Date: Jan 22, 2013 2:22 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: ZFC and God

On 22 Jan., 17:48, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 22 Jan., 15:49, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > FIS: Finite Initial Segment
> > FISON: Finite Initial Segment Of Naturals (or indices)

>
> >> Well, it's not a "union" in the usual sense, but let's let it pass.
>
> > It is a union in that sense that every FISON {1, 2, ..., n+1} added
> > contains all smaler FISONs {1, 2, ..., n}.

>
> We were speaking about the real number d, not the set N.


The set N is required to index the digits of the decimal fractions of
the real numbers.
>
> > This does never change. In particular the set is always finite. If
> > you add with always doubling frequence, you can add all FISONs in
> > finite time - given that an "all" is meaningfull here. But at the
> > end, you think, we cannot follow so quickly, and abracadabra we get
> > something larger than every FISON? Not in mathematics!

>
> Look, you need to offer an actual proof that
>
>   U_n=1^oo {1,...,n} is finite.
>
> I'll give you a hint, by showing you the proof that
>
>   U_n=1^oo {1,...,n} is infinite.


Unfortunately you don't seem to understand what infinite means, in
particular that it has two different meanings.
>
> Just so you know what a proof looks like.
>
> Here's the proof.  First, let me be clear what I mean by infinite.  I
> mean that there is no natural k such that |U_n=1^oo {1,...,n}| = k.


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.
>
> Hence, by definition, U_n=1^oo {1,...,n} is infinite.


Correct, there is no finite threshold.
>
> Now, I'm sure you're familiar with that argument.  I'd like to see
> your claim to the contrary proved with just as much detail and I'd
> like for you to be willing to answer any questions I raise about any
> steps I don't understand.
>
> Until then, I simply cannot see your reasoning that the union is
> finite.


You have not proved that the cardinal number of the union surpasses
all natural numbers. And even if you had done so, it would be useless,
because the union of FISONs cannot surpass itself.
>
>

> > From this "for every n" you claim "for all". And that is wrong for
> > infinite sets.

>
> I have no idea what difference between "for every" and "for all" you
> have in mind, but feel free to read the above as "for every n".


I know that. But as I already mentioned, actual infinity requres "for
all n".

Regards, WM