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

Posted: Tue Jul 18, 2006 6:41 pm Post subject:
Re: An uncountable countable set



In article <1153222834.202538.101020@h48g2000cwc.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Virgil schrieb:
WM's version of logic is quite different from the sort used in ZF of NBG
or any other part of mathematics.
No the logic used in mathematics is the same as mine.

That is WM's perpeetual delusion.
Quote:  It is only in
modern set theory that such silly things as the rejection of the binary
tree occur.

It is WM's sillyness which rejects the infinite binary tree. 

Back to top 


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

Posted: Tue Jul 18, 2006 6:47 pm Post subject:
Re: An uncountable countable set



In article <1153223055.199107.315230@m79g2000cwm.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Virgil schrieb:
The difference is that our "traditions and folklore", which we chose to
call axioms and definitions, are logically consistent,
Only if you decide *arbitrarily* which infinite set can be exhausted
and which cannot.
Regards, WM

On the contrary. Whether one can "exhaust"( deal with every member of )
such a set depends on one's methodology.
If one attempts to exhaust ( deal with every member of ) any infinite
set by operating on its members one at a time, it is not possible.
If one attempts to exhaust ( deal with every member of ) any infinite
set by simultaneously dealing with all its members, it is quite easy.
At least for everyone except WM. 

Back to top 


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

Posted: Tue Jul 18, 2006 6:48 pm Post subject:
Re: An uncountable countable set



In article <1153223462.224954.305750@m73g2000cwd.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Virgil schrieb:
In article <1153148551.942037.97110@s13g2000cwa.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
This is the deep dilemma of
set theory: There is no actually infinite set of finite numbers.
But the existence of this "dilemma" can only be established by assuming
it.
So for those who do not chose to assume it, it does not exist.
Those who choose to close their eyes enjoy always the mercy of not
being forced to see the sheer misery of mathematics.

If it makes you so miserable why do you keep doing it?
You remind one of the small boy who kept hitting himself on the head
with a hammer.
Because it felt so good when he stopped. 

Back to top 


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

Posted: Tue Jul 18, 2006 6:51 pm Post subject:
Re: An uncountable countable set



In article <1153224254.540565.150600@s13g2000cwa.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:
aleph_0 def=  omega 
0.111... =def (a_i) having a_i = 1 A i e omega
a_ij means the well known matrix of figures
IF aleph_0 does exist, THEN 0.111... covers aleph_0 columns.
This is as meaningful as
If i exists then sqrt(1) is i.
IF 0.111... covers aleph_0 columns, THEN aleph_0 columns do exist.
This is as meaningful as
If sqrt(1) is i then i exists.
IF aleph_0 columns do exist THEN we can consider their contents.
This is as meaningful as
If i exists then we can consider its value.
IF we can consider the contents of each column, THEN we can ask how
many 1's are therein.
Lotta questions.
Unknown word.
IF we can ask how many 1's are in each one, THEN the answer can be
"zero 1's" or "not zero 1's".
We can.
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.
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.

Not unless there IS a largest number in the list. An in your list of
terminating strings there is no such thing. 

Back to top 


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

Posted: Tue Jul 18, 2006 6:59 pm Post subject:
Re: An uncountable countable set



In article <1153224623.508718.110530@i42g2000cwa.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Virgil schrieb:
In article <1153168957.805313.57460@p79g2000cwp.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Dik T. Winter schrieb:
Now you use an entirely new term: "can be exhausted". I think you mean
that you can take out elements one by one and doing this at some stage
the infinite set becomes empty. Howver, I think, that if that can be
done, that there is a last element you can take out. And, according
to the axiom of infinity, that is not possible, so infinite sets can
not be exhausted in this sense.
But in another sense?
In the sense of having a set of all of them, as per the axiom of
infinity, an axiom does it.
The axiom is not responsible for the wellorder of *all* elements of
the set, because wellorder is a stepwise process, you see?

What I see is that well ordering is a property of a set as a whole and
not a property of tis members separately from the set.
A well ordering of a set requires the existence of an order relation on
the set >as a whole< with the special property that the induced order
on every nonempty subset identifies a first element of that nonempty
subset.
Any total ordering of a finite set is automatically a well ordering, but
not so for infinite sets. 

Back to top 


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

Posted: Tue Jul 18, 2006 7:04 pm Post subject:
Re: An uncountable countable set



In article <1153225087.356970.296750@h48g2000cwc.googlegroups.com>,
mueckenh@rz.fhaugsburg.de wrote:
Quote:  Dik T. Winter schrieb:
In article <1153145345.281803.129520@35g2000cwc.googlegroups.com
mueckenh@rz.fhaugsburg.de writes:
Franziska Neugebauer schrieb:
mueckenh@rz.fhaugsburg.de wrote:
...
Indeed. I recall. Therefore the linearity of the list numbers
enforces
a column with zeros.
Nice try, but non sequitur. There is no such column. Otherwise show
one
or prove its existence.
The proof requires logic. Therefore I am afraid you will not accept it.
It reads: Either there is a column with only zeros, or there is at
least one 1 in each column spanned by the digit positions of 0.111... ,
isn't it?
Yes, right. And now the remainder of the proof, please?
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. This is so whether you pretend it would not
be so or not.

For every finite list of finite strings of WM's, the so called "*+ sum"
is the last member, so that "*+" is equivalent to "last of".
But what if there is no "last of"? WM still says there still is one even
when there isn't one. 

Back to top 


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

Posted: Wed Jul 19, 2006 12:04 am Post subject:
Re: An uncountable countable set



In article <1153242093.641744.4400@b28g2000cwb.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Quote:  Dik T. Winter schrieb:
Sorry, we are now talking about 0.999..., please remain with the argument.
You said that the definition is silly. Why is the definition above silly?

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?
Quote:  And that there are undefined digits in irrational numbers does not bother
me in the least. I have no idea why you have a problem with that.
In this case not the digits are undefined but the indices are
undefined. See the discussion of the *+ sum of my list of unary
numbers.

what indices in the limit above are undefined?
Quote:  But to go on:
And therefore we have *always* undefined digits in any
irrational number and in the diagonal of Cantor's list. There are
*always* most of its digits unknown, i.e., always there are more digits
unknown than are known. No matter how small an epsilon you select.
For Cantor's diagonal no epsilon is needed, neither is their for irrational
numbers. The only thing we need to know is that the digits are all in the
range [0, 9],
The only thing we need to know for the binary tree is that no path can
split without two additional edges. Why don't you use this kind of
abstract mathematics in that case too?

Because, and I state it again, there is a difference between sequential
processes and simultaneous processes. Four your count of edges and
paths you need limits, but those limits do not exist.
Quote:  Hence you cannot prove that the digits of the diagonal are all
different from those of the line numbers.
Why not? For the proof no exhaustion is needed. Just like (in another
thread) the proposition
For the binary tree no exhaustion is requird either. It suffices to see
that

o
/\
is the ever repeating element which cannot be outwitted.

Again, you need something like a limit here, because adding nodes is a
sequential process.
Quote:  For all p there is an n such that An[p] = K[p]
does not need exhaustion. In the case of the list and the diagonal the
similar statement is
For all p there is an n such that An[p] != D[p]
proving that D is different from all An. BTW, in the first case also the
proposition
For all p there is an n such that An[p] != K[p]
proving that K is also different from all An.
In the first and second case take n = p, in the third case take n = p + 1.
This is *not* a proof by exhaustion. It is simply stating a fact, with
an easy proof.
You can prove it for each one
but you cannot prove it for all.
If something is true for each one, it is true for all.
It is true for each natural, that it cannot index all 1's of 0.111... .

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. 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.
Quote:  Let's reason
with the excluded middle (because that would complicate matters). Assume
it is not true for all, then there must be some for which it is not
true (say z). On the other hand, it is true for each one, so also
true for z. A contradiction, hence the assumption is false. And in
the cases above you can proof it for each one in one sweep, because
you prove it for an arbitrary element.
It is true for each natural, that it cannot index all 1's of 0.111... .

Again, you mean cover here. 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.
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.

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. 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.
Quote:  2) The same happens as with Cantor's diagonal. There is absolutely no
difference.

And indeed, if I write it like I did above, there is absolutely no difference.
K is properly defined, and it is easily shown that it is 0.111..., and in the
same way the diagonal is properly defined.
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 expressen by the three "..."

I do not believe them, unless there is a proper definition or common
usage (that can be given a proper definition).
Quote:  That is why the function f(n) = n + 1 is a bijective mapping from
{1, 2, ...} to {2, 3, ...},
sic!

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. 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.
Quote:  every definition works at once, you need
not to count to 100 to know what f(100) is. And the same is the case
with Cantor's diagonal number. You need not look at the first 100
elements to find the 101st element to know that the 101st digit
of the diagonal is. That is what the definition of a list is. A
mapping from N to the members of the list.
You need not look at the transposition number 100 to see what it does.

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. In all cases you are left
with a sequential process.
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.

(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? 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.
Quote:  you
need sequencing (transpositions are in general noncommutative). In the
case of a sequence you need limiting procedures to show what would be
the "final" result (like lim{n > oo} 1/n = 0). But the problem with
limiting procedures is that the "final" result has not necessarily the
same properties as each and every finite result.
Airyfairy. My transpositions converge, i.e., a partial order by size
can never become worse but can always only become better. I think there
is no proof equired to see that.

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. 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.
Quote: 
Let me give an example with transpositions. Let's say that (a, b)
means that we interchange the ath and bth element in an ordering.
Let us start with the ordered set of natural numbers {1, 2, 3, ...}
(yes, not Bourbaki this time). Let's define a sequence of transpositions:
(1, 2)(2, 3)(3, 4)(4, 5)...
with a proper measure and a proper definition of limit, I think that we
can show that that leads to the ordered set {2, 3, ..., 1}. Now
interchange the first two elements of the transpositions, in this case
we get {1, 3, ..., 2}. So sequencing plays a crucial role.
But that is completely uninteresting, because my transpositions do
never lead off an ordering by size. What is achieved will and can never
again be destroyed. Your example fails.

Yes, your example can possibly (with proper definitions; note my text: "with
a proper measure and a proper definition of limit, I think...") been shown to
converge to the set of rationals in standard ordering. What *my* example
shows is that an infinite sequence of transpositions can destroy the
ordinal number of a set (contradicting Cantor's statement). What your
example shows is that an infinite sequence of transpositions can destroy
wellorderedness. 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. I have found already a few instances where
Cantor's statements are in conflict with modern set theory. 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 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. 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.

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

Back to top 


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

Posted: Wed Jul 19, 2006 12:23 am Post subject:
Re: An uncountable countable set



In article <1153242462.523960.281130@p79g2000cwp.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Quote:  Dik T. Winter schrieb:
In article <1153148551.942037.97110@s13g2000cwa.googlegroups.com> muecken=
h@rz.fhaugsburg.de writes:
Dik T. Winter schrieb:
...
(1): 0.111... is not *the* successor of anything, we may sloppily
say that it is *a* successor of all naturals, just like 10 is
*a* successor of 2
It was Cantor who coined this expression saying " daß omega die erste
ganze Zahl sein soll, welche auf alle Zahlen nu folgt." (Works, p.
195) So it must be larger anyhow.
Yes. Not *the* successor, but *a* successor.
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">:
Quote:  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. It is not *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?
Quote:  is false, or are you arguing that the statement
For all p, there is an n such that An[p] != K[p]
is false?
That statement in the form An != K is correct. The seqence An[p] does
not exist for all p which are in K.

Again, you do not answer the question. Do you disagree with the statement
For all p, Ap1[p] != K[p]?
I know that Ap1[p] = 0 != 1 = K[p] (I see now that p>1 is a requirement).
Now chose in the above n=p1 and you get this statement. What is false
about this statement?
Quote:  There is no actually infinite set of finite numbers.
Axiom of infinity.
In contradiction with mathematics, obviously.

You have not yet shown it. Only in contradiction with *your view* of
mathematics. That is something entirely different.

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

Back to top 


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

Posted: Wed Jul 19, 2006 12:49 am Post subject:
Re: An uncountable countable set



In article <1153243255.293378.172400@i42g2000cwa.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Quote:  Dik T. Winter schrieb:
In article <1153169497.380974.32190@i42g2000cwa.googlegroups.com> mueckenh@rz.fhaugsburg.de writes:
Dik T. Winter schrieb:
I don't like definitions which define nonsense like the corner of a
circle.
Check the manhattanmeasure and you will find square circles.
Then I'll better do without.

Yes, you dislike definitions that do not conform with your views. You are
not interested in the possible usability.
Quote:  That is not an answer to my question. But I will take it in good faith,
so
lim{n > oo} SUM{i = 1 .. n} A_i = 0.111...
tell me where I have gone wrong.
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.
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 "...". What was the reason again?
You apparently have another definition in mind. Pray give *your*
definition.
Quote:  Define: If any case includes at least one 1 the the *+ sum is 1.
Again, in the finite case. You have not defined what you mean with the
infinite sum.
Hell and devil! Can't you read? The definition for *+ sum = 1 is: at
least one 1 must be encountered. That is enough in any case, finite or
not.
Damnation leaving aside, you defined what SUM{i = 1 .. n} An is. You did
*not* define what happens when n grows without bound. Now you state that
in each column, whenever there is a 1, the final result should be 1.
Here is another example for the *+ sum:

I do not ask for examples, I ask for definitions. And those you refuse to
supply. Now clearly above I did think about definitions, and used them
and you state I was wrong. Without even giving the correct definitions.
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.
Quote:  So the statement
For all n, An[n] = K[n]
is true? As is the statement
For all p there is an n such that An[p] = K[p]
also true?
No. Then 0.111... would not differ from every n.
No to which question? Is the first statement false? And if so, why?
Is the second question false, and if so why? And how do you come at
your conclusion?
The reason is that

I ask questions. Again you refrain from giving answers. Are you unwilling
to give answers?
Quote:  0.1
0.11
0.111
...

As my interpretation of your "..." was apparently wrong above, pray, finally
supply a definition of that notation. If you do not supply that there is
no reason to discuss any further, as you are not willing to provide enough
information about what you are writing.
Quote:  You are arguing on two different lines at the same time, confusing one
with the other. When we consider the An as unary representations of
natural numbers the result is aleph_0, not a natural number. When we
consider them as being decimal numbers, the result is 1/9.
How many digits has 1/9 in decimal representation?

That is not a natural number.
Quote:  0.1
0.11
0.111
...
is *not* 0.111...
With your definition (finally extracted) above (if in any column there is
a 1, there is a 1 in the final result), it is. If you think that is false,
please show me a column where there are only 0's.
That column which cannot be covered by a natural number.
(f all couldcovered then 0.111... was a natural.)

There is no such column. And 0.111... is not a natural. What is in *your*
opinion the result of *+ing all natural numbers togther?

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

Back to top 


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

Posted: Wed Jul 19, 2006 9:27 am Post subject:
Re: An uncountable countable set



Franziska Neugebauer schrieb:
Quote:  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.
Quote: 
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.
Quote: 
++
 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.
Quote: 
summing up:
a_i = 1 if i < n (1a)
a_i = 0 if n <= i (1b)

false.
Quote: 
A tabulation may help you to understand what I mean:

It is not important what you mean, but what is. xxx is a set of three
x, whatever you may mean.
Quote: 
With (1a) and (1b) it is even not necessary to define unary(omega)
specially.
This is the deep dilemma of set theory: There is no actually
infinite set of finite numbers.
Non sequitur. Ever considered _your_ representation theory broken?
If you insist that 0.111 represents 4,
I never did and will not.

See above.
Quote:  Please explain in detail what the *problem* is. I can't see any problem.

That is the problem!
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 is not different
from the sum excluding it. Therefore all your corresponding assertions
break down.
Regards, WM 

Back to top 


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

Posted: Wed Jul 19, 2006 9:38 am Post subject:
Re: An uncountable countable set



Dik T. Winter schrieb:
Quote:  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.
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. Actual
infinity or, in other words, finished infinity is the necessary
consequence of the settheoretical theorem that infinity can be
surpassed.
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. The assumption
is false that irrational numbers and non terminating fractions have as
many digits as there are natural numbers *and* that this is a number
aleph_0.
Regards, WM 

Back to top 


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

Posted: Wed Jul 19, 2006 9:49 am Post subject:
Re: An uncountable countable set



Virgil schrieb:
Quote:  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. And if it is a number, then it must e lager or equal or
less than a natural number. As he latter is impossible, it can only be
larger.
Quote: 
There is no number following all naturals and, hence, there is no
number omega at all.
Perhaps not in your philosophy, but you are not God, to command what is
true or false.

But I can see what is contradictory.
Quote: 
There is no such 'natural' but there is such a cardinal or ordinal. Both
are generically 'numbers' not all cardinals or ordinals are naturals.
Why do you emphasize this? Of course it cannot be a natural if it
follows after all naturals.
I emphasize it because WM needs to learn it.

You are a liar. Therefore I cease my correpondence with you.
Last regards, WM 

Back to top 


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

Posted: Wed Jul 19, 2006 10:18 am Post subject:
Re: An uncountable countable set



mueckenh@rz.fhaugsburg.de wrote:
Quote:  Franziska Neugebauer schrieb:
[...]
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!) 
 ++
`
Quote:  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, ... }?
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".
Quote:  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.
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?
Quote:  With (1a) and (1b) it is even not necessary to define unary(omega)
specially.
This is the deep dilemma of set theory: There is no actually
infinite set of finite numbers.
Non sequitur. Ever considered _your_ representation theory broken?
If you insist that 0.111 represents 4,
I never did and will not.
See above.

I suspect you are desperately trying to play the game "how many of FN's
text must I delete to get her wrong".
Quote:  Please explain in detail what the *problem* is. I can't see any
problem.
That is the problem!

Posit your claim clearly then there will be no problem.
context restored:
,[ <44bcd393$0$17968$892e7fe2@authen.yellow.readfreenews.net> ]
 > This is a contradiction,

 Once again: I can't see a "problem" or a "contradiction". Would you
 like to explain, what exactly is the problem?

 > and it is solved only by the observation that the *+ sum of all
 > naturals is not omega,

 You are trying to "solve" a yet nonexisting "problem". BTW: This can
 not be "ovserved" since every "observation" of the present *sum of
 all unaries of naturals is obviously representing omega. You confuse
 observation by hallucination.

 > because a *+ sum of finite numbers alone cannot result in an
 > infinite number.
`
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)?
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!
Quote:  Therefore all your corresponding assertions break down.

Wishful thinking.
F. N.

xyz 

Back to top 


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

Posted: Wed Jul 19, 2006 1:16 pm Post subject:
Re: An uncountable countable set



Franziska Neugebauer schrieb:
Quote:  mueckenh@rz.fhaugsburg.de wrote:
Franziska Neugebauer schrieb:
[...]
IF we can ask how many 1's are in each one, THEN the answer can be
"zero 1's" or "not zero 1's".
We can.
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.
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.
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.
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. Its existence is proven in
set theory.
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.
Regards, WM 

Back to top 


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

Posted: Wed Jul 19, 2006 1:30 pm Post subject:
Re: An uncountable countable set



Franziska Neugebauer schrieb:
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.
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.
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 because the largest number of he list is
finite, although we don't know it, we know that it is not infinite. You
always mix up infinitely many with infinitely large. Look here: The *+
sum 1 *+ 1 *+ 1 *+ ... = 1. This sum does not depend on "how many"but
only on "how large". Therefore it demasks your wishful thinking
infinitely many finite numbers could give an infinite *+ sum.
Regards, WM 

Back to top 


Google


Back to top 



The time now is Sun Jan 20, 2019 3:41 pm  All times are GMT

