# Two results of set geometry

4 views

### WM

Aug 28, 2007, 7:10:11 AM8/28/07
to
Two results of set geometry

Set geometry, a new branch of mathematics, is devised to investigate
relations of finite and infinite sets by means of geometrical
representations. Up to now there have been two important results.

The first result is that the ordinal number of the set of natural
numbers is not unique. It has been shown that if {1, 2, 3, ...} =
omega, then {1, 2, 3, ...} = omega + 1 too.

The second result shows that the set of real numbers is countable.

Regards, WM

### Robert J. Kolker

Aug 28, 2007, 11:13:56 AM8/28/07
to

Totally bogus.

Bob Kolker

>

### Virgil

Aug 28, 2007, 12:18:28 PM8/28/07
to
WM <muec...@rz.fh-augsburg.de> wrote:

> Two results of set geometry
>
> Set geometry, a new branch of mathematics, is devised to investigate
> relations of finite and infinite sets by means of geometrical
> representations. Up to now there have been two important results.
>
> The first result is that the ordinal number of the set of natural
> numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> omega, then {1, 2, 3, ...} = omega + 1 too.
>
> urce&hl=de

Except that there is no proof here at all. Unless 2 = 1.

>
> The second result shows that the set of real numbers is countable.
>
> e

That alleged "showing" has been debunked severally, both by the flaws
in its arguments and by counterproofs.

That WM has succeeded in posting claims he has not proved, is not
evidence of their validity.

### WM

Aug 28, 2007, 1:06:33 PM8/28/07
to
On 28 Aug., 01:41, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> Sorry to break in again, I have been away a bit too long.
>
> In article <1188077979.547053.115...@50g2000hsm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 25 Aug., 14:13, William Hughes <wpihug...@hotmail.com> wrote:
> ...
> > > M' does not contain a second index omega. A does
> > > not contain an element omega. A+1 does not contain
> > > omega + 1.
> >
> > But by extending all rows of M' we get a set of order omega + 1. Don't
> > we get at least one non-natural index, necessarily?
>
> As you do not define what *you* mean with "extend" (and you do apparently
> not use the standard definition), it is even unclear what this means.
> Normally with "extend" is meant "adding something at the end" when talking
> about ordered sets. But as a set with order type omega does not have an
> end, it is unclear what is meant in that case. But if you mean "add an
> element with index greater than all other indices", we clearly get from
> an ordered set with order type omega a set with order type omega+1.

I mean simply: append an element as Cantor does define it. There is no
index to be considered. You can see this if you append 1 to the set
{2,3,4,...} to get {2,3,4,...,1}.

>
> > So you don't believe in the unique bijection between digits and second
> > indexes?
> >
> > 1 1 1 1 ... and
> > 1,2,3,4,...
>
> But this is again thoroughly unmathematical. Can you show a bijection
> between the two sets? From {1, 1, 1, ...} to {1, 2, 3, 4, ...}, what
> is f(1)? You are talking about order preserving bijections between
> ordered multi-sets. Argh.

Look at the matrix. It should be obvious:

1
11
111
1111
...

>
> > > The elements of the sets are not second indexes, but
> > > sequences of ones. The set of sequences of ones
> > > in M' is not the same as the set of sequences of ones
> > > in M.
> >
> > As above bijection shows, then also the set of second indexes must
> > differ.
>
> That is an order preserving bijection between multi-sets.

You misunderstand. The bijection is between n and {1,2,3,...,n} but
the n's are written in unary representation by sets of 1's in the rows
and the initial segments are those of the first column.

1 <--> 1

1
1 <--> 11

1
1
1 <--> 111

...

If the set of natural numbers has omega elements, then the set of rows
is in bijection with the complete first column. If a bijection exists,
then appending (in the manner of Cantor) one element to each partner
of the bijection, then the bijection persists. But this leads to omega
+ 1 being in bijection with the set of natural numbers.

>
> > > Appending a one the the all elements of M'
> > > does not give the same result as appending a one
> > > to all elements of M.
> >
> > That is because set theory is blurry.
>
> No, it is because you are blurry. Each '1' in M and M' actually
> represents a pair of natural numbers (r,c), where r is the row number
> and c the column number. The elements of the triangular matrix M are
> defined as (r,c) as above and non-existing when c > r. The matrix M'
> (of some strange format) is defined as M except when r = 1, in that
> row *all* elements do exist. When defined such we can talk about
> bijections. Now let us try a few actions:
> (1) append an element to each row.
> starting with M we get M itself with the first line omitted (each
> line terminates, so there is a clear new position where we can append).
> starting with M' we get a new matrix M'' (we have to define how we
> append an element to the first line, but let that be with column
> index 'omega+1'.
> (2) append an element to each column.
> starting with M we get the same problem as before with M', but if we
> allow a row 'omega+1', we get an element with row number 'omega+1' in
> each column.
> starting with M' something similar happens (although each column has
> a gap).
> (3) prepend an element to each column.
> starting with M we get M itself (when we look at the 1's only).
> starting with M' we get a matrix with the first line duplicated.
> (4) prepend an element to each row.
> starting with M we get a matrix similar to M with the first row omitted.
> starting with M' we get a matrix similar to M' with the second row
> omitted.
> Now explain why (1) and (2) should provide an inconsistency, while (3) and
> (4) do not.

You said it already yourself: Starting with (1) M, the rows of M
remain (except the first line), while in (2) M we get omega + 1
columns. Your points (3) an d (4) are uninteresting.

>
> > The set of second indexes of M' (which is containing a completed
> > infinity) cannot be the same as the set of second indexes of M (which
> > is not containing a completed infinity).
>
> Nobody says so, when you properly define what you mean with appending.

Cantor has defined how an element is appended to omega. Just this is
what I am doing and what you should know - and what forces W.H. to
assume the same set of second indices in M aa in M'.

Regards, WM

### G. Frege

Aug 28, 2007, 1:06:13 PM8/28/07
to
On Tue, 28 Aug 2007 10:18:28 -0600, Virgil <vir...@comcast.net> wrote:

>
> That WM has succeeded in posting claims he has not proved, is not
> evidence of their validity.
>

Actually,

"Unproven statements carry little weight in the world of
mathematics." - Amir D. Aczel

F.

--

E-mail: info<at>simple-line<dot>de

Aug 28, 2007, 3:21:22 PM8/28/07
to
WM wrote:

The first and foremost result of your new idiocy is this:

From: "Prof. Dr. W. Mueckenheim" <muec...@rz.fh-augsburg.de>
Newsgroups: sci.math.research
Subject: thread closed: "A problem of set geometry"
Date: 17 Aug 2007 13:49:44 -0400
Organization: World Wide Maths
Lines: 122
Approved: G A Edgar <ed...@math.ohio-state.edu> moderator for
sci.math.research
Message-ID: <fa4n3o\$c8\$1...@dizzy.math.ohio-state.edu>
[...]
[
Moderator's note:
This thread is closing. Further discussion of "trijection"
may be done in sci.math, for example.
]
[...]

In other words, you have been thrown out of sci.math.research.
Just in case you didn't notice, Mr. "Prof. Dr. W. Mueckenheim".

### WM

Aug 28, 2007, 4:23:03 PM8/28/07
to
On 28 Aug., 18:18, Virgil <vir...@comcast.net> wrote:

>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > Two results of set geometry
>
> > Set geometry, a new branch of mathematics, is devised to investigate
> > relations of finite and infinite sets by means of geometrical
> > representations. Up to now there have been two important results.
>
> > The first result is that the ordinal number of the set of natural
> > numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> > omega, then {1, 2, 3, ...} = omega + 1 too.
>

> > urce&hl=de
>
> Except that there is no proof here at all. Unless 2 = 1.
>
>
>
> > The second result shows that the set of real numbers is countable.
>

> > e
>
> That alleged "showing" has been debunked severally, both by the flaws
> in its arguments and by counterproofs.

is again. (Just an opportunity for you to learn something. I do not
intend to further discuss your uneducated assertions like infinite
sets being implied by Peano etc.)

____________________________________________________________________

> If you mean that the unbounded initial segments are inclused in the
> "trijection" along with the bounded ones, that is quite obvious, but
> equally irrelevant.

I beg your pardon, why should it be obvious that the unbounded
initial
segments, i.e., the complete sequences be included in my trichotomy? I
think
that need only be so if I defined it so.

> > > > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)

> > > > such that all elements belonging to an initial segment of
> > column,
> > > > diagonal, and line are 1's.

> > > > But there is no such trijection for n in omega including omega
between
> > > > the first column, the diagonal and the n-th line

> > > But there is no need for one either, since for all m, n IN omega, and
> > > all m <= n, a_mn = a_nn = a_nm.

It is not important though, but for the matrix given above we have
0 = a_23 =/= a_32 = 1, for example.

> > > WM's problem is that he wants endless sequences to have two ends, when
> > > they don't.-

> > The complete set of natural numbers has omega elements. It is a valid
> > statement in set theory that a set with omega elements (like the set
> > of all positions of the first column) can be extended to a set of
> > order omega + 1 by adding one element.

> Not at all.

What precisely do you deny?

> It is only true that a well ordered set of omega elements
> after having a new element appended to its ordering is a new set of of
> order type (omega+ 1). Anything stated less precisely need not be hold.

So let us consider the first column, i.e., the set of positions {a_k1
| k in
N} and append one element to obtain a set of order omega + 1.

- Zitierten Text ausblenden -
- Zitierten Text anzeigen -

> > This feature can be used to
> > investigate the trijection between the positions of 1's of the first
> > column, the diagonal, and the lines in the matrix M

> > 1000...
> > 11000...
> > 111000...
> > ...

> > If there is a trijection, then it should remain a trijection after
> > extending the sets involved by one element each. Extending the first
> > column of M by one element we obtain a sequence of order type omega +
> > 1 (because it has the order type omega). After adding one 1 to each
> > sequence of 1's in the lines of M no sequence of 1's has order type
> > omega + 1 (because each one has a finite order type). This can be
> > repeated infinitely often, such that the order type of the first
> > column becomes omega*2 (notation from Cantor, 1895) while the order
> > type of the set of lines cannot surpass omega.

> It is as easy to append a 1 to any endless line (as filled in to
> endlessness by spaces or zeros) as to any endless column or diagonal.

That is correct, but it does not concern my argument. If you would
like to
discuss another topic, it would be desirable to open a new thread in
order
to avoid mixing up the facts. In my argument the 1 should not be
appended to
the endless line but to the finite sequence of 1's in each line
(because
only finite sequences can represent natural numbers - not infinite
sequences). In my argument it is essential that those natural numbers
are to
be put in trijection with the initial segments of column and
diagonal.

Regards, WM

### WM

Aug 28, 2007, 4:25:56 PM8/28/07
to
On 28 Aug., 17:13, "Robert J. Kolker" <bobkol...@comcast.net> wrote:
> WM wrote:
> > Two results of set geometry
>
> > Set geometry, a new branch of mathematics, is devised to investigate
> > relations of finite and infinite sets by means of geometrical
> > representations. Up to now there have been two important results.
>
> > The first result is that the ordinal number of the set of natural
> > numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> > omega, then {1, 2, 3, ...} = omega + 1 too.
>

>
> > The second result shows that the set of real numbers is countable.
>
>
> > Regards, WM
>
> Totally bogus.

Utterings of your taste are appreciated but of no value concerning set
geometry. The arguments of set geometry stand unrefuted.

Regards, WM

### Peter Webb

Aug 28, 2007, 6:04:21 PM8/28/07
to
I haven't bothered to read your proofs in detail. Its basically like being
given a proof that 1=2 and told it is correct. It has been proved that N
cannot be bijected with R, so unless you have proved that set theory is
inconsistent your claim is obviously wrong. The chances of you doing that
are so small as to make reading your "proofs" a waste of time.

I was however struck by the following:

> If the set of natural numbers has omega elements, then the set of rows
> is in bijection with the complete first column. If a bijection exists,
> then appending (in the manner of Cantor) one element to each partner
> of the bijection, then the bijection persists. But this leads to omega
> + 1 being in bijection with the set of natural numbers.
>

You can easily biject omega+1 with N. Here is one example:

1 <-> w+1
2 <-> 1
3 <-> 2
4 <-> 3
.
.
.

If you think that this is paradoxical and worth creating a proof of, you are
about 100 years too late. Cantor himself proved that N can be bijected with
w+1, w+w, and w^127 + 93w + 17 for that matter.

This makes your first result akin to proving 1+1=2 (everybody knows its true
and nobody disputes it) and your second result akin to proving 1+1=3
(everybody knows its false, because it has been proven to be so).

### Virgil

Aug 28, 2007, 7:04:43 PM8/28/07
to
WM <muec...@rz.fh-augsburg.de> wrote:

> On 28 Aug., 01:41, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > Sorry to break in again, I have been away a bit too long.
> >
> > In article <1188077979.547053.115...@50g2000hsm.googlegroups.com> WM
> > <mueck...@rz.fh-augsburg.de> writes:
> > > On 25 Aug., 14:13, William Hughes <wpihug...@hotmail.com> wrote:
> > ...
> > > > M' does not contain a second index omega. A does
> > > > not contain an element omega. A+1 does not contain
> > > > omega + 1.
> > >
> > > But by extending all rows of M' we get a set of order omega + 1. Don't
> > > we get at least one non-natural index, necessarily?
> >
> > As you do not define what *you* mean with "extend" (and you do apparently
> > not use the standard definition), it is even unclear what this means.
> > Normally with "extend" is meant "adding something at the end" when talking
> > about ordered sets. But as a set with order type omega does not have an
> > end, it is unclear what is meant in that case. But if you mean "add an
> > element with index greater than all other indices", we clearly get from
> > an ordered set with order type omega a set with order type omega+1.
>
> I mean simply: append an element as Cantor does define it. There is no
> index to be considered. You can see this if you append 1 to the set
> {2,3,4,...} to get {2,3,4,...,1}.

As sets, {2,3,4,...,1} = {1,2,3,4,...}.
As ordered sets, (2,3,...;1) =/= (1,2,3,...).
Absent indexing, or some equivalent establishment of ordering,
one has merely {2,3,4,...,1} = {1,2,3,4,...}.

> Look at the matrix. It should be obvious:
>
> 1
> 11
> 111
> 1111
> ...

>

> You misunderstand. The bijection is between n and {1,2,3,...,n} but
> the n's are written in unary representation by sets of 1's in the rows
> and the initial segments are those of the first column.
>
> 1 <--> 1
>
> 1
> 1 <--> 11
>
> 1
> 1
> 1 <--> 111
>
> ...
>
> If the set of natural numbers has omega elements, then the set of rows
> is in bijection with the complete first column. If a bijection exists,
> then appending (in the manner of Cantor) one element to each partner
> of the bijection, then the bijection persists.

The partners being paired are a row from the set of rows with the lead
entries of that row as an entry in the first column.

So appending an entry at the end of that column requires appending a new
row, longer than all the present rows.

Which leads to the order type of each, the extended set of column
entries and the extended set of rows (including a row longer than any of
the infinitely many finite rows), being of order type omega + 1.

When WM tries to switch from members to sets or vice versa, as he does
in his "extending" of order isomorphisms, his alleged proofs become
invalid.

### Virgil

Aug 28, 2007, 7:19:19 PM8/28/07
to
WM <muec...@rz.fh-augsburg.de> wrote:

> On 28 Aug., 18:18, Virgil <vir...@comcast.net> wrote:
> >

> I beg your pardon, why should it be obvious that the unbounded

> initial segments, i.e., the complete sequences be included in my
> trichotomy? I think that need only be so if I defined it so.
>
>
> > > > > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)

If it does not hold for N, then there must be a first n in N for which
it doe not hold. But induction proves otherwise.

>
>
>
> It is not important though, but for the matrix given above we have 0
> = a_23 =/= a_32 = 1, for example.
>
>
> > > > WM's problem is that he wants endless sequences to have two
> > > > ends, when they don't.-
>
>
> > > The complete set of natural numbers has omega elements. It is a
> > > valid statement in set theory that a set with omega elements
> > > (like the set of all positions of the first column) can be
> > > extended to a set of order omega + 1 by adding one element.
>
>
> > Not at all.
>
>
>
> What precisely do you deny?

That prepending or inserting a new element, both forms of addition, into
an endless but well-ordered set effects a change its order type.

>
>
> > It is only true that a well ordered set of omega elements after
> > having a new element appended to its ordering is a new set of of
> > order type (omega+ 1). Anything stated less precisely need not be
> > hold.
>
>
> So let us consider the first column, i.e., the set of positions {a_k1
> | k in
> N} and append one element to obtain a set of order omega + 1.
>
>
>
> - Zitierten Text ausblenden - - Zitierten Text anzeigen -
>
> > > This feature can be used to investigate the trijection between
> > > the positions of 1's of the first column, the diagonal, and the
> > > lines in the matrix M
>
>
> > > 1000... 11000... 111000... ...
>
>
> > > If there is a trijection, then it should remain a trijection
> > > after extending the sets involved by one element each. Extending
> > > the first column of M by one element we obtain a sequence of
> > > order type omega + 1 (because it has the order type omega). After
> > > adding one 1 to each sequence of 1's in the lines of M no
> > > sequence of 1's has order type omega + 1 (because each one has a
> > > finite order type).

Apples and oranges. When one associates an element of some column with
some row, adding an element to that column equates with adding an entire
row, not adding an element to a row.

>
> > It is as easy to append a 1 to any endless line (as filled in to
> > endlessness by spaces or zeros) as to any endless column or
> > diagonal.
>
>
>
> That is correct, but it does not concern my argument.

But it scuttles your argument as you are adding apples to oranges.

> If you would
> like to discuss another topic

That WM is too dense to realize that it is the proper topic to reveal
the faults in his arguments is his own personal problem. And an
irrelevance to the proof he is wrong.

### Virgil

Aug 28, 2007, 7:21:12 PM8/28/07
to
WM <muec...@rz.fh-augsburg.de> wrote:

They certainly stand unestablished.

At least by any standards mathematically acceptable.

### Dik T. Winter

Aug 28, 2007, 8:27:39 PM8/28/07
to
In article <1188299411.9...@d55g2000hsg.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
> Two results of set geometry
>
> Set geometry, a new branch of mathematics, is devised to investigate
> relations of finite and infinite sets by means of geometrical
> representations. Up to now there have been two important results.
>
> The first result is that the ordinal number of the set of natural
> numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> omega, then {1, 2, 3, ...} = omega + 1 too.
>

You like posting nonsense? The above has not been shown in that article
at all! To repeat from that article:

> The complete set of natural numbers has omega elements. It is a valid
> statement in set theory that a set with omega elements (like the set
> of all positions of the first column) can be extended to a set of

> order omega + 1 by adding one element. This feature can be used to

> investigate the trijection between the positions of 1's of the first
> column, the diagonal, and the lines in the matrix M
>
> 1000...
> 11000...
> 111000...
> ...
>
> If there is a trijection,

(You actually mean mutiple bijections between sets of indices, but I will
What bijection is there between the ones of the first column and of *any*
line of that matrix? There is a bijection between the first column and
each and every complete line of the matrix.

> then it should remain a trijection after
> extending the sets involved by one element each. Extending the first
> column of M by one element we obtain a sequence of order type omega +
> 1 (because it has the order type omega). After adding one 1 to each
> sequence of 1's in the lines of M no sequence of 1's has order type
> omega + 1 (because each one has a finite order type).

You are doing different additions to the first column and the first line.
You add an element at the end of the first column, but in the middle of
the first line. So why do you expect a bijection between the first line
and the first column? (Note that "trijection" is just obfuscation, there
are simply multiple bijections.)

> This can be
> repeated infinitely often, such that the order type of the first
> column becomes omega*2 (notation from Cantor, 1895) while the order
> type of the set of lines cannot surpass omega.

They *can* surpass omega, if you add elements at the end, which you do
not. This is most properly seen when you use indices. At the first
line you add a new element at position (1,2), shifting *all* 0's up
one place. When you add a new element to the first column, you do *not*
do it at position (2,1), shifting *all* 1's up one place, but you do it
at a new position (w,1).

> For the diagonal the
> case remains undecided. But whatever the diagonal may be, there is no
> trijection between the sequences of 1's in the first column, the
> diagonal, and the lines - not finally and, therefore, also not
> initially.

Bogus. When you add an element after the first column, what is the
diagonal? But we *do* know that there does exist a bijection between
a set of order w and a set of order w+1. Let's get the new matrix with
a new line added at position (w,1). I get the following bijections,
from first column (c_i1) to first line (l_1i) and to diagonal (d_ii):
if i = w: (c_w1) <-> (l_11) <-> (d_11)
if i != w: (c_i1) <-> (l_[i+1]1) <-> (d_[i+1][i+1]).

> The second result shows that the set of real numbers is countable.
>

And this is just as wrong.
> Each number is given by a path
> stretching over infinitely many nodes (bits). All nodes (bits) of the
> tree are countable. The paths are not, according to Cantor's famous
> diagonal proof.
>
> But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> whereas 2^n different paths arrive at and 2^(n+1) different paths
> spring off from line number n.

How you can state that couple of sentences with a straight face escapes
me. As *all* paths are infinite in length, even at the root node your
statements is wrong, because *all* paths spring off from the root node
and not only two different paths. And more general, at each node arrive
*all* the paths that do emanate from that node. So you should explain
how it is possible that at a single node arrives a single path while
two do emanate from it (all paths start at the root). In the second
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

### Dik T. Winter

Aug 28, 2007, 9:02:04 PM8/28/07
to
In article <1188320793.3...@g4g2000hsf.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
> On 28 Aug., 01:41, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > Sorry to break in again, I have been away a bit too long.
> >
> > In article <1188077979.547053.115...@50g2000hsm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > On 25 Aug., 14:13, William Hughes <wpihug...@hotmail.com> wrote:
> > ...
> > > > M' does not contain a second index omega. A does
> > > > not contain an element omega. A+1 does not contain
> > > > omega + 1.
> > >
> > > But by extending all rows of M' we get a set of order omega + 1. Don't
> > > we get at least one non-natural index, necessarily?
> >
> > As you do not define what *you* mean with "extend" (and you do apparently
> > not use the standard definition), it is even unclear what this means.
> > Normally with "extend" is meant "adding something at the end" when talking
> > about ordered sets. But as a set with order type omega does not have an
> > end, it is unclear what is meant in that case. But if you mean "add an
> > element with index greater than all other indices", we clearly get from
> > an ordered set with order type omega a set with order type omega+1.
>
> I mean simply: append an element as Cantor does define it. There is no
> index to be considered. You can see this if you append 1 to the set
> {2,3,4,...} to get {2,3,4,...,1}.

There *are* indices to be considered. All your elements are '1'!

> > > So you don't believe in the unique bijection between digits and second
> > > indexes?
> > >
> > > 1 1 1 1 ... and
> > > 1,2,3,4,...
> >
> > But this is again thoroughly unmathematical. Can you show a bijection
> > between the two sets? From {1, 1, 1, ...} to {1, 2, 3, 4, ...}, what
> > is f(1)? You are talking about order preserving bijections between
> > ordered multi-sets. Argh.
>
> Look at the matrix. It should be obvious:
>
> 1
> 11
> 111
> 1111
> ...

What is f(1)? Darn, you *say* you are talking about bijections between
elements when actually you are talking about bijections between index
positions (and so, yes, index positions *are* to be considered).

> > > > The elements of the sets are not second indexes, but
> > > > sequences of ones. The set of sequences of ones
> > > > in M' is not the same as the set of sequences of ones
> > > > in M.
> > >
> > > As above bijection shows, then also the set of second indexes must
> > > differ.
> >
> > That is an order preserving bijection between multi-sets.
>
> You misunderstand. The bijection is between n and {1,2,3,...,n} but
> the n's are written in unary representation by sets of 1's in the rows
> and the initial segments are those of the first column.

Now, this is again utterly confused. With this explanation I see only
a single set: the set of natural numbers written out in unary. But if
you consider it that way, there is only a single column, namely the
list of natural numbers.

> If the set of natural numbers has omega elements, then the set of rows
> is in bijection with the complete first column.

Yes, of course, because the set of rows is the set of natural numbers
(because each element is a natural number). And it is in bijection with
the first column, which is the set of rows. Quite trivial I would think.

> If a bijection exists,
> then appending (in the manner of Cantor) one element to each partner
> of the bijection, then the bijection persists.

Obvious. As each partner in the bijection is the set of natural numbers.

> But this leads to omega
> + 1 being in bijection with the set of natural numbers.

Not at all. Why do you think so? What you actually *do* when you append
a 1 to each row is changing each element of the set {1,2,3,...} to its
successor, so you change it to {2,3,4,...}. You do not add something to
the set the column is bijected with.

> > > > Appending a one the the all elements of M'
> > > > does not give the same result as appending a one
> > > > to all elements of M.
> > >
> > > That is because set theory is blurry.
> >
> > No, it is because you are blurry. Each '1' in M and M' actually
> > represents a pair of natural numbers (r,c), where r is the row number
> > and c the column number. The elements of the triangular matrix M are
> > defined as (r,c) as above and non-existing when c > r. The matrix M'
> > (of some strange format) is defined as M except when r = 1, in that
> > row *all* elements do exist. When defined such we can talk about
> > bijections. Now let us try a few actions:
> > (1) append an element to each row.
> > starting with M we get M itself with the first line omitted (each
> > line terminates, so there is a clear new position where we can append).
> > starting with M' we get a new matrix M'' (we have to define how we
> > append an element to the first line, but let that be with column
> > index 'omega+1'.
> > (2) append an element to each column.
> > starting with M we get the same problem as before with M', but if we
> > allow a row 'omega+1', we get an element with row number 'omega+1' in
> > each column.
> > starting with M' something similar happens (although each column has
> > a gap).

> > Now explain why (1) and (2) should provide an inconsistency, while (3) and
> > (4) do not.
>
> You said it already yourself: Starting with (1) M, the rows of M
> remain (except the first line), while in (2) M we get omega + 1
> columns. Your points (3) an d (4) are uninteresting.

Still I maintain, why is that an inconsistency? Note that here we are

> > > The set of second indexes of M' (which is containing a completed
> > > infinity) cannot be the same as the set of second indexes of M (which
> > > is not containing a completed infinity).

Note that M' violates your writing above that the rows represent natural
numbers written in unary notation! So either go one way or the other, do
not mix them.

> > Nobody says so, when you properly define what you mean with appending.
>
> Cantor has defined how an element is appended to omega. Just this is
> what I am doing and what you should know - and what forces W.H. to
> assume the same set of second indices in M aa in M'.

But let us see. When defined as indices, your M looks like:
a_ij is defined for every j in N and every i <= j
and your M' looks like (I assume here replacing the first line, I disremember
which it was, but it makes no difference):
a_ij is defined for every j in N and whenever i = 1 or i <= j
where is the difference between sets of second indices?

So M and M' have the same set of second indices. Now we add elements to
rows. Let us define that we add an element by inserting a new element at
a second column index position that is larger than all other index positions
in that same line. With M we can add all elements at finite positions
and get M_app:
a_ij is defined for every j in N and every i <= j - 1
for M' there is difficulty with the first line, and we get:
a_ij is defined for every j in N and whenever i = 1 or i <= j - 1,
and for i = 1 and j = w.
I still do not see a problem.

### WM

Aug 29, 2007, 1:23:24 PM8/29/07
to
On 29 Aug., 00:04, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
wrote:

> I haven't bothered to read your proofs in detail. Its basically like being
> given a proof that 1=2 and told it is correct. It has been proved that N
> cannot be bijected with R, so unless you have proved that set theory is
> inconsistent your claim is obviously wrong. The chances of you doing that
> are so small as to make reading your "proofs" a waste of time.

That attitude may by the reason why set theory has not yet been
recognized as inconsistent by many mathematicians.

>
> I was however struck by the following:
>
> > If the set of natural numbers has omega elements, then the set of rows
> > is in bijection with the complete first column. If a bijection exists,
> > then appending (in the manner of Cantor) one element to each partner
> > of the bijection, then the bijection persists. But this leads to omega
> > + 1 being in bijection with the set of natural numbers.
>
> You can easily biject omega+1 with N. Here is one example:
>
> 1 <-> w+1
> 2 <-> 1
> 3 <-> 2
> 4 <-> 3
> .
> .
> .
>
> If you think that this is paradoxical and worth creating a proof of, you are
> about 100 years too late. Cantor himself proved that N can be bijected with
> w+1, w+w, and w^127 + 93w + 17 for that matter.

Sorry, what you say is correct but not to the point. Of course N can
be bijected in this way to ordered sets (at least it was assumed so
for a long while; already Cantor gave examples like {1, 3, 5, ..., 2,
4, 6, ...}). What I found is simply that if the set of finite natural
numbers in the normal order can be bijected to omega (represented by
the infinite first column of the matrix

1
11
111
...

then the N *in the normal order* can also be bijected to omega + 1.
This is shown by appending just one 1 to every partner of the
bijection.

First, according to the matrix given above, but in more detail, we
have the bijection:

1 <--> 1

1
1 <--> 11

1
1
1 <--> 111

...

where also the infinite first column belongs to the bijection
(although there is no infinite natural number).

If this bijection exists, then we can, without destroying it, append
one 1 to every partner. This leaves every natural number finite (only
the first line of the bijection disappears) but it introduces as the
new first column the ordinal omega + 1. Therefore the set {2, 3,
4, ...} of all but one natural numbers is in bijection with omega + 1.

>
> This makes your first result akin to proving 1+1=2 (everybody knows its true
> and nobody disputes it) and your second result akin to proving 1+1=3
> (everybody knows its false, because it has been proven to be so).

It has been proven so under the assumption that it is possible to
attach a number omega to infinity. This is shown inconsistent. My
second proof is also very simple:

In the binary tree there are all real numbers of the interval [0, 1]
represented by paths from 0.000... to 0.111... .

0.
/ \
0 1
/ \ / \
0 1 0 1
...

It can easily be seen that there cannot be more separate paths than
nodes, because every path needs a node to separate itself from the
other paths. Therefore there are not more different real numbers in
the interval [0, 1] than nodes. But the number of nodes is countable
with no doubt. And, by construction, the tree contains all separable
real numbers of the interval. There is no chance for a "diagonal
number".

Regards, WM

### WM

Aug 29, 2007, 1:27:06 PM8/29/07
to
On 29 Aug., 01:04, Virgil <vir...@comcast.net> wrote:

The ordering is established by my matrix, which after appending one 1
to every partner of the bijection reads

1
1 <--> 11

1
1
1 <--> 111

...

1
1
1
.
.
. <--> ?

1
1
1
.
.
.
1 <--> ?

>
>
>
>
> > Look at the matrix. It should be obvious:
>
> > 1
> > 11
> > 111
> > 1111
> > ...
>
> > You misunderstand. The bijection is between n and {1,2,3,...,n} but
> > the n's are written in unary representation by sets of 1's in the rows
> > and the initial segments are those of the first column.
>
> > 1 <--> 1
>
> > 1
> > 1 <--> 11
>
> > 1
> > 1
> > 1 <--> 111
>
> > ...
>
> > If the set of natural numbers has omega elements, then the set of rows
> > is in bijection with the complete first column. If a bijection exists,
> > then appending (in the manner of Cantor) one element to each partner
> > of the bijection, then the bijection persists.
>
> The partners being paired are a row from the set of rows with the lead
> entries of that row as an entry in the first column.
>
> So appending an entry at the end of that column requires appending a new
> row, longer than all the present rows.

That would be the case indeed *if the complete column existed*.
However, it is not so.

>
> Which leads to the order type of each, the extended set of column
> entries and the extended set of rows (including a row longer than any of
> the infinitely many finite rows), being of order type omega + 1.

The set of rows is extended but not by an infinite number. All numbers
n + 1 are finite - for n in N.

>
> When WM tries to switch from members to sets or vice versa, as he does
> in his "extending" of order isomorphisms, his alleged proofs become
> invalid

It is said that the natural numbers are the members of a set.

Regards, WM

### WM

Aug 29, 2007, 1:32:32 PM8/29/07
to
On 29 Aug., 01:19, Virgil <vir...@comcast.net> wrote:

>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 28 Aug., 18:18, Virgil <vir...@comcast.net> wrote:
> > > In article <1188299411.985907.123...@d55g2000hsg.googlegroups.com>,
>
> > I beg your pardon, why should it be obvious that the unbounded
> > initial segments, i.e., the complete sequences be included in my
> > trichotomy? I think that need only be so if I defined it so.
>
> > > > > > (a_11, .., a_n1) <--> (a_11, .., a_nn) <--> (a_11, .., a_1n)
>
> If it does not hold for N, then there must be a first n in N for which
> it doe not hold. But induction proves otherwise.

First: What holds for N need not hold or not hold for its elements.
Second: You asserted just the opposite, namely: If you mean that the
unbounded initial segments are included in the "trijection" along with

the bounded ones, that is quite obvious, but equally irrelevant.

> > It is not important though, but for the matrix given above we have 0
> > = a_23 =/= a_32 = 1, for example.
>
> > > > > WM's problem is that he wants endless sequences to have two
> > > > > ends, when they don't.-
>
> > > > The complete set of natural numbers has omega elements. It is a
> > > > valid statement in set theory that a set with omega elements
> > > > (like the set of all positions of the first column) can be
> > > > extended to a set of order omega + 1 by adding one element.
>
> > > Not at all.
>
> > What precisely do you deny?
>
> That prepending or inserting a new element, both forms of addition, into
> an endless but well-ordered set effects a change its order type.
>

As I did not consider _prepending_ at any time, why then did you say
"not at all"? That _appending_ an element can change the order type is
undisputed by mathematicians, however.

> > > It is only true that a well ordered set of omega elements after
> > > having a new element appended to its ordering is a new set of of
> > > order type (omega+ 1). Anything stated less precisely need not be
> > > hold.
>
> > So let us consider the first column, i.e., the set of positions {a_k1
> > | k in
> > N} and append one element to obtain a set of order omega + 1.
>
> > - Zitierten Text ausblenden - - Zitierten Text anzeigen -
>
> > > > This feature can be used to investigate the trijection between
> > > > the positions of 1's of the first column, the diagonal, and the
> > > > lines in the matrix M
>
> > > > 1000... 11000... 111000... ...
>
> > > > If there is a trijection, then it should remain a trijection
> > > > after extending the sets involved by one element each. Extending
> > > > the first column of M by one element we obtain a sequence of
> > > > order type omega + 1 (because it has the order type omega). After
> > > > adding one 1 to each sequence of 1's in the lines of M no
> > > > sequence of 1's has order type omega + 1 (because each one has a
> > > > finite order type).
>
> Apples and oranges.

In set theory it does not matter whether sets consist of apples or
oranges. Important is whether they can be put in bijection.

> When one associates an element of some column with
> some row, adding an element to that column equates with adding an entire
> row, not adding an element to a row.
>

The bijection

1 <--> 1

1
1 <--> 11

1
1
1 <--> 111

remains a bijection after appending one 1 to every partner

1
1 <--> 11

1
1
1 <--> 111

1
1
1
1 <--> 1111

>
> > > It is as easy to append a 1 to any endless line (as filled in to
> > > endlessness by spaces or zeros) as to any endless column or
> > > diagonal.
>
> > That is correct, but it does not concern my argument.
>
> But it scuttles your argument as you are adding apples to oranges.
>
> > If you would
> > like to discuss another topic
>
> That WM is too dense to realize that it is the proper topic to reveal
> the faults in his arguments is his own personal problem.

It is refreshing to hear this from a five-star-"mathematician" who
does not know that Peano does not provide actually infinite sets and
who describes the matrix

1000...
11000...
111000...
...

by: " for all m, n IN omega, and all m <= n, a_mn = a_nn = a_nm."

Why m <= n but not n <= m?

Regards, WM

### WM

Aug 29, 2007, 1:33:43 PM8/29/07
to
On 29 Aug., 01:21, Virgil <vir...@comcast.net> wrote:

> > Utterings of your taste are appreciated but of no value concerning set
> > geometry. The arguments of set geometry stand unrefuted.
>
> They certainly stand unestablished.
>

> At least by any standards mathematically acceptable.-

By the standards which assume the existence of nameless and
undefinable numbers?

Regards, WM

### WM

Aug 29, 2007, 4:00:06 PM8/29/07
to
On 29 Aug., 02:27, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> In article <1188299411.985907.123...@d55g2000hsg.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > Two results of set geometry
> >
> > Set geometry, a new branch of mathematics, is devised to investigate
> > relations of finite and infinite sets by means of geometrical
> > representations. Up to now there have been two important results.
> >
> > The first result is that the ordinal number of the set of natural
> > numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> > omega, then {1, 2, 3, ...} = omega + 1 too.
> >

>
> You like posting nonsense? The above has not been shown in that article
> at all! To repeat from that article:
>
> > The complete set of natural numbers has omega elements. It is a valid
> > statement in set theory that a set with omega elements (like the set
> > of all positions of the first column) can be extended to a set of
> > order omega + 1 by adding one element. This feature can be used to
> > investigate the trijection between the positions of 1's of the first
> > column, the diagonal, and the lines in the matrix M
> >
> > 1000...
> > 11000...
> > 111000...
> > ...
> >
> > If there is a trijection,
>
> (You actually mean mutiple bijections between sets of indices, but I will

I gave an exact definition of trijection in the thread and explicated
it in detail to MoeBlee.

> What bijection is there between the ones of the first column and of *any*
> line of that matrix? There is a bijection between the first column and
> each and every complete line of the matrix.
>
> > then it should remain a trijection after
> > extending the sets involved by one element each. Extending the first
> > column of M by one element we obtain a sequence of order type omega +
> > 1 (because it has the order type omega). After adding one 1 to each
> > sequence of 1's in the lines of M no sequence of 1's has order type
> > omega + 1 (because each one has a finite order type).
>
> You are doing different additions to the first column and the first line.
> You add an element at the end of the first column, but in the middle of
> the first line. So why do you expect a bijection between the first line
> and the first column? (Note that "trijection" is just obfuscation, there
> are simply multiple bijections.)

In fact the trijection can be understood in this way. If you are
unable to comprehend the principle of trijection, we can set up two
bijections. There is one bijection between the initial segments
(including the complete sequences) of the first column and the
diagonal:

1 <--> 1

1 1
1 <--> 1

1 1
1 1
1 <--> 1

...

1 1
1 1
1 1
. .
. .
. <--> .

There is another bijection between the first column and the lines (if
there are infinitely many finite natural numbers)

1 <--> 1

1
1 <--> 11

1
1
1 <--> 111

...

1
1
1
.
.
. <--> ?

Third, there is the bijection between rows and initial segments of the
diagonal. You will be able to illustrate it yourself.

>
> > This can be
> > repeated infinitely often, such that the order type of the first
> > column becomes omega*2 (notation from Cantor, 1895) while the order
> > type of the set of lines cannot surpass omega.
>
> They *can* surpass omega, if you add elements at the end, which you do
> not. This is most properly seen when you use indices. At the first
> line you add a new element at position (1,2), shifting *all* 0's up
> one place. When you add a new element to the first column, you do *not*
> do it at position (2,1), shifting *all* 1's up one place, but you do it
> at a new position (w,1).

Wrong description. The second bijection after appending one element

1
1 <--> 11

1
1
1 <--> 111

1
1
1
1 <--> 1111

...

Of course appending one element to omega yields omega + 1.

>
> > For the diagonal the
> > case remains undecided. But whatever the diagonal may be, there is no
> > trijection between the sequences of 1's in the first column, the
> > diagonal, and the lines - not finally and, therefore, also not
> > initially.
>
> Bogus. When you add an element after the first column, what is the
> diagonal? But we *do* know that there does exist a bijection between
> a set of order w and a set of order w+1. Let's get the new matrix with
> a new line added at position (w,1). I get the following bijections,
> from first column (c_i1) to first line (l_1i) and to diagonal (d_ii):
> if i = w: (c_w1) <-> (l_11) <-> (d_11)
> if i != w: (c_i1) <-> (l_[i+1]1) <-> (d_[i+1][i+1]).

The clue however is, that in normal order N is omega + 1.

>
> > The second result shows that the set of real numbers is countable.
> >

>
> And this is just as wrong.
> > Each number is given by a path
> > stretching over infinitely many nodes (bits). All nodes (bits) of the
> > tree are countable. The paths are not, according to Cantor's famous
> > diagonal proof.
> >
> > But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> > whereas 2^n different paths arrive at and 2^(n+1) different paths
> > spring off from line number n.
>
> How you can state that couple of sentences with a straight face escapes
> me. As *all* paths are infinite in length, even at the root node your
> statements is wrong, because *all* paths spring off from the root node
> and not only two different paths. And more general, at each node arrive
> *all* the paths that do emanate from that node. So you should explain
> how it is possible that at a single node arrives a single path while
> two do emanate from it (all paths start at the root). In the second

Only the number of separated paths is interesting, i.e., the number of
different real numbers. In the binary tree there are all real numbers

of the interval [0, 1] represented by paths from 0.000... to
0.111... .

0.
/ \
0 1
/ \ / \
0 1 0 1
...

It can easily be seen that there cannot be more *separate* paths than

### WM

Aug 29, 2007, 4:00:50 PM8/29/07
to
On 29 Aug., 03:02, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> In article <1188320793.372430.104...@g4g2000hsf.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 28 Aug., 01:41, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > Sorry to break in again, I have been away a bit too long.
> > >
> > > In article <1188077979.547053.115...@50g2000hsm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > > On 25 Aug., 14:13, William Hughes <wpihug...@hotmail.com> wrote:
> > > ...
> > > > > M' does not contain a second index omega. A does
> > > > > not contain an element omega. A+1 does not contain
> > > > > omega + 1.
> > > >
> > > > But by extending all rows of M' we get a set of order omega + 1. Don't
> > > > we get at least one non-natural index, necessarily?
> > >
> > > As you do not define what *you* mean with "extend" (and you do apparently
> > > not use the standard definition), it is even unclear what this means.
> > > Normally with "extend" is meant "adding something at the end" when talking
> > > about ordered sets. But as a set with order type omega does not have an
> > > end, it is unclear what is meant in that case. But if you mean "add an
> > > element with index greater than all other indices", we clearly get from
> > > an ordered set with order type omega a set with order type omega+1.
> >
> > I mean simply: append an element as Cantor does define it. There is no
> > index to be considered. You can see this if you append 1 to the set
> > {2,3,4,...} to get {2,3,4,...,1}.
>
> There *are* indices to be considered. All your elements are '1'!

No, the elements are sequences of 1's.

>
> > > > So you don't believe in the unique bijection between digits and second
> > > > indexes?
> > > >
> > > > 1 1 1 1 ... and
> > > > 1,2,3,4,...
> > >
> > > But this is again thoroughly unmathematical. Can you show a bijection
> > > between the two sets? From {1, 1, 1, ...} to {1, 2, 3, 4, ...}, what
> > > is f(1)? You are talking about order preserving bijections between
> > > ordered multi-sets. Argh.
> >
> > Look at the matrix. It should be obvious:
> >
> > 1
> > 11
> > 111
> > 1111
> > ...
>
> What is f(1)? Darn, you *say* you are talking about bijections between
> elements when actually you are talking about bijections between index
> positions (and so, yes, index positions *are* to be considered).

See my last contribution, where I illustrated in detail the trijection
which I had precisely defined earlier.

>
> > > > > The elements of the sets are not second indexes, but
> > > > > sequences of ones. The set of sequences of ones
> > > > > in M' is not the same as the set of sequences of ones
> > > > > in M.
> > > >
> > > > As above bijection shows, then also the set of second indexes must
> > > > differ.
> > >
> > > That is an order preserving bijection between multi-sets.
> >
> > You misunderstand. The bijection is between n and {1,2,3,...,n} but
> > the n's are written in unary representation by sets of 1's in the rows
> > and the initial segments are those of the first column.
>
> Now, this is again utterly confused. With this explanation I see only
> a single set: the set of natural numbers written out in unary. But if
> you consider it that way, there is only a single column, namely the
> list of natural numbers.

You see only one column, but don't know what has to be understood by
the expression "initial segments of the first column"?

>
> > If the set of natural numbers has omega elements, then the set of rows
> > is in bijection with the complete first column.
>
> Yes, of course, because the set of rows is the set of natural numbers
> (because each element is a natural number). And it is in bijection with
> the first column, which is the set of rows. Quite trivial I would think.

Only the fact that the first column has a lot of finite initial
segments and an infinite initial segment while the rows are all finite
may be raising some suspicion.

>
> > If a bijection exists,
> > then appending (in the manner of Cantor) one element to each partner
> > of the bijection, then the bijection persists.
>
> Obvious. As each partner in the bijection is the set of natural numbers.

That is incorrect. Each partner on the one side is a number of natural
numbers and on the other side each partner is a natural number. I
tried to express that by the variables {1,2,3,...,n} <--> n.

>
> > But this leads to omega
> > + 1 being in bijection with the set of natural numbers.
>
> Not at all. Why do you think so? What you actually *do* when you append
> a 1 to each row is changing each element of the set {1,2,3,...} to its
> successor, so you change it to {2,3,4,...}. You do not add something to
> the set the column is bijected with.

In fact I do not add something to the set the column is bijected witd,
but I do neither add something to the column. The initial segments on
the "left hand side" change from
1, 11, 111, ..., 111... to
11, 111, ..., 111..., 111...1.
If they rermain in bijection with the sequences in the rows, then the
set {2, 3, 4, ...} has ordinal omega + 1.

We have a bijection between natural numbers and ordinal numbers of
sets of natural numbers.

>
> > > > The set of second indexes of M' (which is containing a completed
> > > > infinity) cannot be the same as the set of second indexes of M (which
> > > > is not containing a completed infinity).
>
> Note that M' violates your writing above that the rows represent natural
> numbers written in unary notation! So either go one way or the other, do
> not mix them.

M' is mere an example of a possible trijection.

>
> > > Nobody says so, when you properly define what you mean with appending.
> >
> > Cantor has defined how an element is appended to omega. Just this is
> > what I am doing and what you should know - and what forces W.H. to

> > assume the same set of second indices in M and in M'.

>
> But let us see. When defined as indices, your M looks like:
> a_ij is defined for every j in N and every i <= j
> and your M' looks like (I assume here replacing the first line, I disremember
> which it was, but it makes no difference):
> a_ij is defined for every j in N and whenever i = 1 or i <= j
> where is the difference between sets of second indices?
>
> So M and M' have the same set of second indices.

That means there is no difference in the set of numbers represented by
the sequences of 1's in the rows of M and M'. The set of natural
numbers has the same set of second indices as that set with an

> Now we add elements to
> rows. Let us define that we add an element by inserting a new element at
> a second column index position that is larger than all other index positions
> in that same line. With M we can add all elements at finite positions
> and get M_app:
> a_ij is defined for every j in N and every i <= j - 1
> for M' there is difficulty with the first line, and we get:
> a_ij is defined for every j in N and whenever i = 1 or i <= j - 1,
> and for i = 1 and j = w.
> I still do not see a problem.

I must say, I am not really surprised. (The problem is that a set of
finite natural numbers cannot be complete, i.e., its ordinal number
cannot exist as a number which is in trichotomy with the elements of
the set and which is different from omega + 1.)

Regards, WM

### lwa...@lausd.net

Aug 29, 2007, 5:40:19 PM8/29/07
to
On Aug 28, 4:10 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> Two results of set geometry
>
> Set geometry, a new branch of mathematics, is devised to investigate
> relations of finite and infinite sets by means of geometrical
> representations. Up to now there have been two important results.
>
> The first result is that the ordinal number of the set of natural
> numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> omega, then {1, 2, 3, ...} = omega + 1 too.

I believe the confusion here is that by themselves, _sets_ do not
have "ordinal numbers" at all. One has to define an _order_ on
the set, and if the order happens to be a wellorder, then it is
isomorphic to an ordinal. For the set N, if the order is <, then
it is isomorphic to omega, and if the order happens to be <<,
defined as:

m << n iff (1 < m < n or 1 = n < m)

then it is order-isomorphic to omega + 1. Indeed, in ZFC or any
standard set theory, N is isomorphic to any countable ordinal,
depending on the wellorder.

> The second result shows that the set of real numbers is countable.

already pointed out that WM desires to have the properties of
finite trees transfer to infinite trees.

Both Robinson's hyperreals and Conway's surreals have been
mentioned several times in these Cantor threads. In this case
I believe that the surreals are more relevant, because they
are often represented by a binary tree:

http://www.valdostamuseum.org/hamsmith/surreal.html

As you can see in the link above, the binary tree of surreals
has every path, even the infinite ones, ending in a leaf node,
with such surreals as omega, pi, e, sqrt(2), 2/3, iota, -iota,
and -omega being generated after infinitely many steps -- what
Conway calls "birthday omega." Thus every path, whether finite
or infinite, has a leaf node, and thus there cannot be more
paths than nodes in the binary tree of surreals.

Of course, the surreal binary tree still doesn't have every
property that WM wants. In this case, there are at least as
many nodes as standard reals, and so both the set of nodes
and the set of paths are uncountable.

Also, ironically, once one allows surreals or hyperreals, WM's
"first result" no longer holds. When one considers the set of
positive omnific integers (or hyperintegers) rather than N,
several things happen. First of all, < is no longer a well
ordering, because omega - 1, omega - 2, omega - 3, etc. (or
the hyperreal equivalent) is an infinitely descending set of
surreals (respectively hyperreals). But if we consider the
set of positive hyperintegers less than a given infinite
integer W, then < and << (defined above) are order isomorphic
because in {1, 2, 3, ..., W - 3, W - 2, W - 1} maps to
{2, 3, 4, ..., W - 2, W - 1, 1} by mapping each hypernatural
less than W - 1 to its successor and W - 1 to 1. The surreals
are more problematic because they form a proper class and not
a set, but something similar happens here.

### Peter Webb

Aug 29, 2007, 7:33:16 PM8/29/07
to

"WM" <muec...@rz.fh-augsburg.de> wrote in message

I will put aside the issue of what a bijection "in the normal order" is
supposed to mean, by assuming that you mean an order preserving bijection.

What element of N is bijected with w?

<SNIP>

> If this bijection exists, then we can, without destroying it, append
> one 1 to every partner. This leaves every natural number finite (only
> the first line of the bijection disappears) but it introduces as the
> new first column the ordinal omega + 1. Therefore the set {2, 3,
> 4, ...} of all but one natural numbers is in bijection with omega + 1.
>

Yes, a subset of N can be bijected with w, w+1, w+2 or any other countable
ordinal, though not in the manner you describe.

If you want to find a bijection between N and w+1, you have to find a value
for "1" to map to. If you want this to be order preserving, then you are
going to have some difficulty I suspect. If its not order preserving, all
you have done is shown that the cardinality of N is the same as the
cardinality of w+1, which has been known ever since w has been defined.

<SNIP>

>
> It can easily be seen that there cannot be more separate paths than
> nodes, because every path needs a node to separate itself from the
> other paths. Therefore there are not more different real numbers in
> the interval [0, 1] than nodes. But the number of nodes is countable
> with no doubt.

"With no doubt", heh?

Prove it.

If you accept the fact there are a countable number of nodes, you have
accepted the fact that there are a countable number of Reals. You don't need
any arguments concerning paths or edges to "prove" this; you have simply
assumed the result you are trying to prove.

Can you prove there is only a countable number of nodes (Reals) without
first assuming there is a countable number of nodes (Reals)? Somehow I doubt
it, but feel free to try.

### Fuckwit

Aug 29, 2007, 7:50:54 PM8/29/07
to
On Thu, 30 Aug 2007 09:33:16 +1000, "Peter Webb"
<webbf...@DIESPAMDIEoptusnet.com.au> wrote:

>>
>> [...] the number of nodes is countable with no doubt.

>>
> "With no doubt", heh?
>

Well, yes, (since) that's true.

>
> Prove it.
>
Look at the tree:

(level)

(0) o
/ \
(1) o o
/ \ / \
(2) o o o o
: : : :

To every level there corresponds is a countable set of notes, and the
number of levels is countable. Now it can be proved in ZFC that the
union of countable many countable sets is countable. Hence the set of
_all_ notes is countable.

F.

### Peter Webb

Aug 29, 2007, 11:40:42 PM8/29/07
to

"Fuckwit" <nomail@invalid> wrote in message
news:ev0cd3lv638jishfu...@4ax.com...

> On Thu, 30 Aug 2007 09:33:16 +1000, "Peter Webb"
> <webbf...@DIESPAMDIEoptusnet.com.au> wrote:
>
>>>
>>> [...] the number of nodes is countable with no doubt.
>>>
>> "With no doubt", heh?
>>
> Well, yes, (since) that's true.
>
>>
>> Prove it.
>>
> Look at the tree:
>

Whoops, sorry, I stand corrected.

Can I just fall back on the oldest argument in the book - that there is no
node corresponding to 1/3?

### WM

Aug 30, 2007, 1:31:57 PM8/30/07
to
On 29 Aug., 23:40, lwal...@lausd.net wrote:
> On Aug 28, 4:10 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Two results of set geometry
>
> > Set geometry, a new branch of mathematics, is devised to investigate
> > relations of finite and infinite sets by means of geometrical
> > representations. Up to now there have been two important results.
>
> > The first result is that the ordinal number of the set of natural
> > numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> > omega, then {1, 2, 3, ...} = omega + 1 too.
>
> I believe the confusion here is that by themselves, _sets_ do not
> have "ordinal numbers" at all.

There is no confusion. But your further discussion is not to the
point, because the set N which I consider is the set of sequences of
1's in the rows. It is well-ordered, as is easily seen:

1
11
111
...

>
> > The second result shows that the set of real numbers is countable.
>
> already pointed out that WM desires to have the properties of
> finite trees transfer to infinite trees.

We cannot argue other than by finite arguments. We simply don't know
non-finitely defined infinities.

question: Do you believe that there can be more *separated* paths in
the tree than nodes?

Regards, WM

### WM

Aug 30, 2007, 1:35:16 PM8/30/07
to
On 30 Aug., 01:33, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
wrote:

> > 1
> > 11
> > 111
> > ...
>
> > then the N *in the normal order* can also be bijected to omega + 1.
> > This is shown by appending just one 1 to every partner of the
> > bijection.
>
> I will put aside the issue of what a bijection "in the normal order" is
> supposed to mean, by assuming that you mean an order preserving bijection.
>
> What element of N is bijected with w?

I don't know. I was told however, that there should be omega natural
numbers. So every initial segment of omega including omega (the
complete first column) must be in bijection with a natural number.

>
> <SNIP>
>
> > If this bijection exists, then we can, without destroying it, append
> > one 1 to every partner. This leaves every natural number finite (only
> > the first line of the bijection disappears) but it introduces as the
> > new first column the ordinal omega + 1. Therefore the set {2, 3,
> > 4, ...} of all but one natural numbers is in bijection with omega + 1.
>
> Yes, a subset of N can be bijected with w, w+1, w+2 or any other countable
> ordinal, though not in the manner you describe.

What I describe is not an arbitrary choice but is forced by the
assumption of omega natural numbers. As it cannot be, as you confess,
the assumption is wrong.

>
> If you want to find a bijection between N and w+1, you have to find a value
> for "1" to map to. If you want this to be order preserving, then you are
> going to have some difficulty I suspect. If its not order preserving, all
> you have done is shown that the cardinality of N is the same as the
> cardinality of w+1, which has been known ever since w has been defined.
>

The set of sequences in the rows is well ordered. And the mapping is
order preserving.

> <SNIP>
>
>
>
> > It can easily be seen that there cannot be more separate paths than
> > nodes, because every path needs a node to separate itself from the
> > other paths. Therefore there are not more different real numbers in
> > the interval [0, 1] than nodes. But the number of nodes is countable
> > with no doubt.
>
> "With no doubt", heh?
>
> Prove it.
>
> If you accept the fact there are a countable number of nodes, you have
> accepted the fact that there are a countable number of Reals.

You are right. Every sensible mathematician not yet spoilt by
matheology is able to see that.

> You don't need
> any arguments concerning paths or edges to "prove" this; you have simply
> assumed the result you are trying to prove.

Yes, I am glad to see you understand the clarity of the argument. The
set of nodes and the set of separated paths are irrevocably connected.

Regards, WM

### WM

Aug 30, 2007, 1:36:37 PM8/30/07
to
On 30 Aug., 05:40, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
wrote:

No. The first node from the left hand side in the second row is mapped
on 1/3. You can be sure that I can map a node on every real you can
ask for. But in order to shorten the discussion: It is impossible to
define or construct a bijection between the set of constructible reals
and N. Nevertheless the set of constructible real numbers is
countable. Hence, why should I be able to construct or define a
bijection between the countable set of paths and N?

Regards, WM

### WM

Aug 30, 2007, 1:57:24 PM8/30/07
to
On 29 Aug., 23:40, lwal...@lausd.net wrote:
> On Aug 28, 4:10 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Two results of set geometry
>
> > Set geometry, a new branch of mathematics, is devised to investigate
> > relations of finite and infinite sets by means of geometrical
> > representations. Up to now there have been two important results.
>
> > The first result is that the ordinal number of the set of natural
> > numbers is not unique. It has been shown that if {1, 2, 3, ...} =
> > omega, then {1, 2, 3, ...} = omega + 1 too.
>
> I believe the confusion here is that by themselves, _sets_ do not
> have "ordinal numbers" at all.

There is no confusion. But your further discussion is not to the

point, because the set N which I consider is the set of sequences of
1's in the rows. It is well-ordered, as is easily seen:

1
11
111
...

>

> > The second result shows that the set of real numbers is countable.
>
> already pointed out that WM desires to have the properties of
> finite trees transfer to infinite trees.

We cannot argue other than by finite arguments. We simply don't know

### WM

Aug 30, 2007, 2:04:18 PM8/30/07
to
On 30 Aug., 01:33, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
>
> > 1
> > 11
> > 111
> > ...
>
> > then the N *in the normal order* can also be bijected to omega + 1.
> > This is shown by appending just one 1 to every partner of the
> > bijection.
>
> I will put aside the issue of what a bijection "in the normal order" is
> supposed to mean, by assuming that you mean an order preserving bijection.
>
> What element of N is bijected with w?

I don't know. I was told however, that there should be omega natural

numbers. So every initial segment of omega including omega (the
complete first column) must be in bijection with a natural number.

>

> <SNIP>
>
> > If this bijection exists, then we can, without destroying it, append
> > one 1 to every partner. This leaves every natural number finite (only
> > the first line of the bijection disappears) but it introduces as the
> > new first column the ordinal omega + 1. Therefore the set {2, 3,
> > 4, ...} of all but one natural numbers is in bijection with omega + 1.
>
> Yes, a subset of N can be bijected with w, w+1, w+2 or any other countable
> ordinal, though not in the manner you describe.

What I describe is not an arbitrary choice but is forced by the

assumption of omega natural numbers. As it cannot be, as you confess,
the assumption is wrong.
>

> If you want to find a bijection between N and w+1, you have to find a value
> for "1" to map to. If you want this to be order preserving, then you are
> going to have some difficulty I suspect.

The set of sequences in the rows is well ordered. And the mapping is
order preserving.

> If its not order preserving, all

> you have done is shown that the cardinality of N is the same as the
> cardinality of w+1, which has been known ever since w has been defined.

No. My proof concerns ordinal numbers.

>
> <SNIP>
>
>
>
> > It can easily be seen that there cannot be more separate paths than
> > nodes, because every path needs a node to separate itself from the
> > other paths. Therefore there are not more different real numbers in
> > the interval [0, 1] than nodes. But the number of nodes is countable
> > with no doubt.
>
> "With no doubt", heh?
>
> Prove it.
>
> If you accept the fact there are a countable number of nodes, you have
> accepted the fact that there are a countable number of Reals.

You are right. Every sensible mathematician not yet spoilt by

matheology is able to see that.

> You don't need

> any arguments concerning paths or edges to "prove" this; you have simply
> assumed the result you are trying to prove.

Yes, I am glad to see you understand the clarity of the argument. The

### WM

Aug 30, 2007, 2:05:28 PM8/30/07
to
On 30 Aug., 05:40, "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au>
wrote:

No. The first node from the left hand side in the second row is mapped
on 1/3.n You can be sure that I can map a node on every real you can

Aug 30, 2007, 2:44:58 PM8/30/07
to
WM wrote:

Mückenheim, you should first learn the simple things before you try to be
intelligent. For example, that two is one more than one, and sometimes one
too much. In other words, it suffices if you post your garbage once.

### tommy1729

Aug 30, 2007, 2:43:30 PM8/30/07
to
where is neilist now ??

and david C. Ullrich ??

and denis feldman ??

it seems they accepted this set geometry (snicker)

yeah sure this is better than my set theory or cantor

lol

why not be rude now neilist ??

i guess you accept this set geometry :-)

and why cant "professor" (snicker) david see this set geometry is bogus ?

i guess the perfect spelling maybe (lol)

and where is denis feldmann ?

math bruder ?

666 ?

and all the other big critics ??

nowhere , they accepted this

yeah really this " set geometry" is the best ever.

way better than any post i ever made (snicker)

the OP does not even accepts the reals as an uncountable set (lol)

yet doesnt seem to bother neilist , or david c ullrich or denis feldmann ... ( the others in that category )

btw just as i predicted denis feldmann left

the pattern described in my post of critics towards me sure is a rule !

to be honest the only decent geometry post done recently is by galathaea

maybe the OP wants to compare notes with denis feldmann , neilist , the "prof" and JSH.

tommy1729

ps : one thing i do have to admit : cantor's math was better than this :-)

### G. Frege

Aug 30, 2007, 4:20:24 PM8/30/07
to
On Thu, 30 Aug 2007 13:40:42 +1000, "Peter Webb"
<webbf...@DIESPAMDIEoptusnet.com.au> wrote:

>>>>
>>>> [...] the number of nodes is countable with no doubt.
>>>>
>>> "With no doubt", heh?
>>>
>> Well, yes, (since) that's true.
>>

>> [...]

>>
> Whoops, sorry, I stand corrected.
>

Ok.

>
> Can I just fall back on the oldest argument in the book
> - that there is no node corresponding to 1/3?
>

No. Since we are interested in _paths_ not nodes. And of course there
is a path which corresponds to the real number 1/3. Actually, there is
a "natural" correspondence between the set of paths and the set of all
real numbers in [0,1].*) This means that the number of paths is
uncountable (since the number of real numbers is uncountable).

F.

_____________________

*) Note that some real numbers then correspond to _two_ paths.

--

E-mail: info<at>simple-line<dot>de

### Dik T. Winter

Aug 30, 2007, 9:24:22 PM8/30/07
to
Please do not respond to this two or even three times. Once is enough.

In article <1188417606.4...@r34g2000hsd.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
> On 29 Aug., 02:27, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > In article <1188299411.985907.123...@d55g2000hsg.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > 1000...
> > > 11000...
> > > 111000...
> > > ...
> > >
> > > If there is a trijection,
> >
> > (You actually mean mutiple bijections between sets of indices, but I will
>
> I gave an exact definition of trijection in the thread and explicated
> it in detail to MoeBlee.

Yes. What I understand is that there is a trijection between sets A, B and
C so that we can form triples (a_i, b_i, c_i) over some index range. That
is nothing more, nor less, than bijections between A and B, A and C and
B and C.

But my major point was that your "trijection" was *not* between elements
but between sets of indices.

> > > then it should remain a trijection after
> > > extending the sets involved by one element each. Extending the first
> > > column of M by one element we obtain a sequence of order type omega +
> > > 1 (because it has the order type omega). After adding one 1 to each
> > > sequence of 1's in the lines of M no sequence of 1's has order type
> > > omega + 1 (because each one has a finite order type).
> >
> > You are doing different additions to the first column and the first line.
> > You add an element at the end of the first column, but in the middle of
> > the first line. So why do you expect a bijection between the first line
> > and the first column? (Note that "trijection" is just obfuscation, there
> > are simply multiple bijections.)
>
> In fact the trijection can be understood in this way. If you are
> unable to comprehend the principle of trijection, we can set up two
> bijections. There is one bijection between the initial segments
> (including the complete sequences) of the first column and the
> diagonal:

Yes, I understand that one. And you *know* I do, so why you elaborate
escapes me. The initial segments are the rows. But you have not a
bijection between the initial segments and the *elements* of the first
column, but between the initial segments and the *indices* of the elements
of the first column.

> 1 <--> 1

Trivial, the identity mapping.

> 1 1
> 1 <--> 1

Trivial, again the identity mapping.

> 1 1
> 1 1
> 1 <--> 1

Trivial, again the identity mapping.

> ...

etc.

> There is another bijection between the first column and the lines (if
> there are infinitely many finite natural numbers)

Is it?

> 1 <--> 1
>
> 1
> 1 <--> 11

This violates the first mapping. Mappings are from sets to sets. So
does '1' map to '1' or to '11'? It can't be both.

> Third, there is the bijection between rows and initial segments of the
> diagonal. You will be able to illustrate it yourself.

With your above aid, nothing can be illustrated because the illustrations
above are invalid to show mappings. When do you start to properly do
mathematics rather than these haphazard ramblings? If I understand what
you are saying (but I do fail on occasion because you are not clear at
all), the matrix M is defined as follows:
m_ij is defined (and 1) when i in N and j <= i.
The first column is the sequence (so not a set):
(m_11, m_21, m_31, ...)
The n-th initial segment of the diagonal is:
(m_11, m_22, m_33, ..., m_nn).
There is a clear bijection between the set of indices of the elements
of the first column and the set of initial segments of the diagonal.
What is the problem?

> > > This can be
> > > repeated infinitely often, such that the order type of the first
> > > column becomes omega*2 (notation from Cantor, 1895) while the order
> > > type of the set of lines cannot surpass omega.
> >
> > They *can* surpass omega, if you add elements at the end, which you do
> > not. This is most properly seen when you use indices. At the first
> > line you add a new element at position (1,2), shifting *all* 0's up
> > one place. When you add a new element to the first column, you do *not*
> > do it at position (2,1), shifting *all* 1's up one place, but you do it
> > at a new position (w,1).
>
> Wrong description. The second bijection after appending one element

And what you write is *again* not a bijection between sets.

> 1
> 1 <--> 11
>
> 1
> 1
> 1 <--> 111

Violates the first equivalence. Is '1' mapped to '11' or to '111'?
You contiuously ignore that the set {1, 1, 1, 1, 1, ...} is the same
as the set {1} and so has only one element.

> Of course appending one element to omega yields omega + 1.

Now you are talking about ordered sets. And actually ordered multi-sets.
Nevertheless when I wrote that you were talking about order-preserving
bijections between ordered multi-sets you said I was wrong.

> > > For the diagonal the
> > > case remains undecided. But whatever the diagonal may be, there is no
> > > trijection between the sequences of 1's in the first column, the
> > > diagonal, and the lines - not finally and, therefore, also not
> > > initially.
> >
> > Bogus. When you add an element after the first column, what is the
> > diagonal? But we *do* know that there does exist a bijection between
> > a set of order w and a set of order w+1. Let's get the new matrix with
> > a new line added at position (w,1). I get the following bijections,
> > from first column (c_i1) to first line (l_1i) and to diagonal (d_ii):
> > if i = w: (c_w1) <-> (l_11) <-> (d_11)
> > if i != w: (c_i1) <-> (l_[i+1]1) <-> (d_[i+1][i+1]).
>
> The clue however is, that in normal order N is omega + 1.

Was it wrong what I wrote? If so, where. And for your statement,

> > > Each number is given by a path
> > > stretching over infinitely many nodes (bits). All nodes (bits) of the
> > > tree are countable. The paths are not, according to Cantor's famous
> > > diagonal proof.
> > >
> > > But we find that, up to line number n, there are -1 + 2^(n+1) nodes
> > > whereas 2^n different paths arrive at and 2^(n+1) different paths
> > > spring off from line number n.
> >
> > How you can state that couple of sentences with a straight face escapes
> > me. As *all* paths are infinite in length, even at the root node your
> > statements is wrong, because *all* paths spring off from the root node
> > and not only two different paths. And more general, at each node arrive
> > *all* the paths that do emanate from that node. So you should explain
> > how it is possible that at a single node arrives a single path while
> > two do emanate from it (all paths start at the root). In the second
> > paragraph you are not talking about paths but about edges.
>
> Only the number of separated paths is interesting, i.e., the number of
> different real numbers. In the binary tree there are all real numbers
> of the interval [0, 1] represented by paths from 0.000... to
> 0.111... .

Yes, I know that picture well enough. You have posted it so many times
that the novelty is wearing off.

> It can easily be seen that there cannot be more *separate* paths than
> nodes, because every path needs a node to separate itself from the
> other paths.

You state that only the number of separated paths is interesting. As I
state at each node, infinitely many paths come in and infinitely many
paths come out. You stated that at each node one more path comes out
than did come in. I am still wondering how you came at that amazing
discovery. But you are again back to your quantifier dislexia:
for each pair of paths there is a node where the two paths separate
does *not* mean:
for each path there is a node where it is separated from all other
paths.
You have to *prove* the second from the first. The first is true, the
second is not.

### Dik T. Winter

Aug 30, 2007, 10:07:26 PM8/30/07
to
...

> > > I mean simply: append an element as Cantor does define it. There is no
> > > index to be considered. You can see this if you append 1 to the set
> > > {2,3,4,...} to get {2,3,4,...,1}.
> >
> > There *are* indices to be considered. All your elements are '1'!
>
> No, the elements are sequences of 1's.

Oh. So the elements of the first column are sequences of '1's? If not,
how can you talk about a bijection between the elements of the first
column and anything else?

> > > > > 1 1 1 1 ... and
> > > > > 1,2,3,4,...
> > > >
> > > > But this is again thoroughly unmathematical. Can you show a bijection
> > > > between the two sets? From {1, 1, 1, ...} to {1, 2, 3, 4, ...}, what
> > > > is f(1)? You are talking about order preserving bijections between
> > > > ordered multi-sets. Argh.
> > >
> > > Look at the matrix. It should be obvious:
> > >
> > > 1
> > > 11
> > > 111
> > > 1111
> > > ...
> >
> > What is f(1)? Darn, you *say* you are talking about bijections between
> > elements when actually you are talking about bijections between index
> > positions (and so, yes, index positions *are* to be considered).
>
> See my last contribution, where I illustrated in detail the trijection
> which I had precisely defined earlier.

It failed to clarify what f(1) is. And illustrations I do not need, I need
*definitions*.

> > > > That is an order preserving bijection between multi-sets.
> > >
> > > You misunderstand. The bijection is between n and {1,2,3,...,n} but
> > > the n's are written in unary representation by sets of 1's in the rows
> > > and the initial segments are those of the first column.
> >
> > Now, this is again utterly confused. With this explanation I see only
> > a single set: the set of natural numbers written out in unary. But if
> > you consider it that way, there is only a single column, namely the
> > list of natural numbers.
>
> You see only one column, but don't know what has to be understood by
> the expression "initial segments of the first column"?

No. I would understand under it the various sets of the first k elements
of the first column.

> > > If the set of natural numbers has omega elements, then the set of rows

> > > is in bijection with the complete first column.
> >
> > Yes, of course, because the set of rows is the set of natural numbers
> > (because each element is a natural number). And it is in bijection with
> > the first column, which is the set of rows. Quite trivial I would think.
>
> Only the fact that the first column has a lot of finite initial
> segments and an infinite initial segment while the rows are all finite
> may be raising some suspicion.

What this is stating, precisely is:
That the ordered set of natural numbers has a lot of finite initial
segments and one infinite initial segment, while the set of natural
numbers is infinite may be raising some suspicion.
Yes, I knew all along that you distrust the axiom of infinity. Well,
reject it, and go on. But do not bother to argue that that axiom is
inconsistent using methods that *negate* that axiom.

> > > If a bijection exists,
> > > then appending (in the manner of Cantor) one element to each partner
> > > of the bijection, then the bijection persists.
> >
> > Obvious. As each partner in the bijection is the set of natural numbers.
>
> That is incorrect. Each partner on the one side is a number of natural
> numbers and on the other side each partner is a natural number. I
> tried to express that by the variables {1,2,3,...,n} <--> n.

Again, you lost me here. The partners where both the set of natural
numbers.

> > > But this leads to omega

> > > + 1 being in bijection with the set of natural numbers.
> >
> > Not at all. Why do you think so? What you actually *do* when you append
> > a 1 to each row is changing each element of the set {1,2,3,...} to its
> > successor, so you change it to {2,3,4,...}. You do not add something to
> > the set the column is bijected with.
>
> In fact I do not add something to the set the column is bijected witd,
> but I do neither add something to the column. The initial segments on
> the "left hand side" change from
> 1, 11, 111, ..., 111... to
> 11, 111, ..., 111..., 111...1.

There is no last row, so where does the 111...1 come from?

> If they rermain in bijection with the sequences in the rows, then the
> set {2, 3, 4, ...} has ordinal omega + 1.

It is based on your assumption that there is a last row. There is none,
so now what?

> > > You said it already yourself: Starting with (1) M, the rows of M
> > > remain (except the first line), while in (2) M we get omega + 1
> > > columns. Your points (3) an d (4) are uninteresting.
> >
> > Still I maintain, why is that an inconsistency? Note that here we are
>
> We have a bijection between natural numbers and ordinal numbers of
> sets of natural numbers.

Oh. I did not know that. As far as I know, in this part we were talking
about matrices and of bijections between lines and columns. You switch
point of view from one to another without consistency.

> > > > > The set of second indexes of M' (which is containing a completed
> > > > > infinity) cannot be the same as the set of second indexes of M (which
> > > > > is not containing a completed infinity).
> >
> > Note that M' violates your writing above that the rows represent natural
> > numbers written in unary notation! So either go one way or the other, do
> > not mix them.
>
> M' is mere an example of a possible trijection.

But it violates your view that the rows represent natural numbers. So
when you go to M' you should lose the view that each row represents a
natural number.

> > But let us see. When defined as indices, your M looks like:
> > a_ij is defined for every j in N and every i <= j
> > and your M' looks like (I assume here replacing the first line, I
> > disremember which it was, but it makes no difference):
> > a_ij is defined for every j in N and whenever i = 1 or i <= j
> > where is the difference between sets of second indices?
> >
> > So M and M' have the same set of second indices.
>
> That means there is no difference in the set of numbers represented by
> the sequences of 1's in the rows of M and M'.

That means that there is no difference in the *size* of those sets of
numbers.

> The set of natural
> numbers has the same set of second indices as that set with an

I said *replaced*. But let's see whether I do understand you:
a natural number n has the set of second indices {1, 2, 3, ..., n}
a set of natural numbers {n1, ..., nk} has the set of second indices
U{l = 1, ..., k} {1, 2, 3, ..., l} = {1, 2, 3, ..., k}. The set of
natural numbers has as second indices {1, 2, 3, ...} = N. If we add
that "infinite number" 111..., we have that that number has the set
of indices {1, 2, 3, ...} = N. So, yes, the sets of second indices
is the same. I do not see why that should be a problem.

> I must say, I am not really surprised. (The problem is that a set of
> finite natural numbers cannot be complete, i.e., its ordinal number
> cannot exist as a number which is in trichotomy with the elements of
> the set and which is different from omega + 1.)

Again the negation of the axiom of infinity. There is no problem once
you are able to distinguish between natural numbers and ordinal numbers.
Properly speaking, the first ordinal number is 0, not 1. So each
ordinal number is the order number of the ordered set of all preceding
ordinal numbers. And once you see that, there is no problem at all.

### Dik T. Winter

Aug 30, 2007, 10:18:23 PM8/30/07
to
In article <1188495397.9...@50g2000hsm.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
...

> But in order to shorten the discussion: It is impossible to
> define or construct a bijection between the set of constructible reals
> and N.

Again you are guilty of abuse of terminology. Using common mathematical
terminology, it is possible to construct a bijection between N and the
constructible numbers. I even gave the construction in an article in
response to your garbage. But you mean "finitely defined". State so
when you mean that. And be aware that that notion holds a lot of
problems.

> Nevertheless the set of constructible real numbers is
> countable.

Yup, as is the set of finitely defined numbers (when a proper mathematical
definition has been given).

> Hence, why should I be able to construct or define a
> bijection between the countable set of paths and N?

It is sufficient when you show an *injection* between the paths and N.
The other way is clear, and there is a theorem that if there exists
an injection from A to B and an injection from B to A, there is also
a bijection. So when you state that an bijection exists you have to
prove it. In this case by proving an injection from the set of paths
to N. Otherwise your claim is unproven. (Note that in the cases above,
"constructable numbers" and "finitely defined numbers" an injection from
that set to N can be defined.) And do not come up (as you did a previous
time) with an injection from N to the set of paths. That one is known.
The other way around is needed.

### neilist

Aug 31, 2007, 12:19:06 PM8/31/07
to

Ah, wouldn't some like silence, perhaps from yours truly? :-)

And to not hear Tommy's sloppy drivel. And for Tommy to actually
believe his ridiculous proposition that silence=acceptance of his
pompous declarations of greatness ala James Harris.

In the realm of law, silence may be imputed to acceptance or
acquiescence. See "A Man For All Seasons" with Paul Scofield. An
excellent movie.

But in the realm of ordinary people, such as this newsgroup, silence
can be a sign of NON-acceptance of either the posting OR the POSTER.

Now, if someone else stated some of these propositions, then maybe I'd
pay attention, or give it a bid of credibility.

But from YOU?

You don't even know the definition of "integral" as used in normal and
you're WRONG, and can't accept the fact that YOU'RE PWNED!!!

And Tommy, if you can't source your WRONG definition, don't even
attempt to respond. It'll just put you in deeper and deeper Pwn-dom-
ness, Tommy.

("pwn-dom-ness"? I should submit that to the Urban Dictionary)

So go join James Harris and Bassam Ass-King with your grandiose
"theories" and "mathematics", you Pwn-boy Tommy-typo.

And have a great weekend everyone!!

### David R Tribble

Aug 31, 2007, 4:11:36 PM8/31/07
to
W Mueckenheim wrote:
>> It can easily be seen that there cannot be more separate paths than
>> nodes, because every path needs a node to separate itself from the
>> other paths. Therefore there are not more different real numbers in
>> the interval [0, 1] than nodes. But the number of nodes is countable
>> with no doubt.
>

Peter Webb wrote:
> "With no doubt", heh?
>
> Prove it.
>
> If you accept the fact there are a countable number of nodes, you have
> accepted the fact that there are a countable number of Reals. You don't need
> any arguments concerning paths or edges to "prove" this; you have simply
> assumed the result you are trying to prove.
>
> Can you prove there is only a countable number of nodes (Reals) without
> first assuming there is a countable number of nodes (Reals)? Somehow I doubt
> it, but feel free to try.

I'm more interested in WM showing us how the reals are
well-ordered, since he thinks they are countable. I'd like
to see how he bijects N to [0,1], which immediately
makes the reals well-ordered in that interval. In particular,
I'd like to know which reals get mapped to 1 and 2, or
more generally, to k and k+1.

### WM

Aug 31, 2007, 5:04:14 PM8/31/07
to
On 31 Aug., 03:24, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> Please do not respond to this two or even three times. Once is enough.
>
> In article <1188417606.434017.149...@r34g2000hsd.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > On 29 Aug., 02:27, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > In article <1188299411.985907.123...@d55g2000hsg.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > > > 1000...
> > > > 11000...
> > > > 111000...
> > > > ...
> > > >
> > > > If there is a trijection,
> > >
> > > (You actually mean mutiple bijections between sets of indices, but I will
> >
> > I gave an exact definition of trijection in the thread and explicated
> > it in detail to MoeBlee.
>
> Yes. What I understand is that there is a trijection between sets A, B and
> C so that we can form triples (a_i, b_i, c_i) over some index range. That
> is nothing more, nor less, than bijections between A and B, A and C and
> B and C.
>
> But my major point was that your "trijection" was *not* between elements
> but between sets of indices.

And why should that be a problem? These sets of indices are the
elements of the bijection.

>
> > > > then it should remain a trijection after
> > > > extending the sets involved by one element each. Extending the first
> > > > column of M by one element we obtain a sequence of order type omega +
> > > > 1 (because it has the order type omega). After adding one 1 to each
> > > > sequence of 1's in the lines of M no sequence of 1's has order type
> > > > omega + 1 (because each one has a finite order type).
> > >
> > > You are doing different additions to the first column and the first line.
> > > You add an element at the end of the first column, but in the middle of
> > > the first line. So why do you expect a bijection between the first line
> > > and the first column? (Note that "trijection" is just obfuscation, there
> > > are simply multiple bijections.)
> >
> > In fact the trijection can be understood in this way. If you are
> > unable to comprehend the principle of trijection, we can set up two
> > bijections. There is one bijection between the initial segments
> > (including the complete sequences) of the first column and the
> > diagonal:
>
> Yes, I understand that one. And you *know* I do, so why you elaborate
> escapes me. The initial segments are the rows. But you have not a
> bijection between the initial segments and the *elements* of the first
> column, but between the initial segments and the *indices* of the elements
> of the first column.
>
> > 1 <--> 1
>
> Trivial, the identity mapping.

No! The mapping between initial segments of column and diagonal.

>
> > 1 1
> > 1 <--> 1
>
> Trivial, again the identity mapping.

No! The mapping between initial segments of column and diagonal.

>
> > 1 1
> > 1 1
> > 1 <--> 1
>
> Trivial, again the identity mapping.

No! The mapping between initial segments of column and diagonal.

>
> > ...
>
> etc.
>
> > There is another bijection between the first column and the lines (if
> > there are infinitely many finite natural numbers)
>
> Is it?
>
> > 1 <--> 1
> >
> > 1
> > 1 <--> 11
>
> This violates the first mapping. Mappings are from sets to sets. So
> does '1' map to '1' or to '11'? It can't be both.

Between column and diagonal you understood and said "trivial". Here we
have the same. A mapping between sets.

1
1

maps on

11

etc.

> > And what you write is *again* not a bijection between sets.
>
> > 1
> > 1 <--> 11
> >
> > 1
> > 1
> > 1 <--> 111
>
> Violates the first equivalence. Is '1' mapped to '11' or to '111'?
> You contiuously ignore that the set {1, 1, 1, 1, 1, ...} is the same
> as the set {1} and so has only one element.

Why then did you understand

> 1 1
> 1 1
> 1 <--> 1

as "trivial, again the identity mapping"?

But you did not yet understand what Peter Webb for instance understoof
immediately.

>
> > It can easily be seen that there cannot be more *separate* paths than
> > nodes, because every path needs a node to separate itself from the
> > other paths.
>
> You state that only the number of separated paths is interesting.

Yes. Paths which never separate are of no interest.

> As I
> state at each node, infinitely many paths come in and infinitely many
> paths come out. You stated that at each node one more path comes out
> than did come in.

One more separated path bunch to be as clear as possible.

> I am still wondering how you came at that amazing
> discovery. But you are again back to your quantifier dislexia:

actually completely existing sets.

> for each pair of paths there is a node where the two paths separate

That does not prevent the number of nodes to be a limit for the number
of separations. There are not more separated paths than separations (=
nodes).

> does *not* mean:
> for each path there is a node where it is separated from all other
> paths.
> You have to *prove* the second from the first. The first is true, the
> second is not.

The proof is very simple. Most mathematicians will understand it. You
are not able to do so. That is because you are spoilt once and
forever.

Ich glaube mein möglichstes gethan zu haben, um Sie von Ihren
traurigen Irrthümern abzubringen. Es ist mir nicht gelungen und ich
kann nur wünschen, daß spätere reiflichere Ueberlegung Sie zur
Rückkehr in vernunftmäßige Bahnen bewegen wird. Denn ich zweifle nicht
daran, daß bei Ihnen die Liebe zur Wahrheit grösser sein wird, als der
unglückliche Hang zur Rechthaberei und zum Streite, dem sich unsere
Zeit so gern und allgemein hingiebt. (G.C.)

Regards, WM

### WM

Aug 31, 2007, 5:11:35 PM8/31/07
to
On 31 Aug., 22:11, David R Tribble <da...@tribble.com> wrote:
> W Mueckenheim wrote:
> >> It can easily be seen that there cannot be more separate paths than
> >> nodes, because every path needs a node to separate itself from the
> >> other paths. Therefore there are not more different real numbers in
> >> the interval [0, 1] than nodes. But the number of nodes is countable
> >> with no doubt.
>
> Peter Webb wrote:
> > "With no doubt", heh?
>
> > Prove it.
>
> > If you accept the fact there are a countable number of nodes, you have
> > accepted the fact that there are a countable number of Reals. You don't need
> > any arguments concerning paths or edges to "prove" this; you have simply
> > assumed the result you are trying to prove.
>
> > Can you prove there is only a countable number of nodes (Reals) without
> > first assuming there is a countable number of nodes (Reals)? Somehow I doubt
> > it, but feel free to try.
>
> I'm more interested in WM showing us how the reals are
> well-ordered, since he thinks they are countable.

My binary tree shows that they are not uncountable. There is no room
for a diagonal argument.
My matrix shows that there is no actual infinity at all, so there are
neither "all naturals" nor "all reals". Hence "all reals" cannot be
well-ordered

> I'd like
> to see how he bijects N to [0,1], which immediately
> makes the reals well-ordered in that interval. In particular,
> I'd like to know which reals get mapped to 1 and 2, or
> more generally, to k and k+1

Ask for any real you can ask and map it on the set 1,2,3,.... You will
not run out of naturals.

Regards, WM

### WM

Sep 1, 2007, 11:10:14 AM9/1/07
to
On 31 Aug., 04:18, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> In article <1188495397.998255.260...@50g2000hsm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> ...
> > But in order to shorten the discussion: It is impossible to
> > define or construct a bijection between the set of constructible reals
> > and N.
>
> Again you are guilty of abuse of terminology. Using common mathematical
> terminology, it is possible to construct a bijection between N and the
> constructible numbers. I even gave the construction in an article in

I did not see it. (By the way, do I have the point?) But if you can,
then define it and take the diagonal number which then is defined too.

> But you mean "finitely defined". State so
> when you mean that. And be aware that that notion holds a lot of
> problems.
>

Perhaps there are problems in matheology but not in mathematics. Any
number which can be defined, i.e., which can be addressed as an
individuum, is a finitely definable number. (Becausen non-finite
definitions are not definitions.)

> It is sufficient when you show an *injection* between the paths and N.

Every separation requires a node. There cannot be more separated paths
than nodes. The bijection between nodes and natural numbers is
obvious. So what do you not understand?

Regards, WM

### WM

Sep 1, 2007, 11:24:01 AM9/1/07
to
On 31 Aug., 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

> > Only the fact that the first column has a lot of finite initial
> > segments and an infinite initial segment while the rows are all finite
> > may be raising some suspicion.
>
> What this is stating, precisely is:
> That the ordered set of natural numbers has a lot of finite initial
> segments and one infinite initial segment, while the set of natural
> numbers is infinite may be raising some suspicion.
> Yes, I knew all along that you distrust the axiom of infinity. Well,
> reject it, and go on.

I do not reject the axiom of infinity. I assume that every initial
segment of the first column (including the complete first column)
corresponds to (is in bijection with) a natural number.

>
> > > > If a bijection exists,
> > > > then appending (in the manner of Cantor) one element to each partner
> > > > of the bijection, then the bijection persists.
> > >
> > > Obvious. As each partner in the bijection is the set of natural numbers.
> >
> > That is incorrect. Each partner on the one side is a number of natural
> > numbers and on the other side each partner is a natural number. I
> > tried to express that by the variables {1,2,3,...,n} <--> n.
>
> Again, you lost me here. The partners where both the set of natural
> numbers.

The partners include {1,2,3,...}. That is not a natural number.

>
> > > > But this leads to omega
> > > > + 1 being in bijection with the set of natural numbers.
> > >
> > > Not at all. Why do you think so? What you actually *do* when you append
> > > a 1 to each row is changing each element of the set {1,2,3,...} to its
> > > successor, so you change it to {2,3,4,...}. You do not add something to
> > > the set the column is bijected with.
> >
> > In fact I do not add something to the set the column is bijected witd,
> > but I do neither add something to the column. The initial segments on
> > the "left hand side" change from
> > 1, 11, 111, ..., 111... to
> > 11, 111, ..., 111..., 111...1.
>
> There is no last row, so where does the 111...1 come from?

There are omega initial segments of the first *column*. Adding one 1
to every initial segment supplies 11, 111, ..., 111..., 111...1, the
last one having ordinal omega + 1.

>
> > If they rermain in bijection with the sequences in the rows, then the
> > set {2, 3, 4, ...} has ordinal omega + 1.
>
> It is based on your assumption that there is a last row. There is none,
> so now what?

No. There is no last row. But every initial segment of the first
column is said to have a partner in the set of sequences in the rows.

>
> > > > You said it already yourself: Starting with (1) M, the rows of M
> > > > remain (except the first line), while in (2) M we get omega + 1
> > > > columns. Your points (3) an d (4) are uninteresting.
> > >
> > > Still I maintain, why is that an inconsistency? Note that here we are
> > > talking about indices!
> >
> > We have a bijection between natural numbers and ordinal numbers of
> > sets of natural numbers.
>
> Oh. I did not know that. As far as I know, in this part we were talking
> about matrices and of bijections between lines and columns. You switch
> point of view from one to another without consistency.

The natural numbers are represented by the sequences in the rows. The
ordinal numbers are represented by the initial segments of the first
column.

>
> > > > > > The set of second indexes of M' (which is containing a completed
> > > > > > infinity) cannot be the same as the set of second indexes of M (which
> > > > > > is not containing a completed infinity).
> > >
> > > Note that M' violates your writing above that the rows represent natural
> > > numbers written in unary notation! So either go one way or the other, do
> > > not mix them.
> >
> > M' is mere an example of a possible trijection.
>
> But it violates your view that the rows represent natural numbers. So
> when you go to M' you should lose the view that each row represents a
> natural number.

Of course. I did never state the opposite.

>
> > > But let us see. When defined as indices, your M looks like:
> > > a_ij is defined for every j in N and every i <= j
> > > and your M' looks like (I assume here replacing the first line, I
> > > disremember which it was, but it makes no difference):
> > > a_ij is defined for every j in N and whenever i = 1 or i <= j
> > > where is the difference between sets of second indices?
> > >
> > > So M and M' have the same set of second indices.
> >
> > That means there is no difference in the set of numbers represented by
> > the sequences of 1's in the rows of M and M'.
>
> That means that there is no difference in the *size* of those sets of
> numbers.
>
> > The set of natural
> > numbers has the same set of second indices as that set with an

>
> I said *replaced*. But let's see whether I do understand you:
> a natural number n has the set of second indices {1, 2, 3, ..., n}
> a set of natural numbers {n1, ..., nk} has the set of second indices
> U{l = 1, ..., k} {1, 2, 3, ..., l} = {1, 2, 3, ..., k}. The set of
> natural numbers has as second indices {1, 2, 3, ...} = N. If we add
> that "infinite number" 111..., we have that that number has the set
> of indices {1, 2, 3, ...} = N. So, yes, the sets of second indices
> is the same.

Initially you were of opposite opinion.

> I do not see why that should be a problem.

If you add a sequence which is larger than any finite sequence, then
you do not add any index? That hints to the fact that you add nothing.
The infinite sequence does not provably existt, because there is no
indication (no piece of circumstantial evidence) that you have done
anything at all.

> > I must say, I am not really surprised. (The problem is that a set of
> > finite natural numbers cannot be complete, i.e., its ordinal number
> > cannot exist as a number which is in trichotomy with the elements of
> > the set and which is different from omega + 1.)
>
> Again the negation of the axiom of infinity. There is no problem once
> you are able to distinguish between natural numbers and ordinal numbers.
> Properly speaking, the first ordinal number is 0, not 1. So each
> ordinal number is the order number of the ordered set of all preceding
> ordinal numbers. And once you see that, there is no problem at all.

Wrong. Repeat twice adding a 1 to all partners of the bijection. Then
you see the problem again.

Regards, WM

### lwa...@lausd.net

Sep 1, 2007, 4:59:02 PM9/1/07
to
I've been trying to figure out the best axioms to reconcile what
WM is trying to post. We can try the same trick I used earlier
with tommy1729, by denying the Axiom of Infinity and allowing
proper classes. We notice that since V_omega is a model for
ZF-Infinity, it follows that V_(omega+1) is a model for
NBG-Infinity, and the infinite sets of V_(omega+1) correspond
to infinite proper classes in NBG-Infinity. And they are all
countable, so it follows that every proper class has the same
countable cardinality.

But what about the reals, and the binary tree? If we were to
identify a path by its set of nodes, then, since all _sets_
(as opposed to proper classes) are finite, it follows that the
proper class of all paths has only finite paths as elements,
since the elements of any class must be _sets_. The class of
all paths is indeed countable, but then that class only counts
the finite paths, which are indeed countable. But then you
don't have the class of reals but only the dyadic rationals
(i.e., the ring formed by appending 1/2 to Z).

> question: Do you believe that there can be more *separated* paths in
> the tree than nodes?

I'm not sure what you mean by a _separated_ path. Depending
on your definition of "separated," there may indeed be only
countably many "separated" paths.

If every path has a leaf node, then there is trivally a
bijection between the class of paths and the class of
nodes, but if some paths lack leaf nodes -- as the infinite
paths in ZFC do -- then it is possible that there can be
more paths than nodes.

The reason I mentioned Conway is that in the binary tree of
surreals, every path -- including the infinite paths -- do
have leaf nodes. But then again, I'm not sure what would
happen if one tried to construct the surreals without
the Axiom of Infinity.

### tommy1729

Sep 1, 2007, 5:53:14 PM9/1/07
to

BOGUS neilist , you just dont offend the OP because accept the post or because the OP is not me ...

there was no silence towards me , so dont pretend to be a wise silent man , you just forgot to insult once , thats not silence !!

and as for the integral look that up decently:

integral x dx = x^2 /2 + C

and every sain man on this forum will accept that integral as written above ...

yet ullrich , you and a few others dont accept and dont know the integral of x.

so the jokes on you actually.

go ahead look it up in topic integral zeta : ullrich computed the integral different !!

but i assume you will snip my defense and simply dribble and insult instead , as usual ?

tommy1729

### Dik T. Winter

Sep 1, 2007, 11:15:09 PM9/1/07
to
In article <1188659414.8...@o80g2000hse.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
> On 31 Aug., 04:18, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > In article <1188495397.998255.260...@50g2000hsm.googlegroups.com> WM <mueck...@rz.fh-augsburg.de> writes:
> > ...
> > > But in order to shorten the discussion: It is impossible to
> > > define or construct a bijection between the set of constructible reals
> > > and N.
> >
> > Again you are guilty of abuse of terminology. Using common mathematical
> > terminology, it is possible to construct a bijection between N and the
> > constructible numbers. I even gave the construction in an article in
> > response to your garbage.
>
> I did not see it. (By the way, do I have the point?) But if you can,
> then define it and take the diagonal number which then is defined too.

No, you did not see it, but commented on it. Using common mathematical
terminology a constructible number is a number that can be constructed
using finitely many operations of a certain class. The diagonal number
can not be constructed with finitely many operations, and so is not
inherently constructible.

As I said: abuse of terminology. You do not mean constructible numbers.

> > But you mean "finitely defined". State so
> > when you mean that. And be aware that that notion holds a lot of
> > problems.
>
> Perhaps there are problems in matheology but not in mathematics. Any
> number which can be defined, i.e., which can be addressed as an
> individuum, is a finitely definable number. (Becausen non-finite
> definitions are not definitions.)

Which makes clear that you do *not* intend constructible numbers.
cbrt(2) is finitely definable but not constructible. So why do you
persist to write "constructible" when you mean "finitely definable"?

> > It is sufficient when you show an *injection* between the paths and N.
>
> Every separation requires a node. There cannot be more separated paths
> than nodes. The bijection between nodes and natural numbers is
> obvious. So what do you not understand?

separated paths than nodes" is in question. And again (just like you,
I am getting repetitive):
(1) For each two pair of paths there is a node that separates them.
(2) There is a level where all paths are separated.
You state (2) which does not follow from (1).

You argue with the number of separated paths at each node, but as at each
not infinitely many paths come in and infinitely many paths go out at each
edge, that is not an argument.

### Dik T. Winter

Sep 1, 2007, 11:53:28 PM9/1/07
to
In article <1188660241.1...@22g2000hsm.googlegroups.com> WM <muec...@rz.fh-augsburg.de> writes:
> On 31 Aug., 04:07, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:
> > > Only the fact that the first column has a lot of finite initial
> > > segments and an infinite initial segment while the rows are all finite
> > > may be raising some suspicion.
> >
> > What this is stating, precisely is:
> > That the ordered set of natural numbers has a lot of finite initial
> > segments and one infinite initial segment, while the set of natural
> > numbers is infinite may be raising some suspicion.
> > Yes, I knew all along that you distrust the axiom of infinity. Well,
> > reject it, and go on.
>
> I do not reject the axiom of infinity. I assume that every initial
> segment of the first column (including the complete first column)
> corresponds to (is in bijection with) a natural number.

Now you are using (again) your very own terminology. There is a bijection
between the initial segments (including the complete first column) with the
natural numbers. There are lots of them. But you are not talking about
bijections, but about oder-preserving bijections between ordered sets. And,
no, there is no order-preserving bijection between the ordered set of initial
segments and the ordered set of natural numbers.

> > > That is incorrect. Each partner on the one side is a number of natural
> > > numbers and on the other side each partner is a natural number. I
> > > tried to express that by the variables {1,2,3,...,n} <--> n.
> >
> > Again, you lost me here. The partners where both the set of natural
> > numbers.
>
> The partners include {1,2,3,...}. That is not a natural number.

I disrember the start. As far as I remember the partners where the set of
natural numbers. {1, 2, 3, ...} is not in the set of {1, 2, 3, ..., n}.

> > > In fact I do not add something to the set the column is bijected witd,
> > > but I do neither add something to the column. The initial segments on
> > > the "left hand side" change from
> > > 1, 11, 111, ..., 111... to
> > > 11, 111, ..., 111..., 111...1.
> >
> > There is no last row, so where does the 111...1 come from?
>
> There are omega initial segments of the first *column*. Adding one 1
> to every initial segment supplies 11, 111, ..., 111..., 111...1, the
> last one having ordinal omega + 1.

Again, adding again a one to the initial segment that is the first column.
Which row has that complete initial segment?

> > > If they rermain in bijection with the sequences in the rows, then the
> > > set {2, 3, 4, ...} has ordinal omega + 1.
> >
> > It is based on your assumption that there is a last row. There is none,
> > so now what?
>
> No. There is no last row. But every initial segment of the first
> column is said to have a partner in the set of sequences in the rows.

Who said that? There is not such partner for the *complete* first column.
That could only be true it there *was* a last row.

> > > We have a bijection between natural numbers and ordinal numbers of
> > > sets of natural numbers.
> >
> > Oh. I did not know that. As far as I know, in this part we were talking
> > about matrices and of bijections between lines and columns. You switch
> > point of view from one to another without consistency.
>
> The natural numbers are represented by the sequences in the rows. The
> ordinal numbers are represented by the initial segments of the first
> column.

O. Which initial segment represents 0? And to which line does it
correspond?

> > I said *replaced*. But let's see whether I do understand you:
> > a natural number n has the set of second indices {1, 2, 3, ..., n}
> > a set of natural numbers {n1, ..., nk} has the set of second indices
> > U{l = 1, ..., k} {1, 2, 3, ..., l} = {1, 2, 3, ..., k}. The set of
> > natural numbers has as second indices {1, 2, 3, ...} = N. If we add
> > that "infinite number" 111..., we have that that number has the set
> > of indices {1, 2, 3, ...} = N. So, yes, the sets of second indices
> > is the same.
>
> Initially you were of opposite opinion.

Perhaps. But as you do not clearly state what you mean it is easy
to be set on a wrong footing.

> > I do not see why that should be a problem.
>
> If you add a sequence which is larger than any finite sequence, then
> you do not add any index?

Why is that a problem?

> That hints to the fact that you add nothing.
> The infinite sequence does not provably existt, because there is no
> indication (no piece of circumstantial evidence) that you have done
> anything at all.

You have done something. Namely adding a line with *all* indices.
But of course you can not prove that an infinite line exists. It is
there due to the axiom of infinity (which you consistently reject,
although you state you do not).

> > Again the negation of the axiom of infinity. There is no problem once
> > you are able to distinguish between natural numbers and ordinal numbers.
> > Properly speaking, the first ordinal number is 0, not 1. So each
> > ordinal number is the order number of the ordered set of all preceding
> > ordinal numbers. And once you see that, there is no problem at all.
>
> Wrong. Repeat twice adding a 1 to all partners of the bijection. Then
> you see the problem again.

Your context is so far away that I do not even understand what you mean.

### neilist

Sep 2, 2007, 12:30:21 AM9/2/07
to
On Sep 1, 5:53 pm, tommy1729 <tommy1...@gmail.com> wrote:

<snip>

> and every sain man

You're insain!

Get out of this threat, you looser!

:-)

### WM

Sep 2, 2007, 3:54:59 PM9/2/07
to
On 1 Sep., 22:59, lwal...@lausd.net wrote:

> I'm not sure what you mean by a _separated_ path. Depending
> on your definition of "separated," there may indeed be only
> countably many "separated" paths.

Try to define what you understand by a real number. Must it differ
from every other real number?

>
> If every path has a leaf node, then there is trivally a
> bijection between the class of paths and the class of
> nodes, but if some paths lack leaf nodes -- as the infinite
> paths in ZFC do -- then it is possible that there can be
> more paths than nodes.

The paths of my tree are not finite and, therefore, have no leaf
nodes. But although they extend in infinity, there cannot be more
separate paths than nodes, because every separation yields only two
separated paths - in infinity.
>
Regards, WM

### WM

Sep 2, 2007, 4:00:37 PM9/2/07
to
On 2 Sep., 05:15, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote:

Do you have a bijection between the decimal representations of the
finitely definable numbers and N?

>
> > > It is sufficient when you show an *injection* between the paths and N.
> >
> > Every separation requires a node. There cannot be more separated paths
> > than nodes. The bijection between nodes and natural numbers is
> > obvious. So what do you not understand?
>
> separated paths than nodes" is in question.

It is not in question, because every node accomplishes exactly one
separation.

> And again (just like you,
> I am getting repetitive):
> (1) For each two pair of paths there is a node that separates them.

I agree. And it separates not more than these two real numbers,
because any other real number is separated by an earlier or a later
node.

> (2) There is a level where all paths are separated.

I do not say so. Only if the existence of all natural numbers is
assumed, then there must obviously be all levels. Or: if the existence
of all real numbers is assumed, then there must be an environment
where this existence takes place.

> You state (2) which does not follow from (1).

No.

>
> You argue with the number of separated paths at each node, but as at each
> not infinitely many paths come in and infinitely many paths go out at each
> edge, that is not an argument.

On the contrary, your saying that infinitely many paths go in and go
out, is not an argument. What we can count are only those real numbers
which differ - that means only those paths which are separated. You
can define these two as you like (in agreement with the nodes
belonging to them until they separate). Every other path is separated
by an earlier or a later node.

Regards, WM