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WM · Guest

On 10 Okt., 20:30, Virgil <Vir...@gmale.com> wrote:
[quote]In article
5f06517f-bb41-4a02-b04d-130f16f4a...@i76g2000hsf.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
In article
871188cc-f053-4ab5-870f-0b57ce13e...@n33g2000pri.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:

o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \ ...

And so on.

Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.

Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.

But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".

Not according to the rule of my game. Consider the figure above. Node
x buys you one of the path leaving the path
0.111... at position x. You may say that there are many paths
deviating from 0.111..., and you are right. But if one of them
deviates from said path, it will get get another node. There is no
escape.

o
|
o
/ \
o \
/ \ \
/ / \
/ \ x
/ / / \
... /
\

With every won path you get infinitely many nodes. Each of them helps
you to get another path, one and only one path per node. In this game
you have the ratio of paths to nodes P/N = 0 at every step. The
infinite binary tree covers every step, but not more.

Similarly the usual decimal expansion of real numbers covers every
digit, but not more. There is no last digit of pi. There is every
digits of pi - unless one looks to close, but that is not in question
here. Set theorists insistuing of uncountably many reals must insist
that there are more than every digit of pi. I am not inclined to
discuss about such folly.

Your "game" overlooks every path with infinitely many branchings in both
left and right directions, which is the vast majority of paths.

My game exhausts the complete infinite tree, i.e., it covers every[/quote]
path that can be formed by an infinitude of digits. In other words,
there is no infinite combination of digits 0 and 1 (or 1,2,3,...,0 in
an extended version) that is off the tree.
[quote]
But if every node of the tree has been used to mark one path

Unless that node is a terminal (leaf) node, it marks more than one path,

Try to understand the rule of the game (there is only one). Try to
understand that there is no leaf node.

It is the absence of any such leaf nodes that makes you game fail to
distinguish each path from all its neighbors.- Zitierten Text ausblenden -
[/quote]
A path that can be distinguished from all its neighbours by means of
digits can be distinguished in the tree as well as in Cantor>s list.

The important point for mathematics is the distinction by digits.

My game shows that the complete infinite binary tree can be exhausted
by a countable number of paths. That implies that the set of all real
numbers that can be distinguished by digits is countable. Not more
than a countable number of infinite digit sequences is possible. That
contradicts the (erroneous) interpretation of Cantor>s proof.

Regards, WM

WM · Guest

On 10 Okt., 20:40, Virgil <Vir...@gmale.com> wrote:
[quote]In article
f0f8ab79-568e-4989-b712-5f5c26080...@u27g2000pro.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:

Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals.

Then your trees are either not infinite or not complete.

The infinite binary tree is as complete as the number system

Then in YOUR world, both are incomplete.
[/quote]
I call the tree complete if every string of digits can be found within
it.
[quote]
, i.e.,

the binary or decimal representation of real numbers can be. There is
no chance to introduce another path. There is no chance to introduce
another real. All combinations of nodes-values, all combinations of
digits are present.

The binary expansions for 1/3 or sqrt(2) are neither represented in your
i complete tree, as they both require infinitely many branchings in both
directions, which you game does not allow.
[/quote]
Please look up the tree. Every infinite path is there.
[quote]

The infinite path 0.111... is covered by the infinite sequence of
paths

0.1000..., 0.11000..., 0.111000..., ... (*)

But then WM is speaking of INFINITE SETS of his "finite" paths, not just
the individual paths themselves, and there are always more such sets of
such paths than there are such paths.
[/quote]
I am speaking of path. Their number is countable.

Here is a finite example:

The path 1111 in a finite 4-level-tree can be covered by the sequence
of paths
1000, 1100, 1110, 1111
or it can be covered by the sequence of paths
1001, 1101, 1111
or it can be covered by simply itself
1111.

Regards, WM

WM · Guest

On 10 Okt., 22:22, george <gree...@email.unc.edu> wrote:
[quote]On Oct 9, 3:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.

You HAVE NOT DONE this!
All you have said is "Take node 1. Give it a path.
Take node 2. Give it a path, it doesn>t matter which.
[/quote]
It must be a path that is not identical with the first one.

[quote]Take node 3, give it a path. It doesn>t matter which".
[/quote]
It must be a path that is not identical with the first two, i.e., the
chosen path must lead out of the conquered region and must yield
another infinitude of nodes.

[quote]THAT IS ALL you have said! That obviously DOES NOT NECESSARILY
use all paths! I could do THAT using ONLY paths of all 1s followed by
all 0>s!
[/quote]
Of course. See for instance
Message-ID: <f0f8ab79-568e-4989-
b712-5f5c2608091b@u27g2000pro.googlegroups.com>

[quote]I take node 0. I give it the all 0>s path.
I take node 1. I give it the one 1 followed by all 0>s path.
I take node 2. I give it the two 1>s followed by all 0>s path.
I can do this for ALL natnums. When I get to the end I will have
used up all the natnums, BUT NOWHERE NEAR all the paths!
In fact, NO MATTER WHAT way you assign the paths, YOU WILL NEVER
COME CLOSE to using all the paths!
[/quote]
I will use all infinite strings consisting of 0 and 1. (I an extended
decimal version will use all real numbers that can be identified by
means of infinitely many digits 1,2,3,...,0.) There is no single
combination that is not covered by my game.
[quote]
The easy way to prove this is just (even though you can>t finish)
to ASSUME THAT YOU HAVE finished and given EVERY node a path!
THEN, WE CAN CONSTRUCT a path that YOU DID NOT include!
[/quote]
You cannot construct it by means of aleph_0 digits. All those
construction have got their nodes.

[quote]It is just the path whose nth step is THE OPPOSITE WAY from the
nth step of the path that YOU mapped to node n!
[/quote]
That path gets another node.

[quote]THAT path IS NOT IN YOUR map!
[/quote]
Try to find a concrete example please. You will see that you are
wrong.

If you agree that the path is in the tree and if you agree that every
node of the tree has been conquered, then every path is in my map.

Regards, WM

WM · Guest

On 10 Okt., 22:05, george <gree...@email.unc.edu> wrote:
[quote]On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

It is very easy to see that there are not more paths than nodes in
the
binary tree.

It CANNOT be "easy to see" SINCE IT IS FALSE!
NOBODY, INCLUDING YOU, HAS EVER seen this!
[/quote]
Have you ever heard of Skolem relativity? There is no absolute
uncountability. But that is not our topic.
[quote]
For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....

o
|
/
/
/
...

This means that any node that turns right at the 0th step will NOT map
to 0
(or to node 0; your mapping to nodes is equivalent to mapping to
natnums
since the nodes are countable, which means they can be exactly matched
1-1 with the natnums).
[/quote]
Correct. And every path different from 0.000... also wil not map to
node number 0. They all will map to nodes at the border of the
conquered domain. Obviously no path can leave the conquered domain
without a nodes at which this happens.
[quote]
The next node

WHICH node is next?
[/quote]
The winning strategy is so clear that it doesn>t matter what you
choose.

[quote]We>ll say it>s the one on the left, the path to which is just 0.
We>ll call that node 1.
[/quote]
I had proposed the node enumeration:

0
1
23
4567
....
where 1 is the root node. Node number 0 was only introduced to
simplify the problem.
[quote]
is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...

This means that any path that turns LEFT at the 1th (second) step will
NOT
be mapped to node 1. So MY path, just to make sure it maps to
neither 0 nor 1,
will turn right on the 0th step and left on the 1st.

o
|
o
/ \
/ \
/ \
...

The next node is mapped on another path, no matter which one, say one
always changing direction, i.e., 0.010101...

The next node, node 2, Has a 0 in position 2 in your example.
So MY path will have a 1 there. So so far, my path is 101.
It differs FROM ALL THREE of the example paths you have thus far
mentioned.
It differs from the path you map to 0 digit-position 0 (the first).
It differs from the
path you map to 1 at digit-position 1 (the second). It differs from
the path you
map to 2 at digit-position 2 (the third).
I CAN keep this up ad infinitum, DUMBASS.
[/quote]
Unless you regain self-control, this will be the last reply from my
side to you.

Simply give me a string of nodes such that at least one of the nodes
is not in a path of the completely conquered tree.
[quote]

o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \
...

And so on.

And so on, my finite path keeps growing, DISAGREEING WITH EVERY
infinite path you have proferred, at a rather early FINITE position
(the same position
as the number of the node you are now up to, in fact).
[/quote]
Map the path on that node where the path leaves the conquered domain.
[quote]
Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But it is clear, that every infinite path that exists
in the infinite binary tree has one node mapped onto it.

Of course we could also have selected the path pi-3 or any other one
to start with. There is no path existing (!) in the tree being left
out by
the mapping.

There IS SO TOO. If you give the mapping, I AM GIVING a path that IS
NOT INCLUDED
in it!
[/quote]
Simply show a state of the game where you can come up with a path that
cannot get a node mapped on it.

Regards, WM

WM · Guest

On 10 Okt., 22:07, george <gree...@email.unc.edu> wrote:
[quote]On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

Anyhow, the binary tree is a crystal-clear construction that avoids
the confusing opinion of undistinguishable numbers. It does never
insert its roots into the mud of the continuum (Kontinuumssauce).

This is just bullshit. The INFINITE part of the binary tree IS
ALWAYS
the continuum.
[/quote]
There is no "infinite part". The levels of the tree are enumerated by
thge natural numbers. None of them is infinite or denotes an infinite
part of the tree. By the way, same holds for the enumeration of
Cantor>s list.

[quote]The part you have investigated thus far is finite;
the part you have not investigated yet IS ALWAYS UNcountabl.
[/quote]
The part I have investigated, with the first path already, contains
all levels enumerated by natural numbers. More levels do not exist.
[quote]
Therefore all its paths are and remain countable,

You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"!
[/quote]
It is the usual saying. But if you do not understand it, then I say
with sufficient precision: The set of all paths of the tree is a
countable set.

[quote]All sometimes means EACH around here!
[/quote]
That is true. Therefore it would be better, to translate the general
quantifier by "for every" instaed of "for all".

[quote]And no INDIVDIDUAL path can be uncountable;
[/quote]
With respect to this platitude I took the liberty of being somewhat
sloppy.

[quote]they are all countably
infinitely LONG! But they are uncountably WIDE -- there are
uncountably
infinitely many OF them!
[/quote]
At every level denoted by a natural number n, I can distinguish 2^n. I
do not see where you find infinitely many.
[quote]
I realize your native language is not English but that is simply not
an excuse.
[/quote]
As I said, with respect to this platitude I took the liberty of being
somewhat sloppy. In German this is not different. So there is no
excuse required.
[quote]
It is very easy to see that there are not more paths than nodes in
the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....

o
|
/
/
/
...

The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...

You CANNOT SAY "no matter which one". It MATTERS which one.
[/quote]
No. It matters that all are included. The order may be as you like.

[quote]Here is the short version of your proof that the reals are countable:
1) pick a 1st real, no matter which one.
2) pick a 2nd real, no matter which one, as long as it is different
from all the previous
ones.
3) repeat ad infinitum.
4) This clearly maps every natural to a real
5) therefore (unsound) it clearly maps EVERY real to a natural.
[/quote]
No. The mapping is a bijection. Every path gets a node. Every node is
mapped on one and only one path.
[quote]
THIS IS ALL you have said. It is completely unsupported.
[/quote]
Try to understand the principle, why the tree can be exhausted by a
countable set of paths. It is simple. For numbers (or paths) that
consist only of digits (or nodes) enumerated by a natural numbers n in
N, we have

The sequence
0.1000..., 0.11000..., 0.111000..., ... (*)
contains
0.111... = SUM{n in N} 10^-n (**)
Every position with a natural index n shows digit 1.

It is without interest what might follow behind all positions indexed
by naturals (at a "position oo" or even behind it). That may be of
interest to people who believe in divine or unreal reals. For me and
for realistic readers of Cantor-lists, only the digits indexed by
naturals are of concern.

Regards, WM

WM · Guest

On 10 Okt., 22:14, george <gree...@email.unc.edu> wrote:
[quote]On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

Every node in your infinite binary tree occurs at the
END OF A FINITE path.
[/quote]
Every digit in every number occurs at the end of a finite string.
Every natural number is finite!

[quote]Since that finite path to the node IS A FINITE BIT-STRING,
it uniquely represents A NATURAL NUMBER for the node.
Obviously, the nodes can be ordered:
the root is 1, the (node at the end of the)
terminating path 0 is 2, the terminating path 1 is 3,
the path 00 terminates at the node numbered 4, the path 01
termiantes at node#5,
the path 10 terminates at the node numbered 6, and the terminating
path 11 is 7, etc.
You just order them left-to-right across each row, with 0000 on the
left and 1111 on
the right. If you want to know the number corresponding to a path,
just put a 1
in front of the path-as-a-bit-string and evaluate the resulting string
as a natnum in binary.

If you want to know node n>s number n, represent the path
(from the root) to the node in binary, and replace the root with a 1
in front (left).
Evaluate that bit-string as a natnum. That is node n>s number (also
n).

So you say you have a mapping of every path to a node.
Well, that node has a number (from above), so you also have a mapping
from every path TO A NATURAL NUMBER.
We diagonalize this mapping by constructing the following denumerably
long
infinite path that YOU DO NOT map TO ANY natnum:

Call your mapping Y(.).
Y(p)=n, where p is some path and n is a node
in the infinite binary tree. Let the i>th bit of path p be called
p_i: if the i>th step in the path is to the left on the binary tree,
then p_i=0; if it is to the right then p_i=1.

It is easy to design a path, MY path, M, that PROVABLY, for EVERY n,
is NOT the path
that You map to n: Y(M)=/=n FOR ANY n.
MY PATH M IS NOT INCLUDED in your mapping, so your mapping DOES NOT
include ALL the infinite paths.
[/quote]
Is your path included in the tree? Yes. Does it leave the conquered
domain for any possible domain not yet including it? Yes.

Or is your problem simply that you distinguish real numbers that
cannot be distinguished by digits (and that, therefore, cannot be
distinguished in the tree or in Cantor>s list)?
[quote]
First of all, Y(M)=/=0 since no path maps to 0 in this scenario.
You do not map My path to 1, because whichEVER path YOU mapped to 1,
MY path takes THE OPPOSITE direction at step 1.
[/quote]
The infinite path 0.000... is covered by the infinite sequence of
paths

0.111..., 0.0111..., 0.00111... , 0.000111, ...

[quote]
You see, your problem is, if you want to prove that there are NO MORE
infinite paths
than nodes, you have to not only map each path to 1 node, you have to
ALSO
map each NODE to at MOST ONE path.
[/quote]
That is what I do. It is a bijection between nodes and paths.

[quote]Otherwise you could map EVERY path
to 0,
which wouldn>t prove anything.

THIS MEANS YOUR MAPPING Y IS INVERTIBLE.
Y(p)=n ( you falsely allege), for every path p, but if it>s 1-1
and if you>re not leaving out any numbers, then Y has an inverse, call
it y
(it>ll be clear not only from case but from type of the arguments as
well,
which Y we mean)) satisfying
y(n)=p. In other words, you ALSO map every NODE to some PATH.

Third of all, my path does not map to 2, because whatever path
YOU mapped to 2, MY path goes THE OTHER WAY
in its 2nd step. Do you get it yet? MY PATH IS NOT IN your map.

For all n, M_n=1-[y(n)]_n,
where y(n) is the path that you mapped to n.

This, unlike all your handwaving about the picture,
IS A PROOF.
[/quote]
No.

[quote]And it does NOT DISAGREE with your picture in ANY way (although
it does contradict it). It BEGINS BY ASSUMING THAT YOUR PICTURE IS
CORRECT AND THAT YOU DO HAVE a list.
But it constructs, FROM YOUR mapping, a path that IS NOT IN YOUR
mapping.
[/quote]
No. Not for nodes in paths (or positions in numbers) indexed by
natural numbers.

Regards, WM

george · Guest

On Oct 11, 3:59 am, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]The easy way to prove this is just (even though you can>t finish)
to ASSUME THAT YOU HAVE finished and given EVERY node a path!
THEN, WE CAN CONSTRUCT a path that YOU DID NOT include!

You cannot construct it by means of aleph_0 digits.
[/quote]
I CAN SO TOO, dumbass!
We are ONLY TALKING ABOUT strings of aleph_0 bits!
The string I construct OBVIOUSLY HAS to have aleph_0 digits
because I give it ONE bit for EACH NODE IN THE TREE, and
the tree only has aleph_0 nodes! I give it 1 bit for each natural
number,
and there are only aleph_0 natural numbers!
I TELL YOU EXACTLY HOW TO COMPUTE the nth step of MY path,
the nth-bit of MY bit-string, FOR ALL n! I tell you EXACTLY what my
path
is! All YOU have to do is FIRST tell me EXACTLY what YOUR mapping
of the paths to the nodes (and therefore to the natural numbers) is!
Except, actually, YOU DON>T have to tell me what your mapping is,
because I DON>T CARE what your mapping is: MY path can be created
from it REGARDLESS of what it is. My path is a function of your
mapping,
and that function generates a path FOR ANY mapping, INCLUDING yours.
The one it generates from yours IS PROVABLY a path NOT MAPPED TO
ANY node by your mapping.

[quote]All those construction have got their nodes.
[/quote]
No, they haven>t. At any given time, you have only given nodes to a
finite
number of paths. You can eventually get to all the nodes but you
never come
close to using all the paths.

[quote]It is just the path whose nth step is THE OPPOSITE WAY from the
nth step of the path that YOU mapped to node n!

That path gets another node.
[/quote]
No, it gets the node that YOU ALREADY mapped it to.
You claim that your mapping ALREADY exists. You CAN>T CHANGE
your mapping after the fact.
It wouldn>t matter if you could; I would just then have a DIFFERENT
path
that wasn>t mapped by your NEW mapping EITHER.

george · Guest

On Oct 11, 4:44 am, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]It is very easy to see that there are not more paths than nodes in
the
binary tree.

It CANNOT be "easy to see" SINCE IT IS FALSE!
NOBODY, INCLUDING YOU, HAS EVER seen this!

Have you ever heard of Skolem relativity?
[/quote]
Have you? If you have, you have obviously not UNDERSTOOD it.

[quote]There is no absolute uncountability.
[/quote]
Of course there is. And this is an example.

[quote]But that is not our topic.
Thank god. You were far enough in over your head already.[/quote]

But if you want to start something about how Skolem>s paradox
relates to this, this is the message to reply under, unless you are
going to start a whole new thread.

george · Guest

On Oct 11, 4:44 am, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]The next node

WHICH node is next?

The winning strategy is so clear that it doesn>t matter what you
choose.
[/quote]
Then let>s always choose to go from left to right as we go down,
so that the nodes will be numbered easily.
The root is 1 (not 0; there is no 0); the first down to the left
(along direction 0) is 2, the first down to the right (direction 1) is
3,
etc. THAT IS HOW WE ARE GOING.
[quote]
We>ll say it>s the one on the left, the path to which is just 0.
We>ll call that node 1.

I had proposed the node enumeration:

0
1
23
4567
...
where 1 is the root node. Node number 0 was only introduced to
simplify the problem.
[/quote]
Fine, we agree, except that I didn>t have a node 0.
It doesn>t make any difference.
Node 1 also corresponds to an empty path (to get to it, not to map to
it).

[quote]Unless you regain self-control, this will be the last reply from my
side to you.
[/quote]
Oh, EAT SHIT AND DIE.
You are in a long line of insufferable idiots here.
Nobody gives a shit if you quit replying.
EVERBODY"S GOAL IS TO FORCE you to quit replying!
With anything other than "My God, how could I have been so stupid
for so long?"

george · Guest

On Oct 11, 5:03 am, WM <mueck...@rz.fh-augsburg.de> wrote:
[quote]On 10 Okt., 22:07, george <gree...@email.unc.edu> wrote:

This is just bullshit. The INFINITE part of the binary tree IS
ALWAYS
the continuum.

On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
There is no "infinite part".
[/quote]
Damn, you>re stupid.

[quote]The levels of the tree are enumerated by
the natural numbers.
[/quote]
OF which there are INFINITELY many.

[quote]None of them is infinite
[/quote]
Yet DESPITE that, THERE ARE INFINITELY MANY of them.

[quote]or denotes an infinite part of the tree.
[/quote]
No INDIVIDUAL LEVEL of the tree is infinite.
No INDIVIDUAL NATNUM is infinite.
But there are infinitely MANY natnums, and therefore the tree
has infinitely MANY levels, and therefore THE TREE is infinite.

[quote]By the way, same holds for the enumeration of Cantor>s list.
[/quote]
Since Cantor is proving that THERE IS NO list, that is sort of silly.
Cantor>s list is square and denumerable IN BOTH dimensions.
His point is that it is necessarily INCOMPLETE. There is a
conceivable
(constructible, even) row that IS NOT ON it.

[quote]The part you have investigated thus far is finite;
the part you have not investigated yet IS ALWAYS UNcountabl.

The part I have investigated, with the first path already, contains
all levels enumerated by natural numbers. More levels do not exist.
[/quote]
OK, I didn>t realize you were investigating "vertically".
But you only get countably many (denumerably many)
nodes for each path. There are always an uncountable number left.

[quote]Therefore all its paths are and remain countable,

You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"!

It is the usual saying.
[/quote]
This is sci.logic. You can>t always use natural language.

[quote]But if you do not understand it,
[/quote]
THAT is NOT the issue! YOU do not understand it!
When I say you can>t say something, I don>t mean say it anyway
and then tell me what it allegedly means! I may say something ELSE!

[quote]The set of all paths of the tree is a countable set.
[/quote]
Which is obviously bullshit, since I have just disproved it,
and since Cantor did long before.

[quote] But they are uncountably WIDE -- there are
uncountably
infinitely many OF them!

At every level denoted by a natural number n, I can distinguish 2^n.
[/quote]

Idiot, PLEASE! We are talking ABOUT INFINITELY LONG paths!
At level n, you have ZERO infinitely long paths!

I
[quote]You CANNOT SAY "no matter which one". It MATTERS which one.

No. It matters that all are included.
[/quote]
you CANNOT include ALL the paths!
So you can>t tell me it matters that all are included WHEN NOBODY CAN
DO
that ! If THAT matters then YOUR WHOLE ENTERPRISE FAILS!

[quote]The order may be as you like.
[/quote]
No matter WHAT order you like, IF you are only mapping them to natural
numbers,
YOU WILL NEVER include ALL of them! You will RUN OUT OF NUMBERS
FIRST!

[quote]Here is the short version of your proof that the reals are countable:
1) pick a 1st real, no matter which one.
2) pick a 2nd real, no matter which one, as long as it is different
from all the previous
ones.
3) repeat ad infinitum.
4) This clearly maps every natural to a real
5) therefore (unsound) it clearly maps EVERY real to a natural.

No. The mapping is a bijection. Every path gets a node. Every node is
mapped on one and only one path.
[/quote]
Then since the nodes also correspond to numbers,
THEN OBVIOUSLY, the path whose nth step goes the opposite way from
the path that you map to the nth node IS NOT IN YOUR BIJECTION,
DUMBASS.

Virgil · Guest

In article
<ebbaa099-4385-437d-a026-48fb96982c05@s50g2000hsb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 20:30, Virgil <Vir...@gmale.com> wrote:
In article
5f06517f-bb41-4a02-b04d-130f16f4a...@i76g2000hsf.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:
In article
871188cc-f053-4ab5-870f-0b57ce13e...@n33g2000pri.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@yahoo.es> wrote:

o
|
o
/ \
o \
/ \ \
/ / \
/ \ \
/ / \ ...

And so on.

Your game is just a countable sequence of choices, a countable
choice function. We have no guarantee that such a device will
'finally' give us the entire tree.

Of course there is no "finally". All we can say is that "iff there
was an infinite countable set of nodes" and "iff infinite sets could
be worked through completely", then every node would get its turn.

But every node occurs in infinitely many paths, as many paths as in the
entire infinite complete binary tree. So that every node gets infinitely
many
"turns".

Not according to the rule of my game. Consider the figure above. Node
x buys you one of the path leaving the path
0.111... at position x. You may say that there are many paths
deviating from 0.111..., and you are right. But if one of them
deviates from said path, it will get get another node. There is no
escape.

o
|
o
/ \
o \
/ \ \
/ / \
/ \ x
/ / / \
... /
\

With every won path you get infinitely many nodes. Each of them helps
you to get another path, one and only one path per node. In this game
you have the ratio of paths to nodes P/N = 0 at every step. The
infinite binary tree covers every step, but not more.

Similarly the usual decimal expansion of real numbers covers every
digit, but not more. There is no last digit of pi. There is every
digits of pi - unless one looks to close, but that is not in question
here. Set theorists insistuing of uncountably many reals must insist
that there are more than every digit of pi. I am not inclined to
discuss about such folly.

Your "game" overlooks every path with infinitely many branchings in both
left and right directions, which is the vast majority of paths.

My game exhausts the complete infinite tree
[/quote]
It only exhausts the patience of those not blinded by their own
ideology.

As you described it, there are paths with infinitely many changes of
direction which your "counting" method overlooks.

[quote]i.e., it covers every
path that can be formed by an infinitude of digits. In other words,
there is no infinite combination of digits 0 and 1 (or 1,2,3,...,0 in
an extended version) that is off the tree.
[/quote]
If your method of counting paths were valid, you should be able to give
the position of any path in that count.

So what is the position of the path that alternately branches left then
right? Or the path that branches oppositely?

Your so called counting rule does not account for such paths, nor for
uncountably many others.
[quote]
But if every node of the tree has been used to mark one path

Unless that node is a terminal (leaf) node, it marks more than one path,

Try to understand the rule of the game (there is only one). Try to
understand that there is no leaf node.

It is the absence of any such leaf nodes that makes you game fail to
distinguish each path from all its neighbors.- Zitierten Text ausblenden -

A path that can be distinguished from all its neighbours by means of
digits can be distinguished in the tree as well as in Cantor>s list.
[/quote]
But that would take an infinite string of digits, and the set of all
such strings is equally uncountable.

[quote]
The important point for mathematics is the distinction by digits.
[/quote]
The important points for mathematics is that the set of paths in an
infinite complete binary tree bijects nicely with the set of all
infinite binary strings, and that neither injects into any countable
set.
[quote]
My game shows that the complete infinite binary tree can be exhausted
by a countable number of paths.
[/quote]
Not any mathematical infinite complete binary tree.

http://en.wikipedia.org/wiki/Binary_tree

<quote>
An INFINITE COMPLETE BINARY TREE is a tree with ALEPH_0 levels, where
for each level d the number of existing nodes at level d is equal to
2^d. The cardinal number of the set of all nodes is ALEPH_0. The
cardinal number of the set of all paths is 2^ ALEPH_0.
<end quote>

So that, unless WM>s INFINITE COMPLETE BINARY TREE differs in some
significant way from Wikipedia>s, WM is wrong again.

Virgil · Guest

In article
<f43eca7d-3f27-4e42-9736-b700455115eb@m44g2000hsc.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 20:40, Virgil <Vir...@gmale.com> wrote:
In article
f0f8ab79-568e-4989-b712-5f5c26080...@u27g2000pro.googlegroups.com>,

WM <mueck...@rz.fh-augsburg.de> wrote:
On 10 Okt., 06:55, Virgil <Vir...@gmale.com> wrote:

Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals.

Then your trees are either not infinite or not complete.

The infinite binary tree is as complete as the number system

Then in YOUR world, both are incomplete.

I call the tree complete if every string of digits can be found within
it.

, i.e.,

the binary or decimal representation of real numbers can be. There is
no chance to introduce another path. There is no chance to introduce
another real. All combinations of nodes-values, all combinations of
digits are present.

The binary expansions for 1/3 or sqrt(2) are neither represented in your
i complete tree, as they both require infinitely many branchings in both
directions, which you game does not allow.

Please look up the tree. Every infinite path is there.
[/quote]
You mean like at http://en.wikipedia.org/wiki/Binary_tree "

Where Wiki says that the number of paths is not aleph_0 but 2^aleph_0,
and elsewhere says aleph_0 < 2^aleph_0.

And I challenge WM to find any site, other than his own, at which it
states otherwise.
[quote]

The infinite path 0.111... is covered by the infinite sequence of
paths

0.1000..., 0.11000..., 0.111000..., ... (*)

But then WM is speaking of INFINITE SETS of his "finite" paths, not just
the individual paths themselves, and there are always more such sets of
such paths than there are such paths.

I am speaking of path. Their number is countable.
[/quote]
Which single path of your list of paths, 0.1000..., 0.11000...,
0.111000..., ... , do you claim covers the infinite path 0.111...?

So you are really speaking about LISTS of paths, not individual paths.

And Wiki is still correct that the set of all paths in the infinite
complete binary tree has cardinality 2^aleph_0, not aleph_0.

Virgil · Guest

In article
<2b1f1e53-cc59-4b22-b1a5-89d77f4e748f@v53g2000hsa.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 22:22, george <gree...@email.unc.edu> wrote:
On Oct 9, 3:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.

You HAVE NOT DONE this!
All you have said is "Take node 1. Give it a path.
Take node 2. Give it a path, it doesn>t matter which.

It must be a path that is not identical with the first one.
[/quote]
Even so, you have not shown that for each path there is one and only one
node.

And it is not so, since through every node pass infinitely many paths,
so for each of your chosen nodes there are infinitely many paths passing
through it thus identified only by that node.

Or do you nave two or more of your selected nodes on one path?

Either way, your arguments fall flat.

Virgil · Guest

In article
<7b94707b-529c-4cc2-9807-cc2e69e735c5@x41g2000hsb.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 22:05, george <gree...@email.unc.edu> wrote:
On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

It is very easy to see that there are not more paths than nodes in
the
binary tree.

It CANNOT be "easy to see" SINCE IT IS FALSE!
NOBODY, INCLUDING YOU, HAS EVER seen this!

Have you ever heard of Skolem relativity? There is no absolute
uncountability. But that is not our topic.
[/quote]
Our topic is your false claims that a complete binary tree in which no
path has a terminal node has only countably many paths.
[quote]
For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....

o
|
/
/
/
...

This means that any node that turns right at the 0th step will NOT map
to 0
(or to node 0; your mapping to nodes is equivalent to mapping to
natnums
since the nodes are countable, which means they can be exactly matched
1-1 with the natnums).

Correct. And every path different from 0.000... also wil not map to
node number 0. They all will map to nodes at the border of the
conquered domain. Obviously no path can leave the conquered domain
without a nodes at which this happens.

The next node

WHICH node is next?

The winning strategy is so clear that it doesn>t matter what you
choose.
[/quote]
The winning strategy is so obscure that whichever node you choose the
strategy will fail.

Since through each node pass infinitely many paths (in fact uncountably
many), and only one of these can be paired with that node, there are
always too many paths to pair off with a countable set of nodes.

[quote]
Unless you regain self-control, this will be the last reply from my
side to you.
[/quote]
You do not have a "side", only a hole into which you keep digging
yourself deeper.
[quote]
Simply give me a string of nodes such that at least one of the nodes
is not in a path of the completely conquered tree.
[/quote]
That is not the challenge. The challenge is to show that for each path
there is a node which your rules associate with no other path.

And your rules fail miserably to meet that challenge.

[quote]Simply show a state of the game where you can come up with a path that
cannot get a node mapped on it.
[/quote]
Each path having a node mapped on it is not enough unless one can also
show that no other path also gets the same node mapped on it.

But that is impossible.

And WM fails again.

Virgil · Guest

In article
<63c5c5b6-2fc7-4d22-8f61-de92dc2f1d28@m73g2000hsh.googlegroups.com>,
WM <mueckenh@rz.fh-augsburg.de> wrote:

[quote]On 10 Okt., 22:07, george <gree...@email.unc.edu> wrote:
On Oct 8, 12:24 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

Anyhow, the binary tree is a crystal-clear construction that avoids
the confusing opinion of undistinguishable numbers. It does never
insert its roots into the mud of the continuum (Kontinuumssauce).

This is just bullshit. The INFINITE part of the binary tree IS
ALWAYS
the continuum.

There is no "infinite part". The levels of the tree are enumerated by
thge natural numbers.
[/quote]
If one ennumerates the nodes by naturals (which is possible) then the
set of nodes in any path will be an infinite subset of the naturals.

So WM is trying to surject the set of naturals to a set of infinite
subsets of the naturals. But he will always fail.

[quote]
You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"!

It is the usual saying. But if you do not understand it, then I say
with sufficient precision: The set of all paths of the tree is a
countable set.
[/quote]
If that tree is an infinite complete binary tree as defined in Wiki, then
Wiki says that its set of oaths in UNcountable.
http://en.wikipedia.org/wiki/Binary_tree

[quote]
At every level denoted by a natural number n, I can distinguish 2^n. I
do not see where you find infinitely many.
[/quote]
Then WM is admitting his own blindness.

[quote]
I realize your native language is not English but that is simply not
an excuse.

As I said, with respect to this platitude I took the liberty of being
somewhat sloppy. In German this is not different. So there is no
excuse required.
[/quote]
There is certainly no excuse acceptable for WM>s falsehoods.

[quote]It is very easy to see that there are not more paths than nodes in
the
binary tree.
[/quote]
It is even easier to see that there are more paths than nodes, at least
to anyone with normal vision.

One may, if one choses, deny the very existence of any infinite complete
binary tree, as constructionists would do, but once one allows its
existence, the uncountability of its set of paths is not logically
deniable.

But then, WM never allows logic to constrain his arguments


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