P(N). (All but the first obviously give it for sets smaller than or > comparable to P(N).) I was in a terrible hurry when I wrote that post - sorry for that. Forget about all the baloney. The question is: Is it known whether CH implies CC, DC, or any other weak forms of AC? -- Cheers, Herman Jurjus
From: Herman Rubin on 18 Nov 2009 12:25
In article <hdusrq$udh$1(a)news.eternal-september.org>, Herman Jurjus <hjmotz(a)hetnet.nl> wrote: >Subject says it (CC stands for 'countable choice'). >More generally, the question is: 'how much' of AxC is implied by CH? >(As AC is equivalent to the trichotomy of cardinals, and CH imply >trichotomy for sets not larger than P(N), someone might expect that >ZF+CH perhaps implies some weak forms of choice.) Essentially nothing. In all of the standard Fraenkel-Mostovski models, the classical sets are unaffected. So the GCH can hold for all ordinals, but not necessarily for other objects, and there can even be infinite sets which are not comaparable even with N. I am not sure how much of this goes over to Cohen models; the Fraenkel-Mostovski models are for ZFU, where elements are allowed which are not sets, but at least some of this works. -- This address is for information only. I do not claim that these views are those of the Statistics Department or of Purdue University. Herman Rubin, Department of Statistics, Purdue University hrubin(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558 |