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Article #847

Re: A Guide for Cantor Cranks

#847
From: Graham Cooper
Date: Mon, 21 Mar 2011 11:19
124 lines
5349 bytes
On Mar 21, 9:09 pm, William Hughes <wpihug...@gmail.com> wrote:
> On Mar 21, 4:09 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
>
>
>
>
> > On Mar 21, 4:54 pm, William Hughes <wpihug...@gmail.com> wrote:
>
> > > On Mar 21, 3:33 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > > On Mar 21, 2:30 pm, William Hughes <wpihug...@gmail.com> wrote:
>
> > > > > > My point is you can't support your claim with 2 given diagonals.
>
> > > > > My claim is that given a list of binary sequences I can find a
> > > > > sequence
> > > > > not on the list.
>
> > > > > To support this claim I need only one diagonal.  When you gave me
> > > > > two diagonals I chose one and supported the claim.
>
> > > > > Your claim that you can get a generalized diagonal with any finite
> > > > > prefix is true, but irrelevant.  You cannot set the generalized
> > > > > diagonal to an arbitrary sequence.
> > > > >                                      - William Hughes
>
> > > > You haven't proven your sequence is not on the list in the 2 diagonal
> > > > version.
>
> > > > AD1 = 1110...
> > > > AD2 = 0110...
>
> > > > Can you prove EITHER real is missing from my (secret) list given only
> > > > the above?
>
> > > > IF YES, then by all means do so!
>
> > > There is a sequence not on your secret list that starts
> > > 1100.  Proof: the n'th digit means that sequence is not
> > > the n'th line of the list.  There may be other sequences that
> > > starts 1110 on your secret list.
> > >                                  -William Hughes-
>
> > Well that was concise!
>
> > But it's invalid due to the pigeon hole principle.
>
> > My similar claim to yours is: no matter what numbers I put into the
> > pigeon holes, there are always more empty pigeon holes than filled
> > pigeon holes.
>
> The pigeon hole principle does not work
> for infinite sets.
>
>                  - William Hughes-


I'm not slotting infinite sequences into infinite pigeon holes.


Digit 1 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[ ][ ][ ][ ][ ][ ][P][ ][ ][ ]

.

digit 2 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[ ][ ][ ][P][ ][ ][ ][ ][ ][ ]


.

digit 3 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[ ][ ][ ][ ][ ][ ][P][ ][ ][ ]


....



Your argument is to use the missing pigeon holes to form a 'different'
number.


Digit 1 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[p][p][p][p][p][p][ ][p][p][p]

.

digit 2 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[p][p][p][ ][p][p][p][p][p][p]


.

digit 3 of diagonal

-1--2--3--4--5--6--7--8--9--0-
[p][p][p][p][p][p][ ][p][p][p]


....


These are your options for all the anti-diagonals right?

p = Stealth pigeon
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From: Graham Cooper <grahamcooper7@gmail.com>
Newsgroups: sci.logic,sci.math,comp.theory,sci.physics,aus.science
Subject: Re: A Guide for Cantor Cranks
Date: Mon, 21 Mar 2011 04:19:33 -0700 (PDT)
Organization: http://groups.google.com
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