You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
On Jul 12, 8:54 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:
> Q1
> What are 2 missing reals from this List using Cantor's method?
>
> LIST
> 0.100..
> 0.000..
> 0.001..
A real that begins 0.010...
(also missing from LIST')
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
Indeed (in the sense that at least two real numbers
one starting .010 and one starting .110 are missing)
Yes, (although you have not shown this
for all cases, eg. 0.100...)
Note however that there are reals
starting 0.100... 0.000... and 0.001..
that are not missing and are not the antidiagonal
of the main diagonal of any *permutation* of the list.
Graham Cooper
unread,
Aug 17, 2012, 11:05:02 PM8/17/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
what list?
This is an infinite list
LIST
0.100..
0.000..
0.001..
..
with more than 3 prefixes.
So if we examine the list further
LIST
0.100..
0.000..
0.001..
0.111..
0.000..
0.000..
..
all 8 prefixes are DIAGONALS of some permutation and ANTI-DIAGONALS of
some permutation.
0.000... IS A DIAGONAL
0.000... IS AN ANTIDIAGONAL
0.001... IS A DIAGONAL
0.001... IS AN ANTIDIAGONAL
...
0.111... IS A DIAGONAL
0.111... IS AN ANTIDIAGONAL
That can be deduced just by inspecting the list.
LIST
0.100..
0.000..
0.001..
0.111..
0.000..
0.000..
..
> all 8 prefixes are DIAGONALS of some permutation and ANTI-DIAGONALS of
> some permutation.
>
> 0.000... IS A DIAGONAL
Yes there is a real number starting 0.000... which is a diagonal
of some permutation.
> 0.000... IS AN ANTIDIAGONAL
Yes there is a real number starting 0.000... which is an antidiagonal
of some permutation.
Note there is also a real number starting 0.000 which is not the
antidiagonal
of any permutation of the list.
Graham Cooper
unread,
Aug 17, 2012, 11:30:46 PM8/17/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
On Aug 18, 1:26 pm, William Hughes <wpihug...@gmail.com> wrote:
> Note there is also a real number starting 0.000 which is not the
> antidiagonal
> of any permutation of the list.
Is that true for every digit of this mystery number 0.000..?
LIST
0.100..
0.000..
0.001..
0.111..
0.000.. could be this one?
0.000.. might be this one??
..
Herc
William Hughes
unread,
Aug 17, 2012, 11:44:08 PM8/17/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
On Aug 18, 12:30 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Aug 18, 1:26 pm, William Hughes <wpihug...@gmail.com> wrote:
>
> > Note there is also a real number starting 0.000 which is not the
> > antidiagonal
> > of any permutation of the list.
>
> Is that true for every digit of this mystery number 0.000..?
Certainly, for any digit, x_i, of the "mystery number"
you can find a permutation, p(x_i), such that the digit
is in the correct place in the antidiagonal of the
permutation p(x_i).
However, you cannot find one permutation, q, such that *all*
the digits of the mystery number are in the correct place
in the antidiagonal of the permutation q.
Virgil
unread,
Aug 17, 2012, 11:57:19 PM8/17/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
> Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > On Aug 18, 1:26 pm, William Hughes <wpihug...@gmail.com> wrote:
> > > Note there is also a real number starting 0.000 which is not the
> > > antidiagonal
> > > of any permutation of the list.
>
> > Is that true for every digit of this mystery number 0.000..?
>
> > LIST
> > 0.100..
> > 0.000..
> > 0.001..
> > 0.111..
> > 0.000.. could be this one?
> > 0.000.. might be this one??
> > ..
>
> > Herc
>
> For any n in N, any list of n n-place binaries is incomplete.
>
> And, as it happens, any list of |N| |N|-place binaries is incomplete.
> --
You made 2 statements when only 1 was required.
Herc
Graham Cooper
unread,
Aug 18, 2012, 12:10:56 AM8/18/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
So after seeing 0.000.. *IS* in the list..
there is a real number starting 0.000 which is not the
antidiagonal of any permutation of the list.
so for ANY-REAL-IN-THE-LIST
THAT-REAL is not the antidiagonal
Why is that?
Self referential evident truth?
Extra reals that cannot be listed?
Herc
William Hughes
unread,
Aug 18, 2012, 12:47:49 AM8/18/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
> So after seeing 0.000.. *IS* in the list..
>
> there is a real number starting 0.000 which is not the
> antidiagonal of any permutation of the list.
>
> so for ANY-REAL-IN-THE-LIST
> THAT-REAL is not the antidiagonal
>
> Why is that?
Because ANY-REAL-IN-THE-LIST must be at some
row, say i, and any permutation, q, must put
row i somewhere, thus ANY-REAL-IN-THE-LIST
is not the antidiagonal of permutation q,
Virgil
unread,
Aug 18, 2012, 1:40:20 AM8/18/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
So ANY-REAL-IN-THE-LIST
is not the anti-diagonal of the SET.
William Hughes
unread,
Aug 18, 2012, 8:12:35 AM8/18/12
Delete
You do not have permission to delete messages in this group
Copy link
Report message
Sign in to report message
Show original message
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
On Aug 18, 5:27 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
> On Aug 18, 2:47 pm, William Hughes <wpihug...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Aug 18, 1:10 am, Graham Cooper <grahamcoop...@gmail.com> wrote:
>
> > > So after seeing 0.000.. *IS* in the list..
>
> > > there is a real number starting 0.000 which is not the
> > > antidiagonal of any permutation of the list.
>
> > > so for ANY-REAL-IN-THE-LIST
> > > THAT-REAL is not the antidiagonal
>
> > > Why is that?
>
> > Because ANY-REAL-IN-THE-LIST must be at some
> > row, say i, and any permutation, q, must put
> > row i somewhere, thus ANY-REAL-IN-THE-LIST
> > is not the antidiagonal of permutation q,
>
> So ANY-REAL-IN-THE-LIST
> is not the anti-diagonal of the SET.
>
> Why is that?
Because any anti-diagonal of the SET
must be the anti-diagonal of some permutation
q.