The uncountability infinite binary tree.

[Zuhair]

This is a continuation of post about the infinite binary tree raised
by WM in this Usenet.

To review this subject in brief. The infinite binary tree is a single
rooted labeled tree. Each node must be exclusively labeled either by 0
or by 1. The root node receives the label 0. Each node have exactly
two children nodes and one is labeled by 0 and the other by 1.

Diagrammatically this is:

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

Now whenever "infinite binary tree" is mentioned then it is to be
understood that we are referring to the above one.

A path shall be defined as a linear sub-graph of the above tree that
begins with 0 in which all edges have the same direction (i.e. from
above downwards)

Example: The following linear graph is a path of the above tree:
0
\
1

Another example is:

0
/
0
\
1

Now Any sub-tree of the above tree that has its root labeled 0 is said
to be countable iff the set of all paths of it is bijective to the set
of all natural numbers.

Now it is obvious that the infinite binary tree spoken about above
have only countably many nodes. But how many paths does it have? is it
countable?

The answer is NO, it is not countable!

The infinite binary tree depicted above have UNCOUNTABLY many paths.

This is a simple Corollary of Cantor's diagonal argument actually.

Proof: Let G be any countable subtree of the infinite binary tree that
is 0 rooted and such that all paths ending by 0-0-0-... or by
1-1-1-...are among the paths of G.

Let f be a bijective function from the domain N of all naturals(except
0) to the set of all infinite paths of G.

Construct the diagonal path d_f in the following manner:

The root (i.e. the 1st) node of d_f is 0 labeled.

Now for n=1,2,3,.. ; The n+1_th node of d_f is labeled by a label that
is opposite to the label of the n+1_th node of the path of G that f
sends n to.

Now clearly the diagonal d_f (actually the anti-diagonal but it shall
be called the diagonal for short) is a path and clearly it is labeled
in a way that is different from labeling of all paths of G. So d_f is
missing from G.

So any countable subtree of the infinite binary tree, that is 0 rooted
and that has all paths ending with 0-0-0-.. or with 1-1-1.. among its
paths; would be missing a path of the infinite binary tree.

Since the infinite binary tree is 0 rooted, and it contains all paths
ending by 0-0-0... or 1-1-1-.. among its paths, then the infinite
binary tree cannot be countable, because otherwise it would be missing
a path of its paths which is absurd.

Thus the Infinite binary tree is UNCOUNTABLE.

QED

Zuhair

[Virgil]

In article
<5a844c9f-e16f-42ca...@u13g2000vbc.googlegroups.com>,

Techincally, contable if it injects to N or N surjects to it, it is
countable, as countable should include finite sets.

What you are calling countable is more properly called countably
infinite.

I agree that in standard mathematics the set of ALL paths of any
complete infinite binary tree is necessarily uncountable by any standard
definition of the distinction between countable and uncountable, but you
will never get anyone whose head is permanently stuck in
Wolkenmuekenheim to agree to anything so reasonable and obvious.
--

[William Elliot]

On Fri, 14 Dec 2012, Zuhair wrote:

> This is a continuation of post about the infinite binary tree raised
> by WM in this Usenet.

Why belabour a dead horse?

> To review this subject in brief. The infinite binary tree is a single
> rooted labeled tree. Each node must be exclusively labeled either by 0
> or by 1. The root node receives the label 0. Each node have exactly
> two children nodes and one is labeled by 0 and the other by 1.
>

Exercise. The infinite undirected binary tree has countably many
non-intersecting finite paths and uncountably many non-intersecting
infinite paths.

Haven beaten that to death, let's go beyond.

Questions. How many finite paths are there?
Are there more paths than there are non-intersecting paths?

[Zuhair]

On Dec 15, 5:38 am, Virgil <vir...@ligriv.com> wrote:
> In article
> <5a844c9f-e16f-42ca-86c3-67ab01558...@u13g2000vbc.googlegroups.com>,

Yes of course, no doubt.

[WM]

The other answer is yes. Hence, there is a contradiction.

Zuhair, you cannot argue against a contradiction between A and B by
proving over and over that A is correct.

>
> The infinite binary tree depicted above have UNCOUNTABLY many paths.
>
> This is a simple Corollary of Cantor's diagonal argument actually.

Since the Binary Tree can be constructed by counatbly many paths such
that noone can discern whether and which paths are lacking, we have a
contradiction.

>
> Proof: Let G be any countable subtree of the infinite binary tree that
> is 0 rooted and such that all paths ending by 0-0-0-... or by
> 1-1-1-...are among the paths of G.
>
> Let f be a bijective function from the domain N of all naturals(except
> 0) to the set of all infinite paths of G.
>
> Construct the diagonal path d_f in the following manner:
>
> The root (i.e. the 1st) node of d_f is 0 labeled.
>
> Now for n=1,2,3,.. ; The n+1_th node of d_f is labeled by a label that
> is opposite to the label of the n+1_th node of the path of G that f
> sends n to.
>
> Now clearly the diagonal d_f (actually the anti-diagonal but it shall
> be called the diagonal for short) is a path and clearly it is labeled
> in a way that is different from labeling of all paths of G. So d_f is
> missing from G.

This is correct for every finite level. Alas for every finite level n
it is also correct that the paths traversing the tree from the root
node to that level n are less than the nodes of this finite tree. If
this argument can immediately change to the opposite for an actually
infinite tree, why then can not your argument als change to the
opposite in an actually infinite tree or list?

>
> So any countable subtree of the infinite binary tree, that is 0 rooted
> and that has all paths ending with 0-0-0-.. or with 1-1-1.. among its
> paths; would be missing a path of the infinite binary tree.
>
> Since the infinite binary tree is 0 rooted, and it contains all paths
> ending by 0-0-0... or 1-1-1-.. among its paths, then the infinite
> binary tree cannot be countable, because otherwise it would be missing
> a path of its paths which is absurd.
>
> Thus the Infinite binary tree is UNCOUNTABLE.

Since it can be proved to be countable, there is a contradiction.

Regards, WM

[Rupert]

How do you prove that it is countable?

> Regards, WM

[WM]

First: A Tree that contains all nodes also contains all reals of the
unit interval.

I constrcut the tree, i.e., all nodes, by countably many infinite
paths. But I don't tell you which paths I use. If you claim that more
paths are available, then you must find out by means of the nodes of
the tree which paths I have not used. Of course you cannot solve this
task.

Hence, what you call infinite paths, is not defined by nodes but only
by finite words like 1/3. But there are only countably many finite
words.

Regards, WM

[Zuhair]

On Dec 15, 2:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > Thus the Infinite binary tree is UNCOUNTABLE.
>
> > > Since it can be proved to be countable, there is a contradiction.
>
> > How do you prove that it is countable?
>
> First: A Tree that contains all nodes also contains all reals of the
> unit interval.
>
> I constrcut the tree, i.e., all nodes, by countably many infinite
> paths. But I don't tell you which paths I use. If you claim that more
> paths are available, then you must find out by means of the nodes of
> the tree which paths I have not used. Of course you cannot solve this
> task.

Of course we can. Actually I already did it in the head post but you
keep ignoring it. You say that you constructed a COUNTABLE tree, this
by DEFINITION means that there must be a bijection between the set of
all paths of that tree and the set N of all naturals. Then use the
above approach and illicit the missing path

[Rupert]

On Dec 15, 12:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > Thus the Infinite binary tree is UNCOUNTABLE.
>
> > > Since it can be proved to be countable, there is a contradiction.
>
> > How do you prove that it is countable?
>
> First: A Tree that contains all nodes also contains all reals of the
> unit interval.
>
> I constrcut the tree, i.e., all nodes, by countably many infinite
> paths.

Tell us more about this construction.

[WM]

On 15 Dez., 13:36, Zuhair <zaljo...@gmail.com> wrote:
> On Dec 15, 2:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
>
>
>
>
> > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > Thus the Infinite binary tree is UNCOUNTABLE.
>
> > > > Since it can be proved to be countable, there is a contradiction.
>
> > > How do you prove that it is countable?
>
> > First: A Tree that contains all nodes also contains all reals of the
> > unit interval.
>
> > I constrcut the tree, i.e., all nodes, by countably many infinite
> > paths. But I don't tell you which paths I use. If you claim that more
> > paths are available, then you must find out by means of the nodes of
> > the tree which paths I have not used. Of course you cannot solve this
> > task.
>
> Of course we can. Actually I already did it in the head post but you
> keep ignoring it. You say that you constructed a COUNTABLE tree,

No, I say I construct the complete Binary Tree. And I do so because no
node is missing.

this
> by DEFINITION means that there must be a bijection between the set of
> all paths of that tree and the set N of all naturals.

There is a bijection of the set (A u B) of all path A that I use and
all paths B that you can identify as missing.

> Then use the
> above approach and illicit the missing path

No dreamworks please. I have constructed the complete Binary Tree
using a countable set A of infinite paths. What path is missing? If
you cannot identify it by nodes, then it is not defined by nodes. Or
do you believe that belief in paths is sufficient in mathematics?

Regards, WM

[WM]

On 15 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote:
> On Dec 15, 12:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > Thus the Infinite binary tree is UNCOUNTABLE.
>
> > > > Since it can be proved to be countable, there is a contradiction.
>
> > > How do you prove that it is countable?
>
> > First: A Tree that contains all nodes also contains all reals of the
> > unit interval.
>
> > I constrcut the tree, i.e., all nodes, by countably many infinite
> > paths.
>
> Tell us more about this construction.

I use all finite paths. Every node of the Binary Tree is the end of a
path. Then I append these countably many paths by a tail according to
my choice. For instance I can use the tail
000...
or
010101...
or
the bit sequence of pi
or anything else, for instance a mix of many tails.

In order to show you that you are dreaming if you think that infinite
paths could be identified by their nodes, I don't tell you what tails
I have used. If you don't sleep to deep, then you will wake up and
recognize that infinite paths are merely defined by finite
definitions, and hence belong to a countable set.

Regards, WM

[netzweltler]

Does "constructing" the infinite binary tree mean the same as
"constructing" the set of all infinite binary sequences?

--
netzweltler

[WM]

On 15 Dez., 19:16, netzweltler <reinhard_fisc...@arcor.de> wrote:
> Does "constructing" the infinite binary tree mean the same as
> "constructing" the set of all infinite binary sequences?

Constructing the Binary Tree means constructing all nodes. It is
impossible to define a real number (of the unit interval, solely by
nodes) that would not be in the Binary Tree.

Regards, WM

[Zuhair]

You are the one who is DREAMING. I told you that neither you nor
anybody else can construct the complete infinite binary tree by
countably many paths, this is just impossible, this is like saying I
can draw the circle that is not a circle.

I showed you the impossibility of that and yet you are not
understanding what is written.

Take any complete infinite binary tree then if you say it is countable
(like in saying you constructed it using countably many paths) then
this literally means that there exist a bijective function from the
set N of all naturals to the set of all paths of it (because this what
countable means), then take any such bijective function f, and
construct the diagonal d_f as it was shown in the head post, and this
diagonal will be a path that is PROVABLY missing from the range of f
which is of course the set of all paths of your tree, which is
contradictory because we've already stipulated that tree to be
complete (no path is missing), thus we cannot say at all that the
complete binary tree is countable nor can we say that we can construct
it in a countable manner no matter what you do, this just cannot be
done!

Zuhair

[WM]

On 15 Dez., 19:28, Zuhair <zaljo...@gmail.com> wrote:

> You are the one who is DREAMING. I told you that neither you nor
> anybody else can construct the complete infinite binary tree by
> countably many paths, this is just impossible, this is like saying I
> can draw the circle that is not a circle.

No. I did it several times. Years ago I did it twice before breakfast
already. And I will show you that my construction does not leave out
one single node. But before i tell you my construction try to find a
path that I did not use in the countably set of all paths that are
sufficient,

>
> I showed you the impossibility of that and yet you are not
> understanding what is written.
>
> Take any complete infinite binary tree then if you say it is countable

The set of paths that I used is countable. And you will see it when I
tell you what I used. But before I unveil my construction let me know
whether the constructed Binary Tree is complete in your opinion. Here
it is:

0.
0 1
0 1 0 1
...

Every level starts with 0, is alternating between 0 an 1, and has
twice as many nodes as the level above.

Regards, WM

[Zuhair]

What do you mean by "append" can you give mean a simple example of how
this can be done for some path. Don't forget in the infinite binary
tree every node have exactly TWO child nodes, so if you are going to
append (I understand that as attach) some infinite path of your choice
to an already existing finite path of the binary tree then you need to
cut the path that is distal to it that is already existing, so the
proper word might be REPLACE instead of append.

However there would be a problem How I'm to be sure that all
Replacements (or appendings) of your choice would preserve the binary
tree structure (where every node must have exactly two children nodes
each having a different label from the other).

See WM one can easily prove that NO Matter what appendings (or
replacements) you are doing, still you will never produce the complete
infinite binary tree in this way, WHY? because Cantor's argument will
be applicable to it as I showed in the head post (EVEN if you don't
tell me what are the tails you've used, it really doesn't matter at
all) and it will show that your tree is NOT the complete infinite
binary tree. All what I need to know is if the tree you are producing
is countable or not, now the tree you are producing by the appending
manner is of course countable, so there must exist a function f from N
to the set of all paths of it (because that's what countable means),
so carry one the diagonal argument outlined in the head post, and the
new path so constructed would be missing from your constructed tree,
so your constructed Tree is NOT the complete binary tree, it doesn't
preserve the structure of the complete binary tree. Since if it does
then simply you will be involved with a contradiction of having a
missing path from a complete tree which cannot be.

Zuhair

[Zuhair]

The complete binary tree that I know is a tree that have its root node
labeled by 0.
and for each node of it there is exactly TWO child nodes each having a
different label from the other. And Every node of it is labeled by
exactly ONE label that is either 0 or 1.
It goes like that

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

...........
.
.
.

This tree has ALL possible binary paths as paths of it.

In other words it is COMPLETE.

Zuhair

[Zuhair]

I will also add the following assertion:

And there is absolutely NO way whatsoever to construct that tree by
countably many paths.

Because simply the tree has UNCOUNTABLY many paths.

Zuhair

[WM]

I constructed this very tree by all finite paths that extend from the
root node to a node at level n. Then I appended the sequence 000...

Regards, WM

[Zuhair]

What do you mean I appended the sequence 000..., Can you explain that
in details. I mean the full detail of how did you construct this tree
by this appending. How do you prove the the resulting constructed tree
is the infinite binary tree. DETAILS please. At least refer me to an
article that has all the details about that alleged construction if
there is any.

Zuhair

[Zuhair]

I'll wait the details of that. But I think the so constructed tree is
NOT the infinite binary tree defined above. It is just a proper
subtree of it that is countable. I can easily show a missing path in
that constructed tree using the Diagonal argument of Cantor that I've
presented in the above head post, which proves that it is not the
infinite binary tree. However I'll wait for the details of that
construction to say my final opinion about it.

Zuhair

[WM]

The finite paths are the following:
0.
0.0
0.1
0.00
0.01
0.10
0.11
...

Each of these paths now is equipped with an infinite tail 000...

0.000...
0.0000...
0.1000...
0.00000...
0.01000...
0.10000...
0.11000...
...

Some paths appear more than once. Some nodes are constructed more than
once. But that does not matter. This set of paths is nevertheless
countable.
And there is no "diagonal" that *at a finite level n* differs from all
paths.

Regards, WM

[WM]

On 15 Dez., 21:50, Zuhair <zaljo...@gmail.com> wrote:

> I'll wait the details of that. But I think the so constructed tree is
> NOT the infinite binary tree defined above. It is just a proper
> subtree of it that is countable. I can easily show a missing path in
> that constructed tree using the Diagonal argument of Cantor that I've
> presented in the above head post, which proves that it is not the
> infinite binary tree.

Not at a finite level! Alas Cantor's diagonal argument is restricted
to that domain. It is only valid for digits with finite indexes.

Regards, WM

[Virgil]

In article
<7ef75ae4-f678-4d0d...@u13g2000vbc.googlegroups.com>,

Since it is only WM who answers "Yes,only WM is in any contradiction.

The only legitimate ways to prove countability of any set is either to
demonstrate a surjection from N to that set or to demonstrate an
injection from that set to N, neither of which can WM do.

And an equally \ legitimate way to prove uncountability of a set is to
prove that no surjection from N to the set is possible, or, equally,
that no injection from that set to N is possible.

>
> Zuhair, you cannot argue against a contradiction between A and B by
> proving over and over that A is correct.

WM you cannot validly claim countability of the set of all paths is a
complete infinite binary tree without a demonstration, as cited above of
that countability.

Which you have not done , and cannot do!

> >
> > The infinite binary tree depicted above have UNCOUNTABLY many paths.
> >
> > This is a simple Corollary of Cantor's diagonal argument actually.
>
> Since the Binary Tree can be constructed by counatbly many paths such
> that noone can discern whether and which paths are lacking, we have a
> contradiction.

WM often claims such countability for the set of all paths of a
complete infinite binary tree, but has never proven it.

> >
> > Proof: Let G be any countable subtree of the infinite binary tree that
> > is 0 rooted and such that all paths ending by 0-0-0-... or by
> > 1-1-1-...are among the paths of G.
> >
> > Let f be a bijective function from the domain N of all naturals(except
> > 0) to the set of all infinite paths of G.
> >
> > Construct the diagonal path d_f in the following manner:
> >
> > The root (i.e. the 1st) node of d_f is 0 labeled.
> >
> > Now for n=1,2,3,.. ; The n+1_th node of d_f is labeled by a label that
> > is opposite to the label of the n+1_th node of the path of G that f
> > sends n to.
> >
> > Now clearly the diagonal d_f (actually the anti-diagonal but it shall
> > be called the diagonal for short) is a path and clearly it is labeled
> > in a way that is different from labeling of all paths of G. So d_f is
> > missing from G.
>
> This is correct for every finite level.

Every node in every path is at a finite level, even though each path has
infinitely many levels, just as every natural number in the infinite set
of natural numbers is finite number.

WM has goofed again in assuming a property of the set must also be a
property of some of its members.

> Alas for every finite level n
> it is also correct that the paths traversing the tree from the root
> node to that level n are less than the nodes of this finite tree.

True but irrelevant, All of the finite initial sets of naturals are
finite but the set of all of them is not.

> If this argument can immediately change to the opposite for an actually
> infinite tree, why then can not your argument als change to the
> opposite in an actually infinite tree or list?

Because we are outside Wolkenmuekenheim, not inside it.

> >
> > So any countable subtree of the infinite binary tree, that is 0 rooted
> > and that has all paths ending with 0-0-0-.. or with 1-1-1.. among its
> > paths; would be missing a path of the infinite binary tree.
> >
> > Since the infinite binary tree is 0 rooted, and it contains all paths
> > ending by 0-0-0... or 1-1-1-.. among its paths, then the infinite
> > binary tree cannot be countable, because otherwise it would be missing
> > a path of its paths which is absurd.
> >
> > Thus the Infinite binary tree is UNCOUNTABLE.

> o

> Since it can be proved to be countable, there is a contradiction.

No one but WM claims proof of countability, and even he has never
demonstrated it by an explicit surjection from N onto his alleged set of
all paths or an injection from that alleged set of all paths into N,
or with any set bijectable with N.

Thus WM has never presented us with anything like a valid proof of his
claim that the set of all paths in a complete infinite binary tree is
really countable.
--

[Virgil]

In article
<a7d09a3e-bad7-41f2...@m13g2000vbd.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
> > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > > Thus the Infinite binary tree is UNCOUNTABLE.
> >
> > > Since it can be proved to be countable, there is a contradiction.
> >
> > How do you prove that it is countable?
>
> First: A Tree that contains all nodes also contains all reals of the
> unit interval.
>
> I constrcut the tree, i.e., all nodes, by countably many infinite
> paths. But I don't tell you which paths I use.

The we have no way of verifying that you have done what you said.

> If you claim that more
> paths are available, then you must find out by means of the nodes of
> the tree which paths I have not used. Of course you cannot solve this
> task.

Since you hide your work, how can we ever tell that you have actually
done what you say you have done.

Experience has shown us that we cannot take your mere word for
anything, so we need overt proofs that you have done what you say you
have done, for example by having you list your paths and publish that
list, or otherwise proving that you actually have a list.

But you know that if you ever allow us to see your list your paths, we
can find as many paths missing from it as in it.

>
> Hence, what you call infinite paths, is not defined by nodes but only
> by finite words like 1/3. But there are only countably many finite
> words.

We defined the set of paths by establishing a rule to distinguish which
objects are paths and which are not. If that rule does not require that
a path have a finite definition, then paths without any finite
definition are members.

Our definition of a member of the set of paths is a set of nodes from a
complete infinite binary tree which contains the root node and exactly
one child of each of its nodes.

Note that that does not require any path to have any finite definition
distinguishing it form other paths.
That our definition of the set of paths creates an uncountable set of
them is beyond WM's capability to mess up.
--

[Virgil]

In article
<17037f55-dd1d-459b...@c14g2000vbd.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 15 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote:
> > On Dec 15, 12:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
> >
> > > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > Thus the Infinite binary tree is UNCOUNTABLE.
> >
> > > > > Since it can be proved to be countable, there is a contradiction.
> >
> > > > How do you prove that it is countable?
> >
> > > First: A Tree that contains all nodes also contains all reals of the
> > > unit interval.
> >
> > > I constrcut the tree, i.e., all nodes, by countably many infinite
> > > paths.
> >
> > Tell us more about this construction.
>
> I use all finite paths. Every node of the Binary Tree is the end of a
> path. Then I append these countably many paths by a tail according to
> my choice. For instance I can use the tail
> 000...
> or
> 010101...
> or
> the bit sequence of pi
> or anything else, for instance a mix of many tails.

There are, in fact, as many possible tails as there are paths,
uncounltably many, and unless WM has used every tail at least once, he
has provably omitted all the paths with those missing tails.

Nice try, sucker, but no cigar!!

>
> In order to show you that you are dreaming if you think that infinite
> paths could be identified by their nodes, I don't tell you what tails
> I have used. If you don't sleep to deep, then you will wake up and
> recognize that infinite paths are merely defined by finite
> definitions, and hence belong to a countable set.

Every set , including uncountable ones , is defined by a finite
definition, but that does not limit any of them to mere countability.

N is finitely defined, 2^N, or the set of all subse of N is finitely
defined. N is countably infinite, 2^n is uncountably infinite.
--

[Sam Sung]

Virgil write:

> Every set , including uncountable ones , is defined by a finite
> definition, but that does not limit any of them to mere countability.
>

> N is finitely defined, 2^N, or the set of all subsets of N is finitely

> defined. N is countably infinite, 2^n is uncountably infinite.

Yep.

[Virgil]

In article
<731db2df-5c0c-4970...@i1g2000vbp.googlegroups.com>,

But it is quite easy to define, say, the set of all binary sequences of
elements from {0,1}, most of whose elements are, at least in WM's sense,
undefinable.

Similarly, most of the paths of any complete infinite binary tree are
undefinable, but still members of the set of all that trees paths.

The trouble with WM here is that as soon as anything past finiteness
gets involved, it blows all sorts of circuits in WM brain.
--

[Sam Sung]

Virgil schrieb:

> In article
> <731db2df-5c0c-4970...@i1g2000vbp.googlegroups.com>,
> WM <muec...@rz.fh-augsburg.de> wrote:
>
>> On 15 Dez., 19:16, netzweltler <reinhard_fisc...@arcor.de> wrote:
>>> Does "constructing" the infinite binary tree mean the same as
>>> "constructing" the set of all infinite binary sequences?
>>
>> Constructing the Binary Tree means constructing all nodes. It is
>> impossible to define a real number (of the unit interval, solely by
>> nodes) that would not be in the Binary Tree.
>
> But it is quite easy to define, say, the set of all binary sequences of
> elements from {0,1}, most of whose elements are, at least in WM's sense,
> undefinable.
>
> Similarly, most of the paths of any complete infinite binary tree are
> undefinable, but still members of the set of all that trees paths.

The WM WorldMaster "reason" is "the complement" which means there must
be undefinable many paths in order for "the numbers of elements" to be
limited within the one half of the full infinity (about 10^80 Fermions).

[Nam Nguyen]

On 15/12/2012 3:23 PM, Sam Sung wrote:
> Virgil write:
>
>> Every set , including uncountable ones , is defined by a finite
>> definition,

Of course. One wouldn't dare to define a mathematical concept using
an infinite description, naturally.

> but that does not limit any of them to mere countability.

But all that isn't a blank check for us to claim that we know how
to define exactly what the natural numbers be, however countable they
might collectively be.

>>
>> N is finitely defined,

Fundamentally, that doesn't mean much though!

_One is free_ do define the naturals as that in which there are
finitely may counter examples of the Goldbach Conjecture.

Or the other way around! :-(

> 2^N, or the set of all subsets of N is finitely
>> defined. N is countably infinite, 2^n is uncountably infinite.
>
> Yep.

Exactly. That's what I'd say too: "So what"?

All that "knowledge" and "definition" and in the end it's still
impossible to know if there are finitely may counter examples of
the Goldbach Conjecture!

We indeed have been walking on the roof, so to speak, as far as
mathematics and mathematical "logic" is concerned.

--
----------------------------------------------------
There is no remainder in the mathematics of infinity.

NYOGEN SENZAKI
----------------------------------------------------

[Virgil]

In article
<cbb9d0e3-0a22-426c...@p17g2000vbn.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 15 Dez., 13:36, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 15, 2:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
> >
> > > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > Thus the Infinite binary tree is UNCOUNTABLE.
> >
> > > > > Since it can be proved to be countable, there is a contradiction.
> >
> > > > How do you prove that it is countable?
> >
> > > First: A Tree that contains all nodes also contains all reals of the
> > > unit interval.
> >
> > > I constrcut the tree, i.e., all nodes, by countably many infinite
> > > paths. But I don't tell you which paths I use. If you claim that more
> > > paths are available, then you must find out by means of the nodes of
> > > the tree which paths I have not used. Of course you cannot solve this
> > > task.
> >
> > Of course we can. Actually I already did it in the head post but you
> > keep ignoring it. You say that you constructed a COUNTABLE tree,
>
> No, I say I construct the complete Binary Tree. And I do so because no
> node is missing.

I can construct a set which includes root based sequences of
parent-child linked nodes including of every one of the nodes but
having absolutely no paths from the infinite tree at all

I can also construct all sorts of sets of paths collectively containing
every node, but no set containing all paths of the tree.

Their existence in no way affects a complete set of paths.

>
> this
> > by DEFINITION means that there must be a bijection between the set of
> > all paths of that tree and the set N of all naturals.

Not until you have proved, which you have not done, that you have ALL
POSSIBLE paths.

>
> There is a bijection of the set (A u B) of all path A that I use and
> all paths B that you can identify as missing.

Until you sow us the set of all paths you have included, which according
to your claim of countability should be listable, we cannot tell which
ones are missing.

But any listing of them instantly would instantly prove that you do not
have all of them.

>
> > Then use the
> > above approach and illicit the missing path
>
> No dreamworks please. I have constructed the complete Binary Tree
> using a countable set A of infinite paths.

Such a countable set must be, at last in principle, listable to be
countable, but necessity of a complete list existing instantly proves
that set to be incomplete!



> What path is missing?

Until we know which paths are included we would only be guessing, but if
their are only countably many they can in principle at least, be listed.
But any such listing proves the incompleteness of that list.

And being not even in principle listable would mean uncountable.

> If
> you cannot identify it by nodes, then it is not defined by nodes. Or
> do you believe that belief in paths is sufficient in mathematics?

I believe that WM has short circuit in his mental wiring which is
triggered by any situation suggesting the existence of uncountable sets.
--

[Sam Sung]

Nam Nguyen schrieb:

> On 15/12/2012 3:23 PM, Sam Sung wrote:
>> Virgil write:
>>
>>> Every set , including uncountable ones , is defined by a finite
>>> definition,
>
> Of course. One wouldn't dare to define a mathematical concept using
> an infinite description, naturally.

No, thinking is not a matter on daring.

>> but that does not limit any of them to mere countability.
>
> But all that isn't a blank check for us to claim that we know how
> to define exactly what the natural numbers be, however countable they
> might collectively be.

Same as the day of week did not change much as of today.

>>> N is finitely defined,
>
> Fundamentally, that doesn't mean much though!

Oh! Doh...

> _One is free_

Me is not.

> do define the naturals as that in which there are
> finitely may counter examples of the Goldbach Conjecture.

For: one is not to define all (including the gone) mondays
to be some other day.

> Or the other way around! :-(

?

>> 2^N, or the set of all subsets of N is finitely
>>> defined. N is countably infinite, 2^n is uncountably infinite.
>>
>> Yep.
>
> Exactly. That's what I'd say too: "So what"?

Just say it.

> All that "knowledge" and "definition" and in the end

Aaah, in the end...

> it's still impossible to know

Math is thinking MORE than observing.

> if there are finitely many counter examples of
> the Goldbach Conjecture!

Did you drink proper amounts of water today?

> We indeed have been walking on the roof, so to speak, as far as
> mathematics and mathematical "logic" is concerned.

Bullshit.

[Virgil]

In article
<40446e76-b057-4c82...@gu9g2000vbb.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 15 Dez., 19:28, Zuhair <zaljo...@gmail.com> wrote:
>
> > You are the one who is DREAMING. I told you that neither you nor
> > anybody else can construct the complete infinite binary tree by
> > countably many paths, this is just impossible, this is like saying I
> > can draw the circle that is not a circle.
>
> No. I did it several times. Years ago I did it twice before breakfast
> already.

Like the RED QUEENs ability to believe six impossible things before
breakfast?

> And I will show you that my construction does not leave out
> one single node.

It is not the overlooking of nodes which is at issue, but the
overlooking of paths.

> But before i tell you my construction try to find a
> path that I did not use in the countably set of all paths that are
> sufficient,

This is not a game of hide and seek.

In order for us to be able to tell you which paths you are missing, you
must first tell us which ones are not.

And this you coyly refuse ever to do.

> >
> > I showed you the impossibility of that and yet you are not
> > understanding what is written.
> >
> > Take any complete infinite binary tree then if you say it is countable
>
> The set of paths that I used is countable.

Quite possible, but you have never presented any adequate proof that you
have all of them.

> And you will see it when I
> tell you what I used.

But you never do tell us. You only try to tease us.

In mathematics, it does not do to hide things the way you do and then
expect anyone to believe what you say about what you are so carefully
keeping in hiding.

> But before I unveil my construction let me know
> whether the constructed Binary Tree is complete in your opinion. Here
> it is:
>
> 0.
> 0 1
> 0 1 0 1
> ...
>
> Every level starts with 0, is alternating between 0 an 1, and has
> twice as many nodes as the level above.
>

That does not tell us how to identify a path in yiur incomplete diagram,
and since paths are the issue, your incomplete diagram is essentially a
diversion and an irrelevancy.

Others have proved, well past WM's ability to disprove, that a complete
infinite binary tree has more paths than can be listed, i.e., the set of
all such ch paths is uncountable.

WM continues to argue ineffectively that his infinite binary tres do not
work that way.

To which we can only say that WM must be living in a different world,
such as the one that has been called Wolkenmuekenheim by several people
other than WM.
--

[Virgil]

In article
<e1f74dc5-ba35-4418...@f19g2000vbv.googlegroups.com>,

Thus you have no path that eventually becomes '010101...', while every
TRULY complete such tree has infinitely many of them, along with
uncountably many other "tails".


>
> Regards, WM
--

[Virgil]

In article
<2f0912e5-e54e-47d3...@u19g2000yqj.googlegroups.com>,

Or nodes with finite indices, when speaking of paths.

But, as it transpires, every node in every path HAS a finite index, so
it is perfectly valid for infinite strings of nodes, and thus also for
paths.

WM loses again!
--

[Sam Sung]

Virgil schrieb:

...

> WM loses again!

But he is NOT to be intending 'to play' the good...

[Nam Nguyen]

[Virgil]

In article
<b2c064a1-86d5-46ba...@hf3g2000vbb.googlegroups.com>,

There may be none that differs from all your listed paths at the same
level but there are lots that differ from the nth path in at least the
n+m-th position, for various m, which is quite enough.


Another trouble with WM's argument above is that not all paths in any
COMPLETE infinite binary tree can have the same tail.

In fact, every possible path must occur as a tail infinitely often for
other paths:
Let t be the tail
0.t
0.0t
0.1t
0.00t
0.01t
0.10t
0.11t
and so on ad infinitum are all paths.
--

[Sam Sung]

Oooh...

>>--
----------------------------------------------------
There is no remainder in the mathematics of infinity.

NYOGEN SENZAKI
----------------------------------------------------

No more mind no more remind...

[Virgil]

On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 14 Dez., 23:25, Zuhair <zaljo...@gmail.com> wrote:
>
>
>
>
>
>
>
>
>
> > This is a continuation of post about the infinite binary tree raised
> > by WM in this Usenet.
>
> > To review this subject in brief. The infinite binary tree is a single
> > rooted labeled tree. Each node must be exclusively labeled either by 0
> > or by 1. The root node receives the label 0. Each node have exactly
> > two children nodes and one is labeled by 0 and the other by 1.
>
> > Diagrammatically this is:
>

> >     0
> >    /  \
> >   0   1
> >  / \  /  \
> > 0 1 0  1

> > .. .. ..  ..
> > .
> > .
>
> > Now whenever "infinite binary tree" is mentioned then it is to be
> > understood that we are referring to the above one.
>
> > A path shall be defined as a linear sub-graph of the above tree that
> > begins with 0 in which all edges have the same direction (i.e. from
> > above downwards)
>
> > Example: The following linear graph is a path of the above tree:
> > 0
> >  \
> >   1
>
> > Another example is:
>
> >   0
> >  /
> > 0
> >  \
> >   1
>
> > Now Any sub-tree of the above tree that has its root labeled 0 is said
> > to be countable iff the set of all paths of it is bijective to the set
> > of all natural numbers.
>
> > Now it is obvious that the infinite binary tree spoken about above
> > have only countably many nodes. But how many paths does it have? is it
> > countable?
>
> > The answer is NO, it is not countable!
>
> The other answer is yes. Hence, there is a contradiction.
>

> Zuhair, you cannot argue against a contradiction between A and B by
> proving over and over that A is correct.

The contradiction is obvious, so that only one can be correct.
So proving either automatically disproves the other.

And so Zuhairs proof is indeed relevant as a disproof of the other claim.

>
>
>
> > The infinite binary tree depicted above have UNCOUNTABLY many paths.
>
> > This is a simple Corollary of Cantor's diagonal argument actually.
>
> Since the Binary Tree can be constructed by counatbly many paths such
> that noone can discern whether and which paths are lacking, we have a
> contradiction.

Until WM, or someone, has LISTED ALL those allegedly countable paths of
WM, or shown that they can be listed, there is no evidence of their
being only countably many of them.

But as soon as such an argument is made, it will establish the
incompleteness of any such list.

Thus Wm is caught between a rock and a hard place in his eternal
prevarication.

>
> Since it can be proved to be countable, there is a contradiction.
>

How do you prove that it is countable?

The only unequivocal proof wousd be a listing of all of them.

but WM cannot afford to try listing them, as any such listing would
instantly prove the incompleteness of his set of paths.

Countability implies listablity, but listability of the set of all such
paths is disprovable by the diagonal argument.

Thus non-listability proves non-countablity.

And WM's case is dead in the water.
--

[Virgil]

In article
<c3682253-48ed-4061...@c16g2000yqi.googlegroups.com>,
Rupert <rupertm...@yahoo.com> wrote:

> On Dec 15, 12:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
> >

> > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:

> > > > > Thus the Infinite binary tree is UNCOUNTABLE.
> >

> > > > Since it can be proved to be countable, there is a contradiction.
> >
> > > How do you prove that it is countable?
> >

> > First: A Tree that contains all nodes also contains all reals of the
> > unit interval.
> >
> > I constrcut the tree, i.e., all nodes, by countably many infinite
> > paths.
>

> Tell us more about this construction.
>

> > But I don't tell you which paths I use. If you claim that more
> > paths are available, then you must find out by means of the nodes of
> > the tree which paths I have not used. Of course you cannot solve this
> > task.
> >

> > Hence, what you call infinite paths, is not defined by nodes but only
> > by finite words like 1/3. But there are only countably many finite
> > words.
> >

> > Regards, WM

WM explaining his CONstruction ( and CON is the operative bit) would
lead to WM's DIStruction.
--

[Sam Sung]

Virgil writes:

> ... WM explaining ..

WM is not much required to understand worlds...

So what... we love it but cannot help (n)either...

[Zuhair]

Ok but the resulting construction is not the INFINITE binary tree
we've already defined. For example the path representing the decimal
expansion of pi is not among the paths of your constructed tree. Of
course the tree you've constructed that way is COUNTABLE, I agree, but
it is not a complete infinite binary tree, it is not even near. Now
even if you add any tail of your choice, I mean suppose you have a
countable set of tails T of your choice, and you append those tails to
countably many finite stumps of the infinite binary tree, still the
resulting construction is countable, but it is not the complete
infinite binary tree, why because simply and I say simply apply the
method I've showed you in the head post on your so constructed tree
and you will recover a path of the complete binary tree that is not
among the paths of your constructed tree. I showed you that, but you
keep refusing to understand that part.

There is no way, and I say absolutely no way whatsoever to append
countably many tails to a countable tree of finite paths of the
complete infinite binary tree to get a complete infinite binary tree.
This is just impossible.

Look WM, I'm trying to tell you NO MATTER what construction you use if
the result of the construction is a countable tree, then use the above
methodology which is of course basically Cantor's and you will recover
a path of the complete finite binary tree that is not among the paths
of the tree you've constructed. I can always show that WHATEVER
countable binary tree you construct, no matter how you construct it,
it doesn't really matter, at the end Cantor's diagonal argument will
be applicable to it and an it will be shown to have a missing path.

You are just trying to do the impossible.

Zuhair

[WM]

> Ok but the resulting construction is not the INFINITE binary tree
> we've already defined.

Liar! The resulting construction is what you accepted as the complete
infinite Binary Tree CIBT. Should I

The set of paths that I used is countable. And you will see it when I
tell you what I used. But before I unveil my construction let me know
whether the constructed Binary Tree is complete in your opinion. Here
it is:

0.
0 1
0 1 0 1
...


Every level starts with 0, is alternating between 0 an 1, and has
twice as many nodes as the level above.

You said: This tree has ALL possible binary paths as paths of it.

In other words it is COMPLETE.

> For example the path representing the decimal
> expansion of pi is not among the paths of your constructed tree.

First: pi is not a real number of the unit interval
Second: Every finite initial segment of pi - 3 is in the CIBT.

And even pi - 3 is there because I cheated. I did not use the tails
000... but the tails with the bit-string of pi-3.

>Of
> course the tree you've constructed that way is COUNTABLE, I agree, but
> it is not a complete infinite binary tree, it is not even near.

You are dreaming. But let us play again. I construct a CIBT by using
tails that I do not publish. You have to find out what paths are
missing.

> Now
> even if you add any tail of your choice, I mean suppose you have a
> countable set of tails T of your choice, and you append those tails to
> countably many finite stumps of the infinite binary tree, still the
> resulting construction is countable, but it is not the complete
> infinite binary tree, why because simply and I say simply apply the
> method I've showed you in the head post on your so constructed tree
> and you will recover a path of the complete binary tree that is not
> among the paths of your constructed tree. I showed you that, but you
> keep refusing to understand that part.

If you are really unable to recognize it yourself, then listen to
others like George Greene: The CIBT is complete. It is impossible to
find a digit sequence defined by nodes that is not contained in it.

Regards, WM

[WM]

On 16 Dez., 00:04, Virgil <vir...@ligriv.com> wrote:
> In article
> <40446e76-b057-4c82-95c6-ccb75fce0...@gu9g2000vbb.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 15 Dez., 19:28, Zuhair <zaljo...@gmail.com> wrote:
>
> > > You are the one who is DREAMING. I told you that neither you nor
> > > anybody else can construct the complete infinite binary tree by
> > > countably many paths, this is just impossible, this is like saying I
> > > can draw the circle that is not a circle.
>
> > No. I did it several times. Years ago I did it twice before breakfast
> > already.
>
> Like the RED QUEENs ability to believe six impossible things before
> breakfast?

Like her, but my case is quite different: I *proved* *possible*
things.

>
> > But before  i tell you my construction try to find a
> > path that I did not use in the countably set of all paths that are
> > sufficient,
>
> This is not a game of hide and seek.

This seems to be the only chance to rouse you from your infintary
dreams.
If you have a rest of scientific honesty, then you have to confess
that you cannot find which tails I used but also you cannot find any
path, - defined by nodes only(!), not by a finite definition(!!) -
that is missing in my CIBT.

Regards, WM

[WM]

On 16 Dez., 00:08, Virgil <vir...@ligriv.com> wrote:

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

> > I constructed this very tree by all finite paths that extend from the
> > root node to a node at level n. Then I appended the sequence 000...
>
> Thus you have no path that eventually becomes '010101...',

In mathematics such a claim must be proved by nodes. Which is the
first bit of 0.010101... that is missing in my CIBT?

Regards, WM

[Zuhair]

The CIBT (complete infinite binary tree) is defined in the following
manner:

CIBT is a Tree
The root node of CIBT is labeled 0
Each node of CIBT has two child nodes each having a different label
from the other
Any node of ClBT receives only ONE label that is either 0 or 1.

The ClBT is COMPLETE, in the sense that all possible binary paths
belong to it. I already said that.


But YOUR construction which is done by appending infinite paths to
countably many finite stumps of the of the ClBT is NOT the CIBT
itself, it is another tree, at best it is a proper substree of the
CIBT, actually it can even be not a subtree of CIBT, however I'll take
that the appending process is respective of the binary structure of
CIBT, so this constructed tree of yours would be as I said just a
PROPER subtree of the CIBT. And the countability of your constructed
tree is not important since it is not the CIBT itself. It is very easy
to show a missing path from your constructed tree that of course
belongs to the CIBT. As I said again and again just apply the diagonal
argument on your so constructed tree and you'll see the missing path.

Can you tell me what is your proof that the tree you've constructed by
the appending process is the CIBT??? try to prove it, and you will see
that you cannot.

All you have done up till now is to provide a blind assertion that the
tree constructed by the appending process is the CIBT itself? I didn't
see any proof of that, you didn't present any? so if you have a proof
of that then show it, so that one can see if it is valid or not?

YOU (and anyone who agree to your claims) is the one who must supply a
proof that the tree constructed by the appending process is the
complete infinite binary tree itself. But do you have such a proof? I
simply think you CANNOT have any. But lets see if you have.

Again, supply the proof that your constructed tree is the CIBT?

Zuhair

[Zuhair]

I told you it doesn't matter as long as you are using a COUNTABLE set
of tails, then I can always find a path that is missing from your
CONSTRUCTED tree (please don't confuse your constructed tree with the
CIBT, your constructed tree is NOT the CIBT itself).

Lets say you have the set T of all tails that you will append to a
countably many finite stumps of the CIBT.

Now T is countable? as you say.

So just simply apply the diagonal argument on T.

You have a function f from N to T such that f is bijective (because
this what saying that T is countable means)

Run the diagonal argument of Cantor on f, and you'll derive the anti-
diagonal path d_f which is MISSING from T definitely.

Now either d_f starts with 0, and by then it will be the missing path
from your constructed tree.
Or d_f do not start with 0, by then just construct d_f* by putting 0
in front of d_f, and we'll get a missing path from your constructed
tree.

WE don't need to know the details of the members of T, that's what you
are not understanding. All what we need to know is if T is countable
or not.

There is also a very big problem with your claim, if you don't know
the details of members of T, then how you come to know that the tree
constructed by appending countably many finite stumps of CIBT with
members of T is itself the CIBT??? how you can prove that?

The reality of that matter is simply that: NO matter what construction
you use if the result is a countable tree, then this tree is NOT the
CIBT. Why because the CIBT is UNCOUNTABLE! try to hammer that into
your head.

Zuhair

[Zuhair]

Do you think that calling me a Liar, is a valid way of proving that
the tree you've constructed is the CIBT itself?

You need to prove that the tree constructed by your choice appendings
of infinite tails to countably many finite stumps of the CIBT, is
itself the CIBT.

So far you didn't supply any. So your claim is just a blind assertion
nothing more nothing less.

Zuhair

[WM]

On 16 Dez., 10:37, Zuhair <zaljo...@gmail.com> wrote:

> > Here it is:
>
> >      0.
> >   0   1
> > 0 1 0 1
> > ...
>
> > Every level starts with 0, is alternating between 0 an 1, and has
> > twice as many nodes as the level above.
>
> > You said: This tree has ALL possible binary paths as paths of it.
> > In other words it is COMPLETE.

> The CIBT (complete infinite binary tree) is defined in the following
> manner:
>
> CIBT is a Tree
> The root node of CIBT is labeled 0
> Each node of CIBT has two child nodes each having a different label
> from the other
> Any node of ClBT receives only ONE label that is either 0 or 1.

> Can you tell me what is your proof that the tree you've constructed by
> the appending process is the CIBT??? try to prove it, and you will see
> that you cannot.

My CIBT satisfies exactly what you said above. What is further to
prove?

>
> All you have done up till now is to provide a blind assertion that the
> tree constructed by the appending process is the CIBT itself? I didn't
> see any proof of that, you didn't present any? so if you have a proof
> of that then show it, so that one can see if it is valid or not?

The tree need not be "proved", since it is defined and constructed
according to that definition.

If you doubt that I have correctly constructed it, then find a node
that is missing.

Note: The CIBT is defined by its *nodes* (as you said above). Paths
that are not defined by nodes at finite levels may be or may not be in
it.

Regards, WM

[Rupert]

On Dec 15, 7:07 pm, WM <mueck...@rz.fh-augsburg.de> wrote:

> On 15 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote:
>
>
>
>
>
>
>
>
>
> > On Dec 15, 12:35 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > On 15 Dez., 11:21, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > > On Dec 15, 10:56 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > > > Thus the Infinite binary tree is UNCOUNTABLE.
>
> > > > > Since it can be proved to be countable, there is a contradiction.
>
> > > > How do you prove that it is countable?
>
> > > First: A Tree that contains all nodes also contains all reals of the
> > > unit interval.
>
> > > I constrcut the tree, i.e., all nodes, by countably many infinite
> > > paths.
>
> > Tell us more about this construction.
>

> I use all finite paths. Every node of the Binary Tree is the end of a
> path. Then I append these countably many paths by a tail according to
> my choice. For instance I can use the tail
> 000...
> or
> 010101...
> or
> the bit sequence of pi
> or anything else, for instance a mix of many tails.
>

> In order to show you that you are dreaming if you think that infinite
> paths could be identified by their nodes, I don't tell you what tails
> I have used. If you don't sleep to deep, then you will wake up and
> recognize that infinite paths are merely defined by finite
> definitions, and hence belong to a countable set.
>

> Regards, WM

Okay, so you seem to be saying that you would take all the finite
paths and append an infinite tail to each one, thereby obtaining a
countably infinite collection of infinite paths. Is the claim then
that this would be equal to the collection of all infinite paths? Or
not?

[WM]

> not?-

It is equal to the collection of all paths that are defined solely by
their nodes. It is equal to all finite initial strings of digits that
can be applied in a Cantor-list. (Every digit changed there has a
finite index.)

Actually infinite paths like that of 0.010101... = 1/3 (in binary)
cannot be defined by nodes. No infinite sequence has ever been defined
by its terms. Those things have to be defined by finite definitions
like "0.010101..." or "1/3".

Regards, WM

[Rupert]

Okay. That all sounds fair enough.

So, you were going to show us that there are only countably many
paths. When are you going to do that?

[Zuhair]

so what you are saying is that an actual infinite path like
0.010101... is not among the tree you've constructed. If so then how
do you claim that what you've constructed is the CIBT?

Zuhair

[Zuhair]

No. you are one who must prove that what you've constructed IS the
CIBT. since all of your claim depends on that central issue, you
cannot just make assertions and expect everyone to accept them without
supplying proofs of those assertions. You must have a proof in your
mind that the constructed tree of yours is itself the CIBT, don't you?

> Note: The CIBT is defined by its *nodes* (as you said above). Paths
> that are not defined by nodes at finite levels may be or may not be in
> it.

The CIBT that I've mentioned is COMPLETE, it is an actual infinity
binary tree and it DOES contain ALL actually infinite binary paths. We
know that the path 0.010101.. IS a path of CIBT, and it is definable
by its nodes, why you say it is not? by the way what do YOU exactly
mean by "definable by its nodes" you need to explain that to avoid
being lost in translation.

While your constructed tree is INCOMPLETE. It does miss many paths of
the CIBT from it, and I have showed you how to construct in details
the missing paths from it in the head post, but you keep ignoring it.

Zuhair

[WM]

> paths. When are you going to do that?-

First I showed you that there are only countably many paths that are
defined by nodes. You seemed to agree. Call them the set A.

Well, the other "paths" cannot be defined by nodes. Call them the set
B. They need definitions by finite words. And there are only countably
many finite words.

Now consider the union of two countable sets A and B.

Regards, WM

[WM]

On 16 Dez., 13:02, Zuhair <zaljo...@gmail.com> wrote:

> > Actually infinite paths like that of 0.010101... = 1/3 (in binary)
> > cannot be defined by nodes. No infinite sequence has ever been defined
> > by its terms. Those things have to be defined by finite definitions
> > like "0.010101..." or "1/3".
>
> > Regards, WM
>
> so what you are saying is that an actual infinite path like
> 0.010101... is not among the tree you've constructed.

And if I have not told you the truth? What would you say if I had used
the tails 010101... Is that kind of reasoning reasonable? Is that
mathematics???

> If so then how
> do you claim that what you've constructed is the CIBT?

I do not claim it. I have done it. Try to find a node of that path
that is not in the Binary Tree. You will fail. Can't you think so
complex thoughts?

You have no chance to distinguish the path 1/3 by nodes because there
is no binary fraction of 1/3. All you have been told by matheologians
about completed infinity was and is completed nonsense.

Regards, WM

[WM]

On 16 Dez., 13:10, Zuhair <zaljo...@gmail.com> wrote:

> > If you doubt that I have correctly constructed it, then find a node
> > that is missing.
>
> No.

So you doubt that you are able to find a missing node? You are
correct. There are all nodes by definition in the CIBT.

> you are one who must prove that what you've constructed IS the
> CIBT. since all of your claim depends on that central issue, you
> cannot just make assertions and expect everyone to accept them without
> supplying proofs of those assertions. You must have a proof in your
> mind that the constructed tree of yours is itself the CIBT, don't you?

I have constructed the Binary Tree according to your definition. And
if you don't believe me, then construct it it yourself by all finite
paths. Adding a tail of any kind to every path will not make any of
the nodes disappear.

>
> > Note: The CIBT is defined by its *nodes* (as you said above). Paths
> > that are not defined by nodes at finite levels may be or may not be in
> > it.
>
> The CIBT that I've mentioned is COMPLETE, it is an actual infinity
> binary tree and it DOES contain ALL actually infinite binary paths.

You defined that every node has two child nodes. No word about paths.
Do you refuse to remember what you said? I don't think that such
childish babbling as you like to perform here has to do anything with
mathematics or "discourse".

Regards, WM

[Ross A. Finlayson]

These paths are lexicographically ordered. In that order, their order
type is 2^w (that is countable). In that order, "the" antidiagonal
over the expansions is the same as the 2^w'th element. In that order,
there are no elements for an m'th element, m > 1, between the zeroeth
and first.

How does the diagonal method or nested intervals show these paths
uncountable, directly?

Regards,

Ross Finlayson

[Rupert]

How do you know that there exists a countable language such that every
path can be defined in this language?

> Regards, WM

[George Greene]

On Dec 16, 5:02 am, Rupert <rupertmccal...@yahoo.com> wrote:
> Okay, so you seem to be saying that you would take all the finite
> paths and append an infinite tail to each one, thereby obtaining a
> countably infinite collection of infinite paths. Is the claim then
> that this would be equal to the collection of all infinite paths? Or
> not?

Rupert, please! NOTHING IS GAINED by appending the infinite tails!
The class OF ALL FINITE paths *ALREADY*COVERS* all the nodes!
His claim is that his countable collection of infinite paths would
cover all the nodes, but that, THOUGH TRUE, DOES NOT imply that his
collection
is equal to the collection of all infinite paths! THE COLLECTION OF
ALL *FINITE* PATHS *ALSO*
covers all the nodes, yet IT is OBVIOUSLY NOT equal to the collection
of all finite paths!

I don't see how you are so LATE to this. WM was already doing this
the day you arrived.
He's been at it for a LONG time. He is no less full of shit now than
he was on the day he started.
Zuhair is new and can be forgiven. You, not so much.

[Rupert]

Is my engaging in conversation with WM some kind of problem for you?

[George Greene]

On Dec 16, 7:02 am, Zuhair <zaljo...@gmail.com> wrote:

> so what you are saying is that an actual infinite path like
> 0.010101... is not among the tree you've constructed. If so then how
> do you claim that what you've constructed is the CIBT?

Zuhair, PLEASE, you are asking A STUPID question.
It is the complete infinite binary tree BECAUSE HE SAID it is the
complete infinite
binary tree. THAT IS NOT at issue! WHAT IS at issue is whether the
number of
PATHS THROUGH this tree is countable or Un! He has given a countable
set of paths
and has claimed that it must be all of them. As soon as you exhibit a
path that is
not in his set, THAT SHOULD *SHUT*HIM*UP*, but it DOESN'T because HE
IS A LYING FOOL.
If you are not simply pointing out to him that he is a liar and a
fool, then you are part of the problem.

[George Greene]

On Dec 16, 10:54 am, Rupert <rupertmccal...@yahoo.com> wrote:
> How do you know that there exists a countable language such that every
> path can be defined in this language?

Even if there did, it wouldn't help.
If the language is countable then there are only a countable number of
definitions, so the definitions-of-all-paths would be a countable list
THAT COULD THEREFORE BE ANTI-DIAGONALIZED.
Remember, we usually used INdirect proof for this. We usually BEGAN
by CONCEDING that what "they" claimed, existed. So we DON'T ask them
"how they know" that something exists. We just GRANT it. Their
argument
is SO lame that even CONCEDING it does NOT HELP them!

[Ross A. Finlayson]

Well there aren't any uncountable languages. (Thank you I'd be
interested in as to where there were, with regards to then theoretical
consequences.)

L([01]*)

Basically, does the Kleene (Klayn-ee) star allow infinite words?

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

No, it doesn't. Yet, Turing machines allow infinite programs. As
well, several of the references refer to the closure not being finite.

It's always seemed incongruous that 2^w, here the order type of
BT_oo's domain, is countable.

Regards,

Ross Finlayson

[Ross A. Finlayson]

Hi Greene,

Hey Happy Holidays bud. Your erudite opinion is appreciated, though
of course we can simply read Muckenheim's posts and evaluate the truth
values of their statements, here mathematically, and in mathematics.

[Rupert]

On Dec 16, 5:55 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

There are plenty of uncountable languages. For example, you can have a
first-order language with uncountably many predicate symbols.

>         L([01]*)
>
> Basically, does the Kleene (Klayn-ee) star allow infinite words?
>

You can have infinite words. See for example the infinitary languages
discussed in Chapter 10 of Drake's "Set Theory: An Introduction to
Large Cardinals".

But that wouldn't be covered by the Kleene star operation.

But that's neither here nor there. You don't need infinite words, you
just need to have an uncountable alphabet.

>        http://en.wikipedia.org/wiki/Kleene_star
>
> No, it doesn't.  Yet, Turing machines allow infinite programs.

No.

> As
> well, several of the references refer to the closure not being finite.
>

Closure of what? Closure in what sense?

> It's always seemed incongruous that 2^w, here the order type of
> BT_oo's domain, is countable.
>

The set of all infinite paths in a binary tree does not have an
ordering of order type 2^w (where you take exponentiation in the
ordinal sense, so that 2^w=w).

> Regards,
>
> Ross Finlayson

[WM]

I do know that in every language (as well as in all languages together
since aleph_0 * aleph_0 = aleph_0, and since there are no uncountable
languages) only countably many paths can be defined. What cannot be
defined in any language is not a path. Therefore I know that there are
not more than countably many paths - at least paths that can be
applied in mathematics and can be the result of any diagonalization.

Regards, WM

[WM]

On 16 Dez., 17:27, George Greene <gree...@email.unc.edu> wrote:
> On Dec 16, 5:02 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > Okay, so you seem to be saying that you would take all the finite
> > paths and append an infinite tail to each one, thereby obtaining a
> > countably infinite collection of infinite paths. Is the claim then
> > that this would be equal to the collection of all infinite paths? Or
> > not?
>
> Rupert, please! NOTHING IS GAINED by appending the infinite tails!
> The class OF ALL FINITE paths *ALREADY*COVERS* all the nodes!
> His claim is that his countable collection of infinite paths would
> cover all the nodes, but that, THOUGH TRUE, DOES NOT imply that his
> collection
> is equal to the collection of all infinite paths!  THE COLLECTION OF
> ALL *FINITE* PATHS *ALSO*
> covers all the nodes, yet IT is OBVIOUSLY NOT equal to the collection
> of all finite paths

which obviously are not definable by nodes and therefore cannot be the
result of a diagonalization. They are not connected to anything Cantor
did. Diagonalization does not leave the realm of paths which are
definable by nodes only. That is the narure of diagonalization.

Regards, WM

[WM]

On 16 Dez., 17:29, George Greene <gree...@email.unc.edu> wrote:
> On Dec 16, 7:02 am, Zuhair <zaljo...@gmail.com> wrote:
>
> > so what you are saying is that an actual infinite path like
> > 0.010101... is not among the tree you've constructed. If so then how
> > do you claim that what you've constructed is the CIBT?
>
> Zuhair, PLEASE, you are asking A STUPID question.
> It is the complete infinite binary tree BECAUSE HE SAID it is the
> complete infinite
> binary tree.  THAT IS NOT at issue!  WHAT IS at issue is whether the
> number of
> PATHS THROUGH this tree is countable or Un!  He has given a countable
> set of paths
> and has claimed that it must be all of them. As soon as you exhibit a
> path that is
> not in his set, THAT SHOULD *SHUT*HIM*UP*, but it DOESN'T because HE
> IS A LYING FOOL.

But I need not publish this set. How do you identify a missing path
then?
Remember: Cantor's diagonal is defined by digits or nodes only. But
you cannot identify a path by nodes that is missing in the tree
constructed with countably many paths. You must trust that I am not
lying. How can you do so if you accuse me of lying? Is your
"mathematics" not a little bit inconsistent???

Regards, WM

[Ross A. Finlayson]

The Universal Turing Machine has infinite resources, see for example
the recent discussions on Rice's theorem.

Here, the set 2^w has order type 2^w > w.

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

If the paths, of all of them, from the root through all non-leafs, at
each level n, have order type 2^n, why do they not, at level or day
omega = w, have order type 2^w?

Here that is a passing reference to Conway's surreal numbers.

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

If at day omega as the surreal numbers may well have an initial
construction from the infinite balanced binary tree, the paths have
the same order type as omega, isn't there an order-preserving
bijection between omega and the paths? If they have order type 2^w,
is not that yet countable?

Perhaps simply you could show how the diagonal method or nested
intervals applies directly to the tree.

Regards,

Ross Finlayson

[WM]

On 16 Dez., 17:31, George Greene <gree...@email.unc.edu> wrote:
> On Dec 16, 10:54 am, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > How do you know that there exists a countable language such that every
> > path can be defined in this language?
>
> Even if there did, it wouldn't help.
> If the language is countable then there are only a countable number of
> definitions, so the definitions-of-all-paths would be a countable list
> THAT COULD THEREFORE BE ANTI-DIAGONALIZED.

Nonsense. You can diagonalize a Cantor list. You cannot diagonalize a
list of finite expressions like

The number pi
1/3
Greene's number
...

just because it is never complete. Remember, in order to apply
Cantor's diagonalization you must assume a complete list that
afterwards may not be changed.

> Remember, we usually used INdirect proof for this.  We usually BEGAN
> by CONCEDING that what "they" claimed, existed.  So we DON'T ask them
> "how they know" that something exists.  We just GRANT it.

We just grant that diagonalization is valid for every finite list. To
assume that it holds for an infinite list (and that this infinite list
is somwhow complete, such that it may not be changed afterwards), that
is so much nonsense that you seem to have left your intelligence over
that idea - in case it ever has been with you.

Regards, WM

[WM]

On 16 Dez., 18:01, Rupert <rupertmccal...@yahoo.com> wrote:

>
> There are plenty of uncountable languages.

No languages that could be used for discourse.
An uncoutable language needs uncountably many characters. They cannot
be distinguished by finite sequences of bits. But the latter are
required in order to apply them, for instance here in this discussion.

You may believe that every real number can represent a character - but
not before all these real numbers have been defined by a finite
language and not in order to define the real numbers.

Believing, like GG, that these things are simply existing, is pure
matheology but not mathematics.

Regards, WM

[Rupert]

On Dec 16, 6:18 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com>

The input to any Turing machine has finite length, by definition.

> Here, the set 2^w has order type 2^w > w.
>
> http://en.wikipedia.org/wiki/Order_type
>

Not if you use the standard definition of exponentiation of ordinals.
If you have some other notion in mind, you should specify which one.

> If the paths, of all of them, from the root through all non-leafs, at
> each level n, have order type 2^n, why do they not, at level or day
> omega = w, have order type 2^w?
>

Because it simply doesn't follow. Before we can even sensibly discuss
this you need to give your definition of the order type 2^w, because
it is obviously not the one that I thought you had in mind.

> Here that is a passing reference to Conway's surreal numbers.
>
> http://en.wikipedia.org/wiki/Surreal_number
>
> If at day omega as the surreal numbers may well have an initial
> construction from the infinite balanced binary tree, the paths have
> the same order type as omega, isn't there an order-preserving
> bijection between omega and the paths?

No. There is not.

I don't really know why you think there would be; your argument is not
clear enough.

> If they have order type 2^w,
> is not that yet countable?
>

I don't know what definition of 2^w you have in mind so I can't
comment.

The surreal numbers with birthday omega are not countable.

> Perhaps simply you could show how the diagonal method or nested
> intervals applies directly to the tree.
>

Have you read Zuhair's discussion?

> Regards,
>
> Ross Finlayson

[Ross A. Finlayson]

Al-Jofar writes:
1: "This is a simple Corollary of Cantor's diagonal argument actually.
Proof: Let G be any countable subtree of the infinite binary tree that
is 0 rooted and such that all paths ending by 0-0-0-... or by
1-1-1-...are among the paths of G. "

'Let G be a subtree of the IBBT including all expansions representing
fractions by powers of two (there are two of those for each of those
values).'

2: "Let f be a bijective function from the domain N of all
naturals(except 0) to the set of all infinite paths of G. "

For the countable subtree only containing fractions by powers of two
and their dual representations, that's countable (and not obviously
all the paths).

3: "Construct the diagonal path d_f in the following manner: The root
(i.e. the 1st) node of d_f is 0 labeled. Now for n=1,2,3,.. ; The n
+1_th node of d_f is labeled by a label that is opposite to the label
of the n+1_th node of the path of G that f sends n to. "

Let G be all the paths. In lexicographic order, here as described as
BT _oo (binary tree's breadth-first traversal at infinity), augmented
with .0111... being first to reflect the modification of "the" anti-
diagonal, the generated anti-diagonal path is .0111..., that is the
least element of G in the ordering.

With an un-modified construction of the antidiagonal, the generated
anti-diagonal path is .111... and at the end of any course-of-passage
through the paths in their natural order, i.e., never before the end.
That is, there's a simpler corollary that is the same as the binary
anti-diagonal argument.

4: "Now clearly the diagonal d_f (actually the anti-diagonal but it
shall be called the diagonal for short) is a path and clearly it is
labeled in a way that is different from labeling of all paths of G. So
d_f is missing from G. "

In the course of passage with the natural order of the paths, d_f =
f(0) or here f(1) and is not missing from G.

5: "So any countable subtree of the infinite binary tree, that is 0
rooted and that has all paths ending with 0-0-0-.. or with 1-1-1..
among its paths; would be missing a path of the infinite binary tree.
"

That doesn't follow for the natural ordering of the paths by their
content as expansions. Then though while the paths are totally
ordered, well-ordering them would be as well-ordering the reals.

Then I'll happily accept that the anti-diagonal argument is much the
same for the list of expansions or the tree of expansions, then that
EF's antidiagonal is .111... and at the end of the list and BT's
antidiagonal is .111... and rightmost of the tree.

Thanks,

Ross Finlayson

[Zuhair]

Yes I realized that WM simply doesn't know the difference between a
tree and the set of all nodes of a tree. It is really a wast of time
to discuss those issues. He is way below the level of such discussion.

Zuhair

[WM]

On 16 Dez., 20:01, Zuhair <zaljo...@gmail.com> wrote:

> Yes I realized that WM simply doesn't know the difference between a
> tree and the set of all nodes of a tree.

And you don't know it either unless I tell you what paths I have used.

> It is really a wast of time
> to discuss those issues. He is way below the level of such discussion.
>

A very good argument to stick with matheological nonsense. Taken from
theology? There priest usually recognize that some people are way
below the level of recognizing God (in the correct way, i.e. the
correct God).

Regards, WM

[Rupert]

It is mathematics. Mathematics includes discourse about uncountable
sets. The fact that it would not be possible for a human to use such a
language is irrelevant.

[Rupert]

There are uncountable languages, but that's neither here nor there. We
can restrict attention to countable languages.

However, you haven't justified your claim that there are only
countable many countable languages, and in fact it can be shown that
this isn't true.

> What cannot be
> defined in any language is not a path.

This, too, is something which needs to be justified. A set theorist
would not accept that assumption.

> Therefore I know that there are
> not more than countably many paths - at least paths that can be
> applied in mathematics and can be the result of any diagonalization.
>

No. You don't know that.

> Regards, WM

[Rupert]

On Dec 16, 7:47 pm, "Ross A. Finlayson" <ross.finlay...@gmail.com>
wrote:

> Al-Jofar writes:
>
> 1: "This is a simple Corollary of Cantor's diagonal argument actually.
> Proof: Let G be any countable subtree of the infinite binary tree that
> is 0 rooted and such that all paths ending by 0-0-0-... or by
> 1-1-1-...are among the paths of G. "
>
> 'Let G be a subtree of the IBBT including all expansions representing
> fractions by powers of two (there are two of those for each of those
> values).'
>
> 2: "Let f be a bijective function from the domain N of all
> naturals(except 0) to the set of all infinite paths of G. "
>
> For the countable subtree only containing fractions by powers of two
> and their dual representations, that's countable (and not obviously
> all the paths).
>
> 3: "Construct the diagonal path d_f in the following manner: The root
> (i.e. the 1st) node of d_f is 0 labeled. Now for n=1,2,3,.. ; The n
> +1_th node of d_f is labeled by a label that is opposite to the label
> of the n+1_th node of the path of G that f sends n to. "
>
> Let G be all the paths.  In lexicographic order, here as described as
> BT _oo (binary tree's breadth-first traversal at infinity), augmented
> with .0111... being first to reflect the modification of "the" anti-
> diagonal, the generated anti-diagonal path is .0111..., that is the
> least element of G in the ordering.
>

No, it's not.

> With an un-modified construction of the antidiagonal, the generated
> anti-diagonal path is .111... and at the end of any course-of-passage
> through the paths in their natural order, i.e., never before the end.
> That is, there's a simpler corollary that is the same as the binary
> anti-diagonal argument.
>
> 4:  "Now clearly the diagonal d_f (actually the anti-diagonal but it
> shall be called the diagonal for short) is a path and clearly it is
> labeled in a way that is different from labeling of all paths of G. So
> d_f is missing from G. "
>
> In the course of passage with the natural order of the paths, d_f =
> f(0) or here f(1) and is not missing from G.
>

Nonsense.

> 5:  "So any countable subtree of the infinite binary tree, that is 0
> rooted and that has all paths ending with 0-0-0-.. or with 1-1-1..
> among its paths; would be missing a path of the infinite binary tree.
> "
>
> That doesn't follow for the natural ordering of the paths by their
> content as expansions.  Then though while the paths are totally
> ordered, well-ordering them would be as well-ordering the reals.
>

That's irrelevant.

[WM]

On 16 Dez., 21:34, Rupert <rupertmccal...@yahoo.com> wrote:

>
> It is mathematics. Mathematics includes discourse about uncountable
> sets.

But it does not include discourse using uncountably many characters,
because for that sake infinite strings of bits would be required. By
finite strings of bits only countably many characters could be used.
And if we restrict our conversation to those usable characters and
omit the others, then we discuss in a countable language.

> The fact that it would not be possible for a human to use such a
> language is irrelevant

It would not only be impossible for a human but impossible per se,
because there might be sentences that do never end.

Mathematics may contain many foolish ideas, They can be discussed. But
the language applied to discuss them must be free of foolish items and
must be usable by humans and other intellects. That's the way
mathematics works: It is mainly a discussion with others or with
oneself. Every item (including uncountable sets and inaccessible
cardinals) must have finite definitions. Therefore there is no
uncountable alphabet and there are not uncountably many languages.

And all mathematics that has been done by Cantor can be discussed in
any natural (i.e. countable) language. Further Cantor's diagonal
argument works solely with digits (i.e., nodes). A proof that nobody
can distinguish more than countable many paths in the Binary Tree is a
contradiction of uncountability.

Regards, WM

[Rupert]

On Dec 16, 9:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 16 Dez., 21:34, Rupert <rupertmccal...@yahoo.com> wrote:
>
>
>
> > It is mathematics. Mathematics includes discourse about uncountable
> > sets.
>
> But it does not include discourse using uncountably many characters,
> because for that sake infinite strings of bits would be required. By
> finite strings of bits only countably many characters could be used.
> And if we restrict our conversation to those usable characters and
> omit the others, then we discuss in a countable language.
>

You're missing the distinction between the metatheory and the object
theory. The metatheory in which mathematical discourse takes place, in
which we discuss various object theories, is in a countable language.
But the object theories could be in an uncountable language.

> > The fact that it would not be possible for a human to use such a
> > language is irrelevant
>
> It would not only be impossible for a human but impossible per se,
> because there might be sentences that do never end.
>

I don't know what the distinction is between "impossible for a human
to use" and "impossible per se".

In metamathematics, we can study languages other than the languages
which are in fact possible for humans to use.

> Mathematics may contain many foolish ideas, They can be discussed. But
> the language applied to discuss them must be free of foolish items and
> must be usable by humans and other intellects. That's the way
> mathematics works: It is mainly a discussion with others or with
> oneself. Every item (including uncountable sets and inaccessible
> cardinals) must have finite definitions. Therefore there is no
> uncountable alphabet and there are not uncountably many languages.
>

Not in languages that humans actually use, no. But in the universe of
discourse of metamathematics, there are such languages.

> And all mathematics that has been done by Cantor can be discussed in
> any natural (i.e. countable) language. Further Cantor's diagonal
> argument works solely with digits (i.e., nodes). A proof that nobody
> can distinguish more than countable many paths in the Binary Tree is a
> contradiction of uncountability.
>

It's not.

> Regards, WM

[WM]

On 16 Dez., 21:35, Rupert <rupertmccal...@yahoo.com> wrote:

> However, you haven't justified your claim that there are only
> countable many countable languages, and in fact it can be shown that
> this isn't true.

Then there is another contradiction. I can prove the contrary.

All characters belong to a countable set, because it must be possible
to encode them as finite strings of bits:

0
1
00
01
10
11
000
...

The list is countable. The meanings of words formed by these
characters and used for discourse must be defined by men. There will
never be more than a finite set of definitions. But even an infinite
set of languages would not surpass a countable set of words, because
aleph_0 * aleph_0 = aleph_0. We will never leave the countable domain
unless there are infinite words. But infinite words cannot be used in
discussions with others or with oneself, i.e., in mathematics.

Regards, WM

And languages are practized: A language is something to be defined by
men. There will never be more than a finite set of languages existing.

Further

>
> > What cannot be
> > defined in any language is not a path.
>
> This, too, is something which needs to be justified. A set theorist
> would not accept that assumption.

A set theorist would prefer to accept completed infinity.
But mathematics is mainly a language. And Cantor's proof can be done
in that language. Cantor's proof is limited to paths that can be
identified by their strings of digits or nodes.

Cantor's proof does not prove anything about undefinable numbers. It
proves that the set of definable numbers is uncountable. Otherwise one
should be tempted to insert indefinable numbers in his list. The
result would be: nothing. Not a number, because soe digits were
undefined.

The true result is: There is no diagonal number defined unless the
complete list is defined by a finite definition in everyday language.
But there are only countably many finite definition, hence only
countably many diagonals.

>
> > Therefore I know that there are
> > not more than countably many paths - at least paths that can be
> > applied in mathematics and can be the result of any diagonalization.
>
> No. You don't know that.

I know it. And if it were wrong, if there were in fact uncountably
many words to be used in mathematics, then they must be defined
somewhere (how else would we know what meanings these words would
have?), then there must be a list of them. But that is impossible.

Or do you dpropose the use of words that are completely undefined?

Regards, WM

[WM]

On 16 Dez., 22:01, Rupert <rupertmccal...@yahoo.com> wrote:
> On Dec 16, 9:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > On 16 Dez., 21:34, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > It is mathematics. Mathematics includes discourse about uncountable
> > > sets.
>
> > But it does not include discourse using uncountably many characters,
> > because for that sake infinite strings of bits would be required. By
> > finite strings of bits only countably many characters could be used.
> > And if we restrict our conversation to those usable characters and
> > omit the others, then we discuss in a countable language.
>
> You're missing the distinction between the metatheory and the object
> theory. The metatheory in which mathematical discourse takes place, in
> which we discuss various object theories, is in a countable language.
> But the object theories could be in an uncountable language.

Yes, I agree. We may talk about uncountable languages. But we cannot
use them for discussing and for doing mathematics.

>
> > > The fact that it would not be possible for a human to use such a
> > > language is irrelevant
>
> > It would not only be impossible for a human but impossible per se,
> > because there might be sentences that do never end.
>
> I don't know what the distinction is between "impossible for a human
> to use" and "impossible per se".

Simply impossible.

>
> In metamathematics, we can study languages other than the languages
> which are in fact possible for humans to use.

But we cannot used them in order to define real numbers.

>
> > Mathematics may contain many foolish ideas, They can be discussed. But
> > the language applied to discuss them must be free of foolish items and
> > must be usable by humans and other intellects. That's the way
> > mathematics works: It is mainly a discussion with others or with
> > oneself. Every item (including uncountable sets and inaccessible
> > cardinals) must have finite definitions. Therefore there is no
> > uncountable alphabet and there are not uncountably many languages.
>
> Not in languages that humans actually use, no. But in the universe of
> discourse of metamathematics, there are such languages.

There are ideas about such languages like about unicorns. Assume there
were an uncountable language. Then every word of the language must be
defined. Undefined words do not belong to a language, not even in set
theory, do they? But such a book of definitions would be a list of
uncountably many words, i.e., a list of unlistable elements, i.e., a
contradiction, no?

Regards, WM

[WM]

On 16 Dez., 22:01, Rupert <rupertmccal...@yahoo.com> wrote:

> > And all mathematics that has been done by Cantor can be discussed in
> > any natural (i.e. countable) language. Further Cantor's diagonal
> > argument works solely with digits (i.e., nodes). A proof that nobody
> > can distinguish more than countable many paths in the Binary Tree is a
> > contradiction of uncountability.
>
> It's not.

Why then should we be able to distinguish the diagonal from the
entries of the list? Why then do we need the diagonal argument at all?
If we are ready to believe that there are undefinable and
undistinguishable reals, we need no diagonal argument. We could simply
believe. Cantor would be very sad seing that his famous proof is
completely superfluous.

Regards, WM

[Virgil]

In article
<a97af6af-da7b-4a5d...@n5g2000vbk.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 00:08, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > > 0
> > > > / \
> > > > 0 1
> > > > / \ / \
> > > > 0 1 0 1
> > > > ...........
>
> > > I constructed this very tree by all finite paths that extend from the
> > > root node to a node at level n. Then I appended the sequence 000...
> >
> > Thus you have no path that eventually becomes '010101...',
>
> In mathematics such a claim must be proved by nodes. Which is the
> first bit of 0.010101... that is missing in my CIBT?

Since WM keeps his CIBT under wraps so no one else can know what paths
it does or does not contain, only Wm can say.

But one of the paths missing from any tree each of whose paths are
limited to finitely many 1's. as WM's tree described above is, is
0.0101010... which has too many 1's.
>
> Regards, WM
--

[Rupert]

This is all irrelevant. I am definitely correct in saying that there
are uncountable languages, in the sense that we can define them and
talk about them in some sufficiently strong metatheory such as ZFC.
But that's neither here nor there. The point is that there is no good
reason to think that every real number must be definable in some
language. Also, you could have a situation where there is a countable
language corresponding to every countable ordinal and every real
number could be defined in at least one such language. This
possibility is demonstrably consistent, and it still remains true that
the real numbers are uncountable. But the main point is that there is
no good reason to think that every real number is definable in some
language.

[Rupert]

On Dec 16, 10:20 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 16 Dez., 22:01, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > > And all mathematics that has been done by Cantor can be discussed in
> > > any natural (i.e. countable) language. Further Cantor's diagonal
> > > argument works solely with digits (i.e., nodes). A proof that nobody
> > > can distinguish more than countable many paths in the Binary Tree is a
> > > contradiction of uncountability.
>
> > It's not.
>
> Why then should we be able to distinguish the diagonal from the
> entries of the list?

Suppose that you list all the real numbers that are definable in some
countable language. You must then be working in some metalanguage in
which you can talk about the countable language. The anti-diagonal
number can be defined in this metalanguage.

> Why then do we need the diagonal argument at all?

It is necessary to show that the real numbers are uncountable.

> If we are ready to believe that there are undefinable and
> undistinguishable reals, we need no diagonal argument. We could simply
> believe.

No, the question of whether the real numbers are countable or
uncountable is something which requires a proof one way or the other.
Cantor had to think hard about this question before he got the answer.

> Cantor would be very sad seing that his famous proof is
> completely superfluous.
>

[Virgil]

In article
<82530a8d-6a60-4fa0...@r3g2000vbn.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 00:04, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <40446e76-b057-4c82-95c6-ccb75fce0...@gu9g2000vbb.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 15 Dez., 19:28, Zuhair <zaljo...@gmail.com> wrote:
> >
> > > > You are the one who is DREAMING. I told you that neither you nor
> > > > anybody else can construct the complete infinite binary tree by
> > > > countably many paths, this is just impossible, this is like saying I
> > > > can draw the circle that is not a circle.
> >
> > > No. I did it several times. Years ago I did it twice before breakfast
> > > already.
> >
> > Like the RED QUEENs ability to believe six impossible things before
> > breakfast?
>
> Like her, but my case is quite different: I *proved* *possible*
> things.

WRONG on two counts! Some things you claim are not possible, at least
outsdide of Wolkenmuekenheim, and most of your proof, particularly hof
tose things, are not valid outside of Wolkenmuekenheim, and no one but
WM has access to Wolkenmuekenheim.
> >
> > > But before  i tell you my construction try to find a
> > > path that I did not use in the countably set of all paths that are
> > > sufficient,
> >
> > This is not a game of hide and seek.
>
> This seems to be the only chance to rouse you from your infintary
> dreams.
> If you have a rest of scientific honesty, then you have to confess
> that you cannot find which tails I used but also you cannot find any
> path, - defined by nodes only(!), not by a finite definition(!!) -
> that is missing in my CIBT.

I have often shown that there are more paths in any COMPLETE infinite
binary tree than can exist in your sort of tree, which proof suffices
with the vast majority of mathematicians to show that your tree is not a
complete one.

I certainly do not see many mathematicians, or non-mathematicians for
that matter, supporting your posts here.
>
--

[Rupert]

On Dec 16, 10:03 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 16 Dez., 21:35, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > However, you haven't justified your claim that there are only
> > countable many countable languages, and in fact it can be shown that
> > this isn't true.
>
> Then there is another contradiction. I can prove the contrary.
>
> All characters belong to a countable set, because it must be possible
> to encode them as finite strings of bits:
>
> 0
> 1
> 00
> 01
> 10
> 11
> 000
> ...
>
> The list is countable. The meanings of words formed by these
> characters and used for discourse must be defined by men. There will
> never be more than a finite set of definitions. But even an infinite
> set of languages would not surpass a countable set of words, because
> aleph_0 * aleph_0 = aleph_0. We will never leave the countable domain
> unless there are infinite words. But infinite words cannot be used in
> discussions with others or with oneself, i.e., in mathematics.
>

This is an attempt to reason about humanly constructible languages.
That's different to what I was talking about.

> Regards, WM
>
> And languages are practized: A language is something to be defined by
> men. There will never be more than a finite set of languages existing.
>
> Further
>
>
>
> > > What cannot be
> > > defined in any language is not a path.
>
> > This, too, is something which needs to be justified. A set theorist
> > would not accept that assumption.
>
> A set theorist would prefer to accept completed infinity.
> But mathematics is mainly a language. And Cantor's proof can be done
> in that language. Cantor's proof is limited to paths that can be
> identified by their strings of digits or nodes.
>
> Cantor's proof does not prove anything about undefinable numbers. It
> proves that the set of definable numbers is uncountable.

No, it doesn't. Definable in what language?

> Otherwise one
> should be tempted to insert indefinable numbers in his list. The
> result would be: nothing. Not a number, because soe digits were
> undefined.
>
> The true result is: There is no diagonal number defined unless the
> complete list is defined by a finite definition in everyday language.

What do you mean by "everyday language"?

The point is that you cannot set a limit in advance on which languages
you are going to work with.

Suppose you say "I'm only going to work with the real numbers that are
definable in language L". Then there must be some metalanguage in
which you can talk about the language L. In that metalanguage, you can
define the anti-diagonal for the list of all real numbers definable in
the language L. You're never going to be able to set precise limits to
which real numbers you regard as "definable" without also being able
to show that you can go beyond them. That's the point. There does not
exist any countable language in which all of the real numbers are
definable.

> But there are only countably many finite definition, hence only
> countably many diagonals.
>
>
>
> > > Therefore I know that there are
> > > not more than countably many paths - at least paths that can be
> > > applied in mathematics and can be the result of any diagonalization.
>
> > No. You don't know that.
>
> I know it. And if it were wrong, if there were in fact uncountably
> many words to be used in mathematics, then they must be defined
> somewhere (how else would we know what meanings these words would
> have?), then there must be a list of them. But that is impossible.
>
> Or do you dpropose the use of words that are completely undefined?
>

It's nothing to do with the question of what words we use in
mathematics. The issue is about what mathematical objects exist.

> Regards, WM

[Virgil]

In article
<75598286-1e76-462c...@x20g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 07:18, Zuhair <zaljo...@gmail.com> wrote:
> > On Dec 16, 12:12 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 15 Dez., 21:27, Zuhair <zaljo...@gmail.com> wrote:
> >
> > > > On Dec 15, 10:22 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> > > > What do you mean I appended the sequence 000..., Can you explain that
> > > > in details. I mean the full detail of how did you construct this tree
> > > > by this appending. How do you prove the the resulting constructed tree
> > > > is the infinite binary tree. DETAILS please. At least refer me to an
> > > > article that has all the details about that alleged construction if
> > > > there is any.
> >
> > > The finite paths are the following:
> > > 0.
> > > 0.0
> > > 0.1
> > > 0.00
> > > 0.01
> > > 0.10
> > > 0.11
> > > ...
> >
> > > Each of these paths now is equipped with an infinite tail 000...
> >
> > > 0.000...
> > > 0.0000...
> > > 0.1000...
> > > 0.00000...
> > > 0.01000...
> > > 0.10000...
> > > 0.11000...
> > > ...
> >
> > > Some paths appear more than once. Some nodes are constructed more than
> > > once. But that does not matter. This set of paths is nevertheless
> > > countable.
> > > And there is no "diagonal" that *at a finite level n* differs from all
> > > paths.
>
> >
> > Ok but the resulting construction is not the INFINITE binary tree
> > we've already defined.
>
> Liar! The resulting construction is what you accepted as the complete
> infinite Binary Tree CIBT. Should I

What you should do, but no doubt won't is to quit lying. There are all
sorts of infinite binary trees but only one complete one, at least up to
isomorphism, and yours isn't it.
>
> The set of paths that I used is countable. And you will see it when I
> tell you what I used. But before I unveil my construction let me know
> whether the constructed Binary Tree is complete in your opinion. Here
> it is:
>
> 0.
> 0 1
> 0 1 0 1
> ...
>
One test is to "count" the number of paths, If that number is anything
less that uncountable the tree is incomplete, though not conversely.

> Every level starts with 0, is alternating between 0 an 1, and has
> twice as many nodes as the level above.
>
> You said: This tree has ALL possible binary paths as paths of it.
> In other words it is COMPLETE.
>
>
> > For example the path representing the decimal
> > expansion of pi is not among the paths of your constructed tree.
>
> First: pi is not a real number of the unit interval
> Second: Every finite initial segment of pi - 3 is in the CIBT.
>
> And even pi - 3 is there because I cheated. I did not use the tails
> 000... but the tails with the bit-string of pi-3.

Then 1/sqsrt(2) is not in your tree.

And given any one tail, or even countable set of tails, required of all
paths, such a tree is incomplete.
>
> >Of
> > course the tree you've constructed that way is COUNTABLE, I agree, but
> > it is not a complete infinite binary tree, it is not even near.
>
> You are dreaming. But let us play again. I construct a CIBT by using
> tails that I do not publish. You have to find out what paths are
> missing.

All we need to know is whether your set of tails is countable. It it is
then necessarily your tree is incomplete.
>
> > Now
> > even if you add any tail of your choice, I mean suppose you have a
> > countable set of tails T of your choice, and you append those tails to
> > countably many finite stumps of the infinite binary tree, still the
> > resulting construction is countable, but it is not the complete
> > infinite binary tree, why because simply and I say simply apply the
> > method I've showed you in the head post on your so constructed tree
> > and you will recover a path of the complete binary tree that is not
> > among the paths of your constructed tree. I showed you that, but you
> > keep refusing to understand that part.
>
> If you are really unable to recognize it yourself, then listen to
> others like George Greene: The CIBT is complete. It is impossible to
> find a digit sequence defined by nodes that is not contained in it.

True enough, but that tree is not your tree.
>
> Regards, WM
--

[Virgil]

In article
<0538c9b2-1a91-4a29...@gu9g2000vbb.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 10:37, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Here it is:
> >
> > > � � �0.
> > > � 0 � 1
> > > 0 1 0 1
> > > ...
> >

> > > Every level starts with 0, is alternating between 0 an 1, and has
> > > twice as many nodes as the level above.
> >
> > > You said: This tree has ALL possible binary paths as paths of it.
> > > In other words it is COMPLETE.
>

> > The CIBT (complete infinite binary tree) is defined in the following
> > manner:
> >
> > CIBT is a Tree
> > The root node of CIBT is labeled 0
> > Each node of CIBT has two child nodes each having a different label
> > from the other
> > Any node of ClBT receives only ONE label that is either 0 or 1.
>
> > Can you tell me what is your proof that the tree you've constructed by
> > the appending process is the CIBT??? try to prove it, and you will see
> > that you cannot.
>
> My CIBT satisfies exactly what you said above. What is further to
> prove?

That every possible path occurs in WM's tee, which is not the case.
> >
> > All you have done up till now is to provide a blind assertion that the
> > tree constructed by the appending process is the CIBT itself? I didn't
> > see any proof of that, you didn't present any? so if you have a proof
> > of that then show it, so that one can see if it is valid or not?
>
> The tree need not be "proved", since it is defined and constructed
> according to that definition.

Any tree defined and constructed according to the proper definition of
Complete Infinite Binary Tree will differ from yours, but as you
carefully keep the details of your tree hidden, we cannot tell exactly
which paths you have left out of yours.

Consider the following two binary paths:
0.01010101... which is periodic in the obvious way and
0.01001000100001... in which each 1 is further from the previous 1,
which is obviously non-perodic.

Clearly no tree in which all paths have the same tail can contain both
of these paths, Thus WM's claim to have generated all possible paths
using only one tail is proved absolutely false.
>
> If you doubt that I have correctly constructed it, then find a node
> that is missing.

We have found a pair of paths that cannot both be in WM's tree, which is
an even more satisfactory disproof of his claim than finding a missing
node would be.
--

[Virgil]

In article
<d914fd00-8b58-4513...@x20g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 13:10, Zuhair <zaljo...@gmail.com> wrote:
>
> > > If you doubt that I have correctly constructed it, then find a node
> > > that is missing.
> >

> > No.
>
> So you doubt that you are able to find a missing node? You are
> correct. There are all nodes by definition in the CIBT.

We can do better and find a missing path:
Since by WM's construction all paths have the same tail,
either 0.000... or 0.111... must be missing.
>
> > you are one who must prove that what you've constructed IS the
> > CIBT. since all of your claim depends on that central issue, you
> > cannot just make assertions and expect everyone to accept them without
> > supplying proofs of those assertions. You must have a proof in your
> > mind that the constructed tree of yours is itself the CIBT, don't you?
>
> I have constructed the Binary Tree according to your definition.

Yo have not done so

> And
> if you don't believe me, then construct it it yourself by all finite
> paths. Adding a tail of any kind to every path will not make any of
> the nodes disappear.

And one tail will not complete the Complete Infinite Binary Tree.

Nor will any countable set of them.
> >
> > > Note: The CIBT is defined by its *nodes* (as you said above). Paths
> > > that are not defined by nodes at finite levels may be or may not be in
> > > it.
> >
> > The CIBT that I've mentioned is COMPLETE, it is an actual infinity
> > binary tree and it DOES contain ALL actually infinite binary paths.
>
> You defined that every node has two child nodes. No word about paths.
> Do you refuse to remember what you said? I don't think that such
> childish babbling as you like to perform here has to do anything with
> mathematics or "discourse".

And We, collectively, don't think that such infantile babbling as you
like to perform
here has to do anything with mathematics or "discourse".

If you could produce a Complete Infinite Binary Tree with only ciuntably
may paths in it, you would have no need to hide its details from us s
you keep doing, but wold emblazon those details in every post.

The very fact of your needing to keep such alleged things secret is
evidence that you cannot produce them.
--

[Ross A. Finlayson]

In the breadth-first ordering, after 0 = .000..., we don't
(standardly) have a proof that there's a least element in the normal
ordering (or they'd be well-ordered by their normal ordering), but we
do have a proof that there's a well-ordering for each level of the
tree. So, we can select from G\0 (here being all the paths setminus
zero's) all those less than .1 then all those less than .01 and so on,
where then obviously, from the finite to the infinite, there are never
none left, but, as well, there isn't a level or row such that there
isn't an element starting with that many zeros that would be lesser in
the normal or usual or natural breadth-first ordering. So, as we know
that all the elements of the bread-first traversal would be starting
with zero at each relevant place, the only way for it to be different
at that place and be part of the anti-diagonal, is for the value to be
one.

Paths:

.000... = BT(0)
.000... = BT(1)
.000... = BT(2)
...
.111... = BT(2^w)

Anti-diagonal, from the beginning:

.111... = BT(2^w)


Then it's quite relevant in as to well-ordering the reals, well-
ordering the paths.

Regards,

Ross Finlayson

[Virgil]

In article
<348b10c6-a5e0-4b21...@4g2000yqv.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 11:02, Rupert <rupertmccal...@yahoo.com> wrote:
> > On Dec 15, 7:07�pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> >
> >
> >
> >
> > > On 15 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote:

If that were true than a Complete Infinite Binary Tree could not be
defined purely in terms of its nodes

There is a Complete Infinite Binary Tree defineable by its nodes that
has uncountable many paths.
Each path in it is defined by a subset of the set of positive natuals,
N, by requiring the path to branch left from each level in the subset of
N and branch right at all other levels.

This clearly establishes a bijection between paths and subsets of N.

So that WM must also be claiming that he can list all subsets of N.


It is equal to all finite initial strings of digits that
> can be applied in a Cantor-list.

But every path contains infinitely many branchings, any of which can be
to the laft or to the right, so no finite string of 0-for-left or
1-for-right digits can determine all of them.

(Every digit changed there has a
> finite index.)

Meaning that all infinitely many digits have finite indices.

>
> Actually infinite paths like that of 0.010101... = 1/3 (in binary)
> cannot be defined by nodes.

But they can by infinite sequences of nodes each identified as either a
left child or right child of its parent node.

> No infinite sequence has ever been defined
> by its terms.

How about the sequence of all 0's and the sequence of all 1's?
Or your own example of 0.010101... = 1/3 (in binary) ?

Besides which, there is no necessity for a path to be accessible to be a
memeer of the set of paths any more than for a real number to be
accessible in order to be a real number. Most of them aren't!
--

[Virgil]

In article
<9221f338-cfbf-4e01...@c14g2000vbd.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 13:02, Zuhair <zaljo...@gmail.com> wrote:
>
> > > Actually infinite paths like that of 0.010101... = 1/3 (in binary)

> > > cannot be defined by nodes. No infinite sequence has ever been defined
> > > by its terms. Those things have to be defined by finite definitions
> > > like "0.010101..." or "1/3".
> >

> > > Regards, WM

> >
> > so what you are saying is that an actual infinite path like
> > 0.010101... is not among the tree you've constructed.
>

> And if I have not told you the truth?

You often haven't!


> What would you say if I had used
> the tails 010101... Is that kind of reasoning reasonable? Is that
> mathematics???

It is neither reasonable nor mahematics to claim, as you do, that you
can produce all paths using only one tail pattern.
>
> > If so then how
> > do you claim that what you've constructed is the CIBT?
>
> I do not claim it. I have done it.

There! You have claimed again, but it is easy to prove you wrong.
Some tails will be periodic, like 0,0101010... and some tails well be
non-periodic like 0.10100100010000..., and no one tail can be both.

By a diagonal argument, one may show that no countable set of tails
suffices.



>
> You have no chance to distinguish the path 1/3 by nodes because there
> is no binary fraction of 1/3.

It can still be "distinguished" by the periodic node sequence
corresponding to the binary representation of 1/3, despite WM's apparent
ignorance of that representation.




> All you have been told by matheologians
> about completed infinity was and is completed nonsense.

Not anywhere nearly as nonsensical as what WM has "told" people.
>
> Regards, WM
--

[Virgil]

In article
<d32dca76-9d6b-40a4...@f19g2000vbv.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 11:54, Rupert <rupertmccal...@yahoo.com> wrote:

> > > their nodes. It is equal to all finite initial strings of digits that
> > > can be applied in a Cantor-list. (Every digit changed there has a
> > > finite index.)
> >

> > > Actually infinite paths like that of 0.010101... = 1/3 (in binary)
> > > cannot be defined by nodes. No infinite sequence has ever been defined
> > > by its terms. Those things have to be defined by finite definitions
> > > like "0.010101..." or "1/3".
> >

> > Okay. That all sounds fair enough.
> >
> > So, you were going to show us that there are only countably many
> > paths. When are you going to do that?-
>
> First I showed you that there are only countably many paths that are
> defined by nodes.

You have claimed it, but never established it, and cannot do so until
you can DISPROVE the theorem that sets that the cardinality of any set
is less than that of its power set.

But WM's attempts to establish valid proofs about anything are almost
always cooked.
--

[Virgil]

In article
<d3a3d23d-b0b5-4ffe...@f19g2000vbv.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 16:54, Rupert <rupertmccal...@yahoo.com> wrote:

> > > defined by nodes. You seemed to agree. Call them the set A.
> >
> > > Well, the other "paths" cannot be defined by nodes. Call them the set
> > > B. They need definitions by finite words. And there are only countably
> > > many finite words.
> >
> > > Now consider the union of two countable sets A and B.
> >
> > How do you know that there exists a countable language such that every
> > path can be defined in this language?
>
> I do know that in every language (as well as in all languages together
> since aleph_0 * aleph_0 = aleph_0, and since there are no uncountable
> languages) only countably many paths can be defined.

But it is quite possible to define a set in which some, even most, of it
members are so inaccessible that they cannot be individually defined.

E.g., the set of reals.

Thus requiring the members of a set all to be definable before the set
can be allowed is an invalid objection, at least outside WMytheology.

> What cannot be
> defined in any language is not a path.

It is if it is a member of a set of paths, regardless of either its
accessability or definability.



> Therefore I know

False "knowledge" snipped!
--

[Virgil]

In article
<dfda217d-a624-45fa...@r14g2000vbe.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:

> On 16 Dez., 21:35, Rupert <rupertmccal...@yahoo.com> wrote:
>
> > However, you haven't justified your claim that there are only
> > countable many countable languages, and in fact it can be shown that
> > this isn't true.
>
> Then there is another contradiction. I can prove the contrary.
>
> All characters belong to a countable set, because it must be possible
> to encode them as finite strings of bits:
>
> 0
> 1
> 00
> 01
> 10
> 11
> 000
> ...
>
> The list is countable. The meanings of words formed by these
> characters and used for discourse must be defined by men.

Some nonhuman species are alleged to have languages, though much more
primitive that present human languages.

> There will
> never be more than a finite set of definitions.

But one can define a set whose members need not all be defineable.

> But even an infinite
> set of languages would not surpass a countable set of words

Words terminate, strings need not terminate.
--