Re: Largest Set in ZFC?


"MoeBlee" <jazzmobe@xxxxxxxxxxx> wrote in message news:874b4355-ca2f-4e3b-acfc-e4ceac9397f1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
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.

*******
Well, w is not a model of ZF. There are sets constructible from w which are not within w. Like w+1, for example. w cannot be a model of ZF because you can construct sets in ZF that are not a subset of w.

w is a model of ZF-AoI, excepting for the slight technical problem you can't prove the existence of w without AoI.

In a similar but not identical way, the set of all ordinals would be a model of ZF, but unfortunately we know this doesn't exist. You can have a "proper class" of all ordinals, but unfortunately the definition calls for a "set" X (and quite reasonably; classes are not even defined in 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.

*********
Unless we have a terminology problem, what you are saying is wrong.

I said that there is no known set of sets which which meets the definition of X. If by w and omega we are referring to the smallest infinite ordinal, then clearly w cannot be X, because w doesn't include w U {w) which is one of the explicit requirements of X and indeed of the AoI.

Can you name a single set X that meets the requirements of the definition of X? I can't, and AFAIK nobody has any idea of such a set. That is why the definition of X is left vague - If we could produce an actual set X which satisfied the definition of X and all the other axioms of ZF(C), we would have a model of ZF(C), which by Godel's completeness thereom would prove the consistency of ZF(C) from within ZF(C). This is not impossible, but after a century of looking, nobody has found one, and it looks pretty unlikely such a set X will ever be found.


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.

**************
Show me a proof that there is a set X with the properties described in the OP's textbook. Even better, as the OP was complaining that the definition of X is vague, how about producing a single example of such a set? w doesn't work, because it doesn't include S(w), which is an explicit part of the definition of X.


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.

***************
I can't see this. Any set X construcible within ZF that satisfied all of the axioms of ZF (including AoI) would be a model of ZF. Agreed? A model of ZF would prove the consistency of ZF (by Godel's completeness thereom). Agreed? The consistency of ZF has never been proved using ZF alone. Agreed? Therefore there is no known set X. Agreed?

If you think you have a set X, and can prove it meets all the axioms of ZF, please post it. If you cannot name even a single such set X (as I suspect) the reasons that the definition of X is pretty vague should be obvious - nobody can describe even a single, concrete set X which meets the criteria of ZF.



MoeBlee

.

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