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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 12:31 PM
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On 03.12.2012 17:42, Carsten Schultz wrote:
> 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.
>

Thank you, Carsten. I agree that Z is not a set. If there were a bound,
say, l, on the size of ordinals in Z, then Z would be a subset of l+1,
thus, a set. Thus for each cardinal l there is an ordinal >= l in Z.
Probably by induction you could show that Z has a subset of cardinality
k for any k, right?

> 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.

Do you mean the cardinal sum, i.e.,
card X + k
=
card( (X x {0}) union (k x {1})), where, say, 0={}, 1={0}?
I used a Cartesian product to make the union disjoint.

Here, one has to prove that there is a one-to-one map
k -> k'\X

The existence of such a map sounds plausible, though I don't know
whether one would use the regularity axiom during the proof.

Thank you!

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

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