# [FOM] Why would one prefer ZFC to ZC?

Monroe Eskew meskew at math.uci.edu
Sat Jan 30 21:17:08 EST 2010

On Fri, Jan 29, 2010 at 4:50 PM, Jeremy Bem <jeremy1 at gmail.com> wrote:
>

Now V{\omega+\omega} satisfies the Union axiom, and a fortiori an
axiom of countable unions.  So I think you really mean countable
replacement.  (If you have a formula that defines for each n in omega
a unique object, then there is a set containing all those objects,
e.g.  n |--> V_omega+n.)  But this raises the question: Why not full
replacement?  What's so special about omega in this respect?

>
> I am speaking about V itself.  I am showing that the left hand side of
> the analogy ZC : V_{omega+omega} :: ZFC : V is fundamentally better
> behaved than the right.

I don't disagree that a set is better behaved than a class, but I
don't think you're making a fair analogy.  You are saying that ZC is
good because only a slight extension of it implies the existence of a
natural model of it.  But you seem to suggest that the only similar
thing available for ZFC is the class of all sets.  I say if we compare
apples to apples, then we can likewise get a slight extension of ZFC
that implies the existence of a natural set model of ZFC.

Furthermore since the slight extension that of ZC you have in mind is
merely a fragment of the replacement schema, it seems ironic (if not
inconsistent) to use this to argue against replacement.

-Monroe