On Sep 7, 12:14 am, logic...@comcast.net wrote: > One can show in ZF-I that every set that is provably an ordinal > must have a largest element.
That's a slight mis-location of "in". Showing that *IN* ZF-I requires a lot of fancy encoding of proof-predicates. What you mean is that you can show it ABOUT ZF-I, using ZF+I+C.
> I was told that showing every ordinal has a largest element
Well, that is COMPLETELY irrelevant -- you CAN'T -- not in Z, or ZFC, or Z-I, or anything like that, prove that every ordinal has a largest element. RATHER, you can prove that every set THAT YOU CAN *PROVE* IS AN ORDINAL has a largest element. Something might BE an ordinal withOUT your being able to PROVE it was one!
> is not enough to prove that some set, > a set that can't be proven to be an ordinal in ZF-I, > might be an ordinal without a largest element.
Exactly. Did you not understand what you were told? Not reading ahead, I just made the mistake of telling it to you again.
> I find it incredulous ZF without the
No, you ARE increduLOUS. You find IT incredIBLE.
> Axiom of Infinity is incapable of proving every natural number is finite.
It's not any better WITH the axiom of infinity. JUST WHAT IS a natural number anyway? Just what does "finite" mean anyway? The main reason why this is normally easy to prove is that in the Z context, "natural number" is simply DEFINED as "finite ordinal". If you have any other definition for either term then nobody, not ZFC OR ANYTHING ELSE EITHER, will EVER be able to prove this.
> If there is such a proof in ZF-I, I can easily use it to prove > omega doesn't exist.
As you've already had explained to you, if there is no proof from ZF, then there certainly can't be a proof from any SMALLER axiom set (like Z-I). Since there is no proof in first-order logic PERIOD, though, you have nothing to worry about.