Is ZF + CH + not-CC consistent?
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From: David Libert on 23 Nov 2009 01:08 David Hartley (me9(a)privacy.net) writes: > In message <he58t1$6oi$1(a)theodyn.ncf.ca>, David Libert > <ah170(a)FreeNet.Carleton.CA> writes >>The other possibility David raised above to phrase CH without assuming >>AC is every uncountable cardinal is greater than or equal to c. >> >> I don't know how this version works out. I don't know if Solovay's >>model satisfies this. And I don't even know if the Jech Sochar models >>above satisfy this. >> >> So that version is all unsettled for me for now. > > I threw that one in as a way of getting trichotomy for all cardinals not > larger than c, which Herman had mentioned (though perhaps he really > meant 'smaller than or equal to c'). I too don't know anything about how > it works out (not that I know much about the other versions either). > -- > David Hartley David's third version of CH was every uncountable cardinal is >= c . I realized later, this means aleph_1 >= c, so in particular the reals inject into aleph_1, so the reals can be well-ordered. So c is an aleph. But since alpeh_1 >= c, the only possible aleph for c is c = aleph_1 . So this 3rd reading of CH implies the first version c = aleph_1. But this thurd reading is stroner than that. Not merely is c = aleph_1, but every uncountable cardinal is >= alpeh_1. That last is an extra assumption. If ZF is consistent then there are choiceless ZF models with c = aleph_1 but some uncontable cardinals incomparable to c. So what can we say about this thrid version of CH? c = alpeh_1, so the reals are well-orderable, and in Solovay's model for everything Lebesgue measurable, the reals are not well-orderable. So this version of CH fails in Solovay's model, answering a question I posed quoted above. What about the Jech Sohcar model from my earlier article in this thread? That model took Cohen's excample of an omega sequence of 2 element sets with no choice function, copied it to FM atoms, then pushed those to high rank well founded sets by Jech Sochar. Consider the union of those omega many 2 element sets. If it were countable, then we could get a choice function on the 2 element sets by picking the least of each pair of elements in an omega ordering counting it. That contra the model. So that set is uncountable. So by the assumed property, if it were true in that model, alpeh_1 would inject into that set. If you collapse that injection into the union back to the omega set of pairs, ie send each ordinal alpha < aleph_1 to the least pair set according to the omega ordering on pairs sets containing as member the injection on alpha, you get a function from alpeh_1 into a coubtable set which is at most 2 to 1. This is a contradiction even over ZF. For example, the injection from aleph_1 to the union of pairs has well orderable range, well-ordering induced by map from alpeh_1, and is sujected by omega x omega, so it is countable, jence not injeced from alpeh_1. So we conclude the Jech Sochar model with countable choice failing on pair sets, as from my prevous article, also doesn't satisfy this 3rd CH version. A fairly easy generalization of Cohen's orginal ~AC models, is you can make AC hold for choices below a regular cardinal, and fail at the regular cardinal. Ie making a few choices is ok, more isn't. In fact you can make the innder Ciken model have all short seqwuences from the outer AC model. In these ways we can arrange aleph_1 many choices AC style ok, and AC fails for more choices. By the stronger property actually, inheriting all alpeh_1 sequences from an outer AC model, we can make aleph_1 inject into every uncountble cardinal. Ig our starting AC model also has CH, we can make the ~AC construction add no new reals. So we can get ~AC and c = alpeh_1 and aleph_1 <= every uncountable carindal. So the 3rd version of CH doesn't -> AC, by techniques close to Cohen's original. Ths still doesn't answer the 3rd version CH part of the original questuiobn here: CC. My Jech Sochar ~CC model above didn't work for this 3rd version of CH. What else can be said? Suppose we have an omega swquence of countable sets with no choice fuction. Could that coexist with this CH version? Above I argued that 2 element sets wouldn't work. The point about 2 element sets, is if you inject alpeh_1 into the union, since each part is only size 2, injecting alpeh_1 into the union must spread the injectiion across many parts, contadicting that there are on;y countably many parts. So what if instead you have an omega sequence of countable sets? Could aleph_1 inject onto those omega many pieces, each countable? One difficulty is what if aleph_1 has cofinality omega. This can't happen in AC models but can in ~AC ones. Then aleph_1 is a countable union of countable sets, so inject these in turn maybe onto each part. Actually that itself might require some choice to do, but at least it would show it is hard to rule out. But then I remembered: old theorem, even though aleph_1 might have confinality omega, ZF proves c never has confinatiy omega. And our assumption is c = aleph_1. So I think this allows a generalizatiion of my 2 element set argument above, so say if there is an omega sequence of countable sets then ~ (3rd version CH). And in fact we can similarly argue, if there is any omega sequence of sets with no choice function, then the subsequence of countable sets has choice fuctions restricted to it. So given any such example, you can throw away all the countable sets, and the reduced result still has not choice function. So, all the sets would be uncountable, and by the 3rd version CH if true, would inject aleph_1. So if there are ~CC models with 3rd vereion CH, they have ~CC examples with each set injecting aleph_1. Is that possible? I think so. My quoted article in this thread, do an FM model with aleph_1 support on atoms instead of the more common finite support. Use basic atom sets of ground model size alpeh_2, with aleph_1 support. Redo Cohen's ~CC model that way. The Jech Sochar all that. Do that over ground model AC + CH. Then the FM step with aleph_1 support stil injects aleph_1 to everty uncountable set. And it doesn't add any reals so c = alpeh_1 still. Then Jech Sochar as i review the proof I think preserves c = alpeh_1 and this aleph_1 <= all uncountable cardinals. So I think this finally gives a no answer to 3rd version CH -> CC. -- David Libert ah170(a)FreeNet.Carleton.CA |