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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Wed Jul 19, 2006 1:33 pm Post subject:
Re: An uncountable countable set



Dik T. Winter schrieb:
Sorry, I have 150 examinations these days, so I cannot answer as yet.
But I will come back. (And you meanwhile have time to ponder about the
inconsistencies of set theory.)
Regards, WM 

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Franziska Neugebauer science forum addict
Joined: 23 Apr 2005
Posts: 52

Posted: Wed Jul 19, 2006 1:47 pm Post subject:
Re: An uncountable countable set



mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:
mueckenh@rz.fhaugsburg.de wrote:
Franziska Neugebauer schrieb:
[...]
IF the answer is in each case is "not zero 1's", THEN in each
column at least one 1 must be present.
This is the case, since every a_jj = 1 j e N by definition.
However, there is no natural numbers with this property,
Could you precicely _define_ which /property/ you are talking
about?
To contribute a 1 to each column.
Should be a property of which object?
Should be a property of at least one summand if the *+ sum has this
property.

1. I have shown formally (proved) that there is no single sequence
member having the property to "contribute a 1 to each column of the
sum".
2. If your assertion is a theorem that claim and 1 form a contradiction.
3. To "complete" it you now have to show that your claim "there exists
at least one summand which contributes a 1 to each column" is a
theorem. I. E. you have to _prove_ your assertion.
Quote:  For every column j e N a_mj has the 1 in position m(j) = j, since
a_jj = 1 A j e N. Where exactly lies your problem?
I told you recently: In a list like mine a *+ sum is defined. This
* sum is equal to the largest number of the list.
The notion of "largest number" is misleading. There is no largest
number. Neither in the list nor in omega. If it makes sense to talk
about "a largest number", then please
1) That does not mean that your pretension (sum of all naturals is
0.111...) was correct.2) omega is the largest number less than omega +
1.

1) I have already proven that the *sum of all the unary(i) i e N is
unary (omega).
2) But omega is not the "largest number" < omega. This is what we are
talking about.
Quote:  1. name it (show us its name or position _in_ the list or _in_
omega),
and/or
2. present a _proof_ that it exists.
In
0.1
0.11
0.111
...
0.111...
this element exists , if it does exist at all. Its position can be
selected arbitraily.

This does not imply that (unary(i))_{i e omega} has a member
unary(omega). But this is what we are talking about.
Quote:  Until then we cannot (meaningfully) talk about "a largest number in
the list" or "a largest number in omega". Wittgenstein applies here.
omega is the largest number of he sum above.

Which "sum" do you mean? Please state explicitly.
Quote:  Its existence is proven in set theory.

Omega is neither a sum or a *sum in set theory. It is an axiomatically
defined entity.
Quote:  And if a larger number is included, then the sum grows. You pretend
that this is not valid for 0.111... because without 0.111... the
list has the *+ sum 0.111... and when you include it in the list
(as diagonal number or as the first line or anywhere else) the *+
sum does not grow.
Postponed until existential status of "largest number" is clarified
by
Is clarified.

Not to me.
F. N.

xyz 

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Franziska Neugebauer science forum addict
Joined: 23 Apr 2005
Posts: 52

Posted: Wed Jul 19, 2006 2:16 pm Post subject:
Re: An uncountable countable set



Dear Wolfgang,
I would appreciate if you not cut away that parts of my previous posts
you are trying to argue against. It may also be helpful if you read
before cutting.
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:

context restored:
,[ <44be06f2$0$17972$892e7fe2@authen.yellow.readfreenews.net> ]
 [...]
 >>
 >> > Hence (what means the same as therefore etc.) you should write
 >> > "a_i = 1 if 0 < i <= n".
 >>
 >> If you are confused by the range you may drop "0 <=" entirely,
 >> since
 >> A i (i e omega > 0 <= i).
 >
 > You are confused.
 >>
 >> ++
 >>  a_i = 1 if i < n. (sic!) 
 >> ++
 >
 > How many letters "x" do you see here: xxx. Is 4 the correct answer?
 > If
 > not, then please correct your formula by
 > a_i = 1 if i =< n.
 > Example: = 0.111 = 3 and not 4.

 context restored:

 ,[ <44bcd393$0$17968$892e7fe2@authen.yellow.readfreenews.net> ]
  >> a_i = 0 if n <= i < omega
  > a_i = 1 if i = n. Example: 0.11 is the unary represenation of 2,
  > not
  > of 3.
 
  I presume omega = { 0, 1, 2, ... } and I presume i e omega.
  Therefor unary(2) = 11. The positions 0 and 1 are 1.
 
  If you are confused by the range you may drop "i < omega" entirely,
  since A i (i e omega > i < omega).
 
  ++
   a_i = 0 if n <= i (sic!) 
  ++
 `

 >> summing up:
 >>
 >> a_i = 1 if i < n (1a)
 >> a_i = 0 if n <= i (1b)
 >
 > false.

 What is false unter the abovementioned presumption omega = { 0, 1,
 2, ... }?
You prefered not to read it but to cut away.

 context restored:

 ,[ <44bcd393$0$17968$892e7fe2@authen.yellow.readfreenews.net> ]
  A tabulation may help you to understand what I mean:
 
  unary(n) (only 1's
  n 1allocated i's 0allocated i's written)
  
  0 (none) 0 1 2 ... "" (empty string)
  1 0 1 2 3 ... 1
  2 0 1 2 3 4 ... 11
  3 0 1 2 3 4 5 ... 111
  ... .... ... ....
 
  If you want to drop index and number 0, you can simply use
  omega \ { 0 } instead of omega.
 
  >> omega: unary(omgea) def= (a_i) a_i = 1 A i e omega
 
  >> A tabulation may help you to understand what I mean:
 `

 BTW: The tabulation is generated by (1a) & (1b) so they are neither
 "wrong" nor "false".
Has been ignored, too.

 > It is not important what you mean,

 When it is not important "what I mean", why do you discuss with me?
 This
 is rather insulting.

 > but what is.

 Ontological issues are to be discussed in the philosophical
 newsgroups.
`
Quote:  xxx is a set of three x, whatever you may mean.
Where did I argue against this?
Here: a_i = 1 if i < n.
This means for n = 4: a_i = 1 for i = 1, 2, and 3. This leads to 0.111
= 4 or to xxx = 4. And that is wrong.

Read before cutting!
n = 4 > a_i = 1 for i e {0, 1, 2, 3} > unary(4) = 1111
Quote:  A *sum of _infinitely_ _many_ representation of finite numbers
does in the present case result in a representation of an infinite
number. This is a fact.
That is nonsense. The sum including this very number
What number? unary(omega)?
0.111... with omega 1's, the decimal 1/9 and the unary omega.

unary(omega) _is_ 1111... or 0.1111...
Quote:  is not different from the sum excluding it.
That does neither imply the "nonexistence" of unary(omega) nor does
it imply that unary(omega) is in the list. If you want to argue the
converse, prove it!
I placed omega in the list. I found the *+ sum omega, as it should be:
the *+ sum is the largest number of he list. The *+ sum of the list
without omega is not omega

Wrong. You are arguing against facts. The *sum of (unary(i))_{i e
omega} is _proven_ to be unary (omega).
Quote:  because the largest number of he list is finite,

1. There is no largest member in the sequence (unary(i))_{i e omega}
> Wittgenstein.
2 The *sum of the sequence does not necessarily "inherit" the
properties from their individual sequence members. If this
"inherentance" is kind of a "general principle" you should refer to a
recognized source.
But there is even a finite counterexample:
101
011
110
 *sum
111
The property "does not contain at least one 0" is not inherited by
the *sum. So the "inheritance principle" is not so obvious.
Quote:  although we don't know it, we know that it is not infinite.

Riding a dead horse.
Quote:  You always mix up infinitely many with infinitely large.

unsubstancial.
Quote:  Look here:
The *+ sum 1 *+ 1 *+ 1 *+ ... = 1. This sum does not depend on "how
many"but only on "how large".

Where have I argued against?
Quote:  Therefore it demasks your wishful thinking infinitely many finite
numbers could give an infinite *+ sum.

You forgot that it has already been proven. You are arguing against
facts.
context restored:
,[ <44be06f2$0$17972$892e7fe2@authen.yellow.readfreenews.net> ]
 > Therefore all your corresponding assertions break down.

 Wishful thinking.
`
F. N.

xyz 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 19, 2006 6:47 pm Post subject:
Re: An uncountable countable set



In article <1153301245.706099.39070@m73g2000cwd.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:
My understanding of unary represenations is:
strange!
n e omega: unary(n) def= (a_i)
a_i = 1 if 0 <= i < n
a_i =/= 1 if n = 0.
Of course. unary(0) = 000...
The unary representation of 0 (the empty set) is the empty string
(stiputalting only 1's are written).
correct. If "unary representation" is announced by "0." Then 0. denotes
zero.

Unary representation would be better announced by a "U" for unary or
even a "b1." for base 1. Starting with a "0." is more like announcing
decimal fraction representation.
Quote: 
A *sum of _infinitely_ _many_ representation of finite numbers does in
the present case result in a representation of an infinite number. This
is a fact.
That is nonsense.

Then it is logical nonsense in vivid contrast to your illogical nonsense. 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 19, 2006 6:53 pm Post subject:
Re: An uncountable countable set



In article <1153301881.393720.57330@75g2000cwc.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Dik T. Winter schrieb:
If the *+ sum of the list is a unary representation of some 1's, then,
by the construction of the unary numbers of the list, this sum must be
a unary numer of the list.
As long as your list is finite.
As long as it contains *numbers* which can be distinguised from each
other and which obey trichotomy.
Suppose the list is infinite, and the
*+ sum of all members of the list is a member of the list. That would
mean that the *+ sum of the list is the last element of the list,
contradicting that the list is infinite.
If we assume that nonterminating fractions have aleph_0 digits, then
this case is realized by
0.1
0.11
0.111
...
0.111...
Here the last number of an infinite list is in this list.

Whichever interpretation one puts on the symbols, unary or decimal
fraction, in either case one has the ordinality of (N union {N}) in
that there is a trivial order isomorphism between the ordered members of
your
lists and the members of (N union {N}) 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 19, 2006 6:58 pm Post subject:
Re: An uncountable countable set



In article <1153302541.286243.112340@s13g2000cwa.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Virgil schrieb:
The successor of all naturals is not a natural and, therefore,
must be
larger (because it is not less).
There is no such thing as the successor of all naturals any more
that
there is a single successor common to both 3 a and 6.
This is an expression coined by Cantor: "a number following after all
natural numbers".
But he does not call it a 'natural' number any more than he calls it an
even number or a prime number.
I said above: The successor of all naturals is not a natural.
No one else even says that such a thing exists. There is a set of all
naturals but if is not the successor of any of them.
Cantor said it
.. 
Cantor did not say "successor of all naturals", and if he said anything
at all like that he did not mean successor in the same sense as 2 is the
successor of 1.
Quote:  Perhaps not in your philosophy, but you are not God, to command what is
true or false.
But I can see what is contradictory.

And you can see things as contradictory which are not as well.
Such "visions" are a handicap in logical thinking.
Quote: 
You are a liar. Therefore I cease my correpondence with you.

You are hardly in a position to complain about falsehoods. 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Wed Jul 19, 2006 7:00 pm Post subject:
Re: An uncountable countable set



In article <1153316030.583020.125150@p79g2000cwp.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Dik T. Winter schrieb:
Sorry, I have 150 examinations these days, so I cannot answer as yet.
But I will come back. (And you meanwhile have time to ponder about the
inconsistencies of set theory.)
Regards, WM

I hope that your exams are on set theory. As that might make you see
what it is all about. 

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Dik T. Winter science forum Guru
Joined: 25 Mar 2005
Posts: 1359

Posted: Thu Jul 20, 2006 12:39 am Post subject:
Re: An uncountable countable set



In article <1153301881.393720.57330@75g2000cwc.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Quote:  Dik T. Winter schrieb:
If the *+ sum of the list is a unary representation of some 1's, then,
by the construction of the unary numbers of the list, this sum must be
a unary numer of the list.
As long as your list is finite.
As long as it contains *numbers* which can be distinguised from each
other and which obey trichotomy.

No, as long as your list is finite.
Quote:  Suppose the list is infinite, and the
*+ sum of all members of the list is a member of the list. That would
mean that the *+ sum of the list is the last element of the list,
contradicting that the list is infinite.
If we assume that nonterminating fractions have aleph_0 digits, then
this case is realized by
0.1
0.11
0.111
...
0.111...
Here the last number of an infinite list is in this list.

I thought we were talking about natural numbers. I have not yet seen a
definition that calls the last one a natural number. You should *not*
switch between representations during the process. Either you have a
list of natural numbers (in that case 0.111... does not belong to it)
or you have a list of rational numbers in decimal notation. The latter
is easier to reason about because it can be tackled easier. In that
case your list consists of:
An = sum{i = 1..n} 10^(n) = (1  10^(n))/9.
Also we can easily show what repeated *+ is (notated here as SUM):
SUM{i = 1 .. k} Ai = Ak.
Now we can also get a proper definition about what your notation ... means
in the list:
lim{k > oo} SUM{i = 1 .. k} Ak = 1/9 = K.
And K is (by convention) decimally notated as 0.111..., which makes sense,
in my opinion. K clearly is not equal to any of the An, unless there is
a natural number for which 10^(n) = 0.
Quote:  Here the last number of an infinite list is in this list.

It is not a list when you use the proper definition of list (a mapping
from N to the items of the list). An infinite list does not have a last
element. Because, suppose that list had a last element. By the definition
of list there should be a natural number n that maps to that element. But
n+1 maps to another element.
Quote:  Hence the assumption is false.
Yes, the assumption is clearly false that omega or aleph_0 is a number
larger than any natural but counting all the naturals.

This is insidious misquoting.
You should start working with the definitions. When you want to show an
inconsistency in a system you have to do:
(a) show that within that system you can prove proposition P
(b) show that within that system you can disprove proposition P
But *all* proves should be within the system. So you should also use
the definitions from the system. Otherwise the only thing you show is
that the system is inconsistent with *your* definition when used within
the system.

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ 

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Dik T. Winter science forum Guru
Joined: 25 Mar 2005
Posts: 1359

Posted: Thu Jul 20, 2006 12:45 am Post subject:
Re: An uncountable countable set



A bit late, but I think it requires response:
In article <1153168957.805313.57460@p79g2000cwp.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Quote:  Dik T. Winter schrieb:
....
One should have seen that earlier, then Bourbaki would not have
succedeed to define 0 as natural number, even in political decisions.
Pray explain the last part "even in political decisions".
It is laid down in the guide lines of the European Community that zero
was a natural number. I am indebted to your compatriots that they have
dismissed the constitution of this disastrous association.

You are talking about two different things. First the guide lines and
second the constitution. Where in the guidelines, and where in the
consitution, is it layed down that zero is considered a natural number?
I have access to both, so pray give pointers.
But you fail to understand mathematics. Within mathematics you start with
axioms, use definitions and derive conclusions. When you can derive,
within that system, contradicting conclusions, your axioms are inconsistent.
In principle, no other knowledge than the axioms and the definitions is
needed.

dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ 

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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 10:34 am Post subject:
Re: An uncountable countable set



Dik T. Winter schrieb:
Quote:  Where in the guidelines, and where in the
consitution, is it layed down that zero is considered a natural number?
I have access to both, so pray give pointers.

I do not remember the source, but it was a trustworthy one.
Quote: 
When you can derive,
within that system, contradicting conclusions, your axioms are inconsistent.
In principle, no other knowledge than the axioms and the definitions is
needed.

Within this system I defined a *+ sum and showed that
The * sum of
0.1
0.11
0.111
....
0.111...
is
0.111...
i.e. the result is a number of the list.
The * sum of
0.1
0.11
0.111
....
is
0.111...
The result is not a number of the list.
This is a contradiction, because by tertium non datur only one case can
apply. One of these results is wrong if 0.111... represents a number.
Regards, WM 

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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 10:39 am Post subject:
Re: An uncountable countable set



Dik T. Winter schrieb:
Quote:  In article <1153301881.393720.57330@75g2000cwc.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Dik T. Winter schrieb:
If the *+ sum of the list is a unary representation of some 1's, then,
by the construction of the unary numbers of the list, this sum must be
a unary numer of the list.
As long as your list is finite.
As long as it contains *numbers* which can be distinguised from each
other and which obey trichotomy.
No, as long as your list is finite.

The numbers count the lines. As long as there are only finite numbers,
the list is finite.
But as you have not the understanding, consider the related case:
A finite number covers what it indexes. This does not depend on how
much numbers are in the list.
Quote:  From this discovery you may obtain the result for what you call an
infinity list. 
Quote:  Suppose the list is infinite, and the
*+ sum of all members of the list is a member of the list. That would
mean that the *+ sum of the list is the last element of the list,
contradicting that the list is infinite.
If we assume that nonterminating fractions have aleph_0 digits, then
this case is realized by
0.1
0.11
0.111
...
0.111...
Here the last number of an infinite list is in this list.
I thought we were talking about natural numbers. I have not yet seen a
definition that calls the last one a natural number.

I never said so.
Quote:  You should *not*
switch between representations during the process.

It is important for the discovery.
Quote:  Either you have a
list of natural numbers (in that case 0.111... does not belong to it)
or you have a list of rational numbers in decimal notation. The latter
is easier to reason about because it can be tackled easier. In that
case your list consists of:
An = sum{i = 1..n} 10^(n) = (1  10^(n))/9.
Also we can easily show what repeated *+ is (notated here as SUM):
SUM{i = 1 .. k} Ai = Ak.
Now we can also get a proper definition about what your notation ... means
in the list:
lim{k > oo} SUM{i = 1 .. k} Ak = 1/9 = K.
And K is (by convention) decimally notated as 0.111..., which makes sense,
in my opinion. K clearly is not equal to any of the An, unless there is
a natural number for which 10^(n) = 0.

Nevertheless we have the *+ sum 0.111... from
0.1
0.11
0.111
...
Now we can insert this in the list and we find that there is no change
in the *+ sum.
This is only possible, if it was in the list before already.
Quote: 
Here the last number of an infinite list is in this list.
It is not a list when you use the proper definition of list (a mapping
from N to the items of the list). An infinite list does not have a last
element. Because, suppose that list had a last element. By the definition
of list there should be a natural number n that maps to that element. But
n+1 maps to another element.

Then call
0.1
0.11
0.111
...
0.111...
an extended list or an ordered set or what you like. The proof should
not fail by lack of names.
Regards, WM 

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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 10:47 am Post subject:
Re: An uncountable countable set



Franziska Neugebauer schrieb:
Quote:  Dear Wolfgang,
I would appreciate if you not cut away that parts of my previous posts
you are trying to argue against. It may also be helpful if you read
before cutting.

Dear Franziska, I cut what I regard as unnecessary and not helpful in
order to make reading more comfortable. Your counting starting from
zero is superfluous and does not support your arguing.
Quote: 
2 The *sum of the sequence does not necessarily "inherit" the
properties from their individual sequence members. If this
"inherentance" is kind of a "general principle" you should refer to a
recognized source.
But there is even a finite counterexample:
101
011
110
 *sum
111
The property "does not contain at least one 0" is not inherited by
the *sum. So the "inheritance principle" is not so obvious.

But it is obvious for *all* *finite* *unary* number. And only finite
unary numbers are involved in the second list below:
The * sum of
0.1
0.11
0.111
....
0.111...
is
0.111...
i.e. the result is a number of the list.
The * sum of
0.1
0.11
0.111
....
is
0.111...
The result is allegedly not in the list.
This is a contradiction, because by tertium non datur either the first
case or the second must be true (if aleph_0 is a number with respect to
trichotomy with natural numbers).
Regards, WM 

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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 10:52 am Post subject:
Re: An uncountable countable set



Franziska Neugebauer schrieb:
Quote:  What is false unter the abovementioned presumption omega = { 0, 1,
2, ... }?
I am not willing to discuss this strange counting. 
0.1 means 1 and the first (1) digit after the point is a 1.
0.11 means 2 and the first (1) and the second (2) digit after the point
are 1's.
Quote: 
When it is not important "what I mean", why do you discuss with me? This
is rather insulting.

This remark concerned only the rather strange settheoretic counting.
Other opinions of yours are often welcome, as you know.
Regards, WM 

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W. Mueckenheim science forum Guru
Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 10:55 am Post subject:
Re: An uncountable countable set



Dik T. Winter wrote:
You again refrain from answering questions. The definition of 0.999...
is
as follows:
0.999... = lim{n > oo} sum{k = 1 .. n} 10^(k)
You said that definition is silly. What is silly about it?
_______________
WM:
Sorry, could you help me with "silly". My find function shows with
"silly" only the sentence: "It is only in modern set theory that such
silly things as the rejection of the binary tree occur."
_______________
Dik T. Winter wrote:
what indices in the limit above are undefined?
________________________
WM:
Those which cannot be enumerated by natural numbers.
________________
Dik T. Winter wrote:
Because, and I state it again, there is a difference between sequential
processes and simultaneous processes.
________________
WM:
I know that in the infinite binary tree *every* split is realized by
the presence of two edges and would be impossible without them.

o
/\
Nothing can be more simultaneous than this knowledge which is present
without considering any sequential process and which concerns the whole
tree.
The tree "in its length" is nothing else than a CantorList. The only
difference is that the tree guarantees the presence of every "diagonal
number".
_______________
Dik T. Winter wrote:
Four your count of edges and
paths you need limits, but those limits do not exist.
_________________
WM:
Oh, the limit of 1 + 1/2 + 1/4 + ... = 2 is not existing? Strange.
There is no question that every path can be interpreted as the
representation of a real number. And now "those limits do not exist"?
Very strange.
_____________
Dik T. Winter wrote:
Again, you need something like a limit here, because adding nodes is a
sequential process.
__________________
WM:
Adding lines in Cantor's list is not a sequential process? They are
given by the "higher Being which rules matheology". But adding levels
of the binary tree is a sequential process. Are you really in earnest
about that question? By the way: Notes are not added, they are there.
Take the nodes as digits. There is no question that every path can be
interpreted as the representation of a real number.
______________________
Dik T. Winter wrote:
Also, it is true for each natural that it cannot index all 1's of
0.1111.
That means that for all integers it is impossible to index all 1's of
0.1111. On the other hand, with the *set* if integers it is possible
to index all 1's of 0.1111, and also of 0.111... . That is, when with
"to index" we mean "point to an individual element" as in K[p]. The
reason is that each natural can index one and only one digit.
On the other hand, you on occasion use index to mean something
completely
different. For that I could use the word "cover" where A covers B if
in all positions where B has a 1, A also has a 1. Now in the finite
case we find that the digits of 0.1111 are covered by all naturals
greater than or equal to 4. But the digits of 0.111... are not covered
with any natural.
________________________
WM:
Why then do you expect that all digits are indexed "in the infinite
case"? The corresponding definition is of the same quality as "there
are 10 natural numbers between 0 and 1". There are only finite numbers.
________________________
Dik T. Winter wrote:
But nobody has stated that that is possible, it is
only you who *claims* that that should be possible, without giving any
reasonable proof of it.
________________________
WM:
There is no difference! If a natural indexes the digit n, then it
covers all digits m =< n. This is guaranteed by the construction of the
list of unary representations  for *all finite numbers*!!! How can you
believe you could make someone believe this situation would change?
Therefore I say sometimes each natural n can index all digits m =<
which it covers. Whether you call this to cover or to index is
unimportant. Both definitions imply each othe as long as *finite
numbers* are concerned
_____________________________
Quote:  It is true for each natural, that it cannot index all 1's of
0.111... . 
Dik T. Winter wrote:
Again, you mean cover here.
________________________
WM:
There is no difference between covering the digits 1 to n or to index
the nth digit, because of the construction of the naturals. AND WE
HAVE ONLY NATURALS.
___________________________
Dik T. Winter wrote:
I would like to ask you to refrain from
such confusing use of words. And that is indeed true. But that does
*not* mean that
For all p there is an n such that An[p] = K[p]
is false. That is true, take n = p.
_____________________
WM:
You imply that it is possible to take n = p, but that is impossible for
p being not a natural number like aleph_0.
__________________________________
Quote:  If stepwise exhaustion is impossible, your reordering could be
completed.
But you can with your *+ operation give the proof in one sweep,
but you
failed to answer to my question: "what happens at infinity?".
1) There is o infinity. Ever Transposition has a natural number.

Dik T. Winter wrote:
Sorry, that was loosely speaking. What happens if you add *all* of
them
together? You still fail to tell what happens in that case. (It is
quite easy to tell it, and you need a limit, but you are unwilling, or
perhaps unable to tell it.
___________________
WM:
I defined: The +* sum is 1 if the + sum is 1 or larger. This includes
the infinite case, or would you insist that infinitely many 1's are
less than one 1?
___________________
Dik T. Winter wrote:
When I ask you what the meaning of *+ ...
is you do not give a reasonable answer.) Answer:
for each digit position, when there is an Ap with a one in that
digit position, the identical digit position in K is also a 1.
And that answers the question. The result is:
for all p K[p] = 1
which we notate as K = 0.111..., because:
for all p there is an n such that An[p] = 1.
_____________________
WM:
Yes. And as there is no difference between indexing digit number n and
covering all digits m =< n, there is, according to your arguing, a
natural number which covers all digits of K.
____________________
Quote:  You can exhaust an infinite set if you take out all elements at
once.
You may think that would be possible, but it is not. You always need
the belief expressed by the three "..."

Dik T. Winter wrote:
I do not believe them, unless there is a proper definition or common
usage (that can be given a proper definition).
_____________
WM:
What to you doubt?
1) The tree does contain all binary representations of the reals
between 0 and 1?
2) The set of edges is countable?
3) The geometric series converges for q = 1/2?
_______________
Quote:  That is why the function f(n) = n + 1 is a bijective mapping from
{1, 2, ...} to {2, 3, ...},
sic!

Dik T. Winter wrote:
Yup, there is a common usage for that notation: in this notation it
means
"all subsequent successor included". And by the axiom of infinity
(there
is a set that also contains the successor of each element), the above
sets
do exist.
_________________________
WM:
By this very set all the levels of the tree do exist too.
________________
Dik T. Winter wrote:
In the case of 0.999... there are also proper definitions. And
If you had asked what I meant with "..." here, I would have defined it.
On
the other hand, when you write:
A1 *+ A2 *+ A3 *+ ...
and when I ask you what you *mean* with "..." you do refrain to give a
proper definition.
_______________
WM:
Definition: The *+ sum A1 *+ A2 *+ A3 *+ ... = 1 unless forall i : Ai
= 0, in which case the *+ sum is zero.
_________________
Dik T. Winter wrote:
Your transpositions are *conditional* transpostions. That is when you
write (a, b) you mean that the ath element and the bth element are
interchanged when in standard order the bth element is smaller than
the
ath element. I would like to notate that as R(a, b), as the notation
(a, b) in general means: interchange the ath element and the bth
element without looking at the magnitude. Now when I encounter in your
list of conditional transpositions at some place R(1, 2) I need to know
what previous conditional transpositions have done to be able to
determine
what this conditional transposition is going to do. You may weed out
from
your sequence of conditional transpositions those that do nothing. But
in
order to know whether you can weed out the nth transposition you need
to
know what earlier transpositions have done.
_________________
WM:
I don't have to know anything. This process is defined.
______________
Dik T. Winter wrote:
In all cases you are left with a sequential process.
___________________
WM:
What would it matter? Mathematics happens in a timeless environment.
The axiom of infinity guarantees the existence of all transpositions.
It is even defined by a sequential process: IF the set M contains x
THEN it contains x U {x}. Unless you advanced to x you cannot advance
to x U {x}. Further the sequential process is already implied in every
counting. If you can exhaust M, then you can exhaust the set of my
transpositions. Or put it in other words: If the set M does exist, then
the set of all my transpositions does exist too. If something can be
defined for the whole set simultaneously, then it can be defined for
the set of all my transpositions simultaneously.
___________________________________
Quote:  Your set of transpositions are different. They do not work at
once,
They do work at once. Once they are defined the result cannot be
changed.

Dik T. Winter wrote:
(1, 2)(2, 3) != (2, 3)(1, 2) (still assuming that the numbers are
indices).
So how can you state that they work at once?
______________
WM:
They are defined "at once" and lead to a welldefined result if applied
to a given wellordered set.
Mathematics happens in a timeless environment.
______________
Dik T. Winter wrote:
If I am given the set of
transpositions:
{(1, 2), (2, 3)}
what is the result when it is applied to the ordered set
{a, b, c}
is the ordered set
{b, c, a}
or the ordered set
{c, a, b}
? Doing things at once means that there are no order dependencies
between
the things involved.
______________
WM:
There is an order dependency in Cantor's list too. The nth digit
prevails the digit umber n+1. Mixing them up destroys the proof.
______________
Dik T. Winter wrote:
You have to define in what *sense* the sequence of transpositions
converge.
I think it is possible to give such a definition, others think not, but
let that go, that was not my main critique.
______________
WM:
It is easy to see that a subset which is in order by size can never
shrink. Once the element m < n precedes n by order it can never loose
this property.
______________
Dik T. Winter wrote:
My main critique was that you
have to show that a wellordered set remains wellordered in the limit.
There is not yet any proof forwarded, and a quote from Cantor that does
not contain a proof is not a proof.
____________________
WM:
I have not to prove anything concerning a limit. There is not the
faintest possibility that a wellorder could be destroyed by a
transposition which is enumerated by a natural number. Only such
transpositions do occur. If the set does not remain well ordered "in
the limit" then there is no limit.
Regards, WM 

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Franziska Neugebauer science forum addict
Joined: 23 Apr 2005
Posts: 52

Posted: Fri Jul 21, 2006 11:33 am Post subject:
Re: An uncountable countable set



mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:
Dear Wolfgang,
I would appreciate if you not cut away that parts of my previous
posts you are trying to argue against. It may also be helpful if you
read before cutting.
Dear Franziska, I cut what I regard as unnecessary and not helpful in
order to make reading more comfortable.

lame excuse
Quote:  Your counting starting from zero is superfluous and does not support
your arguing.

In set theory one usually counts from zero. As pointed out it is a quite
simple task to renumber the naturals if you want to insist on counting
from 1.
Quote:  2 The *sum of the sequence does not necessarily "inherit" the
properties from their individual sequence members. If this
"inherentance" is kind of a "general principle" you should refer to a
recognized source.
But there is even a finite counterexample:
101
011
110
 *sum
111
The property "does not contain at least one 0" is not inherited by
the *sum. So the "inheritance principle" is not so obvious.
But it is obvious for *all* *finite* *unary* number.

Can't see that. If it is so obvious then please show (prove!) it.
I consider my 3item list a valid counterexample to an "inheritance
principle".
Quote:  And only finite unary numbers are involved in the second list below:
The * sum of

The following notation I will call "A".
Quote:  0.1
0.11
0.111
...
0.111...
is
0.111...
i.e. the result is a number of the list.

You want to put 0.111... into the sequence. On which position j e N?
What does the sequence'...' in your context
(0.1; 0.11; 0.111; ...; 0.111...)
^^^
mean?
I will call this "B".
Quote:  0.1
0.11
0.111
...
is
0.111...
The result is allegedly not in the list.

We did not put it into the list, so it's not there.
Quote:  This is a contradiction, because by tertium non datur either the first
case or the second must be true (if aleph_0 is a number with respect
to trichotomy with natural numbers).

First of all you produced two notations
A def= (0.1; 0.11; 0.111; ...; 0.111...)
and
B def= (0.1; 0.11; 0.111; ...)
Notation A is (not yet) defined since the first occurence of '...' has
(not yet) a defined meaning. Fomally different is notation B which
simply denotes the original "list".
*Assumed* that A and B denote different sequences the proposition
0.111... is in *A* & 0.111... is not in *B*
is not a contradiction and does not violate the tertium non datur
principle. A contradiction is a proposition of the form
p & ~p
Hence you have (not yet) shown a contradiction as has been pointed out
all too often.
F. N.

xyz 

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