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Topic: Distinguishability argument x Cantor's arguments?
Replies: 15   Last Post: Jan 9, 2013 4:32 PM

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Virgil

Posts: 7,005
Registered: 1/6/11
Re: Distinguishability argument x Cantor's arguments?
Posted: Jan 8, 2013 6:07 PM
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In article
<c0c3f405-f742-430f-b93e-ddbabc667783@r13g2000vbd.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

> On 8 Jan., 17:27, George Greene <gree...@email.unc.edu> wrote:
> > On Jan 3, 9:19 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >

> > > because
> > > no infinite diagonal of a Cantor list can be defined

> >
> > THERE IS *NO*SUCH*THING*AS* "a Cantor list"
> > IN THIS CONTEXT!  Cantor is REFUTING the list!
> > Cantor is REBUTTING the list!  The list comes FROM YOU!

>
> In this context, a Cantor-list is a sequence of real numbers of the
> unit interval, in general without repetitions. But that would not
> disturb the argument.


In order to claim countability for a set , you must also claim the
existence of a list of all it members, or some other form of surjection
from N to that set.

If you object to lists, what other proof of surjection from N to the set
of all reals do you wish to propose?
>
> > IT'S YOUR list!  YOU allege that it is well-defined and that
> > it contains all the reals!

>
> No, I don't, in particular because there is no such thing as "all the
> reals". Why should I claim such a nonsense?


Do you mean that there is no way to tell if something is or is not a
real?
>
> > But you also allege that there are only countably many reals and
> > that they are all definable!

>
> I do not allege but can prove that not more than countably many
> objects can exist a individuals in mathematics and elesewhere.


You just alleged it again! But you did not prove it. And there is no
rule that every member of a set must be defined, only that when
presented with a possible member, one can tell whether it is a member.
>
> > In that case, a definable list of all the definable reals IS YOURS
> > *AND*NOT*
> > Cantor's!

>
> It seems to be yours. It is neither mine nor Cantor's.


It is certainly yours if you claim that there re no more reals than
naturals.
>
> >  If THAT list exists, then THE INFINITE DIAGONAL OF IT
> > *CAN*BE*AND*IS*
> > *TRIVIALLY* defined!  It is just "the real whose nth place is the nth
> > place of the nth-defined real ON YOUR, NOT CANTOR'S, list."
> > It's YOUR list and THAT'S *YOUR* definition of its diagonal!

>
> But I can define a list of all finite strings of digits, that is a
> subset of all rational numbers of the unit interval, where most appear
> more than once. Here it is:
> 0.0
> 0.1
> 0.00
> 0.01
> 0.10
> 0.11
> 0.000
> ...


You ability to define such a list does not prove that you cannot define
other lists.
>
> > And if that diagonal can be defined then THE ANTI-diagonal CAN ALSO be
> > defined!
> > For ANY DEFINED bit-string, its complement IS WELL-defined!
> > The nth digit of the complement is 1 - <the nth digit of the original
> > string>.
> > If the base is 10 then the nth digit of the anti-diagonal is just 9 -
> > the nth digit of the diagonal!
> > THESE ARE TRIVIAL, SIMPLE, STRAIGHTFORWARD definitions!
> > ANYthing defined this way CAN BE *AND*IS*WELL*- defined!

>
> For instance this list is so well defined, that it cannot be objected
> that every finite string of the anti-diagonal is an entry of that
> list. By the way this could also be constructed in decimal. But you
> may be intelligent enough to understand that?


Since it is so obviously irrelevant to the issue of whether the reals
are countable, even your low mentality may be able to grasp its
irrelevance.
> >
> > However, if the list of all definable reals were ITSELF definable,
> > then its anti-diagonal
> > would be definable

>
> Let that strawman rest where it is burried. I am not interested in the
> question whether suchga list was definable.


But that IS the very point to which you are objecting.

By definition, a set is countable ONLY IF there is a surjection from N
to that set. Such a surjection is called a list for the obvious reason
that its members must be ordered, though possibly with repetitions,
like the naturals in any such a surjection.
>
> > and WE WOULD HAVE A CONTRADICTION that the this
> > definition
> > of "the anti-diagonal-of-the-list-of-all-definable-reals" both WAS
> > definable (since the
> > above is a definition) AND WAS NOT definable (since it differs from
> > every row of the
> > list of definable reals).

>
> This contradiction happens already in the above list of all finite
> strings which is certainly definable. The anti-diagonal never deviates
> from every entry (at a finite place - but others do not exist).


That is because the set of all finite strings from any finite alphabet
IS countable ( it can be listed in a sort of alphabetical order),
whereas the set of all infinite strings of an alphabet of two or more
characters is not countable.

This
> contracition shows that it is nonsense to talk about completed
> infinity.


Not at all. It only shows that WM talks nonsense abut it.
>
> > Conclusion: the list of all definable reals
> > IS NOT ITSELF
> > DEFINABLE!  And it most certainly is NOT a CANTOR list!

>
> As I said, definability is not interesting.

> >
> >  IT'S
> >
> >        *  YOUR  *
> >
> >  DELUSION
> >

> You are in error. It is your delusion. I do not believe in completed
> infinity. Not even in the existence of the list defined above.


Your beliefs daily have les to do with reality.
>
> > THAT *YOUR* list of all definable reals is definable, and yoru much
> > deeper
> > delusion that your list of all definable reals lists all the reals!

>
> Do you join me when I say that the list containing all finite strings
> of 0's and 1's given above does not exist?


Since one can finitely define such a list as the list of all binary
natural numerals in natural order, it DOES exist, even in
Wolkenmuekenheim.




> that the notion of countability is nonsense?


Only in Wolkenmuekenheim is it nonsensical.
--





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