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jaakov
Posts:
11
Registered:
12/3/12
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Re: Given a set , is there a disjoint set with an arbitrary cardinality?
Posted:
Dec 3, 2012 11:21 AM
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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?
It's not a homework.
Best regards
Jaakov.
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