Date: Jan 24, 2013 7:50 PM
Author: Jesse F. Hughes
Subject: Re: ZFC and God

WM <mueckenh@rz.fh-augsburg.de> writes:

> On 24 Jan., 14:46, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>

>> It would be swell if you could write it in more or less set-theoretic
>> terms, since, after all, you are allegedly providing a proof in ZF.
>>
>> Thanks much.

>
> A last approach to support your understanding:
> Define the set of all terminating decimals 0 =< x =< 1 in ZF.
> Do all that you want to do (with respect to diagonalization).
> Stop as soon as you encounter a non-terminating decimal


I've no idea what you're talking about.

Let t:N -> [0, 1) be the usual list of non-terminating decimals.

I define

d(j) = 7 if t_j(j) != 7
= 6 else

There. Done. Just as soon as I specified what d is, it is obviously
non-terminating. There's no process here to stop. The variable d was
undefined and then it was defined and once defined, it is clearly a
non-terminating decimal. (I even hate to use temporal talk like "once
defined", but hopefully my meaning is clear enough.)

I patiently await a proof in ZF that d is not non-terminating. Fun
fact: the axioms of ZF don't talk about time or stopping or anything
like that. So, you'll have to rework this "argument" so that it is a
valid argument in ZF.

Let's start with the theorem you want to prove, shall we? What
theorem is it precisely? Is it this:

Let d be defined as above. Then d is a terminating decimal.

Is that your claim? I.e.,

Let d be defined as above. Then there is an i in N such that for
all j > i, d(j) = 0.

Or do you plan on proving something else?

Thanks much. Eagerly awaiting further enlightenment, etc.

--
"Being in the ring of algebraic integers is just kind of being in a
weird place, but it's no different than if you are in an Elk's Lodge
with weird made up rules versus just being out in regular society."
-- James S. Harris, teacher