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Topic: Finitely definable reals.
Replies: 52   Last Post: Jan 18, 2013 2:37 PM

 Messages: [ Previous | Next ]
 Virgil Posts: 8,833 Registered: 1/6/11
Re: Finitely definable reals.
Posted: Jan 15, 2013 2:37 PM

In article
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 15 Jan., 17:05, "Ross A. Finlayson" <ross.finlay...@gmail.com>
> wrote:

> > On Jan 15, 7:53 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >

> > > On 15 Jan., 15:32, forbisga...@gmail.com wrote:
> >
> > > > Only the infinite expansion will
> > > > do where the termial repeating sequense is identified as above.

> >
> > > It will do? It will *never* do! Never ready, that is the meaning of
> > > unfinished work.

> >
> > > > You know this but are just playing games.  You also know that some
> > > > reals
> > > > are irrational and do not have a termial repeating sequence.  Just as
> > > > with the rationals only the infinite expansion will equal itself--all
> > > > elese
> > > > are approximations.  For most human purposes sufficiently close
> > > > approximations
> > > > are good enough

> >
> > > and other decimal approximations do not exist. You may say, they are
> > > never available. Two parallels never cross each other. That is
> > > tantamount to "cross each other in the infinite".

> >
> > > But we can state: If parallels cross each other and irrationals have
> > > complete decimal expansions then Cantor is right. Or better: Only in
> > > the reals where parallels cross each other and irrationals have
> > > complete decimal expansions Cantor is right.
> > >

> > No, there are systems, tools, and perspectives where it is right that
> > parallel lines meet,

>
> Then they are not parallel, i.e., they have not always the same
> distance.

In projective spaces, one does not always have distances between lines.
>
> > and don't, and that irrationals have infinite
> > expansions,

>
> They have infinite expansions, but not actually but potentially. There
> is every d_n you like, but there are not all d_n.

That is sheer nonsense.
>
> >
> > And, as simply noted, ZF's universe of all its sets, would be the
> > Russell set, and contain itself.

Except that "ZF's universe" is not a set in ZF.
>
> If set theory were right, then it should be able to tell the limit
> between existing sets and non-existing sets like the set of all sets.
> In fact there is no infinite set existing, because by the power set
> axiom, the set of all its power sets should exist.

Nonsense! WM seems to be profoundly ignorant of what ZF actually does
say.

The power set axiom for ZF only says that for every set there is another
set containing every subset of the original set, but nothing in ZF
requires any set of all sets. In fact, the axioms of ZF and ZFC are
specifically designed to prohibit any such thing as a set of all sets.
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