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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:42 AM


Am 03.12.12 17:21, schrieb jaakov: > On 03.12.2012 16:07, Carsten Schultz wrote: >> Am 03.12.12 15:21, schrieb jaakov: >>> Given a set X and a cardinal k, is there a set Y such that card(Y)=k and >>> X is disjoint from Y? >>> >>> Is there a proof of this fact that works without the axiom of regularity >>> (= axiom of foundation) and does not assume purity of sets? >>> >> >> I am not sure if I understand all of this. >> >> Consider the class of all ordinals not in X. This is a wellordered >> proper class and hence has an initial segment which is order isomorphic >> to k. Take that initial segment. >> >> hth >> >> Carsten > > Hi Carsten: > > Thank you. To finish this proof, we have to prove that your class, let > us call it Z, contains an initial segment of cardinality k. However, > informally speaking, X may hypothetically contain too many ordinals, > still being a set, such that too few ordinals remain in Z. Could you > please expand on why Z contains an initial segment of cardinality k? >
Z has to be a proper class, otherwise the class of all ordinals would have to be the union of two sets and would therefore be a set.
I you do not like this start with a cardinal number k' greater then card X + k. Then k'\X is a wellordered set of cardinality greater then k and has an initial segment isomorphic to k.



