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W. Mueckenheim
science forum Guru

Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 12:02 pm    Post subject: Re: An uncountable countable set

Dik T. Winter schrieb:

 Quote: You cannot exhaust the naturals. Therefore what you write as 0.111... is not the same as the unary representation which also is denoted by 0.111... but means that N is exhausted and there is at least one 1 which cannot be indexed by a natural number. Makes no sense to me. I see no exhaustion, see, I did use the symbol "lim" there, and I had the hope that there would be a proper definition. Apparently not.

No. You intermingle two different things, all n e N and infinity.
Again: For any *finite* natural n we have a logical equivalence, even
better an inclusion: indexing the n-th digit includes covering at least
all digits from 1 to n.
If each digit of 0.111... is indexed by one natural, then each segment
of 0.111... is covered by at least one natural too.
As long as only natural numbers are concerned there is no outcome.

 Quote: The reason is that 0.1 0.11 0.111 ... gives allegedly the same sum as 0.111... (as representation of aleph_0) 0.1 0.11 0.111 ... Yes, with a proper definition of "...".

But that is not acceptable if aleph_0 is a number.

 Quote: What was the reason again?

If aleph_0 is a number larger than any natural, then it cannot be the
*+ sum of a list containing only naturals excluding itself and
simultaneously the sum of a list including itself.

 Quote: You apparently have another definition in mind. Pray give *your* definition.

1) If "..." means "forall" n, e N then 0.111... is not different from
all n. (Because for naturals indexing includes covering.) But we know
that 0.111... is not ín the list of naturals. Contradiction.
2) If "..." means --> oo including aleph_0 as an index, then the digit
with this index is not defined.

 Quote: But what you are now meaning is that lim{n -> oo} SUM{i = 1 .. n} An = 0.111... . I said above: It is different from n really reaching infinity. But for 1/9 or aleph_0 infinity is reached. With a proper definition of the limit, the sum will *never* reach 1/9 (or aleph_0). But as you still refrain from giving definitions of the notations you are using this discussion is less than fruitfull. Bijection: Every position of a decimal number can be indexed by a

natural number. The set of naturals has cardinality aleph_0. The set of
positions has cardinality aleph_0.

 Quote: I ask questions. Again you refrain from giving answers. Are you unwilling to give answers? I ask questions too which you do not answer. The most important one:

How can a finite natural number index digit number n but not cover all
digits from 1 to n?

 Quote: 0.1 0.11 0.111 ... As my interpretation of your "..." was apparently wrong above, pray, finally supply a definition of that notation.

See above. 1) and 2).

 Quote: How many digits has 1/9 in decimal representation? That is not a natural number.

No, but the set of digits has a cardinality, namely aleph_0.

 Quote: There is no such column. And 0.111... is not a natural.

That is a contradiction. And I told you why:

You said: all digits of 0.111... could be indexed by natural numbers,
but not all could be covered by natural numbers. This is wrong as one
can easily prove: Every natural which indexes a digit covers all digits
up to that one. If all digits of 0.111... can be indexed than all
digits up to every (and that is all digits) can be covered.

What is in *your*
 Quote: opinion the result of *+-ing all natural numbers togther?

As we have seen it is simply impossible to *+ sum all naturals. Every
result yields a contradiction. The set of all naturals does not exist.

Regards, WM
W. Mueckenheim
science forum Guru

Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 11:51 am    Post subject: Re: An uncountable countable set

Dik T. Winter schrieb:

 Quote: The "first" successor. Tere is only one "first" successor! Conveniently snipping your own text again? You wrote in "news:1153052131.852951.273540@b28g2000cwb.googlegroups.com">: Hence, 0.111... as the succesor of all naturals must consist of more 1's, than any natural, if it is to be a number. Now, if you had written "the first successor" I would not have made any remarks.

You would have done better.

 Quote: It is not *the* successor.

That depends on the definition of "successor". 3 is the first successor
of 2 if you consider 4 as another successor. It is usual, however, to
define only the first successor at al. Only so we get the completely
defined succession as required by Cantor. He says "völlig bestimmte
Sukzession" and " zu jedem Elemente E' - mit Ausnahme des letzten, wenn
ein letztes vorhanden ist - ein ihm _nächst_ folgendes E'' vorhanden
ist."(italics by Cantor).Therefore omega ist "das Nächstfolgende",
*the* successor.
 Quote: Are you again arguing that the statement For all p, there is an n such that An[p] = K[p] That statement is false. (Without K including the index p = aleph_0, k cannt be different from all natural numbers.) What has your statement in parentheses to do with the statement I gave? There is no argument at all about K being different from all natural numbers. It is just a plain statement: For all p, there is an n such that An[p] = K[p] you claim that is false, and again without direct proof. Do you disagree with the statement: For all p, Ap[p] = K[p]? I know that Ap[p] = 1 = K[p]. Now chose in the above n = p and you get the new statement. Why would this second statement be false?

You said "for all p". If p is a natural number, then the segment K[p]
is the unary representation of a natural number and, hence, is covered
by a natural number. Remember:

For each natural (i.e. finite) number p we know: to index and to cover
are equivalent.
p covers K[p] <==> p indexes p.

Would K have only natural indexes p, then it would be covered by at
least one natural number.

If you deny then please give an example of a *finite* natural number
which indexes the n-th digit but does not cover the digits number 1 to
n.

Regards, WM
W. Mueckenheim
science forum Guru

Joined: 23 Apr 2005
Posts: 934

Posted: Fri Jul 21, 2006 11:42 am    Post subject: Re: An uncountable countable set

Dik T. Winter schrieb:

 Quote: What your example shows is that an infinite sequence of transpositions can destroy well-orderedness. What is the difference? The only conclusion is that Cantor's statement is incorrect or that the reading of Cantor's statement is incorrect. Take your pick.

Your hinting at any doubt about the clear meaning of Cantor's statement
is outrageous.

 Quote: I have found already a few instances where Cantor's statements are in conflict with modern set theory.

There are many. See for instance for a very simple one: "On Cantor's
proof of continuity-preserving manifolds" on
http://www.fh-augsburg.de/~mueckenh/

 Quote: And I am not surprised. He found the building blocks but was not entirely sure about the way to go. His articles were research in progress, as it happens.

And now it happes that all his way turns out as a deadlock.
 Quote: And now for a debunking of a myth. Nowhere (at least I could not find any place) has Cantor used the diagonal argument to show that the reals are not countable. His proof about the reals shows that a complete, densely ordered set is not countable. That one does not use the diagonal argument at all.

So far we agreed recently.

 Quote: His diagonal proof shows that the set of infinite sequences of two symbols is not countable, and as an extension, that the powerset of a set has cardinality strictly larger than the cardinality of the original set. So all your arguments about 0.999... = 1.000... are *not* directed against Cantor.

Wrong. My arguing is directed against the complete existence of the set
of all naturals, the set of all digits of a real number, the actual
infinity, the first transfinite number. That all is purest Cantor.

Regards, WM
Franziska Neugebauer

Joined: 23 Apr 2005
Posts: 52

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

 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 3-item 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?

 Quote: The *- sum of

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
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 Cantor-List. 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 n-th 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 *+ ...
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 a-th element and the b-th element are
interchanged when in standard order the b-th element is smaller than
the
a-th element. I would like to notate that as R(a, b), as the notation
(a, b) in general means: interchange the a-th element and the b-th
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 n-th 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 well-defined result if applied
to a given well-ordered 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 n-th 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 well-ordered set remains well-ordered 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 well-order 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
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 set-theoretic counting.
Other opinions of yours are often welcome, as you know.

Regards, WM
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
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
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.fh-augsburg.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 non-terminating 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
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
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:

 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?

axioms, use definitions and derive conclusions. When you can derive,
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/
Dik T. Winter
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Joined: 25 Mar 2005
Posts: 1359

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

 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 non-terminating 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
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/
Virgil
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Joined: 24 Mar 2005
Posts: 5536

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

mueckenh@rz.fh-augsburg.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
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

mueckenh@rz.fh-augsburg.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.
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

mueckenh@rz.fh-augsburg.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 non-terminating 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})
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

mueckenh@rz.fh-augsburg.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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