Re: proper classes in ZF

zuhair says...

So what this quotation is telling us, is that all proper classes in
NBG can be reduced in ZFC (or any thoery that doesn't have proper
classes as objects in its universe of discourse) to open formulas.

However the source is mentioning that in MK there are proper classes
that cannot be reduced to open formulas in ZFC and the alike theories.

I want to know which proper classes in MK that cannot be reduced to
open formulas in ZFC, and why?

Let's formulate MK using two different sorts of variables.
Upper case variables such as X, Y, Z range over classes,
and lower case variables such as x,y,z range over sets.
Every set is a class, but not vice-versa.

If Phi is a formula that only involves set variables,
not class variables, then let's call it a "pure" formula.
The pure formulas of MK correspond to the formulas of ZFC.

For any formula Phi(x), MK proves

(EX) (Ax) x e X <-> Phi(x)

So MK proves that the class { x | Phi(x) } exists for any
formula Phi, whether pure or not. (Note that x is lower-case,
which means that this is the collection of all *sets*
satisfying Phi(x). MK does not allow talking about collections
of proper classes)

If Phi is *not* pure (that is, if it contains class variables),
then the corresponding class { x | Phi(x) } does not correspond
to an open formula of ZFC (because ZFC cannot talk about
proper classes).

Okay, there is actually a proof obligation here. If Phi
is not pure, how do you know that there isn't some other
formula Phi' that is pure such that Phi(x) <-> Phi'(x)?
In other words, how do you know that you get any new
classes by allowing impure formulas?

Well, here's a way, using Cantor's theorem. Every
pure formula Phi(x) can be coded up as a set using
Godel coding. So now let's define a function Comp from
sets to classes as follows:

If z is a code for a formula Phi(x), then
Comp(z) = { x | Phi(x) }.
If z is not a code for a formula, then
Comp(z) = the empty set.

Every proper class that is definable by an open formula
in ZFC is equal to Comp(x) for some set x.

Now, let R = { x | x is not an element of Comp(x) }.
Clearly, R cannot be equal to Comp(x) for any x. So
R is a proper class that is not definable by a formula
of ZFC. (It turns out that R *can* be defined in MK.)

--
Daryl McCullough
Ithaca, NY

.

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