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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:42 AM
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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 well-ordered
>> 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 well-ordered set of cardinality greater then
k and has an initial segment isomorphic to k.
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