Date: Jan 24, 2013 4:19 PM
Author: Virgil
Subject: Re: ZFC and God

In article 
WM <> wrote:

> On 24 Jan., 14:16, "Jesse F. Hughes" <> wrote:
> > WM <> writes:
> > > On 24 Jan., 13:36, "Jesse F. Hughes" <> wrote:
> > >> WM <> writes:
> >
> > >> Well, what you present below is *not* a proof of (*).
> >
> > > That is wrong. You have no reason to believe that your definition of
> > > proof is correct or the only one.

> >
> > This argument doesn't involve U_n {1,...,n} *at all*!
> >
> > But let's let it pass.

> If you can't understand that we have (d_1, ..., d_n) with just all the
> indices given by the FIS above, then we should stop for a while.

It is WM who needs to stop and get things straight in his own mind, if

> > > You will have have recognized that here the diagonal argument is
> > > applied. It is obvious that up to every line = column the list is a
> > > square.

> >
> > It is clear that, for all j, d(j) != t_j(j) and hence d != t_j. If
> > that's what you mean by the diagonal argument, great!
> >
> > Once again, however, you say something that has no clear meaning to
> > me. Can you clarify "It is obvious that up to every line = column the
> > list is a square?" I've no clue what it means.

> Then ponder a while about the following sequence
> d
> d1
> 2d
> d11
> 2d2
> 33d
> and so on. In every square there are as many d's as lines. The same
> could be shown for the columns.

But for your only finite sequences, there is no guarantee that line n is
at least n digits long, and, if not, no square.
> >
> >

> > > Neither the set of t_i does have a largest element. Nevertheless there
> > > is no t_i of actually infinite length.

> >
> > Correct, to both claims. So what? The digit d(j) is defined for
> > every j, and is neither 0 nor 9, so d is a real number which has no
> > terminating decimal representation.
> >
> > Do you agree with that (obvious) claim or not?

> No. Pleeze try to understand: Presently we work in the system of
> terminating decimals, by definition. There is no non-terminating
> decimal. If we were so unlucky to met any non-terminating decimal we
> would not know what to do. It would not even be defined. If the
> diagonal gets non-terminating, we have to stop and cut it before it
> becomes non-terminating.

You may have to, but no one else is forced to.

Note that while every terminating decimal can be embedded in a
non-terminating decimal of equal value by appending an infinite string
of zeroes, not every infinite decimal has an equal finite decimal.

So that anything with WM's finite strings finite that WM proposes is
merely a special case of all infinite strings. But the reverse is false.
> >
> > Sorry, I could've sworn you were talking about what ZF proves. Can
> > you state this in the language of ZF and prove it? Much thanks!

> Can't you imagine to define in ZF the set of all terminating decimals?

You are the one who has t do all that define=ing in order t prove your
case. Demanding that others to do your work for you means that your
arguments will continue to fail.

> Is it too hard?

It certainly seems to be too hard for WM to do his own proving.

> > > In particular, what would be changed in the length of d if we admitted
> > > also non-terminating t_i (of infinite length)?

> >
> > Nothing would change. So what?

> Look, presently we work in the system of terminating decimals - by
> definition.

Maybe you do, but the rest of us are perfectly able to see that it is
just a special case of infinite decimals.

> If nothing changes when we switch to the system of non-
> terminating decimals, do we switch then at all?

A lot of things change. For one thing there are provably more
nonterminating decimals than terminating decimals. At least everywhere
except in WMytheology

> How could we recognize
> that we have switched?

Among other things, we regain a lot of rationals that your terminating
decimals lose.