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 youÂ 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