Date: Jan 27, 2013 6:00 PM
Author: Virgil
Subject: Re: ZFC and God

In article 
<d704e43f-79df-40dc-a122-79ec3840a70d@l9g2000yqp.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 27 Jan., 19:56, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 27 Jan., 19:21, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> > >> WM <mueck...@rz.fh-augsburg.de> writes:
> > >> > On 27 Jan., 18:44, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> >
> > >> >> Anyway, you haven't proved that there is a function
> >
> > >> >>   f:{1,...,k} -> {0,...,9}
> >
> > >> >> as required by *your* definition of terminating decimal, so you have
> > >> >> not shown that 0.777... is a terminating decimal.

> >
> > >> > You are wrong. Can't you understand? All natural numbers are finite.
> > >> > Why the heck should I define a single k?

> >
> > >> Because, of course, you accepted the following definition:
> >
> > >>   Let x be a real number in [0,1].  We say that x has a terminating
> > >>   decimal representation iff there is a natural number k and a
> > >>   function f:{1,...,k} -> {0,...,9} such that

> >
> > >>    x = sum_i=1^k f(i) * 10^-i.
> >
> > > I did not fix k but only assumed that it is a natural number.
> >
> > Pardon me?
> >
> > Look, you said that a number has terminating decimal representation
> > iff there is some k and f such that blah blah blah.  Thus, to show
> > that a number *does* have a terminating decimal representation is
> > equivalent to showing that there is, in fact, a k and f satisfying the
> > above.

>
> Of course. But why should we agree on a special k? Every natural
> number will do. So we only have to know that k is one of those natural
> numbers that belong to FISONs. As long as we work in FISONs we cannot
> have a non-terminating decimal.


But there is no need to restrict oneself to FISONs, at least not outside
of WMytheology.
> >
> > What could be more obvious than this?

>
> Remember the Binary Tree to answer your doubts. Do you believe that
> the set of all FISONs is definable in ZF? Is every FISON finite? Is
> there a fixed k limiting all FISONs? No.

> >
> > > You have no cause to claim the contrary, since there is no index
> > > (natural number) infinitely many counts away from the decimal point.

> >
> > Are you asking me to prove that 0.777... does not have a terminating
> > decimal representation?  I would be happy to do so, if needed.

>
> I am doubting that when working in FISONs you can find an index of a
> diagonal number that requires to leave the set of FISONs (with respect
> to the indexes of the digits).


But what is the point of working only in FISONs when we have |N
everywhere outside of WMytheology? ?
>
> > it follows *FROM THE AGREED DEFINITION* that 0.777... has no
> > terminating decimal representation.

>
> Show that in your 0.777..., and in particular in the anti-diagonal of
> a list of terminating decimals, there is an index k that does not
> belong to a FISON {1, 2, ..., n}.


Why? Every such index already belongs to the set of all indices which is
the union of the set of all FISONs. Note that in sane set theories, one
cannot have that all FISONs cannot be collected into a single union.

And in set theories like ZF and ZFC there must exist inductive sets
whose members are essentially FISONs.


> As long as you refuse there is no
> reason to believe you, because every index in a FISON belongs to a
> FISON (finite initial set or sequence of natural numbers). Why do you
> try to deny that?


No one denies it, but why do you deny that in reasonable set theories
like ZF and ZFC, and many others, there exist inductive sets as models
for |N?

>
> Regards, WM

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