[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:
>
> How about countable union?
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
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