Date: Jan 23, 2013 7:33 AM
Subject: Re: ZFC and God

On 23 Jan., 12:47, "Jesse F. Hughes" <> wrote:
> WM <> writes:
> > On 22 Jan., 21:18, "Jesse F. Hughes" <> wrote:
> >> WM <> writes:
> >> > That is potential infinity. That proof is not necessary, because the
> >> > set is obviously potentially infinite. No, you shoudl give a proof,
> >> > that there is a larger k than all finite k.

> >> Er, no.  When I say that the union is infinite, I do not mean that it
> >> contains an infinite number.

> > But you mean that the tree contains infinite paths. And just that is
> > impossible without ...

> > In order to shorten this discussion please have a look at
> No.  It's irrelevant.

You are in error.
> We're talking about whether you can prove that
>  U_n=1^oo {1,...,n}
> is finite.  I'm not switching topics to paths in trees (despite the
> fact that the ignorance of your question is obvious).

The union of FISs is finite. Yes that is my claim. But I cannot give
an upper limit, because the finite numbers have no upper limit. This
is called potentially infinite.
> > There it has meanwhile turned out ... But see it with your own eyes
> > what you would not believe if I told you.

> > The index omega is in reach, it seems.
> You're playing your usual little game of trying to change the topic.
> I won't have it.
> I take it that this new tack is so that you don't have to concede the
> point: there is no mathematical publication which claims that the
> above union contains elements larger than any natural, nor any
> publication which claims that this is what it means to be infinite.

I know. But if you hace read the discussion, you have seen that two
matheologians claim just this. Why do they? Because they cannot answer
the question: What paths are (as subsets of the set of nodes) in a
Binary Tree that is the union of all its levels? Are there only the
finite paths? Or are there also the infinite paths?
Try to answer it, and you will see that you need the omegath level or
must confess that it is impossible to distinguish both cases. Hence,
Cantor's argument applies simultaneously to both or to none.

Regards, WM