Re: Cardinals as Equivalence class?


Rupert wrote:
zuhair wrote:
Hi All,

In a previous Post, Moe Blee showed a proof that Frege's definition of
Cardinality is not in ZFC.

form what I remember his proof was like that.

Let c a set of all injectable sets to set x. were x is not empty

now let z e x.

then let Any v: v !e x replace z in x, then the resulting set is in c

y= ( x \ {z} ) U {v} -> y e c.

from this Moe Blee deduces that Uc is the set of all sets.

Therefore c is not a set is ZFC.
------------------------------------------------------

My reply: I don't know till now how Uc is the set of all sets?


We have v is in Uc, and v was arbitrary, except that it's not in x.
It's easy to prove that everything in x is also in Uc. So every set is
a member of Uc.

What is Uc?

now if c for example is c= { a,b,c}

is Uc= a U b U c.

If this is Uc then Moe Blee argument is clear.

Zuhair

there is an intermediate step that I don't know in this proof.

Second: I personally have something which supports Moe Blee idea though
from another angle.

Since v is any set that is not in x, and since c is a set that is not
in x ( axiom of regularity ), then c itself can be v.

ie c itself can replace z in x and thus can be a member of x, then c is
a member of a member of c. which clearly viloates the axiom of
regularity.


Yes, that's another way.

ie instead of writting v: v !e x , since x in c then c !e x then , we
can take c itself.

y= ( x \ {z} ) U {c} -> y e c which violates the axiom of regularity,
since c is a member of y.

so c is not a set is ZFC, But this can be corrected is we say that c is
a PROPER CLASS

Yes. That's right. Frege's definition works in NBG, which has proper
classes.

since proper classes cannot be members of other sets, ie do not have
supersets.

I don't know whether if we assume that c is a proper class will also
work in the same way on Moe Blee's proof, since Uc ( I think this is
the set union of c, ie the superset of c ) do not exist for
c if c is a proper class.


Yes, it does. As before, it turns out that Uc is the universe.

so an Equivalence class of sets , under equivalence relation
"bijection" is in reality a proper class and it would be better writtin
as.

Cardinal <-> Equivalence proper class of sets , under equivalence
relation "bijection".

Similar thing applies to the definition of ordinals as equivalence
classes, it should be like below:

Ordinal <-> Equivalence proper class of sets , under equivalence
relation "order isomorphism".


Yes, these definitions work fine in NBG.

The only trouble is we would like to have sets of ordinals and
cardinals, and proper classes can't be members of sets.

There are ways around this. Von Neumann showed how to define the
ordinals in ZF. You define them as transitive connected sets. The idea
is that each ordinal is the set of all preceding ordinals. Then for
cardinals you can use Scott's trick. There is a way to define the rank
of a set, and the rank of a set is always an ordinal. In each case the
rank of a set is the first ordinal after all the ranks of the members
of the set. Then you can define the cardinality of a set to be the set
of all sets equipollent to that set of smallest possible rank. You can
prove that's a set. Or, if you're working in ZFC rather than ZF, just
define a cardinal to be an ordinal not equipollent to any smaller
ordinal.

Zuhair

.

Relevant Pages

  • Re: Cardinals as Equivalence class?
    ... ie c itself can replace z in x and thus can be a member of x, ... I don't know whether if we assume that c is a proper class will also ... Similar thing applies to the definition of ordinals as equivalence ... Then you can define the cardinality of a set to be the set ...
    (sci.math)
  • Re: Cardinals as Equivalence class?
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    (sci.math)
  • Re: J class theory with comprehension corrected.
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  • Re: Proper class.Proper class ?
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  • Re: Proper classes
    ... > Even in set theories with an anti-foundation axiom. ... I disagree that "by definition" cardinal numbers are ordinals. ... By definition, cardinal numbers are equivalence classes of sets, ... is not the cardinality of all cardinals, ...
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