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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

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 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 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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Date Subject Author
12/3/12 jaakov
12/3/12 forbisgaryg@gmail.com
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 jaakov
12/3/12 Carsten Schultz
12/3/12 jaakov
12/3/12 Aatu Koskensilta
12/3/12 Butch Malahide
12/3/12 jaakov
12/3/12 Butch Malahide
12/4/12 jaakov
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 forbisgaryg@gmail.com
12/4/12 William Elliot
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 William Elliot
12/4/12 jaakov
12/4/12 Shmuel (Seymour J.) Metz
12/4/12 Spammer
12/4/12 jaakov