Re: Largest Set in ZFC?



On Mar 9, 1:55 am, "Peter Webb" <webbfam...@xxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:

Your [reaste's] argument suggests omega is a model of ZF.

No, that's wrong. In ZF we prove that there exists a unique least
inductive set, which we name 'w'. That doesn't entail that in ZF we
prove w is the universe of a model of ZF. Rather, in ZF, we prove that
IF ZF has a model then w is the universe of a model of ZF.

Yes but No. Omega is not X, because it doesn't include w+1, and hence
doesn't satify the requirement that "and whenever y is in X, so is S(y)".

You're TERRIBLY confused about this. To be successor inductive it is
not required that the set have w as a member so it is not required
that the set have w+ as a member. What is required is that IF w is a
member then w+ is a member; but it is not required that w is a member.

As
I indicated, nobody has ever come up with an actual set X that meets the
requirements of the definition.

What are you talking about?! We DO prove the existence of a unique
defined set that is successor inductive.

If they did, they would prove the
consistency of ZF by providing a "model" of ZF(C), viz the set X itself.

No, that is wrong, as I explained earlier in this post.

MoeBlee
.



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