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Topic:
Given a set , is there a disjoint set with an arbitrary cardinality?
Replies:
28
Last Post:
Dec 4, 2012 5:50 PM



jaakov
Posts:
11
Registered:
12/3/12


Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:21 AM


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?
It's not a homework.
Best regards
Jaakov.
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