Matheology § 233

[WM]

Matheology § 233

The set of all termination decimals is a subset of Q. If the set of
all terminating decimals of the unit interval is arranged as set of
all terminating paths of the decimal tree, unavoidably all irrationals
are written as infinite paths too. But we know that it is impossible
to write the path of even one single irrational number, let alone of
several or infinitely many or uncountably many.

So belief in the above requires strong faith.

A view without faith is this: There is no irrational path at all. But
that would destroy the pet dogma of matheology, namely uncountability.

The question is: How come uncountably many irrational paths into being
during the countable process of constructing the complete decimal tree
by constructing all its countably many nodes. Provably none of the
irrationals is constructed in any step.

And an additional question for skilled matheologians: If we delete all
paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
tree of finite paths: Do all irrationals remain? Is infinity the only
necessary condition of matheological belief?

Regards, WM

[fom]

The "principles" of critique suggested by the idea-bankrupt
author of the above remarks do not relate directly
to the logical principles of construction originally
used to formulate an arithmetical representation
of the linearly ordered continuum.

The source of these criticisms reside with mathematicians
that met their obligations to provide an acceptable
framework for mathematical investigation of their
contrary views.

The willful construction of concatenated crayon marks
would not be generally considered as an acceptable
framework for mathematical investigation.

Thus, WM continues to be the idea-bankrupt author of
this nonsense.

Keep crayoning!!!

[AMeiwes]

"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:a75ff190-b949-40d1...@v8g2000yqe.googlegroups.com...

>Matheology � 233
>
>The set of all termination decimals is a subset of Q. If the set of
>all terminating decimals of the unit interval is arranged as set of
>all terminating paths of the decimal tree, unavoidably all irrationals
>are written as infinite paths too. But we know that it is impossible
>to write the path of even one single irrational number, let alone of
>several or infinitely many or uncountably many.

this changes with the base of the numbering system you are using, you are
assuming base 10, what about base 3 ?

>So belief in the above requires strong faith.
>

faith in what? Your obvious mistake is using the assumption on your part
that it is "unavoidable" to include non-terminating decimals in a set of
terminating decimals.

>A view without faith is this: There is no irrational path at all. But
>that would destroy the pet dogma of matheology, namely uncountability.

my pet dogma ate your pet catma.

>The question is: How come uncountably many irrational paths into being
>during the countable process of constructing the complete decimal tree
>by constructing all its countably many nodes. Provably none of the
>irrationals is constructed in any step.

you are confused, just start writing them down, and report back here.

>And an additional question for skilled matheologians: If we delete all
>paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
>tree of finite paths: Do all irrationals remain? Is infinity the only
>necessary condition of matheological belief?

so you go to binary, and cant any number have a binary representation ?

>Regards, WM

[Virgil]

In article
<a75ff190-b949-40d1...@v8g2000yqe.googlegroups.com>,

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

> Matheology � 233
>
> The set of all termination decimals is a subset of Q. If the set of
> all terminating decimals of the unit interval is arranged as set of
> all terminating paths of the decimal tree, unavoidably all irrationals
> are written as infinite paths too. But we know that it is impossible
> to write the path of even one single irrational number, let alone of
> several or infinitely many or uncountably many.

Then it must equally be impossible to construct a decimal tree
containing every possible finite decimal path and no others.

>
> So belief in the above requires strong faith.

Not so great a faith as yours that there must exist some unfindable
"largesT" OR "last" natural that has no successor.

IF a decimal tree containing a different path for every possible finite
decimal from 0 to 1 exists, it also must contain a path for every real
from 0 to 1.

So that if WM denies existence of paths for those reals, he
automatically also denies the existence of any such trees.

Standard mathematics has no problem with that.

>
> A view without faith is this: There is no irrational path at all.

Then the view, either with or without faith, but using standard logic,
must be that there is no such tree at all.

> But
> that would destroy the pet dogma of matheology, namely uncountability.

Standard mathematics says that if certain trees exist, then they
necessarily have uncountably many paths.

WM is the one claiming such trees exist.

>
> The question is: How come uncountably many irrational paths into being
> during the countable process of constructing the complete decimal tree
> by constructing all its countably many nodes.

Because all sets have more subsets than they have members.

>
> And an additional question for skilled matheologians: If we delete all
> paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
> tree of finite paths: Do all irrationals remain?

Since that will delete everything from 0.12 to 0.21, inclusive, some
rationals and some irrationals vanish,
but the irrational 0.10110111011110... won't.


There are binary trees with only finite paths, but only when having some
finite maximum path length, but there are no binary trees containing
paths of all finite path lengths that do not contain at least one
infinite path as well.
--

[WM]

On 27 Mrz., 18:26, Virgil <vir...@ligriv.com> wrote:

> IF a decimal tree containing a different path for every possible finite
> decimal from 0 to 1 exists, it also must contain a path for every real
> from 0 to 1.

And what says matheology about the existence of the set of every
possible finite decimal path of the unit interval without any tree
structure?

>
> So that if WM denies existence of paths for those reals, he
> automatically also denies the existence of any such trees.

I ask: What can be concluded, IF such a tree exists?

>
> Standard mathematics has no problem with that.
>
>
>
> > A view without faith is this: There is no irrational path at all.
>
> Then the view, either with or without faith, but using standard logic,
> must be that there is no such tree at all.

That is correct. But from the assumption of the actually infinite tree
without irrationals, the proof of its non-existence follows easily.

>
> > But
> > that would destroy the pet dogma of matheology, namely uncountability.
>
> Standard mathematics says that if certain trees exist, then they
> necessarily have uncountably many paths.
>
> WM is the one claiming such trees exist.
>

I assume that the countable set of all rationals exists. If written in
form of a tree, all uncountably many irrationals are created. You say
that certain subsets are considered. But that is definitely wrong,
because only the existing paths are written such that everyone remains
- only some nodes are united. This does not create new sets but only
deletes some nodes.

>
>
> > The question is: How come uncountably many irrational paths into being
> > during the countable process of constructing the complete decimal tree
> > by constructing all its countably many nodes.
>
> Because all sets have more subsets than they have members.
>

Not in case of the tree. Here only special subsets can be formed,
namely only such which are already existing because of inclusion
monotony.

>
>
> > And an additional question for skilled matheologians: If we delete all
> > paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
> > tree of finite paths: Do all irrationals remain?
>
> Since that will delete everything from 0.12 to 0.21, inclusive, some
> rationals and some irrationals vanish,
> but the irrational 0.10110111011110... won't.
>

First it must become existing.

> There are binary trees with only finite paths, but only when having some
> finite maximum path length, but there are no binary trees containing
> paths of all finite path lengths that do not contain at least one
> infinite path as well.

So the structure of the tree is something very special. Uncountability
seems to be a matter of how numbers are written.

Regards, WM

[Virgil]

In article
<32f47105-a873-44a0...@k1g2000yqf.googlegroups.com>,

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

> On 27 Mrz., 18:26, Virgil <vir...@ligriv.com> wrote:
>
> > IF a decimal tree containing a different path for every possible finite
> > decimal from 0 to 1 exists, it also must contain a path for every real
> > from 0 to 1.
>
> And what says matheology about the existence of the set of every
> possible finite decimal path of the unit interval without any tree
> structure?

Since WM is the only one speaking for matheology, he must answer his own
question.

> >
> > So that if WM denies existence of paths for those reals, he
> > automatically also denies the existence of any such trees.
>
> I ask: What can be concluded, IF such a tree exists?
> >
> > Standard mathematics has no problem with that.
> >
> >
> >
> > > A view without faith is this: There is no irrational path at all.
> >
> > Then the view, either with or without faith, but using standard logic,
> > must be that there is no such tree at all.
>
> That is correct. But from the assumption of the actually infinite tree
> without irrationals, the proof of its non-existence follows easily.

WM makes my case for me.

> >
> > > But
> > > that would destroy the pet dogma of matheology, namely uncountability.
> >
> > Standard mathematics says that if certain trees exist, then they
> > necessarily have uncountably many paths.
> >
> > WM is the one claiming such trees exist.
> >
> I assume that the countable set of all rationals exists. If written in
> form of a tree, all uncountably many irrationals are created.

RIGHT!

If an n-ary tree contains all (eventually constant) paths paths
corresponding to n-ary rationals between 0 and 1, it must contain the
path for any limit of a sequence of such rational n-ary (eventually
constant) paths, thus also for all reals between 0 and 1.

> You say
> that certain subsets are considered. But that is definitely wrong,
> because only the existing paths are written such that everyone remains
> - only some nodes are united. This does not create new sets but only
> deletes some nodes.

Which nodes get deleted when one has each binary rational other than 0
and 1 in the infinite binary tree being represented by two infinite
paths, one ending with infinitely many 0's the other ending with
infinitely many 1's.

> >
> >
> > > The question is: How come uncountably many irrational paths into being
> > > during the countable process of constructing the complete decimal tree
> > > by constructing all its countably many nodes.
> >
> > Because all sets have more subsets than they have members.
> >
> Not in case of the tree.

How does the existence of a tree structure prevent any set from having
more subsets than members?

That inequality holds for ALL sets anywhere.

> Here only special subsets can be formed,

But enough of them.

> >
> >
> > > And an additional question for skilled matheologians: If we delete all
> > > paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
> > > tree of finite paths: Do all irrationals remain?
> >
> > Since that will delete everything from 0.12 to 0.21, inclusive, some
> > rationals and some irrationals vanish,
> > but the irrational 0.10110111011110... won't.
> >
> First it must become existing.

If the tree exists, then that path exists within it.

>
> > There are binary trees with only finite paths, but only when having some
> > finite maximum path length, but there are no binary trees containing
> > paths of all finite path lengths that do not contain at least one
> > infinite path as well.
>
> So the structure of the tree is something very special. Uncountability
> seems to be a matter of how numbers are written.

In a binary tree, either the entire set of path lengths is bounded by
some maximum finite number of nodes or the tree has at least one path of
non-finite length.
At least when not incarcerated in Wolkenmuekenheim.
--

[fom]

On 3/27/2013 4:56 PM, WM wrote:
>
> So the structure of the tree is something very special. Uncountability
> seems to be a matter of how numbers are written.
>

No. Uncountability is a logical delineation.

The justification for making that delineation is
an argument scheme that applies to syntactic representations
intended to prove a specific claim.

The fact that syntactic representations cannot faithfully
represent the claim leads to defining systems in which
the substance of the claim may be subjected to mathematical
analysis.

Thus one obtains a definition of countability.

Thus one obtains the logical form of genus and species.

Genus -
number system

Species -
countable number system

Coordinate species -
uncountable number system


"For falsity and truth involve combination and division."

Aristotle


Not that WM could understand a word of the above.

[WM]

On 28 Mrz., 02:54, Virgil <vir...@ligriv.com> wrote:
> In article
> <32f47105-a873-44a0-bbcc-f744d1e71...@k1g2000yqf.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 27 Mrz., 18:26, Virgil <vir...@ligriv.com> wrote:
>
> > > IF a decimal tree containing a different path for every possible finite
> > > decimal from 0 to 1 exists, it also must contain a path for every real
> > > from 0 to 1.
>
> > And what says matheology about the existence of the set of every
> > possible finite decimal path of the unit interval without any tree
> > structure?
>
> Since WM is the only one speaking for matheology, he must answer his own
> question.

Matheology is the teaching of unnameable names and of actually
infinite decimal farctions that nobody can apply (opposite to
countably many names which define potentially infinite decimal
expansions). That religion is not what I adhere to.

>
>
> > > So that if WM denies existence of paths for those reals, he
> > > automatically also denies the existence of any such trees.
>
> > I ask: What can be concluded, IF such a tree exists?

And the foundation of matheology is that such a node-complete tree
exists since the actually infinite set of all finite decimal
expansions exist, namely the set of all rationals that end in a
decimal period 000...

> > That is correct. But from the assumption of the actually infinite tree
> > without irrationals, the proof of its non-existence follows easily.
>
> WM makes my case for me.

For a matheologian, that is inconsequent, because the result is a
contradiction of uncountably sets.

>
>
>
> > > > But
> > > > that would destroy the pet dogma of matheology, namely uncountability.
>
> > > Standard mathematics says that if certain trees exist, then they
> > > necessarily have uncountably many paths.
>
> > > WM is the one claiming such trees exist.
>
> > I assume that the countable set of all rationals exists. If written in
> > form of a tree, all uncountably many irrationals are created.
>
> RIGHT!

That means irrationals cannot be distinguished from terminating
rationals by nodes. They appear and disappear by the form of writing.
So you agree that irrational numbers as decimal fractions cannot be
distinguished from the terminating rationals by nodes. But this is the
principle of Cantor's argument, which, therefore, must fail.

> If an n-ary tree contains all  (eventually constant) paths paths
> corresponding to n-ary rationals between 0 and 1, it must contain the
> path for any limit of a sequence of such rational n-ary (eventually
> constant) paths, thus also for all reals between 0 and 1.

And why should the set of all rationals not contain that limit?
Does the sequence of all natural numbers, when written in the form 1,
2, 3, ... contain its limit, but when written as 7, 5, 11, ... not
contain it? Amazing.

>
> > You say
> > that certain subsets are considered. But that is definitely wrong,
> > because only the existing paths are written such that everyone remains
> > - only some nodes are united. This does not create new sets but only
> > deletes some nodes.
>
> Which nodes get deleted when one has each binary rational other than 0
> and 1 in the infinite binary tree being represented by two infinite
> paths, one ending with infinitely many 0's the other ending with
> infinitely many 1's.

The rationals
0.11000...
0.10000...
when written in the Binary Tree have the first node 1 in common, so
one of the first nodes disappears in the Binary Tree.

>
>
>
> > > > The question is: How come uncountably many irrational paths into being
> > > > during the countable process of constructing the complete decimal tree
> > > > by constructing all its countably many nodes.
>
> > > Because all sets have more subsets than they have members.
>
> > Not in case of the tree.
>
> How does the existence of a tree structure prevent any set from having
> more subsets than members?

In an inclusion monotonic set, written in the Binary Tree, all
included elements belong to the element including them. Therefore the
set can only shrink. It is not possible in the Binary Tree to combine
elements destroying the tree structure.

>
> That inequality holds for ALL sets anywhere.

No. There are aleph_0 elements
0.1
0.11
0.111
...
which in the Binary tree are combined to only one path 0.111...

Since the set of all rationals contains aleph_0 elements, which only
lose nodes when combined to the tree, the Binary tree cannnot contain
more paths because no path is added.

>
> > Here only special subsets can be formed,
>
> But enough of them.

A statement of unjustified faith. See above. Always two rationals are
combined. And there cobination belongs to not more than countably many
paths.

>
>
>
> > > > And an additional question for skilled matheologians: If we delete all
> > > > paths containing digits 2, 3, 4, 5, 6, 7, 8, and 9 from the decimal
> > > > tree of finite paths: Do all irrationals remain?
>
> > > Since that will delete everything from 0.12 to 0.21, inclusive, some
> > > rationals and some irrationals vanish,
> > > but the irrational 0.10110111011110... won't.
>
> > First it must become existing.
>
> If the tree exists, then that path exists within it.
>

If the set of all rationals exists, then that limit exists already in
that set. Combining paths with loss of nodes is not useful to increase
the number of paths.

Regards, WM

[fom]

On 3/28/2013 7:48 AM, WM wrote:
> On 28 Mrz., 02:54, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <32f47105-a873-44a0-bbcc-f744d1e71...@k1g2000yqf.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>> On 27 Mrz., 18:26, Virgil <vir...@ligriv.com> wrote:
>>
>>>> IF a decimal tree containing a different path for every possible finite
>>>> decimal from 0 to 1 exists, it also must contain a path for every real
>>>> from 0 to 1.
>>
>>> And what says matheology about the existence of the set of every
>>> possible finite decimal path of the unit interval without any tree
>>> structure?
>>
>> Since WM is the only one speaking for matheology, he must answer his own
>> question.
>
> Matheology is the teaching of unnameable names and of actually
> infinite decimal farctions that nobody can apply (opposite to
> countably many names which define potentially infinite decimal
> expansions). That religion is not what I adhere to.

And, because your world view is cast in terms of religious
belief, you make yourself incapable to understand the nature
of logic or how it is applied in the profession of mathematics.

[fom]

On 3/28/2013 7:48 AM, WM wrote:

Neither your questions nor your crayon marks constitute
mathematics.

Aristotle is clear about the nature and use of categorical
quantifiers in demonstrations using a deductive calculus.

The development of mathematics is intimately bound with the
not-so-definite theories of science that you pretend would
fix mathematics. Complaints such as yours span the entire
history of mathematics and every repair to the system of
mathematical thought in response to those criticisms breeds
a new generation of loudmouths such as yourself.

As noted before, Berkeley's complaints about Newton's fluxions
provide the fodder for your complaints about Cantor's transfinite
arithmetic.

All of your questions demonstrate the same two things: ignorance
of logic and ignorance of mathematics.

[fom]

On 3/28/2013 7:48 AM, WM wrote:
>>

> If the set of all rationals exists, then that limit exists already in
> that set. Combining paths with loss of nodes is not useful to increase
> the number of paths.

But no one is talking about whether the sequence
of rationals converging to a rational is in the
set of rationals.

The issue is a representation of apparent geometric
completeness within an arithmetical system.

The issue is the logical form of such a construction.

That you believe your crayon marks justify a material
belief in some abstract, non-material objects but deny
a material belief in other abstract, non-material
objects verges on sheer lunacy.

Crayon marks have nothing to do with it.

Belief has nothing to do with it.

And, your ability to call certain crayon marks "names"
has nothing to do with it either. You failed your
science lesson on those days too.

[Virgil]

In article
<e880e3d2-4842-4e0d...@u20g2000yqj.googlegroups.com>,

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

> And the foundation of matheology is that such a node-complete tree
> exists since the actually infinite set of all finite decimal
> expansions exist, namely the set of all rationals that end in a
> decimal period 000...

It is WM who claims that such a tree with a path for each rational must
exist, even though it requires the actualness of an infinity, which WM
claims does not exist..

Standard mathematics only says what must follow IF it exists.
--

[fom]

Religious zealotry precludes understanding such distinctions.

[Gus Gassmann]

On 28/03/2013 3:38 PM, fom wrote:
> On 3/28/2013 7:48 AM, WM wrote:
>>>
>> If the set of all rationals exists, then that limit exists already in
>> that set. Combining paths with loss of nodes is not useful to increase
>> the number of paths.
>
> But no one is talking about whether the sequence
> of rationals converging to a rational is in the
> set of rationals.
>
> The issue is a representation of apparent geometric
> completeness within an arithmetical system.

I don't think so. I think the issue is that Mueckenheim, whom someone
decided to hire as a professor of mathematics at a third-rate
institution, manages to obfuscate just enough the distinction between
repeating and non-repeating decimals when he applies them to paths, and
that he is too dense to comprehend that. Crayon marks, indeed.

[fom]

On 3/28/2013 2:46 PM, Gus Gassmann wrote:
> On 28/03/2013 3:38 PM, fom wrote:
>> On 3/28/2013 7:48 AM, WM wrote:
>>>>
>>> If the set of all rationals exists, then that limit exists already in
>>> that set. Combining paths with loss of nodes is not useful to increase
>>> the number of paths.
>>
>> But no one is talking about whether the sequence
>> of rationals converging to a rational is in the
>> set of rationals.
>>
>> The issue is a representation of apparent geometric
>> completeness within an arithmetical system.
>
> I don't think so. I think the issue is that Mueckenheim, whom someone
> decided to hire as a professor of mathematics at a third-rate
> institution, manages to obfuscate just enough the distinction between
> repeating and non-repeating decimals when he applies them to paths, and
> that he is too dense to comprehend that. Crayon marks, indeed.
>

Well, that is the real-world issue. It is what motivates
Virgil to reply regularly to this nonsense.

He has stated as much.

[fom]

It is *the obfuscation* that motivates Virgil.

[WM]

On 28 Mrz., 20:13, Virgil <vir...@ligriv.com> wrote:
> In article

> <e880e3d2-4842-4e0d-9430-d3b6638d2...@u20g2000yqj.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > And the foundation of matheology is that such a node-complete tree
> > exists since the actually infinite set of all finite decimal
> > expansions exist, namely the set of all rationals that end in a
> > decimal period 000...
>
> It is WM who claims that such a tree with a path for each rational must
> exist, even though it requires the actualness of an infinity, which WM
> claims does not exist..

But which Cantor and you and many others claim to exist.

>
> Standard mathematics only says what must follow IF it exists.

Yes. And what must follwo shows a contradiction.

The set of all rational numbers of the unit interval, when written as
the Binary Tree, contains all irrational numbers of the unit interval.
So we have established the fact that an irrational number has no node
of its own. This knowledge applied to the Cantor-list containing all
rational numbers implies that it is impossible to get an anti-diagonal
that differs from all entries. The first bit causes that the anti-
diagonal differs from 2^-1 of all entries and agrees with 2^-1, the
second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of
the list agreeing with the anti-diagonal. There is no bit that would
leave zero entries agreeing with the diagonal, since there is no bit
at position infinity.

This proves that irrational numbers cannot be distinguished by any
potentially infinite sequence of digits from all rationals. Cantor's
argument would require an actually infinite sequence, but, as is
obviuous from the facts learned at the Binary Tree, this assumption of
Cantor's is false.

Regards, WM

[Virgil]

In article
<46aafcf0-9fef-4ed7...@fn10g2000vbb.googlegroups.com>,

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

> On 28 Mrz., 20:13, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <e880e3d2-4842-4e0d-9430-d3b6638d2...@u20g2000yqj.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > And the foundation of matheology is that such a node-complete tree
> > > exists since the actually infinite set of all finite decimal
> > > expansions exist, namely the set of all rationals that end in a
> > > decimal period 000...
> >
> > It is WM who claims that such a tree with a path for each rational must
> > exist, even though it requires the actualness of an infinity, which WM
> > claims does not exist..
>
> But which Cantor and you and many others claim to exist.
> >
> > Standard mathematics only says what must follow IF it exists.
>
> Yes. And what must follwo shows a contradiction.

WRONG!

>
> The set of all rational numbers of the unit interval, when written as
> the Binary Tree, contains all irrational numbers of the unit interval.

Only binary rationals will have paths that end in all left branchings or
all right branchings in an infinite BINARY tree.

> So we have established the fact that an irrational number has no node
> of its own.

No number in any infinite binary tree has any node "of its own", as
every node has two child nodes belonging to necessarily different
numbers.

> This knowledge applied to the Cantor-list containing all
> rational numbers implies that it is impossible to get an anti-diagonal
> that differs from all entries.

But as what WM claims as "this knowledge" has just shown itself to be
false, it is still possible to have an anti-diagonal different from all

entries.


> The first bit causes that the anti-
> diagonal differs from 2^-1 of all entries and agrees with 2^-1, the
> second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of
> the list agreeing with the anti-diagonal. There is no bit that would
> leave zero entries agreeing with the diagonal, since there is no bit
> at position infinity.

Thus there is always at least one bit of any listed entry disagreeing
with the antidiagonanl, just as the Cantor proof requires.

So that, as ever, WM's claims cannot hold true anywhere outside of the
wild weird world of Wolkenmuekenheim.
--

[fom]

On 3/29/2013 6:00 AM, WM wrote:
> On 28 Mrz., 20:13, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <e880e3d2-4842-4e0d-9430-d3b6638d2...@u20g2000yqj.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>> And the foundation of matheology is that such a node-complete tree
>>> exists since the actually infinite set of all finite decimal
>>> expansions exist, namely the set of all rationals that end in a
>>> decimal period 000...
>>
>> It is WM who claims that such a tree with a path for each rational must
>> exist, even though it requires the actualness of an infinity, which WM
>> claims does not exist..
>
> But which Cantor and you and many others claim to exist.
>>
>> Standard mathematics only says what must follow IF it exists.
>
> Yes. And what must follwo shows a contradiction.
>

But when asked to demonstrate that contradiction,
WM never does so.

He will not use established mathematical conventions.

He refuses to propose non-standard methods that others
may evaluate and choose to accept as an agreed upon
standard.

What WM does do may be called a "no showing" of his
claims.

[Virgil]

In article <ku2dnYMk_OD7fsjM...@giganews.com>,

It is because, however much of a researcher into the creation of
mathematics WM is, he is not himself able create any.
Not even for the "proofs" that he so often claims.
--

[WM]

On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:

>
> > So we have established the fact that an irrational number has no node
> > of its own.
>
> No number in any infinite binary tree has any node "of its own", as
> every node has two child nodes belonging to necessarily different
> numbers.

That is correct, but only establishes the fact that no actually
infinite path can be distinguished from all rational paths as should
be possible in a Cantor-list - but is not.

> > The first bit causes that the anti-
> > diagonal differs from 2^-1 of all entries and agrees with 2^-1, the
> > second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of
> > the list agreeing with the anti-diagonal. There is no bit that would
> > leave zero entries agreeing with the diagonal, since there is no bit
> > at position infinity.
>
> Thus there is always at least one bit of any listed entry disagreeing
> with the antidiagonanl, just as the Cantor proof requires.

In a list containing every rational: Is there always, i.e., up to
every digit, an infinite set of paths identical with the anti-
diagonal? Yes or no?

Regards, WM

[fom]

On 3/30/2013 8:38 AM, WM wrote:
> On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:
>
>>
>>> So we have established the fact that an irrational number has no node
>>> of its own.
>>
>> No number in any infinite binary tree has any node "of its own", as
>> every node has two child nodes belonging to necessarily different
>> numbers.
>
> That is correct, but only establishes the fact that no actually
> infinite path can be distinguished from all rational paths as should
> be possible in a Cantor-list - but is not.
>

WM failed the science lesson again today.

The Cantor argument is an argument scheme.

It presupposes a standard, classical use of
of the quantifier "all".

WM has never defined his non-standard uses
for the word "all". It has no agreed upon
usage. It is meaningless by WM's own standards
of meaning through pragmatic agreements between
language users.

By definition, all paths in the complete infinite
binary tree are infinite whether or not they
become eventually constant.

Any purported countable listing of all the paths
of that tree will result in a successful
defeat of the claim by a Cantor argument.

[Virgil]

In article
<25276bde-3813-4e0a...@t5g2000vbm.googlegroups.com>,

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

> On 29 Mrz., 19:34, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > So we have established the fact that an irrational number has no node
> > > of its own.
> >
> > No number in any infinite binary tree has any node "of its own", as
> > every node has two child nodes belonging to necessarily different
> > numbers.
>
> That is correct, but only establishes the fact that no actually
> infinite path can be distinguished from all rational paths as should
> be possible in a Cantor-list - but is not.

Every rational path in ANY Complete Infinite Binary Tree is an actually
infinite path, as that is the only sort of paths such a tree is allowed
to have.

Thus every binary-rational path in a Complete Infinite Binary Tree is an
actually infinite path. Such binary-rational paths are the ones that are
eventually all left-child nodes or all right-child nodes from some
node onward.

>
> > > The first bit causes that the anti-
> > > diagonal differs from 2^-1 of all entries and agrees with 2^-1, the
> > > second bit leaves 2^2 entries and the n-th bits leaves 2-^n entries of
> > > the list agreeing with the anti-diagonal. There is no bit that would
> > > leave zero entries agreeing with the diagonal, since there is no bit
> > > at position infinity.
> >
> > Thus there is always at least one bit of any listed entry disagreeing
> > with the antidiagonanl, just as the Cantor proof requires.
>
> In a list containing every rational: Is there always, i.e., up to
> every digit, an infinite set of paths identical with the anti-
> diagonal? Yes or no?

The set of paths in any Complete Infinite Binary Tree which agree with
any particular path up to its nth node is equinumerous with the set of
all paths in the entire tree i.e., is uncountably infinite.

Note that in Wolkenmuekenheim, at least according to WM's standard
constraints there, no such thing as a Complete Infinite Binary Tree can
exist.

And WM's ability to think about things outside Wolkenmuekenheim is
seriosly inhibited by his delusions about how things work outside of
Wolkenmuekenheim, .
--

[Ross A. Finlayson]

A breadth-first traversal of the paths, as the level of the tree goes
to infinity, is the same as depth-first of the nodes, for the infinite
tree, for each path, and each node, in reading out that paths as
nodes, as paths. The depth first traversal goes through nodes in a
transfinite sequence, as it were, that is the same as the readout of
the nodes of the paths of the breadth-first traversal of paths, as
modeled from the finite, in exhaustion, in the infinite.

And, it is very similar to n/d, with n, d e N and d -> oo, n->d (or,
"EF"), in this manner: unfilled nested intervals and the antidiagonal
result don't follow from the premises.

And besides that rays through countable ordinal points are dense in
the paths, of the infinite balanced binary tree (and ordinals are well-
ordered). This is similar to the notion of that the rationals are
dense in the reals: that for other sets dense in the reals (countable
or not): the rationals are dense in those, and stronger: for the
ordinal points as well-ordered, each is identified with a distinct
path, here regardless of what its elements as nodes are, except:
first, and last.

Then transitively that 2^w <-> P(N) and P(N) <-/-> N that N <-/-> 2^w,
due the set-theoretic Cantor's theorem of set and powerset (where even
infinite sets are _axiomatized_ to be well-founded and the universe
doesn't exist), well, with ubiquitous ordinals that S of Cantor's
theorem is {} (not containing any elements of the set) for n -> n+ 1 ,
again there are systems with mapping the continuum of the naturals
through the sweep of the continuum of the segment, that it is 1-1,
onto, and furthermore constant monotone: and where it is not onto if
not constant monotone or sweep.

So: draw a line. Cantor's theorem for reals is simply read as that
the drawing of the point is through the stroke, not each mark, for
each mark besides beginning or end as of the others: of the drawing,
of the points, of the line: all of them.

The drawing, the stroke, the sweep of the line, is through all, of
infinitely many: only at once.

There are proofs that irrationals exist without the uncountable. It
would be of tremendous interest to many to find application solely due
transfinite cardinals. To accommodate the results of real analysis
matching geometry: countably additivity of no greater than
infinitesimal reals, to accommodate the existence of a universe: non-
sets in set theory, it's easier to find non-application, and eschewal,
of transfinite cardinals: for results.

Then, with regards to the universal quantifier: sometimes: transfer
carries. That is where: "for each" _is_ "for all". This with
regards to "the" universal quantifier, here over sets.

So: make a science lesson today.

Regards,

Ross Finlayson

[Virgil]

In article <ff6dnb5sjeDL1crM...@giganews.com>,

It would if there were one, which there isn't!
--

[WM]

On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote:

> > > Thus there is always at least one bit of any listed entry disagreeing
> > > with the antidiagonanl, just as the Cantor proof requires.
>
> > In a list containing every rational: Is there always, i.e., up to
> > every digit, an infinite set of paths identical with the anti-
> > diagonal? Yes or no?
>
> The set of paths in any Complete Infinite Binary Tree which agree with
> any particular path up to its nth node is equinumerous with the set of
> all paths in the entire tree i.e., is uncountably infinite.

This was the question: In a list containing every rational: Is there

always, i.e., up to every digit, an infinite set of paths identical

with the anti-diagonal? Yes or no?

Regards, WM

[Virgil]

In article
<2bc13fff-5cbb-43dd...@m9g2000vbc.googlegroups.com>,

Lists and trees are different. And anti-diagonals derive from lists, not
trees.
The entries in list are well ordered.
The entries in a Complete Infinite Binary Tree are densely ordered.
Those order types are incompatible.
So questions, like WM's, which confuse them, are nonsense.
At least outside Wolkenmuekenheim.
--

[Ross A. Finlayson]

On Mar 30, 5:56 pm, Virgil <vir...@ligriv.com> wrote:
> In article

> <2bc13fff-5cbb-43dd-a06a-218c68c99...@m9g2000vbc.googlegroups.com>,

Zuhair simply brought forth an anti-diagonal argument for the infinite
balanced binary tree, and then the breadth-first traversal or sweep
was shown to iterate the paths that it didn't apply.

With f = lim_d->oo n/d, n -> d, the elements of ran(f) are dense in
[0,1] and well-ordered.

Fuse the elements, or, un-fuse them: don't con-fuse them.

Regards,

Ross Finlayson

[fom]

But note that the question also demonstrates WM's
complete lack of understanding of the diagonal
argument.

He has been told time and time again that it is
an argument scheme which only has application
under certain assumptions.

He chooses to believe otherwise for the agenda
of his fanaticism.

Suppose one is given a countable listing of
the rationals (with the appropriate restriction
on double representation) according to the
infinite listing of an expansion.

Suppose one performs a diagonalization on
that listing.

Is the resultant a rational number? No.

What may be concluded? That the rational numbers
do not exhaust the capacity of the algorithm
to generate representations if that algorithm
is to generate a representation for every
rational number.

Although the burden of proof lies with WM
concerning the nature of the diagonal, it
is a simple matter to understand if one
uses a Baire space representation instead.

In the Baire space, rationals are in correspondence
with eventually constant sequences. Since,
by construction, a list of Baire space rationals
would exhaust all of the eventually constant
sequences, the resultant of a diagonal argument
could not have an eventually constant sequence
unless the original premise had been false.

[Virgil]

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


> In a list containing every rational: Is there
> always, i.e., up to every digit, an infinite set of paths identical
> with the anti-diagonal? Yes or no?
>
Lists and trees are different.

Anti-diagonals derive ONLY from lists, never from trees,
and paths of the type referred to exist only in trees, not in lists.


The entries in any�list are WELL-ORDERED.

The paths in any Complete Infinite Binary Tree are DENSELY ORDERED.

Those order types, well-ordering versus dense orderings, are entirely
incompatible.

So questions, like WM's, which confuse and conflate them, are nonsense.

At least everywhere outside Wolkenmuekenheim.
--

[WM]

On 31 Mrz., 02:56, Virgil <vir...@ligriv.com> wrote:
> In article

> <2bc13fff-5cbb-43dd-a06a-218c68c99...@m9g2000vbc.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote:
>
> > > > > Thus there is always at least one bit of any listed entry disagreeing
> > > > > with the antidiagonanl, just as the Cantor proof requires.
>
> > > > In a list containing every rational: Is there always, i.e., up to
> > > > every digit, an infinite set of paths identical with the anti-
> > > > diagonal? Yes or no?
>
> > > The set of paths in any Complete Infinite Binary Tree which agree with
> > > any particular path up to its nth node is equinumerous with the set of
> > > all paths in the entire tree i.e., is uncountably infinite.
>
> > This was the question: In a list containing every rational: Is there
> > always, i.e., up to every digit, an infinite set of paths identical
> > with the anti-diagonal? Yes or no?
>
> Lists and trees are different. And anti-diagonals derive from lists, not
> trees.
> The entries in  list are well ordered.
> The entries in a Complete Infinite Binary Tree are densely ordered.

The entries in form of nodes are well ordered. Anything else is not
added to the list.

> Those order types are incompatible.

This was the question: In a list containing every rational: Is there

always, i.e., up to every digit, an infinite set of paths (rational
numbers) identical with the anti-diagonal? Yes or no?

Regards, WM

[fom]

As noted before, WM's questions often make no sense.

As explained before:

[Virgil]

In article
<17e6599f-5537-449a...@g8g2000vbf.googlegroups.com>,

This is an equally valid question: What's the difference between a duck?
--

[WM]

On 31 Mrz., 17:30, Virgil <vir...@ligriv.com> wrote:
> In article

> <17e6599f-5537-449a-b4aa-bb1764a3c...@g8g2000vbf.googlegroups.com>,

From the standpoint of matheology, perhaps. Did you hitherto respond
in an unreasonable way because you misunderstood the question?

In mathematics, in particular in analysis, we have ficed rules to
answer such an easy question and to calculate (instead of believing)
the limit.

See Matheology § 234.

Regards, WM

[fom]

No. He responded in a unreasonable way because you
repeated a senseless question despite having its
senselessness explained to you.

[Virgil]

In article
<9dc44e27-dced-4191...@a8g2000vbx.googlegroups.com>,

From the standpoint of logic and common sense, undoubtdly.

Your question makes no sense anywhere outside of your own miniscule
mini-nation of Wolkenmuekenheim.

> Did you hitherto respond
> in an unreasonable way because you misunderstood the question?

Your questions assume conditions contrary to fact.

They could only make sense in that wooly world of Wolkenmuekenheim to
which most of us have no entry

>
> In mathematics, in particular in analysis, we

WM claims access to a world he is constitutionally unable to enter.
--

[WM]

On 31 Mrz., 19:15, Virgil <vir...@ligriv.com> wrote:

> > > > This was the question: In a list containing every rational: Is there
> > > > always, i.e., up to every digit, an infinite set of paths (rational
> > > > numbers) identical with the anti-diagonal? Yes or no?
>
> > > This is an equally valid question: What's the difference between a duck?
>
> > From the standpoint of matheology, perhaps.
>
> From the standpoint of logic and common sense, undoubtdly.
>
> Your question makes no sense

>

> > Did you hitherto respond
> > in an unreasonable way because you misunderstood the question?
>
> Your questions assume conditions contrary to fact.

You do not believe that a sequence or list of all rational numbers can
be constructed? Or do you not believe that the first n digits of all
these numbers in decimal representation can be compared with the first
n digits of the anti-diagonal number? Or is there something else you
do not understand?

Regards, WM

[William Hughes]

On Mar 27, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
> Matheology § 233
>
> The set of all termination decimals is a subset of Q. If the set of
> all terminating decimals of the unit interval is arranged as set of
> all terminating paths of the decimal tree,

It is, of course, impossible to write this out
(there number of terminating decimals is infinite).

> unavoidably all irrationals
> are written as infinite paths too. But we know that it is impossible
> to write the path of even one single irrational number, let alone of
> several or infinitely many or uncountably many.

Correct, it is impossible.

[fom]

These questions are not equivalent to the senseless
question above.

Even so, these questions are also senseless because
belief has nothing to do with it.

Then there is...

[Virgil]

In article
<ca9ea385-e6aa-4759...@y14g2000vbk.googlegroups.com>,

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

> On 31 Mrz., 19:15, Virgil <vir...@ligriv.com> wrote:
>
> > > > > This was the question: In a list containing every rational: Is there
> > > > > always, i.e., up to every digit, an infinite set of paths (rational
> > > > > numbers) identical with the anti-diagonal? Yes or no?
> >
> > > > This is an equally valid question: What's the difference between a duck?
> >
> > > From the standpoint of matheology, perhaps.
> >
> > From the standpoint of logic and common sense, undoubtdly.
> >
> > Your question makes no sense
>
> >
> > > Did you hitherto respond
> > > in an unreasonable way because you misunderstood the question?
> >
> > Your questions assume conditions contrary to fact.
>
> You do not believe that a sequence or list of all rational numbers can
> be constructed?

One can "enumerate" the set of all rationals by formula, as has been
quite often done, but not by physically listing all of them.

Note that one cannot ennumerate by listing even sufficiently large
finite sets, so being listable other than by formula is not a relevant
criterion.

> Or is there something else you
> do not understand?

What none of us outside of your Wolkenmuekenheim understand is why you
need its walls to protect you from the sanity of standard mathematical
practice, nor why few, if any, of your alleged proofs survive scrutiny
from outside of Wolkenmuekenheim, .
--

[WM]

On 1 Apr., 17:05, William Hughes <wpihug...@gmail.com> wrote:
> On Mar 27, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Matheology § 233
>
> > The set of all termination decimals is a subset of Q. If the set of
> > all terminating decimals of the unit interval is arranged as set of
> > all terminating paths of the decimal tree,
>
> It is, of course, impossible to write this out
> (there number of terminating decimals is infinite).

Nevertheless Cantor has given a finite formula to construct the list
of all rationals between 0 and 1. From that formula we can find every
entry and the anti-diagonal up to every digit d_n. From that we can
easily prove that for every FIS d_1, d_2, ..., d_n of d there are
infinitely many rationals with the same FISs. For every finite number
n - and there are no other lines than such enumerated with a finite
number!

Regards, WM

[WM]

On 1 Apr., 22:44, Virgil <vir...@ligriv.com> wrote:

> > You do not believe that a sequence or list of all rational numbers can
> > be constructed?
>
> One can "enumerate" the  set of all rationals by formula, as has been
> quite often done, but not by physically listing all of them.

A formula giving every entry is enough.

>
> Note that one cannot ennumerate by listing even sufficiently large
> finite sets, so being listable other than by formula is not a  relevant
> criterion.

Constructing a list by a formula is enough to prove what I said.
That physical listing is impossible is well known.

Can you prove that for every FIS of d there are infinitely many
rationals in the list having the same FIS? This holds up to every line
number n - and there are no further lines.

Regards, WM

[WM]

On 1 Apr., 17:05, William Hughes <wpihug...@gmail.com> wrote:

> On Mar 27, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > Matheology § 233
>
> > The set of all termination decimals is a subset of Q. If the set of
> > all terminating decimals of the unit interval is arranged as set of
> > all terminating paths of the decimal tree,
>
> It is, of course, impossible to write this out
> (there number of terminating decimals is infinite).
>

But it is possible to construct the Binary Tree according to this
method.

> > unavoidably all irrationals
> > are written as infinite paths too. But we know that it is impossible
> > to write the path of even one single irrational number, let alone of
> > several or infinitely many or uncountably many.
>
> Correct, it is impossible.

It is impossible to write them out. Yes. But they are constructed like
the finita paths. It is impossible to prohibit infinite paths (of
rationals and of irrationals) to be constructed when the complete set
of nodes of the Binary Tree is constructed by means of all finite
paths. Therefore it is impossible to distinguish infinite paths by
nodes other than be naming infinite sets of nodes. Alas there are only
countably many names available.

Regards, WM

[William Hughes]

On Apr 1, 11:08 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 1 Apr., 17:05, William Hughes <wpihug...@gmail.com> wrote:> On Mar 27, 8:55 am, WM <mueck...@rz.fh-augsburg.de> wrote:
>
> > > Matheology § 233
>
> > > The set of all termination decimals is a subset of Q. If the set of
> > > all terminating decimals of the unit interval is arranged as set of
> > > all terminating paths of the decimal tree,
>
> > It is, of course, impossible to write this out

> > (the number of terminating decimals is infinite).
>
> But

Nope, no new arguments before you concede that
the bit about not being able to write stuff out
was nonsense

[Virgil]

In article
<c1713766-dbe4-4327...@h9g2000vbk.googlegroups.com>,

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

> On 1 Apr., 17:05, William Hughes <wpihug...@gmail.com> wrote:
> > On Mar 27, 8:55�am, WM <mueck...@rz.fh-augsburg.de> wrote:
> >
> > > Matheology � 233
> >
> > > The set of all termination decimals is a subset of Q. If the set of
> > > all terminating decimals of the unit interval is arranged as set of
> > > all terminating paths of the decimal tree,
> >
> > It is, of course, impossible to write this out
> > (there number of terminating decimals is infinite).
> >
> But it is possible to construct the Binary Tree according to this
> method.

I am not at all sure that it is even possible to build any binary tree
this way but it is clearly impossible to built a COMPLETE INFINITE
BINARY TREE, this way.

For one thing, in a CIBT, every path is by definition maximal in the
sense that no additional node can be added to a path without making the
result not a path, and is also minimal in the sense that no node can be
removed from it without making the result not a path.

In WM's "trees", every FISON (Finite Initial Sequncee Of Nodes) appears
to be a path, which is quite differnt notion of path.

> It is impossible to write them out. Yes. But they are constructed like
> the finita paths. It is impossible to prohibit infinite paths (of
> rationals and of irrationals) to be constructed when the complete set
> of nodes of the Binary Tree is constructed by means of all finite
> paths. Therefore it is impossible to distinguish infinite paths by
> nodes other than be naming infinite sets of nodes. Alas there are only
> countably many names available.

Thus not all CIBT-paths are nameable, just like not all real numbers are
nameable.

Nothing in the vast extent of mathematics outide Wolkenmuekenheim
requires that all things be nameable.
--

[Virgil]

> The set of all termination decimals is a subset of Q. If the set of
> all terminating decimals of the unit interval is arranged as set of
> all terminating paths of the decimal tree,

In a properly defined infinite tree, there are no such things as
terminating paths.
If WM means FISONs (finite initial sequences of nodes}, he should not be
calling them paths.
--

[Virgil]

In article
<1dd2037c-407c-49f6...@w21g2000vbp.googlegroups.com>,

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

> On 1 Apr., 22:44, Virgil <vir...@ligriv.com> wrote:
>
> > > You do not believe that a sequence or list of all rational numbers can
> > > be constructed?
> >
> > One can "enumerate" the �set of all rationals by formula, as has been
> > quite often done, but not by physically listing all of them.
>
> A formula giving every entry is enough.
> >
> > Note that one cannot ennumerate by listing even sufficiently large
> > finite sets, so being listable other than by formula is not a �relevant
> > criterion.
>
> Constructing a list by a formula is enough to prove what I said.

And enough to disprove what WM has said as well.

AS a matter of fact, almost any honest math is enough to disprove much
of want WM has said.
--

[Virgil]

In article
<b87e674d-d8ba-44ff...@a14g2000vbm.googlegroups.com>,

But as there is no end to the set/list of natural/finite numbers, there
is also no end to the set/list of such lines.
--

[WM]

Again you confuse different infinities!
It is nonsense to claim the existence of all bits or digits of an
actually infinite sequence of bits or digits.
It is correct to claim the existence of every bit or digit of an
infinite sequence that is defined by a finite formula such that every
bit or digit in principle can be determined in finite time. (Potential
infinity.)

Regards, WM

[WM]

On 2 Apr., 01:54, Virgil <vir...@ligriv.com> wrote:

>
> I am not at all sure that it is even possible to build  any binary tree
> this way but it is clearly impossible to built a COMPLETE INFINITE
> BINARY TREE, this way.

Given the foundations of matheology, it is possible to construct every
node / every finite path of a Binary Tree that is complete with
respect to its nodes.

>
> For one thing, in a CIBT, every path is by definition maximal in the
> sense that no additional node can be added to a path without making the
> result not a path, and is also minimal in the sense that no node can be
> removed from it without making the result not a path.
>
> In WM's "trees", every FISON (Finite Initial Sequncee Of Nodes) appears
> to be a path, which is quite differnt notion of path.

Call it as you like. I call it finite path as an abbreviation of FIS
of an infinite path.

>
> > It is impossible to write them out. Yes. But they are constructed like

> > the finite paths. It is impossible to prohibit infinite paths (of

> > rationals and of irrationals) to be constructed when the complete set
> > of nodes of the Binary Tree is constructed by means of all finite
> > paths. Therefore it is impossible to distinguish infinite paths by
> > nodes other than be naming infinite sets of nodes. Alas there are only
> > countably many names available.
>
> Thus not all CIBT-paths are nameable, just like not all real numbers are
> nameable.

They are not even distinguishable by nodes. They are purest belief.

Regards, WM

[WM]

On 2 Apr., 02:01, Virgil <vir...@ligriv.com> wrote:
> In article

> <1dd2037c-407c-49f6-afc6-e00c1d853...@w21g2000vbp.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 1 Apr., 22:44, Virgil <vir...@ligriv.com> wrote:
>
> > > > You do not believe that a sequence or list of all rational numbers can
> > > > be constructed?
>
> > > One can "enumerate" the set of all rationals by formula, as has been
> > > quite often done, but not by physically listing all of them.
>
> > A formula giving every entry is enough.
>
> > > Note that one cannot ennumerate by listing even sufficiently large
> > > finite sets, so being listable other than by formula is not a relevant
> > > criterion.
>
> > Constructing a list by a formula is enough to prove what I said.
>
> And enough to disprove what WM has said as well.

Then try it.
What did I say?
This: After every line n of the list of all rational numbers there are
infinitely many rational numbers that up to digit n are identical with
the anti-diagonal up to digit n.

This holds for the digits up to every finite n. And more digits cannot
be expected to exist in any decimal representation of a number.

Regards, WM

[WM]

On 2 Apr., 02:06, Virgil <vir...@ligriv.com> wrote:

>
> > Nevertheless Cantor has given a finite formula to construct the list
> > of all rationals between 0 and 1. From that formula we can find every
> > entry and the anti-diagonal up to every digit d_n. From that we can
> > easily prove that for every FIS d_1, d_2, ..., d_n of d there are
> > infinitely many rationals with the same FISs. For every finite number
> > n - and there are no other lines than such enumerated with a finite
> > number!
>
> But as there is no end to the set/list of natural/finite numbers, there
> is also no end to the set/list of such lines.

What shall that "argument" be good for? Why should there be an end?
If I say every natural number is divisible by 1 without remainder.
Would you doubt that on the grounds that there is no end to the
sequence of natural numbers? Such that possibly a natural number would
follow that is not divisible by 1?

Regards, WM

[Virgil]

In article
<7e6e8c91-ba4b-47fe...@y9g2000vbb.googlegroups.com>,

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

> On 2 Apr., 02:06, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > Nevertheless Cantor has given a finite formula to construct the list
> > > of all rationals between 0 and 1. From that formula we can find every
> > > entry and the anti-diagonal up to every digit d_n. From that we can
> > > easily prove that for every FIS d_1, d_2, ..., d_n of d there are
> > > infinitely many rationals with the same FISs. For every finite number
> > > n - and there are no other lines than such enumerated with a finite
> > > number!
> >
> > But as there is no end to the set/list of natural/finite numbers, there
> > is also no end to the set/list of such lines.
>
> What shall that "argument" be good for?

WM is continually claiming a "last" natural must always exist, even
though it is always also indeterminate, but now any of his arguments
that require such a last natural automatically fail.

> Why should there be an end?

That's what we keep asking WM.

> If I say every natural number is divisible by 1 without remainder.
> Would you doubt that on the grounds that there is no end to the
> sequence of natural numbers? Such that possibly a natural number would
> follow that is not divisible by 1?

WM has long been claiming that every and any set of naturals must have
an inaccessible last member, i.e., that every set of naturals is
necessarily a subset of a FISON (finite initial set of naturals),
just as he always claims that in a Complete Infinite Binary Tree every
path is a FISON (finite initial sequence of nodes), but he is wrong on
both counts..
--

[Virgil]

In article
<7ab07dbc-c9b1-48af...@c15g2000vbl.googlegroups.com>,

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

> On 2 Apr., 02:01, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <1dd2037c-407c-49f6-afc6-e00c1d853...@w21g2000vbp.googlegroups.com>,
> >
> >
> >
> >
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 1 Apr., 22:44, Virgil <vir...@ligriv.com> wrote:
> >
> > > > > You do not believe that a sequence or list of all rational numbers can
> > > > > be constructed?
> >
> > > > One can "enumerate" the set of all rationals by formula, as has been
> > > > quite often done, but not by physically listing all of them.
> >
> > > A formula giving every entry is enough.
> >
> > > > Note that one cannot ennumerate by listing even sufficiently large
> > > > finite sets, so being listable other than by formula is not a relevant
> > > > criterion.
> >
> > > Constructing a list by a formula is enough to prove what I said.
> >
> > And enough to disprove what WM has said as well.
>
> Then try it.
> What did I say?

Among other things, WM has said that in a COMPLETE INFINITE BINARY TREE
there are paths of finite length, but as paths are, by definition, only
MAXIMAL sequences of parent-child linked nodes and EVERY node has child
nodes, there cannot be any such sequence of parent-child linked nodes
which is both finite and maximal.

What WM miscalls paths are merely FISONs (finite initial sequences of
nodes), no one of which fails to be a PROPER subset of some other FISON,
and thus cannot be maximal and thus cannot be a path.
--

[Virgil]

In article
<1b631b44-21cc-4e3a...@m1g2000vbe.googlegroups.com>,

What in mathematics is not a matter of belief? Even small natural
numbers have no physical existence but are merely creations of the mind.

And if other minds than WM's can create what WM does not and cannot
either comprehend or imagine, that is only WM's loss.
--

[Virgil]

In article
<0fa962ac-5f63-4f00...@r8g2000vbj.googlegroups.com>,

They are equally sensible or nonsensical, as all such "existence" occurs
only in the "world" of imagination, and imagination is not hampered by
the constraints of any physical reality.
--

[YBM]

Le 02.04.2013 22:10, WM a �crit :

Oh my God! WM is posting something stupid on the Internet, should
I reply?

[WM]

On 2 Apr., 23:46, Virgil <vir...@ligriv.com> wrote:

> > > > Nevertheless Cantor has given a finite formula to construct the list
> > > > of all rationals between 0 and 1. From that formula we can find every
> > > > entry and the anti-diagonal up to every digit d_n. From that we can
> > > > easily prove that for every FIS d_1, d_2, ..., d_n of d there are
> > > > infinitely many rationals with the same FISs. For every finite number
> > > > n - and there are no other lines than such enumerated with a finite
> > > > number!
>
> > > But as there is no end to the set/list of natural/finite numbers, there
> > > is also no end to the set/list of such lines.
>
> > What shall that "argument" be good for?
>
> WM is continually claiming

Here ordinary set theory is scrutinized.
Is the above observation correct in set theory or not?

Regards, WM

[Virgil]

In article
<f5ec9187-74e4-4017...@j9g2000vbz.googlegroups.com>,

WM <muec...@rz.fh-augsburg.de> wrote:
>
> Here ordinary set theory is scrutinized.

Since that "scrutiny" is by WM, It must be through the obscuring
astigmatic lenses of his WMYTHEOLOGY, and thus presents a false image.
--

[WM]

On 3 Apr., 00:03, Virgil <vir...@ligriv.com> wrote:
> In article

> <1b631b44-21cc-4e3a-8028-1bcfb1cc5...@m1g2000vbe.googlegroups.com>,

Given that an enumeration of all rational numbers is believed to form
a Cantor-list, then the following is not a matter of belief. Beyond
the n-th line there are f(n) rational numbers the first n digits of
which are the same as the first n digits d_1, d_2, ..., d_n of the
anti-diagonal. For every n in |N, f(n) > k for every k in |N. Define
for every n in |N the function g(n) = 1/f(n) = 0. In analysis the
limit of this function is lim[n-->oo] g(n) = 0.

Regards, WM

[Alan Smaill]

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

> What did I say?
> This: After every line n of the list of all rational numbers there are
> infinitely many rational numbers that up to digit n are identical with
> the anti-diagonal up to digit n.
>
> This holds for the digits up to every finite n. And more digits cannot
> be expected to exist in any decimal representation of a number.

So far you are in agreement with the drug-crazed Cantorists.
You must therefore be in error.

> Regards, WM

--
Alan Smaill

[WM]

On 3 Apr., 15:05, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:

If a mathematician had realized this before, he would no longer have
believed in Cantor's argument. Both statements, 1 and 2, are of
similar power:

1) For every n, there is at least one difference between the first n
bits of the bit sequence of line n and the first n bits of the
diagonal.

2) For every n, there is at least one bit sequence beyond line n the
first n bits of which are identical with the first n bits of the
diagonal.

I would be very interested to see statement 2 or a similar one in the
literature before 2013.

Regards, WM

[fom]

So, when did 'oo' become a natural number to which
a function on the natural numbers could be mapped?

[Virgil]

In article
<2da00a34-e8df-4152...@cd3g2000vbb.googlegroups.com>,

The difference being that for Cantor's diagonal (1) must be true but (2)
can easily be false for some sequence of all distinct binary sequences.
For example, a sequence of sequences all of which start with 0 will
never have the first bit containing a 1.

>
> I would be very interested to see statement 2 or a similar one in the
> literature before 2013.

Why bother searching for a statement so obviously false?
>
> Regards, WM
--

[Virgil]

In article
<b0cab414-2254-439e...@i5g2000vbk.googlegroups.com>,

But it is not given.

The only such "ennumeration" that I know of is summarily restricted to
rationals between 0 and 1 inclusive, and is ambiguous in that what I
like to call binary rationals but fom prefers to call dyadic rationals
in that range all have dual representations.


So I certainly do not believe it

> then the following is not a matter of belief.

Since the claim of "given" is false, what follows is irrelevant.
--

[WM]

On 3 Apr., 20:35, Virgil <vir...@ligriv.com> wrote:

> > If a mathematician had realized this before, he would no longer have
> > believed in Cantor's argument. Both statements, 1 and 2, are of
> > similar power:
>
> > 1) For every n, there is at least one difference between the first n
> > bits of the bit sequence of line n and the first n bits of the
> > diagonal.
>
> > 2) For every n, there is at least one bit sequence beyond line n the
> > first n bits of which are identical with the first n bits of the
> > diagonal.
>
> The difference being that for Cantor's diagonal (1) must be true but (2)
> can easily be false for some sequence of all distinct binary sequences.

Consider only sequences that contain all rational numbers of the unit
interval. I hope you don't want to save Cantor's "proof" by leaving
out some of the easily countable numbers. For these Cantor-lists, the
difference you observed is void.

What's next?

Regards, WM

[Virgil]

In article
<6be00bac-7b9c-4985...@m1g2000vbe.googlegroups.com>,

WM makes another claim, and, as usual, provides no proof of that claim.

And we see by my example L that claim was not true as stated.,

Now WM wants to append additional conditions to his claim.

When WM has provided a valid proof of his claim in (2) above, after
having appended whatever additional necessary conditions to make it
actually true, only then need anyone pay any attention to it.

But considering WM's past ability to provide proofs, even for simple
claims, we can merely round file the whole thing.
--

[WM]

On 3 Apr., 21:24, Virgil <vir...@ligriv.com> wrote:
> In article

> <6be00bac-7b9c-4985-bf6c-1a0851891...@m1g2000vbe.googlegroups.com>,

>
>
>
>
>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 3 Apr., 20:35, Virgil <vir...@ligriv.com> wrote:
>
> > > > If a mathematician had realized this before, he would no longer have
> > > > believed in Cantor's argument. Both statements, 1 and 2, are of
> > > > similar power:
>
> > > > 1) For every n, there is at least one difference between the first n
> > > > bits of the bit sequence of line n and the first n bits of the
> > > > diagonal.
>
> > > > 2) For every n, there is at least one bit sequence beyond line n the
> > > > first n bits of which are identical with the first n bits of the
> > > > diagonal.
>
> > > The difference being that for Cantor's diagonal (1) must be true but (2)
> > > can easily be false for some sequence of all distinct binary sequences.
>
> > Consider only sequences that contain all rational numbers of the unit
> > interval. I hope you don't want to save Cantor's "proof" by leaving
> > out some of the easily countable numbers. For these Cantor-lists, the
> > difference you observed is void.
>
> > What's next?
>
> WM makes another claim, and, as usual, provides no proof of that claim.
>
> And we see by my example L that claim was not true as stated.,
>
> Now WM wants to append additional conditions to his claim.

The appropriate conditions were stated on March 31 here:
http://groups.google.com/group/sci.logic/msg/b54791a03419cb66?dmode=source

And on April 1 I asked you here in this thread:
You do not believe that a sequence or list of all rational numbers can
be constructed?

> When WM has provided a valid proof of his claim in (2) above, after
> having appended whatever additional necessary conditions to make it
> actually true, only then need anyone pay any attention to it.

The only condition is that the list contains all rational numbers. I
see that you are embarrassed since nobody has hitherto made this
simple observation. So you need to find an escape. But you will not
succeed.

Regards, WM

[Virgil]

In article
<c503546e-b37e-4d03...@a14g2000vbm.googlegroups.com>,

so where is your proof?


As usual, you claim a mathematical result but do not demonstrate or
prove that claim.

We do not choose to accept your claims on faith.

If you want us to accept them then provide valid proofs for them.

Mathematical claims made without proofs may justifiably be ignored.

Your claims, being again and as usual made without proofs, we
justifiably ignore.



B
--

[WM]

On 3 Apr., 22:58, Virgil <vir...@ligriv.com> wrote:

> > The only condition is that the list contains all rational numbers.
>
> so where is your proof?

You are really not seeing it?
So it is very probable, that my conjecture is true: Nobody has ever
seen this obvious fact.

>
> As usual, you claim a mathematical result but do not demonstrate or
> prove that claim.

This is the proof:
Every sequence or list that contains all rational numbers, also
contains infinitely many rationals that have the first digits d_1,
d_2, d_3, ..., d_n in common with the anti-diagonal.
Since the sequence or list between line 1 and line n contains only a
finite number of rationals, there must follow infinitely many after
line n.

Regards, WM

[Virgil]

In article
<2b4680be-1562-48f2...@h1g2000vbx.googlegroups.com>,

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

> On 3 Apr., 22:58, Virgil <vir...@ligriv.com> wrote:
>
> > > The only condition is that the list contains all rational numbers.
> >
> > so where is your proof?
>
> You are really not seeing it?

I certainly do not se anything like a proof, because you have not
provided any.

> So it is very probable, that my conjecture is true: Nobody has ever
> seen this obvious fact.

Probability is irrelevant. but if we are to attach one, how about to
WM's claim today that the sum from k = 1 to k = m of 1/k equals m(m+1)/2.
I find that claim highly improbable.

> >
> > As usual, you claim a mathematical result but do not demonstrate or
> > prove that claim.
>
> This is the proof:
> Every sequence or list that contains all rational numbers, also
> contains infinitely many rationals that have the first digits d_1,
> d_2, d_3, ..., d_n in common with the anti-diagonal.

Another claim not proven. And stated too ambiguously.

> Since the sequence or list between line 1 and line n contains only a
> finite number of rationals, there must follow infinitely many after
> line n.
>
> Regards, WM

--

[WM]

On 3 Apr., 23:25, Virgil <vir...@ligriv.com> wrote:

> > This is the proof:
> > Every sequence or list that contains all rational numbers, also
> > contains infinitely many rationals that have the first digits d_1,
> > d_2, d_3, ..., d_n in common with the anti-diagonal.
>
> Another claim not proven. And stated too ambiguously.

Sorry, I will reduce the speed of my writing to the speed of your
thinking. Can you understand that there are infinitely many rational
numbers that in decimal representation have the first n digits d_1,
d_2, d_3, ..., d_n?

Regards, WM

[Virgil]

In article
<3f0b8f39-a4f2-4ad2...@r7g2000vbw.googlegroups.com>,

I do not recognize d_1, d_2, d_3,...,d_n as being any particular decimal
digits. Which digits are they?
--

[WM]

On 4 Apr., 10:02, Virgil <vir...@ligriv.com> wrote:
> In article

> <3f0b8f39-a4f2-4ad2-829c-d7523bb1a...@r7g2000vbw.googlegroups.com>,

For all n these are the first n digits of the anti-diagonal.
It does not make any difference what values they may assume. My proof
holds for every anti-diagonal of every possible Cantor-list (that
contains all rational numbers between 0 and 1 or, if you like, between
-oo and oo).

Regards, WM

[fom]

You really just make these delusions up as you go.

[William Hughes]

[Virgil]

In article
<125dbdf6-ce6d-4d26...@g8g2000vbf.googlegroups.com>,

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

> On 4 Apr., 10:02, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <3f0b8f39-a4f2-4ad2-829c-d7523bb1a...@r7g2000vbw.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 3 Apr., 23:25, Virgil <vir...@ligriv.com> wrote:
> >
> > > > > This is the proof:
> > > > > Every sequence or list that contains all rational numbers, also
> > > > > contains infinitely many rationals that have the first digits d_1,
> > > > > d_2, d_3, ..., d_n in common with the anti-diagonal.
> >
> > > > Another claim not proven. And stated too ambiguously.
> >
> > > Sorry, I will reduce the speed of my writing to the speed of your
> > > thinking. Can you understand that there are infinitely many rational
> > > numbers that in decimal representation have the first n digits d_1,
> > > d_2, d_3, ..., d_n?
> >
> > I do not recognize d_1, d_2, d_3,...,d_n as being any particular decimal
> > digits. Which digits are they?
>
> For all n these are the first n digits of the anti-diagonal.

WHICH anti-diagonal? There are at least a countable infinity of them for
any such listing. And the original Cantor argument was about pure
binary sequences, not decimal or binary representations of reals.

> It does not make any difference what values they may assume. My proof
> holds for every anti-diagonal of every possible Cantor-list (that
> contains all rational numbers between 0 and 1 or, if you like, between
> -oo and oo).

I have yet to see any "proof" which invalidates the origianl Cantor
diagonal argument, only your oft repeated but forever insubstantiated
claim of one.
--

[WM]

On 4 Apr., 21:55, Virgil <vir...@ligriv.com> wrote:

>
> > For all n these are the first n digits of the anti-diagonal.
>
> WHICH anti-diagonal?

Every anti-diagonal.

> There are at least a countable infinity of them for
> any such listing.  And the original Cantor argument was about pure
> binary sequences, not decimal or binary representations of reals.

We know that Cantor's original argument fails, when applied to real
numbers.
Nevertheless, my proof covers all possible cases and all possible anti-
diagonals.

>
> > It does not make any difference what values they may assume. My proof
> > holds for every anti-diagonal of every possible Cantor-list (that
> > contains all rational numbers between 0 and 1 or, if you like, between
> > -oo and oo).
>

> I have yet to see any "proof" which invalidates the original Cantor

> diagonal argument, only your oft repeated but forever insubstantiated
> claim of one.

If you are unable to see it, even after repeated explanations, then
you belong to the famous, nearly empty set of those who will never
understand why Cantor failed.

Regards, WM

[Virgil]

In article
<d1e2d90f-8c5e-4bbc...@cm2g2000vbb.googlegroups.com>,

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

> On 4 Apr., 21:55, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > For all n these are the first n digits of the anti-diagonal.
> >
> > WHICH anti-diagonal?
>
> Every anti-diagonal.

But there can be two anti diagonals whose first n digits do not overlap
at all for some n.

>
> > There are at least a countable infinity of them for
> > any such listing.  And the original Cantor argument was about pure
> > binary sequences, not decimal or binary representations of reals.
>
> We know that Cantor's original argument fails, when applied to real
> numbers.

It only needs some minor modification, and the modified versions
succeed.

> Nevertheless, my proof covers all possible cases and all possible anti-
> diagonals.

Your "proof" is not a proof. At least nowhere outside of
Wolkenmuekenheim.

>
> >
> > > It does not make any difference what values they may assume. My proof
> > > holds for every anti-diagonal of every possible Cantor-list (that
> > > contains all rational numbers between 0 and 1 or, if you like, between
> > > -oo and oo).
> >
> > I have yet to see any "proof" which invalidates the original Cantor
> > diagonal argument, only your oft repeated but forever insubstantiated
> > claim of one.
>
> If you are unable to see it, even after repeated explanations

Your "explanations" do not constitute proofs. At least nowhere outside
of Wolkenmuekenheim.
They are not even close to proofs outside of Wolkenmuekenheim.
--

[WM]

On 4 Apr., 23:17, Virgil <vir...@ligriv.com> wrote:

> In article
> <d1e2d90f-8c5e-4bbc-a8b3-c7b537331...@cm2g2000vbb.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 4 Apr., 21:55, Virgil <vir...@ligriv.com> wrote:
>
> > > > For all n these are the first n digits of the anti-diagonal.
>
> > > WHICH anti-diagonal?
>
> > Every anti-diagonal.
>
> But there can be two anti diagonals whose first n digits do not overlap
> at all for some n.

For each of them there are infinitely many entries in the list, that
start with just a FIS of one of these two anti-diagonals. Even for
infinitely many anti-diagonals this is true.

> > > I have yet to see any "proof" which invalidates the original Cantor
> > > diagonal argument, only your oft repeated but forever insubstantiated
> > > claim of one.
>
> > If you are unable to see it, even after repeated explanations
>
> Your "explanations" do not constitute proofs.

Are you really unable to understand that in the set of all rational
numbers there are infinitely many numbers that have the FIS d_1, d_2,
d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
k =< n?

Regards, WM

[Virgil]

In article
<f7237df3-e100-4dba...@z4g2000vbz.googlegroups.com>,

I am certainly unable to see that whether it is true or not makes any
difference in WM's many false claims about the Cantor diagonal argument.

The bottom line is that for any given list of infinite binary sequences
there exist as many binary sequences unlisted as listed.
--

[WM]

On 5 Apr., 10:50, Virgil <vir...@ligriv.com> wrote:

> > Are you really unable to understand that in the set of all rational
> > numbers there are infinitely many numbers that have the FIS d_1, d_2,
> > d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
> > k =< n?
>
> I am certainly unable to see that whether it is true or not makes any
> difference in WM's many false claims about the Cantor diagonal argument.

Of course you are unable to see that.

>
> The bottom line is that for any given list of infinite binary sequences
> there exist as many binary sequences unlisted as listed.

That is one side of the contradiction. The other bottom line is that
*for every n* the Cantor-list beyond n contains infinitely many
rational numbers q_k that have the same sequence of first n digits as
the anti-diagonal d.

Since in FOPL "for every" is the same as "for all", we have:
For all n the Cantor list contains d_1, ..., d_n infinitely often.

So if d is nothing more than all its FISs, there is a contradiction.
And if d is more than all its FISs, then we can remove all FISs
without removing d. That's also a contradiction.

Regards, WM

[fom]

On 4/5/2013 10:53 AM, WM wrote:
> On 5 Apr., 10:50, Virgil <vir...@ligriv.com> wrote:
>
>>> Are you really unable to understand that in the set of all rational
>>> numbers there are infinitely many numbers that have the FIS d_1, d_2,
>>> d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
>>> k =< n?
>>
>> I am certainly unable to see that whether it is true or not makes any
>> difference in WM's many false claims about the Cantor diagonal argument.
>
> Of course you are unable to see that.
>>
>> The bottom line is that for any given list of infinite binary sequences
>> there exist as many binary sequences unlisted as listed.
>
> That is one side of the contradiction. The other bottom line is that
> *for every n* the Cantor-list beyond n contains infinitely many
> rational numbers q_k that have the same sequence of first n digits as
> the anti-diagonal d.
>

And Cantor's argument is such that the constructed diagonal
will differ from each of them at some finite length greater
than n.

> Since in FOPL "for every" is the same as "for all", we have:
> For all n the Cantor list contains d_1, ..., d_n infinitely often.
>

It is strange that when WM proves things, he generally
proves his own misunderstandings. The statement
is correct, but the remarks concerning the quantifiers
is relevant only to WM

> So if d is nothing more than all its FISs, there is a contradiction.

Contradictions have to be proven. Your claim overlooks the
fact that at every n, the finite initial of the same length
as the element of the listing will be purposely formed to
be different from that element from the listing.

If by "more than all its finite initial segments" you mean
that its definition relies on the given list, you are
correct. Then there is no contradiction.

> And if d is more than all its FISs, then we can remove all FISs
> without removing d. That's also a contradiction.

If by "more than all its finite initial segments" you mean
that its definition relies on the given list, you are
correct. Then there is no contradiction. Indeed, were
"removing" a legitimate mathematical practice, removing
the list would remove the definition. Without the definition
there can be no contradiction with respect to non-propositions
involving the non-definition.

[Virgil]

In article
<96d4d6af-163e-4542...@h1g2000vbx.googlegroups.com>,

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

> On 5 Apr., 10:50, Virgil <vir...@ligriv.com> wrote:
>
> > > Are you really unable to understand that in the set of all rational
> > > numbers there are infinitely many numbers that have the FIS d_1, d_2,
> > > d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
> > > k =< n?
> >
> > I am certainly unable to see that whether it is true or not makes any
> > difference in WM's many false claims about the Cantor diagonal argument.
>
> Of course you are unable to see that.
> >
> > The bottom line is that for any given list of infinite binary sequences
> > there exist as many binary sequences unlisted as listed.
>
> That is one side of the contradiction. The other bottom line is that
> *for every n* the Cantor-list beyond n contains infinitely many
> rational numbers q_k that have the same sequence of first n digits as
> the anti-diagonal d.

That is false, since a "Cantor list" may well consist of a list in which
every line begins with a zero but the antidiagonal does not.
"Cantor lists" are arbitrary lists and must be allowed to have any
pattern or avoid any pattern, since ANY list is a "Cantor list".

Thus WM entirely misses the point once more!

>
> Since in FOPL "for every" is the same as "for all", we have:
> For all n the Cantor list contains d_1, ..., d_n infinitely often.

A list in which every line begins with a zero is still a "Cantor list"
since EVERY list is a Cantor list.

>
> So if d is nothing more than all its FISs, there is a contradiction.

Not outside of Wolkenmuekenheim.


While WM may control what lists are "CAntor lists" inside
Wolkenmuekenheim, EVERY infinite list of digit sequences list is a
Cantor list outside of Wolkenmuekenheim.

> And if d is more than all its FISs, then we can remove all FISs

Not without changing the list from which those FIS's are constructed.

WM seems to have forgot that the construction of an antidiagonal is
dependent on the list from which it is constructed, and that any change
in the list may force a change in the antidiagonal.
--

[WM]

On 5 Apr., 22:36, Virgil <vir...@ligriv.com> wrote:

>
> > > > Are you really unable to understand that in the set of all rational
> > > > numbers there are infinitely many numbers that have the FIS d_1, d_2,
> > > > d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
> > > > k =< n?

> > > The bottom line is that for any given list of infinite binary sequences
> > > there exist as many binary sequences unlisted as listed.
>
> > That is one side of the contradiction. The other bottom line is that
> > *for every n* the Cantor-list beyond n contains infinitely many
> > rational numbers q_k that have the same sequence of first n digits as
> > the anti-diagonal d.
>
> That is false, since a "Cantor list"

Read carefully: I did not say "a Cantor-list" but "the Cantor-list"
with reference to that one containing all ratinal numbers (see first
paragraph).

> may well consist of a list in which
> every line begins with a zero but the antidiagonal does not.
> "Cantor lists" are arbitrary lists and must be allowed to have any
> pattern or avoid any pattern, since ANY list is a "Cantor list".

And that is all you can argue?

> WM seems to have forgot that the construction of an antidiagonal is
> dependent on the list from which it is constructed, and that any change
> in the list may force a change in the antidiagonal.

Again you miss the point. See the first paragraph.

Cantor's argues that the constructed diagonal will differ from each
entry at some digit n.
I have proved that this is false, since for every n there are
infinitely many entries with the same FIS d_1, ..., d_n as the anti-
diagonal.
Since this is true for every n, it cannot be false for any n.
Contradiction, if d is nothing else but the sequence of all its FISs.

The situation would be different, if d was the *limit* of the sequence
of all its FISs, not a member - like 0 is the limit of the sequence
(10^-n). Alas Cantor's argument is only valid for finite n. (10^-n)
has its first digit 1 not at a finite index.

Regards, WM

[fom]

On 4/6/2013 4:14 AM, WM wrote:
> On 5 Apr., 22:36, Virgil <vir...@ligriv.com> wrote:
>

<snip>

>
> Cantor's argues that the constructed diagonal will differ from each
> entry at some digit n.

The Cantor argument says that the constructed diagonal will differ
from the n-th entry of a complete fixed list at the n-th symbol of
its concatenation. It is not "some" ambiguous location.

> I have proved that this is false, since for every n there are
> infinitely many entries with the same FIS d_1, ..., d_n as the anti-
> diagonal.

WM has proved nothing. He never does.

The number of entries that share a given finite initial segment
is irrelevant.

> Since this is true for every n, it cannot be false for any n.

Perhaps, but it is still irrelevant and proves nothing.

> Contradiction, if d is nothing else but the sequence of all its FISs.
>

The constructed diagonal is defined by its relation to a fixed
list.

> The situation would be different, if d was the *limit* of the sequence
> of all its FISs, not a member -

The constructed diagonal is not defined with respect to its
initial segments, is not defined as a limit, and is not a
member of any set containing only finite sets as elements.

> like 0 is the limit of the sequence
> (10^-n). Alas Cantor's argument is only valid for finite n.

Cantor's argument applies in any situation where the
premises for the argument have been satisfied. Nothing
WM ever says meets that stipulation.

> (10^-n)
> has its first digit 1 not at a finite index.

If n is finite, 10^n is not zero.

You even get that wrong with your typical
carelessness despite having stated it correctly
10 words prior.

[Virgil]

In article
<428fd1e4-64bb-4142...@m1g2000vbe.googlegroups.com>,

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

> On 5 Apr., 22:36, Virgil <vir...@ligriv.com> wrote:
>
> >
> > > > > Are you really unable to understand that in the set of all rational
> > > > > numbers there are infinitely many numbers that have the FIS d_1, d_2,
> > > > > d_3, ..., d_n irrespective of the values of n in |N and d_k with 1 =<
> > > > > k =< n?
>
> > > > The bottom line is that for any given list of infinite binary sequences
> > > > there exist as many binary sequences unlisted as listed.
> >
> > > That is one side of the contradiction. The other bottom line is that
> > > *for every n* the Cantor-list beyond n contains infinitely many
> > > rational numbers q_k that have the same sequence of first n digits as
> > > the anti-diagonal d.
> >
> > That is false, since a "Cantor list"
>
> Read carefully: I did not say "a Cantor-list" but "the Cantor-list"
> with reference to that one containing all ratinal numbers (see first
> paragraph).

But there is no such "the Cantor-list" which is the only one containing
all rationals, since any permutation of the order of listing produces a
different "the Cantor-list" also containing all the rationals.

So which "the Cantor-list" does WM mean?

>
> > may well consist of a list in which
> > every line begins with a zero but the antidiagonal does not.
> > "Cantor lists" are arbitrary lists and must be allowed to have any
> > pattern or avoid any pattern, since ANY list is a "Cantor list".
>
> And that is all you can argue?

Beats hell out of anything that WM has so far managed to argue!

>
> > WM seems to have forgot that the construction of an antidiagonal is
> > dependent on the list from which it is constructed, and that any change
> > in the list may force a change in the antidiagonal.
>
> Again you miss the point. See the first paragraph.

It is WM who, as usual, deliberately misses every point he can manage to
slide around.

Every permutation of a list containing all rationals is also a list of
all rationals, so that there is no list unique in that respect.

>
> Cantor's argues that the constructed diagonal will differ from each
> entry at some digit n.
> I have proved that this is false

No you have not. Claiming to have done something and actually doing it
are quite different, and most of what WM claims to have done here has
never been done by him anywhere.


Further garbage deleted.
--

[WM]

On 6 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:

> But there is no such "the Cantor-list" which is the only one containing
> all rationals, since any permutation of the order of listing produces a
> different "the Cantor-list" also containing all the rationals.

But "the Cantor list" that I referred to is containing all rational
numbers. Whether it contains anything else is irrelevant for my proof.

>
>
> > Cantor's argues that the constructed diagonal will differ from each
> > entry at some digit n.
> > I have proved that this is false
>
> No you have not.

For all n the sequence d_1, d_2, d_3, ..., d_n is infinitely often in
the list that contains all rational numbers. If you cannot prove it by
yourself, then you cannot prove it and probably cannot understand it.
So much is deplorable but sure.

Regards, WM

[fom]

What WM seems to have proven here is that his own
non-standard notions of quantification confuse his
ability to reason.

He appears to be interpreting "each [particular] entry at
some digit n" as "every entry at some [particualar]
digit n".

But, if Cantor had actually made an argument along
those lines, no one would be discussing Cantor's
ideas at all.

[Virgil]

In article
<f98c9d07-bb2e-4b8c...@he10g2000vbb.googlegroups.com>,

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

> On 6 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
>
> > But there is no such "the Cantor-list" which is the only one containing
> > all rationals, since any permutation of the order of listing produces a
> > different "the Cantor-list" also containing all the rationals.
>
> But "the Cantor list" that I referred to is containing all rational
> numbers. Whether it contains anything else is irrelevant for my proof.

Which of the uncountably many Cantor lists that contains all rationals
are you referring to? And in what notation, fractional with decimal
numerator and denominator for those rationals?

> >
> >
> > > Cantor's argues that the constructed diagonal will differ from each
> > > entry at some digit n.
> > > I have proved that this is false

Wm has claimed to have proved it false, but has not actually proved it,
or much of anything else.
--

[WM]

On 7 Apr., 00:01, Virgil <vir...@ligriv.com> wrote:
> In article

> <f98c9d07-bb2e-4b8c-b0c7-07ddb6e51...@he10g2000vbb.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > On 6 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
>
> > > But there is no such "the Cantor-list" which is the only one containing
> > > all rationals, since any permutation of the order of listing produces a
> > > different "the Cantor-list" also containing all the rationals.
>
> > But "the Cantor list" that I referred to is containing all rational
> > numbers. Whether it contains anything else is irrelevant for my proof.
>
> Which of the uncountably many Cantor lists that contains all rationals
> are you referring to?

Every one. When you will have understood my proof, if that happens,
you will recognize that the special form of the sequence does not
matter.

> And in what notation, fractional  with decimal
> numerator and denominator for those rationals?

Even that does not really matter, but I recommend that you use the
number 10 as base (like I recommend to some weaker students to insert
numbers if they feel uncomfortable with general expressions).

Regards, WM

[fom]

On 4/7/2013 3:51 AM, WM wrote:
> On 7 Apr., 00:01, Virgil <vir...@ligriv.com> wrote:
>> In article
>> <f98c9d07-bb2e-4b8c-b0c7-07ddb6e51...@he10g2000vbb.googlegroups.com>,
>>
>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>> On 6 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
>>
>>>> But there is no such "the Cantor-list" which is the only one containing
>>>> all rationals, since any permutation of the order of listing produces a
>>>> different "the Cantor-list" also containing all the rationals.
>>
>>> But "the Cantor list" that I referred to is containing all rational
>>> numbers. Whether it contains anything else is irrelevant for my proof.
>>
>> Which of the uncountably many Cantor lists that contains all rationals
>> are you referring to?
>
> Every one. When you will have understood my proof, if that happens,
> you will recognize that the special form of the sequence does not
> matter.
>

And, although you will never quite grasp
it, the Cantor argument is a scheme which
only applies when the listing is a fixed
list.

It is sort of interesting that you seem
to have used "every" as an "all".

[WM]

If you are really unable to understand general ideas and proofs then
use the well-known enumeration of the rationals by Cantor himself.

Again and again you prove why you could not but fail to graduate.

Regards, WM

[fom]

chuckle

[Virgil]

In article
<5d700a23-446a-4d3c...@gp5g2000vbb.googlegroups.com>,

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

> On 7 Apr., 00:01, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <f98c9d07-bb2e-4b8c-b0c7-07ddb6e51...@he10g2000vbb.googlegroups.com>,
> >
> >  WM <mueck...@rz.fh-augsburg.de> wrote:
> > > On 6 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
> >
> > > > But there is no such "the Cantor-list" which is the only one containing
> > > > all rationals, since any permutation of the order of listing produces a
> > > > different "the Cantor-list" also containing all the rationals.
> >
> > > But "the Cantor list" that I referred to is containing all rational
> > > numbers. Whether it contains anything else is irrelevant for my proof.
> >
> > Which of the uncountably many Cantor lists that contains all rationals
> > are you referring to?
>
> Every one. When you will have understood my proof, if that happens,
> you will recognize that the special form of the sequence does not
> matter.

The proof of your "theorem" is irrelevant as the truth of your theorem
in no way counter the truth of the CDA (Cantor Diagoanl Argument).

From another of WM;s posts, here is his :theorem":
Consider a Cantor-list that contains a complete sequence (q_k) of all
rational numbers q_k. The first n digits of the anti-diagonal d are
d_1, d_2, d_3, ..., d_n. It can be shown *for every n* that the Cantor-
list beyond line n contains infinitely many rational numbers q_k that

have the same sequence of first n digits as the anti-diagonal d.

Since that argument in no way forces the anti-diagonal to be a rational
number, and since no one claims that the rationals cannot be listed,
WM's present argument is irrelevant to the truth of the CDA
--

[Virgil]

In article
<9ed73b7d-9364-4124...@a3g2000vbr.googlegroups.com>,

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

> If you are really unable to understand general ideas and proofs then

Then you will be on an even level with WM.
--

[Ross A. Finlayson]

On 3/30/2013 6:55 PM, fom wrote:
> On 3/30/2013 7:56 PM, Virgil wrote:
>> In article
>> <2bc13fff-5cbb-43dd...@m9g2000vbc.googlegroups.com>,
>> WM <muec...@rz.fh-augsburg.de> wrote:
>>
>>> On 30 Mrz., 22:11, Virgil <vir...@ligriv.com> wrote:
>>>
>>>>>> Thus there is always at least one bit of any listed entry disagreeing
>>>>>> with the antidiagonanl, just as the Cantor proof requires.
>>>>
>>>>> In a list containing every rational: Is there always, i.e., up to
>>>>> every digit, an infinite set of paths identical with the anti-
>>>>> diagonal? Yes or no?
>>>>
>>>> The set of paths in any Complete Infinite Binary Tree which agree with
>>>> any particular path up to its nth node is equinumerous with the set of
>>>> all paths in the entire tree i.e., is uncountably infinite.
>>>
>>> This was the question: In a list containing every rational: Is there
>>> always, i.e., up to every digit, an infinite set of paths identical
>>> with the anti-diagonal? Yes or no?
>>
>> Lists and trees are different. And anti-diagonals derive from lists, not
>> trees.
>> The entries in list are well ordered.
>> The entries in a Complete Infinite Binary Tree are densely ordered.
>> Those order types are incompatible.
>> So questions, like WM's, which confuse them, are nonsense.
>> At least outside Wolkenmuekenheim.
>>
>
> But note that the question also demonstrates WM's
> complete lack of understanding of the diagonal
> argument.
>
> He has been told time and time again that it is
> an argument scheme which only has application
> under certain assumptions.
>
> He chooses to believe otherwise for the agenda
> of his fanaticism.
>
> Suppose one is given a countable listing of
> the rationals (with the appropriate restriction
> on double representation) according to the
> infinite listing of an expansion.
>
> Suppose one performs a diagonalization on
> that listing.
>
> Is the resultant a rational number? No.
>
> What may be concluded? That the rational numbers
> do not exhaust the capacity of the algorithm
> to generate representations if that algorithm
> is to generate a representation for every
> rational number.
>
> Although the burden of proof lies with WM
> concerning the nature of the diagonal, it
> is a simple matter to understand if one
> uses a Baire space representation instead.
>
> In the Baire space, rationals are in correspondence
> with eventually constant sequences. Since,
> by construction, a list of Baire space rationals
> would exhaust all of the eventually constant
> sequences, the resultant of a diagonal argument
> could not have an eventually constant sequence
> unless the original premise had been false.
>
>
>
>
>
>
>
>
>
>
>
>
>
>

Oh this now.

Well I guess this is from last year but still.