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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 12:31 PM


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 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? >> > > 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 wellordered 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 onetoone 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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