Re: Proper class.Proper class ?


Jonathan Hoyle wrote:
Sorry, there was a mistake, I meant B as defined as the set of all sets
that have cardinality k, is not a set, since this would mean that B is
the set of all sets that are bijective to S when S has cardinality k.

Yes, you are correct. Moe Blee's proof is a good one. It is simple
enough to show just using k = 1. Each set containing a cardinal number
is certainly a set. But the collection of all these sets is not a set,
as its union would be the class of cardinal numbers.

It has been proved that such a set is not in ZFC. In short UB ( the set
union of B) would be the set of all sets, since you can replace any
member of a member of B by any set v, without this member of B being
excluded from B. And since the set of all sets is not a set in ZFC,
then B is not a set. But it can be a proper class, this will safe it
from being inconsistent.

While A and B are each proper classes, it seems to me that A.B should
still be a set, certainly for finite S anyway. I'm not sure why the
class of a transitive sets with cardinality omega should not be a set
either.

Jonathan Hoyle
Eastman Kodak

It is not about whether you are sure or not, it is about proof. what is
the proof that for every S:S=/={}, y={x|x is a transitive set /\ x is
bijectable to S ) -> y is a set.

This is important.

Zuhair

.

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