Matheology § 246

[WM]

Matheology § 246

Cantor's list contains real numbers r as binary or decimal fractions.
Real numbers, however, are /limits/ of binary or decimal fractions.

For every terminating fraction of r, Cantor obtains a difference
between r and the due terminating fraction of the anti-diagonal d:
r_nn =/= d_n. He concludes that this remains true for the limits of
the list numbers r and d by using the argument: different sequences
have different limits. But it is well known that this argument is not
admissible in proofs because it is false.

Regards, WM

[fom]

On 4/12/2013 12:57 AM, WM wrote:
> Matheology � 246
>
> Cantor's list contains real numbers r as binary or decimal fractions.

WM is wrong again.

Cantor's list consists of one representation for
a real number that is not on the given list purported
to consist of representations for every number.

> Real numbers, however, are /limits/ of binary or decimal fractions.
>

Yes. This is why the arithmetic of real numbers is not the
arithmetic of rational numbers, although the latter is
representable within the former.

> For every terminating fraction of r, Cantor obtains a difference
> between r and the due terminating fraction of the anti-diagonal d:
> r_nn =/= d_n.

He does not obtain one. He constructs one based on
syntactic criteria.

> He concludes that this remains true for the limits of
> the list numbers r and d by using the argument: different sequences
> have different limits. But it is well known that this argument is not
> admissible in proofs because it is false.

But, the argument is based on the representation
of real numbers with respect to representation
according to the output of the Euclidean algorithm.

There is no assumption concerning the convergence
of partial sums whatsoever.

If WM's statement were to be given credence, the Euclidean
algorithm of long division could no longer be considered
as providing a faithful representation of distinct real
numbers (or rational numbers for that matter).

[JT]

0.333... and so on is not 1/3 in any digit place, only in the
imagination of infinite actual representation, you can do your long
division in infinity that number series will *never* represent 1/3.
Partition of the reals into bases using longdivision is not lossless
to use computer terms. It is 0.333... is an identity now they also
claim 1=1.000... is an identity and it is laughable there is no such
creature. The natural 1 is discete is does not have any decimal
expansion. The zeros is dreamed up from some noneexistent numberline.

[fom]

On 4/12/2013 2:26 AM, JT wrote:

> Partition of the reals into bases using longdivision is not lossless

I think you will find your fancy "computer term"
discussed in another post I did.

news://news.giganews.com:119/5sidnRpmsrGpz4rM...@giganews.com


Then, if you really want to cut your teeth on
a trying to do mathematical constructions
"losslessly" you can try to make sense
of this one. Words like "impenetrable"
apply to the readability of the exposition.

news://news.giganews.com:119/M5qdncGgG5a-VbvM...@giganews.com

[Ralf Bader]

fom wrote:

What you wrote is much too wonkish. Because what Mückenheim wrote shows that
he is already totally clueless about infinite decimal fractions. He is
obviously too stupid, and willfully so, to understand even the most basic
facts about such fractions, taken as sequences of their finite partial
fractions. Facts that are explained in any decent (so probably not in
Mückenheims "bestseller") introductory calculus book. Must be a crazy
country in which such a person qualifies to teach mathematics at a
so-called "university".

[WM]

On 12 Apr., 09:26, JT <jonas.thornv...@gmail.com> wrote:

>
> > > For every terminating fraction of r, Cantor obtains a difference
> > > between r and the due terminating fraction of the anti-diagonal d:
> > > r_nn =/= d_n.
>
> > He does not obtain one.  He constructs one based on
> > syntactic criteria.

Blablabla. He puts d_n =/= r_nn. And he says so. But he forgets about
the limit.

> > There is no assumption concerning the convergence
> > of partial sums whatsoever.

That's why the argument could survive so long.

>
> > If WM's statement were to be given credence, the Euclidean
> > algorithm of long division could no longer be considered
> > as providing a faithful representation of distinct real
> > numbers (or rational numbers for that matter).
>
> 0.333... and so on is not 1/3 in any digit place, only in the
> imagination of infinite actual representation, you can do your long
> division in infinity that number series will *never* represent 1/3.

Correct. Cantorists have to assume that they will get ready, but in
general are not willing (or not able) to recognize what they have to
assume.

> Partition of the reals into bases using longdivision is not lossless
> to use computer terms. It is 0.333... is an identity now they also
> claim 1=1.000... is an identity and it is laughable there is no such
> creature. The natural 1 is discete is does not have any decimal
> expansion.

But some zeros do not change the numerical value. And nobody can
really use more than some.

Regards, WM

[fom]

On 4/12/2013 2:51 AM, Ralf Bader wrote:
> fom wrote:
>
> What you wrote is much too wonkish. Because what Mückenheim wrote shows that
> he is already totally clueless about infinite decimal fractions.

If I ask myself why Virgil has been correcting matters
of fact (with embellishments) it is not because he thinks
WM is correctable. It is because rhetoric can actually
confuse people into thinking that WM has valid points.

Besides, "wonkish" is my specialty.

> He is
> obviously too stupid, and willfully so, to understand even the most basic
> facts about such fractions, taken as sequences of their finite partial
> fractions. Facts that are explained in any decent (so probably not in
> Mückenheims "bestseller") introductory calculus book. Must be a crazy
> country in which such a person qualifies to teach mathematics at a
> so-called "university".
>

My city is full of "for-profit" universities that do
a great deal of advertising and very little teaching.

The U.S. government is finally considering looking into
their practices more closely because of the apparent
problems of student debt and inadequate instruction.

My country would have to be one of those crazy countries.

[fom]

On 4/12/2013 3:00 AM, WM wrote:
> On 12 Apr., 09:26, JT <jonas.thornv...@gmail.com> wrote:
>
>>
>>>> For every terminating fraction of r, Cantor obtains a difference
>>>> between r and the due terminating fraction of the anti-diagonal d:
>>>> r_nn =/= d_n.
>>
>>> He does not obtain one. He constructs one based on
>>> syntactic criteria.
>
> Blablabla.

WM almost managed an argument.

[Virgil]

In article
<d5b94127-f249-4639...@a6g2000vbm.googlegroups.com>,

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

> Matheology � 246
>
> Cantor's list contains real numbers r as binary or decimal fractions.
> Real numbers, however, are /limits/ of binary or decimal fractions.

Real numbers are all certainly expressable as limits of sequences, but
so are rationals and integers and naturals.

None of which weakens the Cantor argument in any way.

>
> For every terminating fraction of r, Cantor obtains a difference
> between r and the due terminating fraction of the anti-diagonal d:
> r_nn =/= d_n. He concludes that this remains true for the limits of
> the list numbers r and d by using the argument: different sequences
> have different limits. But it is well known that this argument is not
> admissible in proofs because it is false.

The Cantor antidiagonal argument does not use the limit properties of
sequences at all. In fact his binary sequences of members of {m,w} are
not assumed to have any numerical or topological properties at all, much
lesimits. Each of the Cantor sequences is, in fact, merely a function
from |N to {m,w} with no other properties assumed or allowed, except
that two such functions are different if the is some n in |N at which
they have different values.

But that is far too mathematical for WM to grasp.
--

[Virgil]

In article <y-udnc3rbL5NOvrM...@giganews.com>,

fom <fom...@nyms.net> wrote:

> On 4/12/2013 12:57 AM, WM wrote:
> > Matheology � 246
> >
> > Cantor's list contains real numbers r as binary or decimal fractions.
>
> WM is wrong again.
>
> Cantor's list consists of one representation for
> a real number that is not on the given list purported
> to consist of representations for every number.

Actually the lists are not Cantor's, but are challenges to Cantor, each
of which is shown by Cantor to be an incomplete listing of all binary
sequnces of the letters "m" and "w".

>
> > Real numbers, however, are /limits/ of binary or decimal fractions.
> >
>
> Yes. This is why the arithmetic of real numbers is not the
> arithmetic of rational numbers, although the latter is
> representable within the former.
>
> > For every terminating fraction of r, Cantor obtains a difference
> > between r and the due terminating fraction of the anti-diagonal d:
> > r_nn =/= d_n.
>
> He does not obtain one. He constructs one based on
> syntactic criteria.
>
> > He concludes that this remains true for the limits of
> > the list numbers r and d by using the argument: different sequences
> > have different limits. But it is well known that this argument is not
> > admissible in proofs because it is false.
>
> But, the argument is based on the representation
> of real numbers with respect to representation
> according to the output of the Euclidean algorithm.
>
> There is no assumption concerning the convergence
> of partial sums whatsoever.
>
> If WM's statement were to be given credence, the Euclidean
> algorithm of long division could no longer be considered
> as providing a faithful representation of distinct real
> numbers (or rational numbers for that matter).

--

[fom]

On 4/12/2013 3:44 AM, Virgil wrote:
> In article <y-udnc3rbL5NOvrM...@giganews.com>,
> fom <fom...@nyms.net> wrote:
>
>> On 4/12/2013 12:57 AM, WM wrote:
>>> Matheology � 246
>>>
>>> Cantor's list contains real numbers r as binary or decimal fractions.
>>
>> WM is wrong again.
>>
>> Cantor's list consists of one representation for
>> a real number that is not on the given list purported
>> to consist of representations for every number.
>
> Actually the lists are not Cantor's, but are challenges to Cantor, each
> of which is shown by Cantor to be an incomplete listing of all binary
> sequnces of the letters "m" and "w".


"Cantor's list consists of one representation for
a real number"

I guess I was not clear. But we agree.

[Virgil]

In article
<5c9991a6-f874-4850...@gb2g2000vbb.googlegroups.com>,

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

> On 12 Apr., 09:26, JT <jonas.thornv...@gmail.com> wrote:
>
> >
> > > > For every terminating fraction of r, Cantor obtains a difference
> > > > between r and the due terminating fraction of the anti-diagonal d:
> > > > r_nn =/= d_n.
> >
> > > He does not obtain one. �He constructs one based on
> > > syntactic criteria.
>
> Blablabla. He puts d_n =/= r_nn. And he says so. But he forgets about
> the limit.

What is "the limit" for sequences of letters from {m,w}?

>
> > > There is no assumption concerning the convergence
> > > of partial sums whatsoever.
>
> That's why the argument could survive so long.

Since the Cantor diagonal argument is totally independent of any
convergence needs or wants, convergence arguments can have no effect on
it at all.

> >
> > > If WM's statement were to be given credence, the Euclidean
> > > algorithm of long division could no longer be considered
> > > as providing a faithful representation of distinct real
> > > numbers (or rational numbers for that matter).
> >
> > 0.333... and so on is not 1/3 in any digit place, only in the
> > imagination of infinite actual representation, you can do your long
> > division in infinity that number series will *never* represent 1/3.
>
> Correct. Cantorists have to assume that they will get ready, but in
> general are not willing (or not able) to recognize what they have to
> assume.

Which is as nothing to what WM keeps assuming even, nay, especially,
when proven wrong.
--

[WM]

On 12 Apr., 10:41, Virgil <vir...@ligriv.com> wrote:
> In article
> <d5b94127-f249-4639-8f2f-9de6dda34...@a6g2000vbm.googlegroups.com>,

>
>  WM <mueck...@rz.fh-augsburg.de> wrote:
> > Matheology § 246
>
> > Cantor's list contains real numbers r as binary or decimal fractions.
> > Real numbers, however, are /limits/ of binary or decimal fractions.
>
> Real numbers are all certainly expressable as limits of sequences, but
> so are rationals and integers and naturals.

No objection.

> > For every terminating fraction of r, Cantor obtains a difference
> > between r and the due terminating fraction of the anti-diagonal d:
> > r_nn =/= d_n. He concludes that this remains true for the limits of
> > the list numbers r and d by using the argument: different sequences
> > have different limits. But it is well known that this argument is not
> > admissible in proofs because it is false.
>
> The Cantor antidiagonal argument does not use the limit properties of
> sequences at all. In fact his binary sequences of members of {m,w} are
> not assumed to have any numerical or topological properties at all,

They are assumed to have infinitely many bits. But an infinite
sequence of bits is not converging (the factors 2^-n are missing).
Most of them cannot be written by finite expressions. And they cannot
be written as infinite expressions.

> much
> lesimits. Each of the Cantor sequences  is, in fact, merely a function
> from |N to {m,w} with no other properties assumed or allowed, except
> that two such functions are different if the is some n in |N at which
> they have different values.

Most of them are undefined and undefinable. But in order to remedy
that problem we use rationals in binary representation. Cantor's
argument, if valid, has to stand that example.

But it does not.

Regards, WM

[fom]

We use the Euclidean algorithm. We use the fact that
our understanding of that algorithm distinguishes rationals
and that there are sequences with properties that differ
from properties common and distinctive of rational sequences.

We use the same reasoning of the "endlessness" found in finite
constructive mathematics.

In order to distinguish all sequences that potentially
describe rationals it is necessary to admit all
potential sequences.

It must be time to begin drawing crayon marks again.

[WM]

On 12 Apr., 11:41, fom <fomJ...@nyms.net> wrote:

> We use the same reasoning of the "endlessness" found in finite
> constructive mathematics.

And that shows you a digit of endlessness that does distinguish the
irrtional limit from all its rational approximations? Wow, what a
power of mind!

Regards, WM

[dull...@sprynet.com]

On Thu, 11 Apr 2013 22:57:57 -0700 (PDT), WM
<muec...@rz.fh-augsburg.de> wrote:

If you believe what you wrote below you're a remarkably
slow learner. On the other hand, if you're aware that
the most important assertion below is simply false you're
a liar.

>Matheology � 246
>
>Cantor's list contains real numbers r as binary or decimal fractions.
>Real numbers, however, are /limits/ of binary or decimal fractions.

No, it _is_ a list of limits of decimal fractions.

>For every terminating fraction of r, Cantor obtains a difference
>between r and the due terminating fraction of the anti-diagonal d:
>r_nn =/= d_n. He concludes that this remains true for the limits of
>the list numbers r and d by using the argument: different sequences
>have different limits.

No, it does not use that "argument". Of course one can find
incorrect versions of the proof on the internet that do
assume this, but a correct version of the argument uses
this fact:

Fact: If two different infinite decimals represent the
same real number then one ends in all 0's and the
other ends in all 9's.

That's a true fact, easy to prove.

>But it is well known that this argument is not
>admissible in proofs because it is false.

It passes belief that a person could actually
think that such a well known proof could contain
such a simple error, without that error being
noticed by any of the thousands of mathematicians
who've studied the subject for a century or so.

>
>Regards, WM

[WM]

On 12 Apr., 16:14, dullr...@sprynet.com wrote:

> >Cantor's list contains real numbers r as binary or decimal fractions.
> >Real numbers, however, are /limits/ of binary or decimal fractions.
>
> No, it _is_ a list of limits of decimal fractions.

There is no such list possible unless you give finite definitions.
It is impossible to define a number by writing an iinfinite sequence.

> >
> It passes belief that a person could actually
> think that such a well known proof could contain
> such a simple error, without that error being
> noticed by any of the thousands of mathematicians
> who've studied the subject for a century or so.

This assumption may be the reason that Cantor's "proof" has been
believed over 100 years. Try to understand the following. (If you are
not a too slow thinker, this will happen before midnight.)

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.

Proof: There are infinitely many rationals q_k with this property.
All
are in the list by definition. At most n of them are in the first n
lines of the list. Infinitely many must exist in the remaining part
of
the list. So we have obtained:

For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
q_kn.
This theorem it is not less important than Cantor's theorem: For all
k: d =/= q_k.

Both theorems contradict each other with the result that finished
infinity as presumed for transfinite set theory is not a valid
mathematical notion.

Regards, WM

[Dan]

They do not contradict each other :
Cantor's affirmation (in its full form) is :

1) forall k , exist n , d_n =/= q_kn

Your (valid) theorem is :

2) forall n , exists k , d_1 = q_k1 and d_2 = q_k2 and ..... d_n =
q_kn

The negation of Cantor theorem would reverse the quantifiers , that
is :
Cantor negated :

3) exists k , forall n , d_n == q_kn

Now , this sounds similar, but not the same as your theorem .
Let's put side by side a simplified version of your theorem , and the
Cantor negation:

2) forall n , exists k d_n = q_kn //your theorem
3) exists k , forall n , d_n == q_kn //Cantor's negative

They look remarkably the same, but they say different things ,
swapping the order of quantifiers has important consequences .

2) forall children , exists woman , woman is child's mother

//this has the same structure as your theorem .
It says "every child has a woman such that the woman is it's mother"

3) exists woman , forall children , woman is child's mother

//this has the same structure as the Cantor negation .
It says "there exists a (single , unique ) woman, who is the mother
of every child" .

[WM]

On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:

> They do not contradict each other :
> Cantor's affirmation (in its full form) is :
>
> 1) forall k  , exist n , d_n =/= q_kn

Tricky! No, please be careful. Cantor shows exactly:
forall k: d_k =/= q_kk
Not more and not less.

This can be extended to
forall k, exists n =< k: d_n =/= q_kn
No statement about n > k is appropriate or possible from the facts.

> Cantor negated  :
>
> 3) exists k  , forall n , d_n == q_kn

No. That negation is valid only for all n =< k.

> swapping the order of quantifiers has important consequences .

That depends on the structure of the set. In linearly ordered sets
like chains of mother-child we have
>
> 2) forall children , exists woman ,  woman is child's ancestor
> 3) exists woman ,  forall children ,  woman is child's ancestor.

Regards, WM

[Ralf Bader]

WM wrote:

> On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>
>> They do not contradict each other :
>> Cantor's affirmation (in its full form) is :
>>
>> 1) forall k  , exist n , d_n =/= q_kn
>
> Tricky! No, please be careful. Cantor shows exactly:
> forall k: d_k =/= q_kk
> Not more and not less.

And that is precisely what he wanted to show - assertion 1) is true with
taking n=k. And you have completed your day's work by showing that again
you reached a level of stupidity unseen and unheard of so far. That is
really crazy, denying something by asserting it.

[fom]

On 4/12/2013 11:41 AM, WM wrote:
> On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>
>> They do not contradict each other :
>> Cantor's affirmation (in its full form) is :
>>
>> 1) forall k , exist n , d_n =/= q_kn
>
> Tricky! No, please be careful. Cantor shows exactly:
> forall k: d_k =/= q_kk
> Not more and not less.
>
> This can be extended to
> forall k, exists n =< k: d_n =/= q_kn
> No statement about n > k is appropriate or possible from the facts.
>

The facts are that unless WM assumes that
an endless application of long division by
the Euclidean algorithm is constant WM has
no "knowledge", in the sense of his finitistic
claims, upon which to base his denial.

WM has not "proved" that each truncated string
is, in fact, the representation which he is
claiming it to be with an abbreviation.

When WM writes,

"No statement about n > k is appropriate or possible from the facts."

he, in fact, is making a statement about n > k.

Anything that might confuse the facts is appropriate and possible
from WM.

>> Cantor negated :
>>
>> 3) exists k , forall n , d_n == q_kn
>
> No. That negation is valid only for all n =< k.
>

No. The only invalid statements here are WM's

>> swapping the order of quantifiers has important consequences .
>
> That depends on the structure of the set. In linearly ordered sets
> like chains of mother-child we have

In WM's world, procreation seems to involve
immaculate conceptions.

>>
>> 2) forall children , exists woman , woman is child's ancestor
>> 3) exists woman , forall children , woman is child's ancestor.

Oddly, based on mitochondrial DNA, genetic analyses have
made exactly this claim.

[Dan]

On Apr 12, 7:41 pm, WM <mueck...@rz.fh-augsburg.de> wrote:
> On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > They do not contradict each other :
> > Cantor's affirmation (in its full form) is :
>
> > 1) forall k  , exist n , d_n =/= q_kn
>
> Tricky! No, please be careful. Cantor shows exactly:
> forall k: d_k =/= q_kk
> Not more and not less.

I see you are also unfamiliar with the rules of logic, that , among
other things allow us to derive more statements from less.

forall k: d_k =/= q_kk =>
forall k , if n = k , then d_k =/= q_kn =>
forall k ,exists n , d_k =/= q_kn


http://en.wikipedia.org/wiki/First-order_logic

>> 3) exists k , forall n , d_n == q_kn
>No. That negation is valid only for all n =< k.

Why ? This is NOT how logic works . If "=<" or "=<" doesn't appear
in the affirmation , then there is no reason for it to appear in the
negation . Either way, a contradiction doesn't exist .

>That depends on the structure of the set. In linearly ordered sets
>like chains of mother-child we have

The Cantor set doesn't have your required structure . Therefore,
swapping order of quantifiers is an illegal operation .
http://math.stackexchange.com/questions/201051/is-the-order-of-universal-existential-quantifiers-important

Unless you've worked out exactly WHERE a contradiction occurs in
Cantor's argument using a STEP BY STEP procedure in SMALL, VALID
STEPS, you have no right to criticize it .

http://star.psy.ohio-state.edu/coglab/Pictures/miracle.gif

By the same token unless you've worked out WHERE and HOW my argument
is wrong, you have no right to criticize it .
"Your argument is wrong because my argument is right" isn't an
argument.
For all its missing steps, I actually took the time to analyze what
your argument MIGHT say, and show how that is invalid.

[WM]

On 12 Apr., 19:37, Dan <dan.ms.ch...@gmail.com> wrote:

> >> 3) exists k  , forall n , d_n == q_kn
> >No. That negation is valid only for all n =< k.
>
> Why ? This is NOT how logic works .

But it is how mathematics works.

If  "=<" or "=<"  doesn't appear
> in the affirmation , then there is no reason for it to appear in the
> negation . Either way, a contradiction doesn't exist

A contradiction does exist *if* d is not more than all d_n.

Forall n: (d_1, ..., d_n) leaves no digit out.
Assume (sets of nodes = paths in the Binary Tree) that d is the union
of all paths (d_1, ..., d_n).
If forall n: (d_1, ..., d_n) is in the list, then d is in the list.

Therefore this is a contradiction:
Forall n: d =/= q_n. d is not in the list.
Forall n: (d_1, ..., d_n) is in the list,. d is in the list.

.
>
> >That depends on the structure of the set. In linearly ordered sets
> >like chains of mother-child we have
>
> The Cantor set doesn't have your required structure .

It has. Every linearly ordered set has that structure.


Therefore,
> swapping order of quantifiers is an illegal operation .http://math.stackexchange.com/questions/201051/is-the-order-of-univer...

>
> Unless you've worked out exactly WHERE a contradiction occurs in
> Cantor's argument

> By the same token unless you've worked out WHERE and HOW my argument
> is wrong,

Your argument is wrong if d is nothing but the union of all its finite
initial segments (d_1, ..., d_n), since for each of them the diagonal
argument (which is only valid for the first k lines of the list) fails
for all following lines.

If you claim, however, that d is more than all its finite initial
segments, then I will present another argument.

So do you claim that d is not in the list but all its finite initial
segments (d_1, ..., d_n) are in the list? If so, what is the
difference
d \ U(d_1, ..., d_n)
d and (d_1, ..., d_n) understood as sets of nodes of paths in decimal
tree.

Regards, WM

[Dan]

> So do you claim that d is not in the list but all its finite initial
> segments (d_1, ..., d_n) are in the list?

YES . Finally .

> If so, what is the
>difference
> d \ U(d_1, ..., d_n)
> d and (d_1, ..., d_n) understood as sets of nodes of paths in decimal
> tree.

The tree is irrelevant unless you can do a small-step STEP by STEP
proof , using LOGICALLY VALID of how it's relevant and how it relates
to the LIST . Cantor's argument was all about the LIST ,not some made-
up tree .

http://star.psy.ohio-state.edu/coglab/Pictures/miracle.gif

[Dan]

> So do you claim that d is not in the list but all its finite initial
> segments (d_1, ..., d_n) are in the list? If so, what is the
> difference

Any finite initial segment of any number is a rational number .
Therefore, forall m, the list of rational numbers contains any finite
inital segment of m.
Does that mean that the list of rational numbers contains all
numbers?

[AMiews]

"WM" <muec...@rz.fh-augsburg.de> wrote in message
news:55765f2d-ef78-4532...@cd3g2000vbb.googlegroups.com...

On 12 Apr., 10:41, Virgil <vir...@ligriv.com> wrote:
> In article
> <d5b94127-f249-4639-8f2f-9de6dda34...@a6g2000vbm.googlegroups.com>,
>
> WM <mueck...@rz.fh-augsburg.de> wrote:
> > Matheology � 246


>

>> The Cantor antidiagonal argument does not use the limit properties of
>> sequences at all. In fact his binary sequences of members of {m,w} are
>> not assumed to have any numerical or topological properties at all,

>They are assumed to have infinitely many bits. But an infinite
>sequence of bits is not converging (the factors 2^-n are missing).

wrong. repeating sequences of bits in an infinitely long string indicate
representation as a fraction.

>Most of them cannot be written by finite expressions. And they cannot
>be written as infinite expressions.

wrong. you seem ill at ease with infinite representations of numbers

[WM]

On 12 Apr., 20:51, Dan <dan.ms.ch...@gmail.com> wrote:
> > So do you claim that d is not in the list but all its finite initial
> > segments (d_1, ..., d_n) are in the list?
>
> YES . Finally .

What does distinguish d from all its finite initial segments (FISs)?

>
> > If so, what is the
> >difference
> > d \ U(d_1, ..., d_n)
> > d and (d_1, ..., d_n) understood as sets of nodes of paths in decimal
> > tree.
>
> The tree is irrelevant unless you can do a small-step STEP by STEP
> proof , using LOGICALLY VALID of how it's relevant and how it relates
> to the LIST . Cantor's argument was all about the LIST ,not some made-
> up tree .

Nevertheless the decimal tree is a representation of all real numbers
of the unit inerval. And Cantor stated his proof as another and
simpler proof for the uncountability of the real numbers.

Do you reject those parts of mathematics which could spoil his proof?

Regards, WM

[WM]

On 12 Apr., 21:01, Dan <dan.ms.ch...@gmail.com> wrote:
> > So do you claim that d is not in the list but all its finite initial
> > segments (d_1, ..., d_n) are in the list? If so, what is the
> > difference
>
> Any finite initial segment of any number is a rational number .

Yes. Let us call it FIS.

> Therefore, forall m,  the list of rational numbers contains any finite
> inital segment of m.
> Does that mean that the list of rational numbers contains all
> numbers?

No, but it contains all decimal fractions, i.e., all terminating
decimals. There is no chance to represent 1/3 or sqrt(2) by digits.
There is no chance to apply Cantor's argument other than to FISs.
Every d_n is part of a FIS.

In order to show that, I'd like to use the Binary Tree.
Consider the path 0.111...
It is the union of all its FISs
0.1
0.11
0.111
...
This is a strictly incresing infinite sequence that does not contain
its limit 0.111...

However, since every FIS is the union of itself with all its
predecessors, this sequence is also a sequence of unions, and, since
it is not finite, contains unions of more than n FIS (for every n you
can take). Usually this is called infinite uninon. So, if the path
0.111... is the union of all its FIS, then it should also be in list,
since the list contains all finite unions, i.e., an infinite union.
(Compare the union of all finite natural numbers results in an
infinite set.)

So we obtain a contradiction. 0.111... is in the list, if the list is
understood as sequence of unions like
{1} U {1, 2} U {1, 2, 3} U ... = |N,
and it is not in the list, if the list is understood as sequence in
mathematics.

Since the Binary Tree deletes the difference between sequence and
unions, it shows this problem in full clarity.

The Binary Tree is constructed by countably many nodes. So it is
impossible to distinguish uncountably many paths *by nodes*.

Regards, WM

[WM]

On 12 Apr., 21:42, "AMiews" <inva...@invalid.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

>
> news:55765f2d-ef78-4532...@cd3g2000vbb.googlegroups.com...
> On 12 Apr., 10:41, Virgil <vir...@ligriv.com> wrote:
>
> > In article
> > <d5b94127-f249-4639-8f2f-9de6dda34...@a6g2000vbm.googlegroups.com>,
>
> > WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Matheology § 246
>
> >> The Cantor antidiagonal argument does not use the limit properties of
> >> sequences at all. In fact his binary sequences of members of {m,w} are
> >> not assumed to have any numerical or topological properties at all,
> >They are assumed to have infinitely many bits. But an infinite
> >sequence of bits is not converging (the factors 2^-n are missing).
>
> wrong. repeating sequences of bits in an infinitely long string indicate
> representation as a fraction.

Since there is no topology defined for Cantor's binary sequences,
there is no chance to determine a limit of wmwmwmwm...

>
> >Most of them cannot be written by finite expressions. And they cannot
> >be written as infinite expressions.
>
> wrong.    you seem ill at ease with infinite representations of numbers

Have you ever seen an infinite expression? Do you think that 0.111...
is an infinite expression? 1/9 or 0.111... are very finite expressions
for infinite sequences. But those sequences are not available. And
every d_n of a numerical Cantor-list is the last digit of a
terminating decimal.
Never, do you understand, never anybody has seen or used a d_n that
does not belong to a terminating decimal.

Therefore Cantor proves that the countable set of rationals is
uncountable.

Regards, WM

[WM]

On 12 Apr., 19:31, Ralf Bader <ba...@nefkom.net> wrote:
> WM wrote:
> > On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>
> >> They do not contradict each other :
> >> Cantor's affirmation (in its full form) is :
>
> >> 1) forall k  , exist n , d_n =/= q_kn
>
> > Tricky! No, please be careful. Cantor shows exactly:
> > forall k: d_k =/= q_kk
> > Not more and not less.
>
> And that is precisely what he wanted to show - assertion 1) is true with
> taking n=k.

I know that. But Cantor's proof is valid only for n =< k.

This is not related to the fact that there is a contradiction between
my theorem

forall n , exists k , d_1 = q_k1 and d_2 = q_k2 and ..... d_n =
q_kn

and the negation of Cantor theorem :
exists k , forall n =<k , d_n == q_kn

iff
d = SUM{all n} d_n*10^-n
since all n are already in the sequence
(d_1*10^-1), (d_1*10^-1 + d_2*10^-2), (d_1*10^-1 + d_2*10^-2 +
d_3*10^-3) , ...
which contains all n.
Or do you know of any n that is missing?

Regards, WM

[Sam Sung]

The idiot WM:

> Consider the path 0.111...
> It is the union of all its FISs
> 0.1
> 0.11
> 0.111
> ...
> This is a strictly incresing infinite sequence that does not contain
> its limit 0.111...
>
> However, since every FIS is the union of itself with all its
> predecessors, this sequence is also a sequence of unions, and, since
> it is not finite, contains unions of more than n FIS (for every n you
> can take). Usually this is called infinite uninon. So, if the path
> 0.111... is the union of all its FIS, then it should also be in list

It is:

{ // begin of unions
0.1
0.11
0.111
...
// end of unions }
limit of the sequence
.

The idiot WM is unable to get that oo IS NOT A NUMBER, because
by definition there is a bijection of infinite sets to at least
one of its proper subsets - e.g. each proper interval within R
is in bijection to R, i.e. they have the same cardinality, which
is vulgo for SIZE, so each proper interval in R has the SAME SIZE
R has.

[fom]

On 4/12/2013 2:47 PM, WM wrote:
> On 12 Apr., 20:51, Dan <dan.ms.ch...@gmail.com> wrote:
>>> So do you claim that d is not in the list but all its finite initial
>>> segments (d_1, ..., d_n) are in the list?
>>
>> YES . Finally .
>
> What does distinguish d from all its finite initial segments (FISs)?

As numbers???

Let me think...

'='

[fom]

On 4/12/2013 3:04 PM, WM wrote:
> On 12 Apr., 21:42, "AMiews" <inva...@invalid.com> wrote:
>> "WM" <mueck...@rz.fh-augsburg.de> wrote in message
>>
>> news:55765f2d-ef78-4532...@cd3g2000vbb.googlegroups.com...
>> On 12 Apr., 10:41, Virgil <vir...@ligriv.com> wrote:
>>
>>> In article
>>> <d5b94127-f249-4639-8f2f-9de6dda34...@a6g2000vbm.googlegroups.com>,
>>
>>> WM <mueck...@rz.fh-augsburg.de> wrote:
>>>> Matheology � 246
>>
>>>> The Cantor antidiagonal argument does not use the limit properties of
>>>> sequences at all. In fact his binary sequences of members of {m,w} are
>>>> not assumed to have any numerical or topological properties at all,
>>> They are assumed to have infinitely many bits. But an infinite
>>> sequence of bits is not converging (the factors 2^-n are missing).
>>
>> wrong. repeating sequences of bits in an infinitely long string indicate
>> representation as a fraction.
>
> Since there is no topology defined for Cantor's binary sequences,
> there is no chance to determine a limit of wmwmwmwm...

That is humorous.

The sense of real numbers as arithmetical individuals
is understood precisely in terms of topological
considerations.

It is true that Hilbert's formalized mathematics reverses
this distinction. But, then WM would be using the
assumption that the real numbers exist to argue for the
fact that the real numbers do not exist.

Oh, wait a minute.

WM will claim that this is what he has been doing with the
hope of demonstrating a contradiction.

Notice, however, that this is not what WM does.

He is simply denying the formal real numbers he would
claim to assume.

Claiming that such denials constitute proof, he repeats
himself again and again.

The use of the word "topology" appears novel, however.

I wonder if WM knows how a typical topology on such a
tree is defined.

[fom]

On 4/12/2013 3:16 PM, WM wrote:
> On 12 Apr., 19:31, Ralf Bader <ba...@nefkom.net> wrote:
>> WM wrote:
>>> On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>>
>>>> They do not contradict each other :
>>>> Cantor's affirmation (in its full form) is :
>>
>>>> 1) forall k , exist n , d_n =/= q_kn
>>
>>> Tricky! No, please be careful. Cantor shows exactly:
>>> forall k: d_k =/= q_kk
>>> Not more and not less.
>>
>> And that is precisely what he wanted to show - assertion 1) is true with
>> taking n=k.
>
> I know that. But Cantor's proof is valid only for n =< k.
>

No. The Cantor argument only stipulates conditions
on what must be given in order for it to be applied.

The argument is irrelevant if what is given corresponds
to your statement. It cannot be applied in such a case.

WM argues with something that is not there.

[Dan]

> No, but it contains all decimal fractions, i.e., all terminating
> decimals. There is no chance to represent 1/3 or sqrt(2) by digits.
> There is no chance to apply Cantor's argument other than to FISs.
> Every d_n is part of a FIS.

You can represent by an infinite sequence of digits, at least in an
abstract sense. That's why we can talk about pi, and the questions
like "what is the 1000000'th digit in the decimal expansion of pi?"
have a valid unambiguous answer .

> However, since every FIS is the union of itself with all its
> predecessors, this sequence is also a sequence of unions, and, since
> it is not finite, contains unions of more than n FIS (for every n you
> can take). Usually this is called infinite uninon. So, if the path
> 0.111... is the union of all its FIS, then it should also be in list,
> since the list contains all finite unions, i.e., an infinite union.
> (Compare the union of all finite natural numbers results in an
> infinite set.)
>
> So we obtain a contradiction. 0.111... is in the list, if the list is
> understood as sequence of unions like
> {1} U {1, 2} U {1, 2, 3} U ... = |N,
> and it is not in the list, if the list is understood as sequence in
> mathematics.
>
> Since the Binary Tree deletes the difference between sequence and
> unions, it shows this problem in full clarity.
>
> The Binary Tree is constructed by countably many nodes. So it is
> impossible to distinguish uncountably many paths *by nodes*.

Yes . The binary tree has countable nodes , and uncountable paths .
(paths sequences of nodes, ie. subsets of the set of nodes , therefore
members of the power-set P(nodes) ) .
And as Cantor showed , |P(nodes)| > |nodes|

When you talk about an irrational number "in the binary tree" , you
can't talk about it as 'any' FIS to represent it .
You have to use a whole sequence of FIS to represent it . That's
called A PATH . So you need to use UNCOUNTABLE PATHS to represent
your UNCOUNTABLY MANY IRRATIONAL NUMBERS .

To recap :
Your tree has :
-Countable nodes :
eg. 0.11 , 0.1110 , 0101 ....
-Countable 'finite' paths : (paths that are made out of finite amount
of nodes)
eg. {0->0.1 -> 0.10 -> 0.101 -> 0.1011 }
UNCOUNTABLE 'infinite' paths: (paths that are made out of infinite
amount of nodes)
eg. {0->0.1 -> 0.11 -> 0.111 -> 0.1111 -> 0.11111 -> ......} to
represent the number "0.11111....."

Real numbers in general are fully represented only by 'infinite'
paths , so they are uncountable .
There are uncountable ways of combining your "finite-length
paths" (that are countable in number) , by union, into "infinite-
length paths" (that are uncountable in number) .

You say that all paths are unions of the "finite paths" , and because
the "finite paths" are countable , so are all paths .
But that does not follow .

[Virgil]

In article
<dc4d82b3-de81-496e...@w1g2000vbw.googlegroups.com>,

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

> On 12 Apr., 16:14, dullr...@sprynet.com wrote:
>
> > >Cantor's list contains real numbers r as binary or decimal fractions.
> > >Real numbers, however, are /limits/ of binary or decimal fractions.
> >
> > No, it _is_ a list of limits of decimal fractions.
>
> There is no such list possible unless you give finite definitions.
> It is impossible to define a number by writing an iinfinite sequence.

But every real number between 0 and 1 defines an infinite sequence.

> > >
> > It passes belief that a person could actually
> > think that such a well known proof could contain
> > such a simple error, without that error being
> > noticed by any of the thousands of mathematicians
> > who've studied the subject for a century or so.
>
> This assumption may be the reason that Cantor's "proof" has been
> believed over 100 years. Try to understand the following. (If you are
> not a too slow thinker, this will happen before midnight.)
>
> 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.

>

> For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
> q_kn.
> This theorem it is not less important than Cantor's theorem: For all
> k: d =/= q_k.

If it were of any importance at all, many others would have found it and
made use of it long before WM came up with it.

But it in no way counters Cantor. At least not outside Wolkenmuekenheim.

>
> Both theorems contradict each other

Not outside of Wolkenmuekenheim!
And if they do inside Wolkenmuekenheim,
they are far from the only contradictions that flourish there.
--

[Virgil]

In article
<580e7ba0-8170-492a...@r4g2000vbf.googlegroups.com>,

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

> On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > They do not contradict each other :
> > Cantor's affirmation (in its full form) is :
> >
> > 1) forall k �, exist n , d_n =/= q_kn
>
> Tricky! No, please be careful. Cantor shows exactly:
> forall k: d_k =/= q_kk
> Not more and not less.
>
> This can be extended to
> forall k, exists n =< k: d_n =/= q_kn

But does not ever require that there exist any n's less than k for which
d_n =/= q_kn, so does not say any more than that d_k =/= q_kk.

> No statement about n > k is appropriate or possible from the facts.
>
> > Cantor negated �:
> >
> > 3) exists k �, forall n , d_n == q_kn
>
> No. That negation is valid only for all n =< k.

Not even for all n < k, only for n = k.

>
> > swapping the order of quantifiers has important consequences .
>
> That depends on the structure of the set. In linearly ordered sets
> like chains of mother-child we have
>
> 2) forall children , exists woman , �woman is child's ancestor
> 3) exists woman , �forall children , �woman is child's ancestor.

We can certainly have 2, at least back tos Eve, but not 3, as no woman
can be her own ancestor.

Thus we have another case of quantifier dyslexia.
--

[Virgil]

In article
<cfa34ebd-fbe4-45c5...@s4g2000vbr.googlegroups.com>,

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

> On 12 Apr., 19:37, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > >> 3) exists k �, forall n , d_n == q_kn
> > >No. That negation is valid only for all n =< k.
> >
> > Why ? This is NOT how logic works .
>
> But it is how mathematics works.

Only the sort of mathematics that disdains logic, like WMytheology.

>
> If �"=<" or "=<" �doesn't appear
> > in the affirmation , then there is no reason for it to appear in the
> > negation . Either way, a contradiction doesn't exist
>
> A contradiction does exist *if* d is not more than all d_n.

Since d is only a sequence of d_n's, it is, at last in one sense,
is nothing more than "all d_n".
But in any case, WM's "contradictions" exist almost entirely inside
Wolkenmuekenheim, and never get out.

>
> Forall n: (d_1, ..., d_n) leaves no digit out.

Out of what?

> Assume (sets of nodes = paths in the Binary Tree)

Not all sets of nodes are paths in most trees, including any trees
having more than two nodes.

> >
> > The Cantor set doesn't have your required structure .
>

>
>

> > Therefore,
> > swapping order of quantifiers is an illegal operation
> > .http://math.stackexchange.com/questions/201051/is-the-order-of-univer...

NOt necessarily in Wolkenmuekenheim, where WM does it far too regularly.

> >
> > Unless you've worked out exactly WHERE a contradiction occurs in
> > Cantor's argument
>
> > By the same token unless you've worked out WHERE and HOW my argument
> > is wrong,
>
> Your argument is wrong if d is nothing but the union of all its finite
> initial segments (d_1, ..., d_n), since for each of them the diagonal
> argument (which is only valid for the first k lines of the list) fails
> for all following lines.

What is the value of k immediately after which the diagonal argument
fails?

I'll bet WM cannot name it, nor even prove that there is a k for which
it works while for k+1 it does not work.

>
> If you claim, however, that d is more than all its finite initial
> segments, then I will present another argument.

Don't bother, as it will be even more wrong that the ones you have been
boring us with so far.

>
> So do you claim that d is not in the list but all its finite initial
> segments (d_1, ..., d_n) are in the list?

As ALL of the members of a list in the Cantor proof are infinite
sequences, there are NO finite initial segments listed.
--

[Virgil]

In article
<fc112cbf-b975-47f5...@y2g2000vbe.googlegroups.com>,

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

> On 12 Apr., 21:01, Dan <dan.ms.ch...@gmail.com> wrote:
> > > So do you claim that d is not in the list but all its finite initial
> > > segments (d_1, ..., d_n) are in the list? If so, what is the
> > > difference
> >
> > Any finite initial segment of any number is a rational number .

In the original Cantor "diagonal" argument, each infinite sequence
consisted of repetitions of only the letters from the set {"m", "w"}.

So that any argument that relies on those infinite sequences being
composed of digits fails to invalidate Cantor's argument.
--

[Virgil]

In article
<9419a98a-5b08-463b...@a6g2000vbm.googlegroups.com>,

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

> On 12 Apr., 20:51, Dan <dan.ms.ch...@gmail.com> wrote:
> > > So do you claim that d is not in the list but all its finite initial
> > > segments (d_1, ..., d_n) are in the list?
> >
> > YES . Finally .
>
> What does distinguish d from all its finite initial segments (FISs)?

Only that d is the union of the set of all its FISs.

> >
> > > If so, what is the
> > >difference
> > > d \ U(d_1, ..., d_n)
> > > d and (d_1, ..., d_n) understood as sets of nodes of paths in decimal
> > > tree.
> >
> > The tree is irrelevant unless you can do a small-step STEP by STEP
> > proof , using LOGICALLY VALID of how it's relevant and how it relates
> > to the LIST . Cantor's argument was all about the LIST ,not some made-
> > up tree .
>
> Nevertheless the decimal tree is a representation of all real numbers
> of the unit inerval. And Cantor stated his proof as another and
> simpler proof for the uncountability of the real numbers.
>
> Do you reject those parts of mathematics which could spoil his proof?

So far, WM has totally failed to show that ANY parts of mathematics are
capable of "spoiling" Cantor's diagonal argument. What goes on in
Wolkenmuekenheim is not really mathematics.
--

[Virgil]

In article
<164f8e3f-c24a-4e18...@16g2000vbx.googlegroups.com>,

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

> On 12 Apr., 19:31, Ralf Bader <ba...@nefkom.net> wrote:
> > WM wrote:
> > > On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
> >
> > >> They do not contradict each other :
> > >> Cantor's affirmation (in its full form) is :
> >
> > >> 1) forall k �, exist n , d_n =/= q_kn
> >
> > > Tricky! No, please be careful. Cantor shows exactly:
> > > forall k: d_k =/= q_kk
> > > Not more and not less.
> >
> > And that is precisely what he wanted to show - assertion 1) is true with
> > taking n=k.
>
> I know that. But Cantor's proof is valid only for

Cantor's diagonal argument has never been shown to be less that
completely true by WM, or by anyone else.
--

[Virgil]

In article
<55765f2d-ef78-4532...@cd3g2000vbb.googlegroups.com>,

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

> On 12 Apr., 10:41, Virgil <vir...@ligriv.com> wrote:
> > In article
> > <d5b94127-f249-4639-8f2f-9de6dda34...@a6g2000vbm.googlegroups.com>,
> >
> > �WM <mueck...@rz.fh-augsburg.de> wrote:
> > > Matheology � 246
> >

> > > Cantor's list contains real numbers r as binary or decimal fractions.
> > > Real numbers, however, are /limits/ of binary or decimal fractions.
> >

> > Real numbers are all certainly expressable as limits of sequences, but
> > so are rationals and integers and naturals.
>
> No objection.
>
> > > For every terminating fraction of r, Cantor obtains a difference
> > > between r and the due terminating fraction of the anti-diagonal d:
> > > r_nn =/= d_n. He concludes that this remains true for the limits of
> > > the list numbers r and d by using the argument: different sequences
> > > have different limits. But it is well known that this argument is not
> > > admissible in proofs because it is false.

> >
> > The Cantor antidiagonal argument does not use the limit properties of
> > sequences at all. In fact his binary sequences of members of {m,w} are
> > not assumed to have any numerical or topological properties at all,
>
> They are assumed to have infinitely many bits.

No, only infinitely many terms, each being one of "m" or "w".

> But an infinite
> sequence of bits is not converging

Neither is an infinite sequence of letters.

>
> > Each of the Cantor sequences �is, in fact, merely a function
> > from |N to {m,w} with no other properties assumed or allowed, except
> > that two such functions are different if there is some n in |N at which
> > they have different values.
>
> Most of them are undefined and undefinable. But in order to remedy
> that problem we

It is not a problem, since the failure of any such list of sequences
existing does not disprove the conclusion that such a list in
necessarily incomplete.
--

[Virgil]

In article
<b3f7e8ac-f541-46a5...@a3g2000vbr.googlegroups.com>,

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

> Therefore Cantor proves that the countable set of rationals is
> uncountable.

Since Cantor's diagonal argument nowhere involves any rationals, either
in its hypotheses nor in its conclusion, WM is wrong, again, as usual!!!
--

[Virgil]

In article
<a9a1f814-5b2e-4a72...@r4g2000vbf.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 12 Apr., 11:41, fom <fomJ...@nyms.net> wrote:
>
> > We use the same reasoning of the "endlessness" found in finite
> > constructive mathematics.
>
> And that shows you a digit of endlessness that does distinguish the
> irrtional limit from all its rational approximations? Wow, what a
> power of mind!
>
It is far better to go that way than to exclude all possibility of any
infinite sequences at all, as WM does. WM's way cancels all of calculus,
among other things,
--

[fom]

On 4/12/2013 6:54 PM, Virgil wrote:
> In article
> <cfa34ebd-fbe4-45c5...@s4g2000vbr.googlegroups.com>,
> WM <muec...@rz.fh-augsburg.de> wrote:
>
>> On 12 Apr., 19:37, Dan <dan.ms.ch...@gmail.com> wrote:
>>
>>>>> 3) exists k , forall n , d_n == q_kn
>>>> No. That negation is valid only for all n =< k.
>>>
>>> Why ? This is NOT how logic works .
>>
>> But it is how mathematics works.
>
> Only the sort of mathematics that disdains logic, like WMytheology.

Isn't that a fundamental symmetry of WMytheology?

The sort of logic that disdains mathematics seems equally at
sacred.

[AMiews]

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

news:b3f7e8ac-f541-46a5...@a3g2000vbr.googlegroups.com...

On 12 Apr., 21:42, "AMiews" <inva...@invalid.com> wrote:
> "WM" <mueck...@rz.fh-augsburg.de> wrote in message

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

<snip>

>> wrong. repeating sequences of bits in an infinitely long string indicate
>> representation as a fraction.

>Since there is no topology defined for Cantor's binary sequences,
>there is no chance to determine a limit of wmwmwmwm...

balonie,
you are only complaining about the three dots or periods " ... "
indicating repeating in that fashion, so get over it...

>
>> >Most of them cannot be written by finite expressions. And they cannot
>> >be written as infinite expressions.
>>
>> wrong. you seem ill at ease with infinite representations of numbers

>Have you ever seen an infinite expression? Do you think that 0.111...
>is an infinite expression?

it is short hand for one,
the "..." mean repeated, usally one uses a bar over the last repeated
numbers, but cant do that with text.

> 1/9 or 0.111... are very finite expressions

yes and you are fussing over notation convention, meaning you are unfamiliar
with math(s)

>for infinite sequences. But those sequences are not available.

why ? where did they go ? if they were infinite, they would fill up your
room...

>And
>every d_n of a numerical Cantor-list is the last digit of a
>terminating decimal.
>Never, do you understand, never anybody has seen or used a d_n that
>does not belong to a terminating decimal.

you seem confused by standard math notation here. Irrationals no one has
seen the end.

>
>Therefore Cantor proves that the countable set of rationals is
>uncountable.

that is what you say, but study up on common math notation first.

>
>Regards, WM

[fom]

On 4/12/2013 8:46 AM, WM wrote:
> On 12 Apr., 11:41, fom <fomJ...@nyms.net> wrote:
>
>> We use the same reasoning of the "endlessness" found in finite
>> constructive mathematics.
>
> And that shows you a digit of endlessness that does distinguish the
> irrtional limit from all its rational approximations? Wow, what a
> power of mind!

It shows nothing.

That is for children with crayons.

It explains that any discrete characterization of
universal quantification cannot be faithful (in
the algebraic sense of representation) with respect
to identity criteria unless a completed identity
is presupposed.

Cantor's topological arguments merely make clear
that Zeno's finished line is only reached when
imminent infinity is taken as completed infinity.

Whoops! No Cantor argument involved.

[Virgil]

In article <kkao13$bel$1...@news.albasani.net>,

If those }e_n's" are digits, no one has ever seen digit that does not
appear in infinitely many natural numbers.

>
> >
> >Therefore Cantor proves that the countable set of rationals is
> >uncountable.

Only in Wolkenmuekenheim.
--

[WM]

On 12 Apr., 23:29, Dan <dan.ms.ch...@gmail.com> wrote:
> > No, but it contains all decimal fractions, i.e., all terminating
> > decimals. There is no chance to represent 1/3 or sqrt(2) by digits.
> > There is no chance to apply Cantor's argument other than to FISs.
> > Every d_n is part of a FIS.
>
> You can represent by an infinite sequence of digits, at least in an
> abstract sense. That's why we can talk about pi, and the questions
> like "what is the 1000000'th digit in the decimal expansion of pi?"
> have a valid unambiguous answer .

Yes, we hav a formula that tells us the n-th digit of pi. But the
*decimal representation of pi* does never leave the domain of FISs. A
decimal string would not allow us to recognize pi. Therefore there is
no decimal representation of pi and of other actually infinite
sequences. Therefore the anti-diagonal d, when only presenting its
digits d_n, cannot differ from all FISs and therefore cannot prove the
uncountability of irrational numbers.

Do you agree that

{1} U {1, 2} U {1, 2, 3} U ... = |N

but the infinite sequence
{1}, {1, 2}, {1, 2, 3}, ...
does not contain |N?

What is the difference?

>
> > Since the Binary Tree deletes the difference between sequence and
> > unions, it shows this problem in full clarity.
>
> > The Binary Tree is constructed by countably many nodes. So it is
> > impossible to distinguish uncountably many paths *by nodes*.
>
> Yes . The binary tree has countable nodes , and uncountable paths .
> (paths sequences of nodes, ie. subsets of the set of nodes , therefore
> members of the power-set P(nodes) ) .
> And as Cantor showed , |P(nodes)| > |nodes|

All what Cantor showed concerns FISs. On the other hand he showed that
the set of FISs is countable.

>
> When you talk about an irrational number "in the binary tree" , you
> can't talk about it as 'any' FIS to represent it .

In fact there is no decimal representation of any irrational number.

> You have to use a whole sequence of FIS to represent it .

That can only be done by a finite definition - not by gathering
infinitely many FISs, obviously.

> Real numbers in general are fully represented only by 'infinite'
> paths , so they are uncountable .
> There are uncountable ways of combining your "finite-length
> paths" (that are countable in number) , by union, into "infinite-
> length paths" (that are uncountable in number) .

Not in the Binary Tree. All paths that are possible there (and are
represented by sets of nodes) belong to a countable set. It is
impossible to distinguish more than countably many. And it is
impossible to apply the diagonal argument to the Binary Tree, since it
contains all possible paths.

>
> You say that all paths are unions of the "finite paths" , and because
> the "finite paths" are countable , so are all paths .
> But that does not follow .

This construction of the Binary Tree shows that it is impossible to
discern uncountably many paths by nodes. (But Cantor's argument
concerns nodes (like d_n) only!)

The Binary Tree contains only those paths that are constructed step by
step of its initial segments Bk:

B0 = a0.

B1 = a0.
/
a1

B2 = a0.
/ \
a1 a2
...

Bk = a0.
/ \
a1 a2
...
/
... ak

...


Nothing else can be distinguished by nodes (like d_n).

Regards, WM

[Sam Sung]

idiot WM drivels:

> {1} U {1, 2} U {1, 2, 3} U ... = |N
> but the infinite sequence
> {1}, {1, 2}, {1, 2, 3}, ...
> does not contain |N?
>
> What is the difference?

N is not number so it could not be contained in a set of numbers {1, 2, n}

[WM]

On 13 Apr., 01:23, Virgil <vir...@ligriv.com> wrote:

> > 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.
>
> > For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
> > q_kn.
> > This theorem it is not less important than Cantor's theorem: For all
> > k: d =/= q_k.
>

> If it were of any importance at all, many others would have found it.

How many others have found that it is impossible to prove the parallel-
axiom in the 2000 years before Gauss?

If d is nothing but its fis, then there is a contradiction. And if you
claim that d is more than its FISs, then there is a contradiction too.

The first contradiction does not contradict Cantor's proof but what
has been interpreted into his result:
Cantor showed: for all k exists n : d_n =/= q_kn & n =< k
This has been interpreted as

For all k: d =/= q_k.

The second contradiction is this:
If d (that has aleph_0 digits with aleph_0 > n for all n) is not in
the sequence of all its FISs, then the union of all FISONs has not
aleph_0 elements.
Proof: The infinite union
A) {1} U {1, 2} U {1, 2, 3} U ...
is in the sequence of all finite unions
B) {1}, {1, 2}, {1, 2, 3}, ...
since in both cases "all n" are applied.
C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ...
Or would you dare to claim that A is not contained in C?

Regards, WM


Regards, WM


Regards, WM

[Sam Sung]

WM drivels:

> In fact there is no decimal representation of any irrational number.

There is just no no FINITE equivalent of irrational numbers.

[Sam Sung]

idiot WM drivels:

> Binary Tree. All paths that are possible there (and are
> represented by sets of nodes) belong to a countable set

This is idiot-WM-nonsense.

[Sam Sung]

idiot WM drivels:

> The Binary Tree contains only those paths that are constructed step by
> step of its initial segments Bk

Infinite sets exist by axiom and are not to be constructed.

[Sam Sung]

idiot WM drivels:

> Regards, WM
>
>
> Regards, WM
>
>
> Regards, WM

Piss off, asshole.

[Virgil]

In article
<e2110442-bb15-4dc8...@cd3g2000vbb.googlegroups.com>,

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

> On 12 Apr., 23:29, Dan <dan.ms.ch...@gmail.com> wrote:
> > > No, but it contains all decimal fractions, i.e., all terminating
> > > decimals. There is no chance to represent 1/3 or sqrt(2) by digits.
> > > There is no chance to apply Cantor's argument other than to FISs.
> > > Every d_n is part of a FIS.
> >
> > You can represent by an infinite sequence of digits, at least in an
> > abstract sense. That's why we can talk about pi, and the questions
> > like "what is the 1000000'th digit in the decimal expansion of pi?"
> > have a valid unambiguous answer .
>
> Yes, we hav a formula that tells us the n-th digit of pi. But the
> *decimal representation of pi* does never leave the domain of FISs. A
> decimal string would not allow us to recognize pi. Therefore there is
> no decimal representation of pi and of other actually infinite
> sequences. Therefore the anti-diagonal d, when only presenting its
> digits d_n, cannot differ from all FISs and therefore cannot prove the
> uncountability of irrational numbers.

The original "antidiagonal" did not contain any digits, nor did the list
of sequences from which is was to be constructed contain any.

Thus any argument based on the presence of digits is irrelevant.

>
> Do you agree that
> {1} U {1, 2} U {1, 2, 3} U ... = |N
> but the infinite sequence
> {1}, {1, 2}, {1, 2, 3}, ...
> does not contain |N?
>
> What is the difference?

An infinite union contains every member of each of the sets being
unioned, and every natural is in at least one FISON, so the union of all
FISONs MUST contain all naturals.

But a strictly increasing sequence, like {1}, {1, 2}, {1, 2, 3}, ...,
can never contain its limit, |N.

Those not mired in the wilds of Wolkenmuekenheim are already aware of
the difference between the two.

> >
> > > The Binary Tree is constructed by countably many nodes. So it is
> > > impossible to distinguish uncountably many paths *by nodes*.

But quite possible by sets of nodes.

> >
> > Yes . The binary tree has countable nodes , and uncountable paths .
> > (paths sequences of nodes, ie. subsets of the set of nodes , therefore
> > members of the power-set P(nodes) ) .
> > And as Cantor showed , |P(nodes)| > |nodes|
>
> All what Cantor showed concerns FISs.

Where in either the statement of Cantor's diagonal argument, or in his
proof of it, is any FIS even mentioned. The terms of sequence, but never
the FISs of one!

>
> In fact there is no decimal representation of any irrational number.

There are, in fact, formulae that allow the determination any and every
digit of certain transcendental numbers.

>
> > You have to use a whole sequence of FIS to represent it .

Nope! One can use a formula for each digit.

>
> That can only be done by a finite definition - not by gathering
> infinitely many FISs, obviously.

And has been done by finite definition.

>
> > Real numbers in general are fully represented only by 'infinite'
> > paths , so they are uncountable .
> > There are uncountable ways of combining your "finite-length
> > paths" (that are countable in number) , by union, into "infinite-
> > length paths" (that are uncountable in number) .
>
> Not in the Binary Tree. All paths that are possible there (and are
> represented by sets of nodes) belong to a countable set.

What WM is speaking of are properly only FISONs of such a tree, not
paths. A path is usually required to be a maximal sequence of
parent-child linked nodes, which no FISON can be, or could be defined to
be any maximal set of FISONs ordered by inclusion.

> And it is
> impossible to apply the diagonal argument to the Binary Tree

Only someone as messed up as WM would even think of trying to apply a
"diagonal argument" to something that as obviously has no "diagoanal" as
a binary tree. Why does WM not try to apply the diagonal argument to
elliptic integrals as well?

A list of binary sequences, to which the diagaonal argument DOES apply,
and a Complete Infinite Binary Tree are not at all the same thing.

> >
> > You say that all paths are unions of the "finite paths" , and because
> > the "finite paths" are countable , so are all paths .
> > But that does not follow .
>
> This construction of the Binary Tree shows that it is impossible to
> discern uncountably many paths by nodes.

But not as sets of nodes, which s what paths are.

WM keeps ignoring that given any set, it has more subsets then members,
so there are far more sets of nodes than nodes.

> (But Cantor's argument
> concerns nodes (like d_n) only!)

Cantor's diagonal argument only concerns lists of binary sequences

>
> The Binary Tree contains only those paths that are constructed step by
> step of its initial segments

A path in a Complete Infinite Binary Tree can only be constructed from a
nested and infinite set of such initial segments, each being order
isomorphic to the set of all Finite Initial Segments Of Naturals
(FISONs).
--

[WM]

On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:

> >every d_n of a numerical Cantor-list is the last digit of a
> >terminating decimal.
> >Never, do you understand, never anybody has seen or used a d_n that
> >does not belong to a terminating decimal.
>
> you seem confused by standard math notation here.  Irrationals no one has
> seen the end.

But you believe in its existence nevertheless?
There is no end and there is no "all", because every scientific use of
"all" would include to find all. And that includes to prove that all
have been found. And that includes that a last one has been confirmed.

>
> >Therefore Cantor proves that the countable set of rationals is
> >uncountable.
>
> that is what you say, but study up on common math notation first.
>

Common math notation says that the digit d_n has the place number n
and belongs to a finite sequence of digits. Cantor's method is
incapable of dealing wth irrational numbers, because nobody can use
them other than by a finite formula. Alas there are only countably
many finite formulas.

Also the set of all defined Cantor-lists and all possible diagonals
belongs to the countable set of finite definitions. So Cantor proves
that this countable set is uncountable. Not a contradiction of course.
Beware of such devilry in matheology!

Regards, WM

[Virgil]

In article
<ec974e95-ef9c-44fb...@z4g2000vbz.googlegroups.com>,

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

> On 13 Apr., 01:23, Virgil <vir...@ligriv.com> wrote:
>
> > > 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.
> >
> > > For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
> > > q_kn.
> > > This theorem it is not less important than Cantor's theorem: For all
> > > k: d =/= q_k.
> >
> > If it were of any importance at all, many others would have found it.
>
> How many others have found that it is impossible to prove the parallel-
> axiom in the 2000 years before Gauss?

It is still impossible without first assuming something equivalent to it.

>
> If d is nothing but its fis, then there is a contradiction. And if you
> claim that d is more than its FISs, then there is a contradiction too.

But that alleged "contradiction" does not exist outside
Wolkenmuekenheim, no no one but WM will ever be bothered by it.

>
> The first contradiction does not contradict Cantor's proof but what
> has been interpreted into his result:
> Cantor showed: for all k exists n : d_n =/= q_kn & n =< k
> This has been interpreted as
> For all k: d =/= q_k.

What Cantor showed was there is a d such that for all k : d_k =/= q_k_k,
therefore d =/= q_k

Anything beyond that was not part of his proof.

>
> The second contradiction is this:
> If d (that has aleph_0 digits with aleph_0 > n for all n) is not in
> the sequence of all its FISs, then the union of all FISONs has not
> aleph_0 elements.

Again WM's incapacity reveals itself!

The set of all FISONs is not the same thing as the sequence of all
FISONs.

The sequence, being strictly increasing by inclusion, has a limit larger
than any member, and that limit is the union of all FISONs, which cannot
itself be a FISON
--

[Virgil]

In article
<eac9e5da-7dbd-4b1d...@b3g2000vbo.googlegroups.com>,

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

> On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>
> > >every d_n of a numerical Cantor-list is the last digit of a
> > >terminating decimal.
> > >Never, do you understand, never anybody has seen or used a d_n that
> > >does not belong to a terminating decimal.

In Cantor's diagonal argument, no one has ever seen a d_n other than "m"
or "w".

So that none of WM's nonsense about digit sequences is relevant to
Cantor's diagonal argument.
--

[WM]

On 13 Apr., 10:05, Virgil <vir...@ligriv.com> wrote:

> But a strictly increasing sequence, like {1}, {1, 2}, {1, 2, 3}, ...,
> can never contain its limit,

And why is that so? Because it does not exist other than in the
imagination of some matheologians.

The sequence
0.1
0.11
0.111
...
cannot contain its limit because it does not exist other than by a
finite definition like 1/9. It does not exist as a string of digits.
And therefore it cannot appear in any proof as an infinite string d of
digits d_n.

Regards, WM

[fom]

On 4/13/2013 3:16 AM, WM wrote:
> On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>
>>> every d_n of a numerical Cantor-list is the last digit of a
>>> terminating decimal.
>>> Never, do you understand, never anybody has seen or used a d_n that
>>> does not belong to a terminating decimal.
>>
>> you seem confused by standard math notation here. Irrationals no one has
>> seen the end.
>
> But you believe in its existence nevertheless?
> There is no end and there is no "all", because every scientific use of
> "all" would include to find all. And that includes to prove that all
> have been found. And that includes that a last one has been confirmed.

WM's finite existence hypothesis.

It is meaningful in the sense that physics
cannot disentangle anything from anything
else in one interpretation of the Aspect
experiments.

Thus, there is 1.

But, any hypothesis by WM postulating a
plurality is a postulation of a finite
universe.

Perhaps he could offer a fixed bound
so other scientists do not err.

[fom]

On 4/13/2013 3:24 AM, WM wrote:
>
> The sequence
> 0.1
> 0.11
> 0.111
> ...

When all else fails...

...get the crayons.

[fom]

On 4/13/2013 3:16 AM, WM wrote:
>

> scientific use of "all"
>

"The dependence of our systems of kinds on our
theories, and the dependence of these, in turn,
on our interests, values, technology, and the
like, make questionable, at best, the thesis
that our predicates pick out real properties
or natural kinds whose existence, extension,
and metaphysical status are independent of
any contribution of ours. And the claim that
just these kinds or properties are required
to answer scientific questions or provide
scientific explanations supports that thesis
only if backed by an account of why these
questions or forms of explanation have
priority -- an account that does not, in turn,
appeal to the practices or institutions of
which they are a part,...

...else all questions are begged."



Catherine Elgin

"With Reference to Reference"

[WM]

On 13 Apr., 10:42, fom <fomJ...@nyms.net> wrote:
> On 4/13/2013 3:16 AM, WM wrote:
>
> > On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>
> >>> every d_n of a numerical Cantor-list is the last digit of a
> >>> terminating decimal.
> >>> Never, do you understand, never anybody has seen or used a d_n that
> >>> does not belong to a terminating decimal.
>
> >> you seem confused by standard math notation here.  Irrationals no one has
> >> seen the end.
>
> > But you believe in its existence nevertheless?
> > There is no end and there is no "all", because every scientific use of
> > "all" would include to find all. And that includes to prove that all
> > have been found. And that includes that a last one has been confirmed.
>
> WM's finite existence hypothesis.

No, there is the simple fact that mathematics is done by finite means
and that statements concerning actions have to be proved,. It it not
sufficient to prove that if some axioms are given, that then things
have to be so and so. With respect to actions like sending signals or
reading tables, the reality is the proof. This can be seen by
Zermelo's AC. He said well-ordering is possible, but reality says no.
So the axiom proves nonsense. For matheology this may me tolerable,
since it inputs and outputs only nonsense which nobody can use in any
respect other than for set-theoretical masturbation. But in doing
mathematics restrictions put by reality cannot be overcome.

Regards, WM

[fom]

On 4/13/2013 4:06 AM, WM wrote:
> On 13 Apr., 10:42, fom <fomJ...@nyms.net> wrote:
>> On 4/13/2013 3:16 AM, WM wrote:
>>
>>> On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>>
>>>>> every d_n of a numerical Cantor-list is the last digit of a
>>>>> terminating decimal.
>>>>> Never, do you understand, never anybody has seen or used a d_n that
>>>>> does not belong to a terminating decimal.
>>
>>>> you seem confused by standard math notation here. Irrationals no one has
>>>> seen the end.
>>
>>> But you believe in its existence nevertheless?
>>> There is no end and there is no "all", because every scientific use of
>>> "all" would include to find all. And that includes to prove that all
>>> have been found. And that includes that a last one has been confirmed.
>>
>> WM's finite existence hypothesis.
>
> No, there is the simple fact that mathematics is done by finite means
> and that statements concerning actions have to be proved,.

The very notion of determining the precision of
a an approximate value in numerical analysis
presupposes an exact value. The "knowledge" of
those exact values in the mathematical theories
WM rejects involve infinity.

Suppose one takes exactly one of WM's crayon
marks,

|

What makes that an individual? What makes that
an instantiation of "one object"?

There is either a single complement of "not that
crayon mark" or there is an apparently indeterminable
plurality of "other objects" which must satisfy the
predicate "not that crayon mark".

Since the plurality is apparently indeterminable,
WM's claims with regard to science and scientific
explanation reduce to a claim that that apparently
indeterminable plurality is exhaustible through
counting. This must be a falsifiable claim because
it is being presented as a scientific claim.

Why ask WM to prove anything, when we can ask
him to conduct science?

Please, WM. Give us a scientific demonstration of
your claims.

[fom]

On 4/13/2013 4:06 AM, WM wrote:
> On 13 Apr., 10:42, fom <fomJ...@nyms.net> wrote:
>> On 4/13/2013 3:16 AM, WM wrote:
>>
>>> On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>>
>>>>> every d_n of a numerical Cantor-list is the last digit of a
>>>>> terminating decimal.
>>>>> Never, do you understand, never anybody has seen or used a d_n that
>>>>> does not belong to a terminating decimal.
>>
>>>> you seem confused by standard math notation here. Irrationals no one has
>>>> seen the end.
>>
>>> But you believe in its existence nevertheless?
>>> There is no end and there is no "all", because every scientific use of
>>> "all" would include to find all. And that includes to prove that all
>>> have been found. And that includes that a last one has been confirmed.
>>
>> WM's finite existence hypothesis.
>
> No, there is the simple fact that mathematics is done by finite means
> and that statements concerning actions have to be proved,. It it not
> sufficient to prove that if some axioms are given, that then things
> have to be so and so.

Perhaps, but the critics of Hilbert's formalism
actually engaged in proper mathematical discourse
when debating these matters.

Nothing you do could be described as "proper
mathematical discourse" and that has nothing to
do with anyone's beliefs but your own.

[fom]

On 4/13/2013 4:06 AM, WM wrote:
> On 13 Apr., 10:42, fom <fomJ...@nyms.net> wrote:
>> On 4/13/2013 3:16 AM, WM wrote:
>>
>>> On 13 Apr., 06:47, "AMiews" <inva...@invalid.com> wrote:
>>
>>>>> every d_n of a numerical Cantor-list is the last digit of a
>>>>> terminating decimal.
>>>>> Never, do you understand, never anybody has seen or used a d_n that
>>>>> does not belong to a terminating decimal.
>>
>>>> you seem confused by standard math notation here. Irrationals no one has
>>>> seen the end.
>>
>>> But you believe in its existence nevertheless?
>>> There is no end and there is no "all", because every scientific use of
>>> "all" would include to find all. And that includes to prove that all
>>> have been found. And that includes that a last one has been confirmed.
>>
>> WM's finite existence hypothesis.
>
> No, there is the simple fact that mathematics is done by finite means
> and that statements concerning actions have to be proved,. It it not
> sufficient to prove that if some axioms are given, that then things
> have to be so and so. With respect to actions like sending signals or
> reading tables, the reality is the proof. This can be seen by
> Zermelo's AC. He said well-ordering is possible, but reality says no.

The fact that WM needs mathematics to be grounded in a metaphysical
reality merely reflects the poverty of his philosophy.

Hilbert reportedly described mathematical existence as "that which
is not contradictory" with a clearly defined notion of contradiction.

Leibniz defined individuated existents in terms of possible worlds
theory or counterpart theory on the basis that no finite account of
an object could sufficiently describe it uniquely.

WM offers no philosophy concerning ontology, epistemology, semantics,
or logic to account for why his beliefs might have any correspondence
to objective representations of either reality or the nature of
mathematics.

WM has apparently ties his self-worth to the chicken scratch
and crayon marks he calls a reality (Oh, it should be granted
that he found a publisher for his books. There is a sort
of success in that, even if it seems unimaginable from the
statements he makes in this forum.).

[WM]

On 13 Apr., 11:46, fom <fomJ...@nyms.net> wrote:
> On 4/13/2013 4:06 AM, WM wrote:

> >>>>> every d_n of a numerical Cantor-list is the last digit of a
> >>>>> terminating decimal.
> >>>>> Never, do you understand, never anybody has seen or used a d_n that
> >>>>> does not belong to a terminating decimal.
>

> >>> But you believe in its existence nevertheless?
> >>> There is no end and there is no "all", because every scientific use of
> >>> "all" would include to find all. And that includes to prove that all
> >>> have been found. And that includes that a last one has been confirmed.
>
> >> WM's finite existence hypothesis.

Not a hypothesis, but the fundament of science.

>
> > No, there is the simple fact that mathematics is done by finite means
> > and that statements concerning actions have to be proved,. It it not
> > sufficient to prove that if some axioms are given,  that then things
> > have to be so and so. With respect to actions like sending signals or
> > reading tables, the reality is the proof. This can be seen by
> > Zermelo's AC. He said well-ordering is possible, but reality says no.
>
> The fact that WM needs mathematics to be grounded in a metaphysical
> reality merely reflects the poverty of his philosophy.

Mathematics is done in reality without any metaphysical touch.

> >
> WM has apparently ties his self-worth to the chicken scratch
> and crayon marks he calls a reality (Oh, it should be granted
> that he found a publisher for his books.

One of them printed in seventh edition, another one in third, even a
best seller.
http://www.hs-augsburg.de/~mueckenh/Infinity/Bestseller.pdf
of a publisher who is famous for his books including German editions
of Hardy and Feynman, as well as risky isues like my book which tells
students to refrain from actual infinity as nonsense or Rudolf
Diesel's or Hermann Oberth's (that man who inspired Wernher von Braun
without whom the Americans never would have entered the moon). By the
way, Oberth's ideas at his times have been considered as ridiculous as
mine few years ago.

Regards, WM

[Dan]

> The Binary Tree contains only those paths that are constructed step by
> step of its initial segments Bk:
>
> B0 =    a0.
>
> B1 =    a0.
>           /
>         a1
>
> B2 =    a0.
>           /   \
>         a1   a2
> ...
>
> Bk =    a0.
>           /   \
>         a1   a2
>        ...
>           /
>     ... ak
>
> ...
>
> Nothing else can be distinguished by nodes (like d_n).
>
> Regards, WM

Trust me on this one :

1)There are an countable/inFinite
number of paths of 'Finite length' :

Examples of paths of finite length :

0->0.1 -> 0.10->0.101->0.1011 : path of length 5
0->0.0->0.00->0.001 : path of length 4


2) There are an unCountable
number of paths of 'Countable/infinite length' :

0->0.1 -> 0.10->0.101->0.1010->0.10101-> ......... :path of infinite
length , representing 0.(10)
0->0.1 -> 0.11->0.111->0.1111->0.11111-> ......... :path of infinite
length , representing 0.(1)


The fact that "the set of 'all Countable Things' is unCountable" ,
is , ultimately , no more surprising that the fact that
"the set of 'all Finite Things' is inFinite" .

If you start counting one number at a time .... 1 , 2 , 3 .... you'll
never count all the numbers, given the fact that , one day , you'll
die . We'll all die, eventually . There are some things that you
cannot count even if you do live forever . The trick is understanding
and learning to live with your limits.

Now, I want you to read this again , slowly , without giving in to
impulse .
http://groups.google.com/group/sci.math/msg/9e7d3b7c74d89f10

"You can lead a horse to water, but you can't make it drink" .

[WM]

On 13 Apr., 12:13, Dan <dan.ms.ch...@gmail.com> wrote:

>
> Trust me on this one :
>
> 1)There are an countable/inFinite
>      number of paths of 'Finite length' :

I do.

> 2)     There are an unCountable
> number of paths of 'Countable/infinite length' :

I do not. And you cannot show any one of them.

>
> 0->0.1 -> 0.10->0.101->0.1010->0.10101-> .........  :path of infinite
> length , representing 0.(10)
> 0->0.1 -> 0.11->0.111->0.1111->0.11111-> .........  :path of infinite
> length , representing 0.(1)

All your examples are finite definitions. Neither these nor the paths
of the Binary Tree can be used to distinguish more than countably many
numbers. But undistinguishable numbers are not numbers that can be
used to distinguish things (Dedekind).

>
> The fact that "the set of 'all Countable Things' is unCountable" ,
> is , ultimately , no more surprising that the fact that
> "the set of 'all Finite Things' is inFinite" .

Both is wrong with respect to actual infinity.

The set of finite things has larger cardinality m than any finite
number n:
For all n exists m : m > n
But there is no cardinality m that is larger than every finite number
n
Exists m for all n : m > n
is wrong.

The first expression describes potential infinity, the second actual
infinity with m = aleph_0. But it is independent of the name of m and
it is false.

>
> If you start counting one number at a time .... 1 , 2  , 3 .... you'll
> never count all the numbers, given the fact that , one day , you'll
> die . We'll all die, eventually .   There are some things that you
> cannot count even if you do live forever . The trick is understanding
> and learning to live with your limits.

Yes, to understand them! That is important. To understand, for
instance, that independent of lifetime or time at all the sequence
0.1
0.11
0.111
...
does not contain 1/9. Why? Because there is no sequence of 1's
supplying 1/9.

>
> Now, I want you to read this again

I have known that for several years. Nothing new. I know that most
mathematicians do not understand that limits are never decimal
fractions, not even 0. There is no 0.000... other than this very
finite definition "0.000....".

Now I want you to read and possibly answer my question:
A) {1} U {1, 2} U {1, 2, 3} U ...
B) {1}, {1, 2}, {1, 2, 3}, ...
C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ...
Is A contained in C?
Is anything in C that is not in B?

Regards, WM

[Dan]

> The set of finite things has larger cardinality m than any finite
> number n:
> For all n exists m : m > n

True, if m is indeed, a cardinality, and not a finite number .

> But there is no cardinality m that is larger than every finite number
> n
> Exists m for all n : m > n
> is wrong.

http://en.wikipedia.org/wiki/Axiom_of_infinity
The axiom of infinity, one of the basic axioms of set theory, says
precisely that m exists .
If you reject it,not only are we left unable to talk about real
numbers ,banning them as invalid, we are left unable to talk about
your binary tree .

Because it contains ALL Finite initial segments , the binary tree
should not exist . It's wrong .

We've been using 'magic'/infinity in our mathematics up until now .
An important point to make is IT WORKS .
The same way we've been using 'i' in our mathematics .Modern physics
is based on 'i' .
As long as it will CONTINUE TO WORK , we will CONTINUE TO USE IT .

Opinions on infinity are varied, but the paradoxes you keep rambling
about DON'T EXIST FOR ANYONE ELSE .Nor do you have any results to
show. Unless you manage to understand that, you'll make no further
progress .

"Wanting to reform the world without discovering one's true self is
like trying to cover the world with leather to avoid the pain of
walking on stones and thorns. It is much simpler to wear shoes." -
Ramana Maharshi

http://www.webcitation.org/5nZWht6FE

"Leibniz, one of the co-inventors of infinitesimal calculus,
speculated widely about infinite numbers and their use in mathematics.
To Leibniz, both infinitesimals and infinite quantities were ideal
entities, not of the same nature as appreciable quantities, but
enjoying the same properties."

[WM]

On 13 Apr., 13:30, Dan <dan.ms.ch...@gmail.com> wrote:
>  > The set of finite things has larger cardinality m than any finite
>
> > number n:
> > For all n exists m : m > n
>
> True, if m is indeed, a cardinality, and not a finite number .
>
> > But there is no cardinality m that is larger than every finite number
> > n
> > Exists m for all n : m > n
> > is wrong.
>
> http://en.wikipedia.org/wiki/Axiom_of_infinity
> The axiom of infinity, one of the basic axioms of set theory, says
> precisely that m exists .

No. The axiom of infinity has been invented by Zermelo who confessed
that he had taken it from Dedekind, who has it invented independently
from but anticipated by Bolzano. It is explained as follows: I can
think an idea. I can think that I can think an idea. I can think
that ... and so on. Obviously the set of ideas (or, as Bolzano put it,
of theorems) is potentiall infinite. It is not actually infinite. So
the axiom of infinity is based upon potential infinity. Only set
theorists have confused the "there is a set ..." with actual or
finished infinity.

> If you reject it,not only are we left unable to talk about real
> numbers ,banning them as invalid, we are left unable to talk about
> your binary tree .

No. I assume it, when discussing here. Otherwise any discussion would
be in vain.

>
> Because it contains ALL Finite initial segments , the binary tree
> should not exist . It's wrong .
>

In fact "all FISONs" are als an actually infinite set - and as such do
not exist. What exists is a FISON and for that there is a longer one.
But already here we face a problem: The set

1
2,1
3,2,1
...

does not contain its limit |N as a line, but obviously contains its
limit |N in the first (and every other) column. On the other hand we
can prove, by the construction of this set, that everything that is in
the list, is in one line.

> We've been using 'magic'/infinity  in our mathematics up until now .

Euler, Gauss, and Weierstarss did not use it. Their mathematics was
better than the present one.

> Opinions on infinity are varied, but the paradoxes you keep rambling
> about DON'T EXIST FOR ANYONE ELSE .Nor do you have any results to
> show.

Have you overlook my repeated question?

A) {1} U {1, 2} U {1, 2, 3} U ...
B) {1}, {1, 2}, {1, 2, 3}, ...
C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,

3} U ..., ...

Is A contained in C?
Is anything in C that is not in B?
>

> "Leibniz, one of the co-inventors of infinitesimal calculus,
> speculated widely about infinite numbers and their use in mathematics.

He had a split tongue in this respect. There are quotes in favour and
quotes against actual infinity. Cantor himself named him among the
opponents of actual infinity: D'ALEMBERT, ARISTOTELES, CAUCHY,
CAVALIERI, FULLERTON, GALILEI, GAUSS, GERDIL, GOUDIN, GULDIN,
LAGRANGE, LEIBNIZ, MOIGNO, NEWTON, PERERIUS, PESCH, RENOUVIER,
SANSEVERINO, TONGIORGI, TORRICELLI, ZIGLIARA.

Regards, WM

[Bergholt Stuttley Johnson]

Dan wrote:
> your binary tree .


WM's binary tree?
Who invented it?
http://en.wikipedia.org/wiki/Cantor_tree
WM is an idiot.

[Dan]

> He had a split tongue in this respect. There are quotes in favour and
> quotes against actual infinity. Cantor himself named him among the
> opponents of actual infinity:  D'ALEMBERT, ARISTOTELES, CAUCHY,
> CAVALIERI, FULLERTON, GALILEI, GAUSS, GERDIL, GOUDIN, GULDIN,
> LAGRANGE, LEIBNIZ, MOIGNO, NEWTON, PERERIUS, PESCH, RENOUVIER,
> SANSEVERINO, TONGIORGI, TORRICELLI, ZIGLIARA.
>
> Regards, WM

Summing up supporters for or against a cause (and wrongly so) is not
in itself an argument for or against it .

"I am so in favor of the actual infinite that instead of admitting
that Nature abhors
it, as is commonly said, I hold that Nature makes frequent use of it
everywhere,
in order to show more effectively the perfections of its Author. Thus
I believe that
there is no part of matter which is not, I do not say divisible, but
actually divided;
and consequently the least particle ought to be considered as a world
full of an
infinity of different creatures." -Leibniz , the co-inventor of
calculus .

The only reason he seemed to have had 'a split tongue' is his attempt
to make himself understood to those who's minds were , shall we say ,
limited by finitude , and , as always , try and make a unified system
within which all the different perspectives would be integrated .You
make no such attempt, nor could you , given you immense lack of
understanding. "Burn the witches! Ban the infinite!" .

"The continuum actually consists of infinitely many indivisibles" -
Galilei

I could continue, but it's a futile endeavor. The answer to both your
questions is no.

[dull...@sprynet.com]

On Fri, 12 Apr 2013 07:27:50 -0700 (PDT), WM
<muec...@rz.fh-augsburg.de> wrote:

>On 12 Apr., 16:14, dullr...@sprynet.com wrote:
>
>> >Cantor's list contains real numbers r as binary or decimal fractions.
>> >Real numbers, however, are /limits/ of binary or decimal fractions.
>>
>> No, it _is_ a list of limits of decimal fractions.
>
>There is no such list possible unless you give finite definitions.
>It is impossible to define a number by writing an iinfinite sequence.
>> >
>> It passes belief that a person could actually
>> think that such a well known proof could contain
>> such a simple error, without that error being
>> noticed by any of the thousands of mathematicians
>> who've studied the subject for a century or so.
>
>This assumption may be the reason that Cantor's "proof" has been
>believed over 100 years.

Curiously you deleted my statement of exactly what error of
yours I was referring to. Why was that? Wait, I know.
Because you have no intellectual integrity.

>Try to understand the following. (If you are
>not a too slow thinker, this will happen before midnight.)

>
>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.
>

>Proof: There are infinitely many rationals q_k with this property.
>All
>are in the list by definition. At most n of them are in the first n
>lines of the list. Infinitely many must exist in the remaining part
>of
>the list. So we have obtained:

>
>For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
>q_kn.
>This theorem it is not less important than Cantor's theorem: For all
>k: d =/= q_k.
>

>Both theorems contradict each other with the result that finished
>infinity as presumed for transfinite set theory is not a valid
>mathematical notion.
>
>Regards, WM
>

[fom]

On 4/13/2013 5:10 AM, WM wrote:
> By the
> way, Oberth's ideas at his times have been considered as ridiculous as
> mine few years ago.

What is ridiculous are not your ideas.

What is ridiculous is your desire to treat
the issues of mathematical investigation
in terms of "right" and "wrong".

Your ideas with respect to treating counting
as concatenations of symbols are to be found
in the Russian school of constructive
mathematics.

Prattling on about how ideas you do not like
should be abolished is distasteful, at best.
Making claims that sound mathematical, but
are not, is unprofessional. Pretending that
these claims are grounded in the physical
universe as described by empirical evidence
is bad philosophy.

That is what is ridiculous.

The issues of transfinite arithmetic do
have bearing on the mathematics used by
physicists. It may well be that some
confluence of empirical data and set-theoretic
models actually decide a question in the
future. But unless untestable theories
such as string theory or loop quantum
gravity are considered and unless forcing
models on countable transitive models of
set theory examine different modal possibilities,
that confluence will never be recognized.

It is clear that certain physical investigations
do not require any mathematics or physics beyond
the late nineteenth century. This is the same
mathematics and physics whose probabilistic
models depend on a classical probability theory
based on measure theory that uses the axiom of
choice. Your views of what is required from
mathematics fall into this pre-axiomatic era.

The problems arose as all the mathematics
used to coordinate all of the empirical data
became the justification for the truth of the
philosophies developing in that industrial
age. Everyone thought the mathematics would
simply justify the beliefs of the new professionals
called "scientists".

Mathematics has certainly supported empirical
science, and, empirical science has contributed
a great deal to mathematics. But, those philosophical
debates are precisely what created the information
age. You may look to Claude Shannon for the details
of bits and bytes. He was employed by a commercial
enterprise whose unit of measurement had been the
almighty dollar. But, the theory of algorithms
derives from those philosophical debates.

Undoubtedly, as the discipline of mathematics is
practiced primarily in computational environments,
the significance of classical mathematics with its
classical logic will be less prominent. The
"newspeak" of this culture is already felt with
phrases like "common era" to obscure the system of
calendar dates based on the influence of
Christianity in Western culture. One is now hearing
how it is too expensive to change the clocks on
computing systems to keep synchronized with
astronomical calculations. I can see a day when
people speak of "nominal time" disconnected from
"GPS time" or "satellite time" just because
"counting with crayon marks" is the "right way".

Anyway, your narrow views are just your beliefs.

What is ridiculous is that you insist that everyone
should abide by your views. What is not ridiculous
is that you probably feel that the current situation
in the mathematical community is imposing that
exact situation on you. You are fortunate that this
is probably the worst injustice "the system" commits
against you.

[WM]

On 13 Apr., 14:13, Dan <dan.ms.ch...@gmail.com> wrote:

> Summing up supporters for or against a cause (and wrongly so) is not
> in itself an argument for or against it .

No. But the list has been prepared by Cantor. He seems to have been
kind of proud to have defeated all the scholars in the list including
Gauss and Leibniz and Newton - at least in the eyes of matheologians.

>
> "The continuum actually consists of infinitely many indivisibles" -

besides those of geometry and matter also those of the vacuum.
Galilei

I did not count him as an opponent of actual infinity. Cantor did. And
he has with certainty read ten times more than you.

>
> The answer to both your questions is no.

Really???

The first question was:

Is A contained in C?

If you say no, then you should be able to determine something that is
in A but not in C. In case you prefer to argue instead of to believe,
try logic. The last U in (A) is of higher power than the last U in
(C)?

A) {1} U {1, 2} U {1, 2, 3} U ...
B) {1}, {1, 2}, {1, 2, 3}, ...
C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ..., ...

The second question was:

Is anything in C that is not in B?

Here your no is provably correct.

But one falsity can crash a theory. Until you or someone else finds
anything that is in A but not in C, I claim to have contradicted
transfinite set theory.

Regards, WM

[WM]

On 13 Apr., 16:47, dullr...@sprynet.com wrote:

> Curiously you deleted my statement of exactly what error of
> yours I was referring to. Why was that?

What do you allude to? Why don't you repeat it, if it was important? I
delete all phrases which I do not consider important for the further
discussion. Otherwise the post would be too long.
But if you think a bit about the following questions, you will
certainly recognize that your attempts are in vain, to find an error
in my theory:

A) {1} U {1, 2} U {1, 2, 3} U ...
B) {1}, {1, 2}, {1, 2, 3}, ...
C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ..., ...

Is A contained in C?

Is anything in C that is not in B?

In case you tend to answer the first question with a resounding no
(like Feferman answered the question: "Is Cantor necessary"), please
be prepared to find something of A that is not contained in C - as
custom has it in mathematics.

Regards, WM

[fom]

On 4/13/2013 11:58 AM, WM wrote:
> On 13 Apr., 16:47, dullr...@sprynet.com wrote:
>
>> Curiously you deleted my statement of exactly what error of
>> yours I was referring to. Why was that?
>
> What do you allude to? Why don't you repeat it, if it was important? I
> delete all phrases which I do not consider important for the further
> discussion. Otherwise the post would be too long.

chuckle

That is rarely the criterion.

WM usually deletes whatever challenges that
would require an accounting of his statements
more substantial than

|
||
|||
||||

[dull...@sprynet.com]

On Sat, 13 Apr 2013 09:58:39 -0700 (PDT), WM
<muec...@rz.fh-augsburg.de> wrote:

>On 13 Apr., 16:47, dullr...@sprynet.com wrote:
>
>> Curiously you deleted my statement of exactly what error of
>> yours I was referring to. Why was that?
>
>What do you allude to? Why don't you repeat it, if it was important?

You asserted that "He concludes that this remains true for the limits
of
the list numbers r and d by using the argument: different sequences
have different limits."

And that is simply not true.

> I
>delete all phrases which I do not consider important for the further
>discussion. Otherwise the post would be too long.
>But if you think a bit about the following questions, you will
>certainly recognize that your attempts are in vain, to find an error
>in my theory:
>
>A) {1} U {1, 2} U {1, 2, 3} U ...
>B) {1}, {1, 2}, {1, 2, 3}, ...
>C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
>3} U ..., ...
>Is A contained in C?

No, of course not.

>Is anything in C that is not in B?
>
>In case you tend to answer the first question with a resounding no
>(like Feferman answered the question: "Is Cantor necessary"), please
>be prepared to find something of A that is not contained in C - as
>custom has it in mathematics.

A = {1,2,3,...}, and that set is not in B.

>
>Regards, WM

[WM]

On 13 Apr., 19:59, dullr...@sprynet.com wrote:
> On Sat, 13 Apr 2013 09:58:39 -0700 (PDT), WM
>

> <mueck...@rz.fh-augsburg.de> wrote:
> >On 13 Apr., 16:47, dullr...@sprynet.com wrote:
>
> >> Curiously you deleted my statement of exactly what error of
> >> yours I was referring to. Why was that?
>
> >What do you allude to? Why don't you repeat it, if it was important?
>
> You asserted that "He concludes that this remains true for the limits
> of
> the list numbers r and d by using the argument: different sequences
> have different limits."
>
> And that is simply not true.

Remember, what you said: Only cases like 0.999... = 1.000... can lead
to same limits. This holds if you have digits a_n multiplied by powers
of 10^-n. But Cantor simply used bits that he called w and m. Without
a topology, there are no limits at all! There are only strings of
bits. So also the anti-diagonal d is not a limit - it is simply a
string of bits without end. And of that kind you can find behind every
finite string of bits (d_1, ..., d_n) the same string repeated
infinitely often. For every n. Further nothing happens in the list.

>
> > I
> >delete all phrases which I do not consider important for the further
> >discussion. Otherwise the post would be too long.
> >But if you think a bit about the following questions, you will
> >certainly recognize that your attempts are in vain, to find an error
> >in my theory:
>
> >A) {1} U {1, 2} U {1, 2, 3} U ...
> >B) {1}, {1, 2}, {1, 2, 3}, ...
> >C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
> >3} U ..., ...
> >Is A contained in C?
>
> No, of course not.

Then let me know the first natural number that is in A but not in C.
You know that every non-empty set of natural numbers has a first
element. That is even true in common mathematics without any set
theory.

>
> >Is anything in C that is not in B?
>
> >In case you tend to answer the first question with a resounding no
> >(like Feferman answered the question: "Is Cantor necessary"), please
> >be prepared to find something of A that is not contained in C - as
> >custom has it in mathematics.
>
> A = {1,2,3,...}, and that set is not in B.

I appreciate your clear speech. But I do not believe you. So I invite
you to determine a natural number, possibly the first one, that is in
A but not in B.
Please be aware, that I require mathematics, not matheology. In the
latter case the faithful believe that A contains only all finite
natural numbers and that B also contains all finite natural numbers,
but that the U has some magig power like the horseshoe that Bohr hat
nailed above the door of his summer house. And when Heisenberg asked
him, whether he, Bohr, was superstitious believing in good luck spent
by the horseshoe, Bohr answered, no, he did not believe, but the
horseshoe is said to be of help inspite of that.

Well, please do not use such arguments. Speak frankly, what is missing
in B that is present in A? Sets are differing by elements. If all
elements of A are in B and vice versa, then by Zermelo's first axiom,
A = B.

Regards, WM

[WM]

On 13 Apr., 19:34, fom <fomJ...@nyms.net> wrote:

> WM usually deletes whatever challenges that
> would require an accounting of his statements
> more substantial than
>
> |
> ||
> |||
> ||||

You have forgotten the important three points for "and so on". If you
were a mathematician, you would see, that every number that is in the
list

1
2, 1
3, 2, 1
...

is in one line of the list. This is the construction principle. And it
is clear by symmetry considerations, that every attempt to deny this
fact is condemned to fail in a community of sober minds.

Regards, WM

[Dan]

> I appreciate your clear speech. But I do not believe you. So I invite
> you to determine a natural number, possibly the first one, that is in
> A but not in B.

> Well, please do not use such arguments. Speak frankly, what is missing
> in B that is present in A? Sets are differing by elements. If all
> elements of A are in B and vice versa, then by Zermelo's first axiom,
> A = B.

B dose not contain numbers as members . B contains sets of numbers .
A contains numbers .
It's like A is a list of apples , where the apples are of various
kinds : 1, 2 ,3 ....
And B is a list of 'boxes of apples' :
{1}, {1, 2}, {1, 2, 3} ....
You won't find a 'kind of apple' on the B-list , but you'll find a box
that contains that 'kind of apple' , among other kinds .
In all honesty, it's easier to teach these things to a five-year
old .

On the other hand , if you're using 'numbers as sets' representation,
a form commonly used for ordinals , then
{1,2 ...n} = n+1 , in which case 1 is the first , and only natural
number that appears A , and not in C .
In shorthand , A= {1,2,3 .... }
C= {2,3,4......}

[Virgil]

In article
<c536ca1c-9c78-4cb5...@a3g2000vbr.googlegroups.com>,

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

> On 13 Apr., 16:47, dullr...@sprynet.com wrote:
>
> > Curiously you deleted my statement of exactly what error of
> > yours I was referring to. Why was that?
>
> What do you allude to?

It is a bad habit of WM's to pretend that his many past errors simply do
not exist, but outside of his closed world of Wolkenmuekenheim, he
cannot maintain it.
A recent error of his that he will try to ignore is his inability to
distinguish between the union of all FISONs (Finite Iinitial SeTS of
NaturalS), and the sequence of all FISONs.

The union of the set of all FISONs is |N, but the sequence of all FISONs
does not contain |N, even though |N is its limit.

At least everywhere other than in Wolkenmuekenheim,
--

[Virgil]

In article
<6f0d8a55-f965-4934...@a34g2000vbt.googlegroups.com>,

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

> Please be aware, that I require mathematics, not matheology.

it is a shame that WM is unable to provide any of what he reqires!
--

[Virgil]

In article <atrim8pasc6hk9968...@4ax.com>,

dull...@sprynet.com wrote:

> >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.
> >
> >Proof: There are infinitely many rationals q_k with this property.
> >All
> >are in the list by definition. At most n of them are in the first n
> >lines of the list. Infinitely many must exist in the remaining part
> >of
> >the list. So we have obtained:
> >
> >For all n exists k: d_1, d_2, d_3, ..., d_n = q_k1, q_k2, q_k3, ...,
> >q_kn.
> >This theorem it is not less important than Cantor's theorem: For all
> >k: d =/= q_k.
> >
> >Both theorems contradict each other with the result that finished
> >infinity as presumed for transfinite set theory is not a valid
> >mathematical notion.

Outside or Wolkenmuekenheim it is of no importance at all, since what WM
claims follows from it does not follow from it.

A list of all rationals between 0 and 1 need not contain any
non-rationals, and cannot contain all irrationals, of which there are at
least as many as there are rationals, so is clearly an incomplete
listing of all the numbers between 0 and 1, just as Cantor said!
--

[WM]

On 13 Apr., 21:58, Dan <dan.ms.ch...@gmail.com> wrote:

> B dose not contain numbers as members .
> B contains sets of numbers .
> A contains numbers .

My first question to you was this: Is A contained in C?
Therefore your following text is not on topic and has been deleted.

A) {1} U {1, 2} U {1, 2, 3} U ...

C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ...

A is an infinite set of numbers, C is an infinite sequence of finite
sets of numbers.

Is the set of all finite natural numbers missing in the sequence C
that obviously contains all possible finite initial segments of the
set of natural numbers? You said A is not in C. Therefore there must
be something missing in every set of the sequence C. Since we are
dealing with natural numbers, this something can only be a natural
number - at least if we are doing mathematics.

Regards, WM

[Virgil]

On Sat, 13 Apr 2013 09:58:39 -0700 (PDT), WM
<muec...@rz.fh-augsburg.de> wrote:


> I
>delete all phrases which I do not consider important for the further
>discussion. Otherwise the post would be too long.

You also delete any questions you cannot answer and any proofs of your
many errors or false claims.

>A) {1} U {1, 2} U {1, 2, 3} U ...
>B) {1}, {1, 2}, {1, 2, 3}, ...
>C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
>3} U ..., ...
>Is A contained in C?

No. C is a sequence which is term by term equal to sequence B, but it is
a strictly increasing sequence so cannot contain its limit, which is A.

>Is anything in C that is not in A?

More importantly, there is something "in" A that is not in C, the union
of the set of terms of sequence C.
--

[WM]

On 13 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:

> The union of the set of all FISONs is |N, but the sequence of all FISONs
> does not contain |N, even though |N is its limit.

A) {1} U {1, 2} U {1, 2, 3} U ...

C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
3} U ...

The sequence C of all unions of FISONs does not contain |N? Why should
it stop before reaching this goal, if it can be accomplished by A? But
if it stops in fact, then also A cannot reach this goal, because C
contains A . Otherwise there should be a natural number or a FISON
that is in A but not in C. At least when discussing this question in
mathematics, the proponent of your position would be asked to prove
his point. Of course, I know that it would be in vain to ask you to do
so.

Regards, WM

[Sam Sung]

The idiot WM is even incapable to get that a comma is not the same as
a union symbol and the idiot WM does not get that sequences, unlike
sets, are strictly ordered, and that in sequences like B and C each
member has a unique and exaclty specified successor.

[Sam Sung]

The idiot WM is even incapable to get that a comma is not the same as
a union symbol and the idiot WM does not get that sequences, unlike
sets, are strictly ordered, and that in sequences like B and C each

member has a unique and exactlty specified successor.

That's just one reason the idiot WM is even incapable to get that
in the series A and B there is ALWAYS a unique successor which is
a FINITE sized element, whereas the union A includes EACH member of
N, no matter whether it is finitely sized or of infinite cardinality.

[Sam Sung]

Sam Sung wrote:

> The idiot WM is even incapable to get that a comma is not the same as
> a union symbol and the idiot WM does not get that sequences, unlike
> sets, are strictly ordered, and that in sequences like B and C each

> member has a unique and exactly specified successor.

>
> That's just one reason the idiot WM is even incapable to get that
> in the series A and B

Error - read: in the sequences B and C

[fom]

On 4/13/2013 1:58 PM, WM wrote:
> On 13 Apr., 19:34, fom <fomJ...@nyms.net> wrote:
>
>> WM usually deletes whatever challenges that
>> would require an accounting of his statements
>> more substantial than
>>
>> |
>> ||
>> |||
>> ||||
>
> You have forgotten the important three points for "and so on". If you
> were a mathematician, you would see, that every number that is in the
> list
>
> 1
> 2, 1
> 3, 2, 1
> ...
>
> is in one line of the list. This is the construction principle.

Hence, constructive mathematics.

That is only a proper part of mathematics. And, you
embarrass that tradition as much as any other.

[Ralf Bader]

WM wrote:

> On 12 Apr., 19:31, Ralf Bader <ba...@nefkom.net> wrote:
>> WM wrote:
>> > On 12 Apr., 17:24, Dan <dan.ms.ch...@gmail.com> wrote:
>>
>> >> They do not contradict each other :
>> >> Cantor's affirmation (in its full form) is :
>>
>> >> 1) forall k , exist n , d_n =/= q_kn
>>
>> > Tricky! No, please be careful. Cantor shows exactly:
>> > forall k: d_k =/= q_kk
>> > Not more and not less.
>>
>> And that is precisely what he wanted to show - assertion 1) is true with
>> taking n=k.
>
> I know that. But Cantor's proof is valid only for n =< k.
>
> This is not related to the fact that there is a contradiction between
> my theorem
> forall n , exists k , d_1 = q_k1 and d_2 = q_k2 and ..... d_n =
> q_kn
> and the negation of Cantor theorem :
> exists k , forall n =<k , d_n == q_kn
>
> iff
> d = SUM{all n} d_n*10^-n
> since all n are already in the sequence
> (d_1*10^-1), (d_1*10^-1 + d_2*10^-2), (d_1*10^-1 + d_2*10^-2 +
> d_3*10^-3) , ...
> which contains all n.
> Or do you know of any n that is missing?
>
> Regards, WM

There is no theorem. There is only garbage. You are unable to express
anything in a correct manner. You have no idea how quantifiers work. They
are not an abbreviation for colloquial phrases like "for all", in the way
you introduce them in your idiotic "bestseller". Your question is nonsense.

[Virgil]

In article <kkci6d$k4s$1...@speranza.aioe.org>, Sam Sung <n...@mail.invalid>
wrote:

The set N. or |N, while being itself infinite does not contain anything
but finite naturals as members (though infinitely many of them).
--

[Virgil]

In article
<1d60c1b4-9e0e-4471...@c9g2000vbr.googlegroups.com>,

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

> On 13 Apr., 22:32, Virgil <vir...@ligriv.com> wrote:
>
> > The union of the set of all FISONs is |N, but the sequence of all FISONs
> > does not contain |N, even though |N is its limit.
>
> A) {1} U {1, 2} U {1, 2, 3} U ...
> C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
> 3} U ...
>
> The sequence C of all unions of FISONs does not contain |N?

C is no more that the sequence of FISONs which makes it a strictly
increasing ordered sequence, ordered by inclusion, whose LIMIT is |N.

A strictly increasing infinite sequence like cannot have its limit as
any one of its terms as there is always a larger term.

> Why should
> it stop before reaching this goal

It doesn't stop, but if it ever reached its limit, being a strictly
increasing sequence, it would then have to pass that limit.

Which may well happen in Wolkenmuekenheim, but nowhere else.

> if it can be accomplished by A? But
> if it stops in fact, then also A cannot reach this goal, because C
> contains A .

Not outside of Wolkenmuekenheim it doesn't. A is a set which is not
equal to or contained in any term of sequence C, but only in C's limit,
which is not a member of C or a subset of C.

> Otherwise there should be a natural number or a FISON
> that is in A but not in C.

The only way that could be true would be if the limit of a strictly
increasing sequence could be a member of the sequence, but other than in
Wolkenmuekenheim, the limit of a strictly increasing sequence CANNOT be
a member of the sequence

At least when discussing this question in
> mathematics

WM is incapable of discussing any thing in mathematics, as that would
require him to leave his Wolkenmuekenheim.
--

[Virgil]

In article
<1f9ae72b-c720-4a82...@f18g2000vbs.googlegroups.com>,

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

> On 13 Apr., 21:58, Dan <dan.ms.ch...@gmail.com> wrote:
>
> > B dose not contain numbers as members .
> > B contains sets of numbers .
> > A contains numbers .
>
> My first question to you was this: Is A contained in C?

A is the limit of a strictly increasing sequence of sets of naturals
denoted by either B or C, and is therefore not a member of either B or C.

It is only in weird worlds like Wolkenmuekenheim that one can have
strictly increasing infnite sequences which contain their limits as
terms of those sequences.

> Therefore your following text is not on topic and has been deleted.
>
> A) {1} U {1, 2} U {1, 2, 3} U ...
> C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
> 3} U ...
>
> A is an infinite set of numbers, C is an infinite sequence of finite
> sets of numbers.

And is a strictly increasing sequence of sets of natural numbers when
ordered by inclusion, which must, therefore, not contain its limit, |N.

>
> Is the set of all finite natural numbers missing in the sequence C

Yes!

> that obviously contains all possible finite initial segments of the
> set of natural numbers? You said A is not in C. Therefore there must
> be something missing in every set of the sequence C.

Every set in C is missing far more naturals than it contains, as many as
in all of |N.

> Since we are
> dealing with natural numbers,

We are dealing with sets of natural numbers in sequnce, C is a strictly
increasing sequence of such sets, the sequence of all FISONs, taken in
increasing order of size, in fact.

> - at least if we are doing mathematics.

Which WM is not doing!
--

[Virgil]

In article
<0179df02-65cf-4062...@r4g2000vbf.googlegroups.com>,

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

> On 13 Apr., 19:34, fom <fomJ...@nyms.net> wrote:
>
> > WM usually deletes whatever challenges that
> > would require an accounting of his statements
> > more substantial than
> >
> > |
> > ||
> > |||
> > ||||
>
> You have forgotten the important three points

While fom may have forgotten much,
he, at least, at one time knew it,
something that WM fails at.

WM's latest ex cathedra is that a strictly increasing infinite sequence
having a limit may contain that limit as a term of the sequence.
--

[Virgil]

In article
<1cdea071-3cd0-4a77...@c9g2000vbr.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 13 Apr., 12:13, Dan <dan.ms.ch...@gmail.com> wrote:
>
> >
> > Trust me on this one :
> >
> > 1)There are an countable/inFinite
> > � � �number of paths of 'Finite length' :
>
> I do.
>
> > 2) � � There are an unCountable
> > number of paths of 'Countable/infinite length' :
>
> I do not. And you cannot show any one of them.
> >
> > 0->0.1 -> 0.10->0.101->0.1010->0.10101-> ......... �:path of infinite
> > length , representing 0.(10)
> > 0->0.1 -> 0.11->0.111->0.1111->0.11111-> ......... �:path of infinite
> > length , representing 0.(1)
>
> All your examples are finite definitions.

Nothing in either mathematics nor logic prohibits finite definition of
infinite processes or procedures.

> Neither these nor the paths
> of the Binary Tree can be used to distinguish more than countably many
> numbers.

Distinguishability is not a requirement or prerequisit for existence, at
last not outside Wolkenmuekenheim.


> But undistinguishable numbers are not numbers that can be
> used to distinguish things (Dedekind).

It is only their existence, not their distinguishability, that is at
issue.
> >
> > The fact that "the set of 'all Countable Things' is unCountable" ,
> > is , ultimately , no more surprising that the fact that
> > "the set of 'all Finite Things' is inFinite" .
>
> Both is wrong with respect to actual infinity.

Both ARE right with respect to life outside Wolkenmuekenheim.
>
> The set of finite things has larger cardinality m than any finite
> number n:
> For all n exists m : m > n
> But there is no cardinality m that is larger than every finite number
> n

> Exists m for all n : m > n is wrong.

But "For all n exists m: m > n" is right.

WM's quantifier dyslexia strikes again!

> Now I want you to read and possibly answer my question:

> A) {1} U {1, 2} U {1, 2, 3} U ...

> B) {1}, {1, 2}, {1, 2, 3}, ...

> C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,
> 3} U ...

> Is A contained in C?

NO!
> Is anything in C that is not in B?

NO!

As sequences, B and C are the same sequence which is a strictly
increasing sequence ordered by inclusion. and as such cannot contain the
limit set, which is A.
--

[Virgil]

In article
<d6e306ee-ee0b-4ffa...@b3g2000vbo.googlegroups.com>,
WM <muec...@rz.fh-augsburg.de> wrote:

> On 13 Apr., 13:30, Dan <dan.ms.ch...@gmail.com> wrote:
> > �> The set of finite things has larger cardinality m than any finite

> >
> > > number n:
> > > For all n exists m : m > n
> >

> > True, if m is indeed, a cardinality, and not a finite number .

> >
> > > But there is no cardinality m that is larger than every finite number
> > > n
> > > Exists m for all n : m > n
> > > is wrong.
> >

> > http://en.wikipedia.org/wiki/Axiom_of_infinity
> > The axiom of infinity, one of the basic axioms of set theory, says
> > precisely that m exists .
>
> No.

Yes!
>
> > If you reject it,not only are we left unable to talk about real
> > numbers ,banning them as invalid, we are left unable to talk about
> > your binary tree .
>
> No. I assume it, when discussing here.

Then you assume an actual infinity, and cannot then impose its
non-existence on what follows.


> Otherwise any discussion would
> be in vain.

What makes you think the yours, at least, are not both vain and in vain?
> >
.
> >
>
> In fact "all FISONs" are als an actually infinite set - and as such do
> not exist.

Once you allow an actual infinitey, you can no longer deny it.

What exists is a FISON and for that there is a longer one.

And so on ad infinitum.
> But already here we face a problem: The set
>
> 1
> 2,1
> 3,2,1
> ...
>
> does not contain its limit |N as a line

Again WM does not comprehend that a strictly increasing sequence, like
the above, is NEVER able to contain its limit.

That is the natural order of things which should hold even in the
Stygian darknesses of Wolkenmuekenheim.





, but obviously contains its
> limit |N in the first (and every other) column. On the other hand we
> can prove, by the construction of this set, that everything that is in
> the list, is in one line.

Thus again WM claims that a strictly increasing sequence, can contain
(have a term equal to ) its limit.

That claim is false everywhere outside Wolkenmuekenheim,

>
> > Opinions on infinity are varied, but the paradoxes you keep rambling
> > about DON'T EXIST FOR ANYONE ELSE .Nor do you have any results to
> > show.
>
> Have you overlook my repeated question?

> A) {1} U {1, 2} U {1, 2, 3} U ...
> B) {1}, {1, 2}, {1, 2, 3}, ...
> C) {1}, {1} U {1, 2}, {1} U {1, 2} U {1, 2, 3}, {1} U {1, 2} U {1, 2,

> 3} U ..., ...

> Is A contained in C?

NO! At least unless a strictly increasing sequence can have a term equal
to its limit, which does not happen outside Wolkenmuekenheim. .
--