infinity ...
albs...@gmx.de
(Hint: The sketches would not looks good in proportional fond)
Let's start with a representation of the natural numbers in unitary
(1-adic) system as follows:
O O O O O O O O O ...
O O O O O O O O ...
O O O O O O O ...
O O O O O O ...
O O O O O ...
O O O O ...
O O O ...
O O ...
O ...
.
.
.
1 2 3 4 5 6 7 8 9 ...
Each vertical row shows a natural number. Horizontally, it is the
sequence of the natural numbers. Since the Os, the elements, are local
distinguished from each other, we can also look at the rows as sets. A
set may contain the coordinates of the elements as their representation
or may look like this: S3 = {O1, O2, O3} e.g. .
>From the Peano axiomes follows that the set of all naturals is
infinite. So, the set of the elements of the first horizontal row is
infinite. Actually the set of the elements of every horizontal row is
infinite. And the set of all the elements in this representation is
also infinite.
But there is no vertical row with infinite many elements since there is
no infinite natural.
Now let's fill the horizontal rows or lines with other symbols. We have
to take into account that only lines should be filled with #s which
containes Os.
# O O O O O O O O O ... 1
# # O O O O O O O O ... 2
# # # O O O O O O O ... 3
# # # # O O O O O O ... 4
# # # # # O O O O O ... 5
# # # # # # O O O O ... 6
# # # # # # # O O O ... 7
# # # # # # # # O O ... 8
# # # # # # # # # O ... 9
. . . . . . . . . . .
. . . . . . . . . . .
. . . . . . . . . . .
The vertical sequence of sets of #s fullfill the peano axiomes exactly
as the horizontal sequence of the sets of Os does.
But there is a slight difference. Since there is no infinite natural in
form of a set of Os and since after every set of #s there should be a
O, the size of the set of the naturals as sets of #s could not extend
the "biggest" number of the naturals in form of sets of Os.
Since there is no biggest number and since there is no infinite number,
the size of the set of numbers in form of sets of #s is undefined as
the biggest natural number is undefined.
But the sequence of the sets of # fullfill the peano axiomes. So this
set must be infinite.
The cardinality of a set is not able to be infinite and "not defined"
at the same time.
This is the contradiction.
Or let's say it in another form: The first vertical row of #s could not
exceed the biggest vertical row of Os (and could not be smaller). So,
the cardinality of this set is undefined like the biggest natural
number. But the set of the elements of the first vertical row of #s has
the same cardinality like the set of the natural numbers.
--> Contradiction.
Or did I construct a monster set which cardinality is subtransfinite?
Comments?
Best regards
Albrecht S. Storz, Germany
ste...@nomail.com
> But there is a slight difference. Since there is no infinite natural in
> form of a set of Os and since after every set of #s there should be a
> O, the size of the set of the naturals as sets of #s could not extend
> the "biggest" number of the naturals in form of sets of Os.
> Since there is no biggest number and since there is no infinite number,
> the size of the set of numbers in form of sets of #s is undefined as
> the biggest natural number is undefined.
Whoever said the size of a set has anything to do with the "biggest"
element?
> But the sequence of the sets of # fullfill the peano axiomes. So this
> set must be infinite.
> The cardinality of a set is not able to be infinite and "not defined"
> at the same time.
> This is the contradiction.
No, the contradiction is assuming that cardinality has
anything to do with the "biggest" element. Cardinality
is not defined in terms of the largest element. It
is defined in terms of bijections. Your post says
nothing about bijections, so it says nothing about
cardinality.
Stephen
Tony Orlow
I think your diagram is very nice, and your point pretty clear. That is a good
graphic illustration of the equality between element value and element count
for the natural numbers. It would seem very hard to argue that the array with
its diagonal is somehow longer than it is wide, using this unary notation. I
believe that you have constructed a representation of a set which is
transfinite, but not infinite, unless the strings of 0's and #'s are allowed to
become infinite in both directions. Good job! Danke!
>
>
> Best regards
>
> Albrecht S. Storz, Germany
>
>
--
Smiles,
Tony
Tony Orlow
> albs...@gmx.de wrote:
>
> > But there is a slight difference. Since there is no infinite natural in
> > form of a set of Os and since after every set of #s there should be a
> > O, the size of the set of the naturals as sets of #s could not extend
> > the "biggest" number of the naturals in form of sets of Os.
> > Since there is no biggest number and since there is no infinite number,
> > the size of the set of numbers in form of sets of #s is undefined as
> > the biggest natural number is undefined.
>
> Whoever said the size of a set has anything to do with the "biggest"
> element?
the width of the square and the height are the same. Do you not see that the
width is the count of naturals and the height is the value? The picture said
so, that's who.
>
> > But the sequence of the sets of # fullfill the peano axiomes. So this
> > set must be infinite.
>
> > The cardinality of a set is not able to be infinite and "not defined"
> > at the same time.
>
> > This is the contradiction.
>
> No, the contradiction is assuming that cardinality has
> anything to do with the "biggest" element. Cardinality
> is not defined in terms of the largest element. It
> is defined in terms of bijections. Your post says
> nothing about bijections, so it says nothing about
> cardinality.
values in the set. Can the square be wider than it is tall? No? Well, then,
what does that say to you? (probably nothing, of course)
>
> Stephen
>
>
--
Smiles,
Tony
ste...@nomail.com
> ste...@nomail.com said:
>> albs...@gmx.de wrote:
>>
>> > But there is a slight difference. Since there is no infinite natural in
>> > form of a set of Os and since after every set of #s there should be a
>> > O, the size of the set of the naturals as sets of #s could not extend
>> > the "biggest" number of the naturals in form of sets of Os.
>> > Since there is no biggest number and since there is no infinite number,
>> > the size of the set of numbers in form of sets of #s is undefined as
>> > the biggest natural number is undefined.
>>
>> Whoever said the size of a set has anything to do with the "biggest"
>> element?
> Stephen, did you even look at the diagrams he presented? Do you not see that
> the width of the square and the height are the same. Do you not see that the
> width is the count of naturals and the height is the value? The picture said
> so, that's who.
What square? The sides of a square are line segments.
The four corners of the square are defined by the ends
of those line segments. If your lines extend indefinitely,
then there is no square.
This is not a square:
+-----------.....
|
|
.
.
.
A square has four corners. This only has one "corner".
Remember, infinite lines do not end. Not even "at infinity".
<snip>
> It says everything about the size of the set of naturals compared with the
> values in the set. Can the square be wider than it is tall? No? Well, then,
> what does that say to you? (probably nothing, of course)
As there is no square in the first place, it is irrelevant
if a square can be wider than it is tall.
Stephen
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> Can the square be wider than it is tall?
TO seems to be a square much wider than he is tall.
albs...@gmx.de
The idea results from the understanding, that every number is a set. A
number is the unchanged aspect of a simultanity of endlessly many
objects which only and absolutly only has common aspects in the number
of their elements.
That's the only, or one possible definition of a number.
The similarity of my sketches with the usual representation of the
Cantor diagonal argument is not an accidently happend effect.
We could interpret the struktures in the sketches alternatively. In one
sense, the lines or columns are natural numbers, in the form of the
1-adic system or in the other sense as sets which completeness follows
the peano axiomes.
Numbers are sets. A set without number isn't a set. Since there is no
infinite numeral, there is no set with an infinite number of elements.
But if someone don't like this interpretation, he may think, that
infinity is something as "undefined". Infinity, endlessly,
uncountability means the same aspect in different "dimensiones".
Endlessly is the aspect of infinity in space and time, uncountability
the aspcet of infinity of sets of discret things.
If we accept the uncountability as a form of infinity, this leads to
the paradoxon that the natural numbers are not countable. That's
paradox since the natural numbers count themself.
The most mathematics shurely say, that the word "uncountability" is
just a word. It's accidentally another word for infinity. Since
infinity is defined. Infinity is that, what could be biject to a part
of itself.
With infinity you can do very interesting things.
You can find two of them: potential infinity and actual infinity.
Perhaps three? Undefined?
Infinity is just a "facon de parler". In this sense it's the strongest
tool of math.
I know, the reactions of the most other posters will be the usual one.
You know them. A dream can be stronger than every argument. So I thank
you for your kindness and your understanding. Glückauf.
AS
albs...@gmx.de
> albs...@gmx.de wrote:
>
> > But there is a slight difference. Since there is no infinite natural in
> > form of a set of Os and since after every set of #s there should be a
> > O, the size of the set of the naturals as sets of #s could not extend
> > the "biggest" number of the naturals in form of sets of Os.
> > Since there is no biggest number and since there is no infinite number,
> > the size of the set of numbers in form of sets of #s is undefined as
> > the biggest natural number is undefined.
>
> Whoever said the size of a set has anything to do with the "biggest"
> element?
You are unable to recognize the primitivest form of bijection?
>
> > But the sequence of the sets of # fullfill the peano axiomes. So this
> > set must be infinite.
>
> > The cardinality of a set is not able to be infinite and "not defined"
> > at the same time.
>
> > This is the contradiction.
>
> No, the contradiction is assuming that cardinality has
> anything to do with the "biggest" element. Cardinality
> is not defined in terms of the largest element. It
> is defined in terms of bijections. Your post says
> nothing about bijections, so it says nothing about
> cardinality.
>
> Stephen
I did not use the word "bijection"? This must be really the flaw of my
proof.
AS
albs...@gmx.de
Image a rectangular triangle with a = b. Than c = sqrt(2) * a.
Now expand the side a to infinity. What is the lenght of of the side b?
Since a = b, b must be equal infinity.
Some may argue, that there is no triangle with infinite sides.
Consider the angle between the straight lines a and c. What is the
rectangular distance between a and c in infinity? Infinity or
undefined?
If this real value is infinite, there must be a numeral, which is
infinite. Or there is no infinity.
To made this concept connected to my argument in the starting posting.
Consider the rectangular triangle with the side c as a staircaise with
steps 1 in length and hight. If side a has the length 10, the length of
the x-components of the side c is equally 10. The lenght of the
y-components of the side c is also 10 like the lenght of side b.
The lenghts of the sides of the sequence of this triangles follows the
peano axiomes. Consequently there are infinitely many steps if a is
infinite. So the sum of the x-components of c is infinite. But the sum
of the y-components of c isn't infinite. It is undefined. A really
logical concept.
Regards
Albrecht Storz
zuhair
>
> Albrecht S. Storz, Germany
I don't know were is the problem. May be I am not the first one who
advocated aproaching the infinite by unary approach , but I don't know
another one who advocated that .
I made a lot of viusal trainning on using the unitary system in
infinity and I find it a very useful way in understanding the infinite
and I have some experience with that, yet I don't see any problem in
the figure you just said, in reality I made such a figure of yours four
years ago and I saw no problems then , let me review some of what you
say.
You said: The first vertical row of #s could not
> exceed the biggest vertical row of Os.
Why it couldn't exceed? if you want the reality it exceeds it by
one.because the last vertical column contains # at its end and it is
not full of zeros as you imagined . lets see how that works on finites
and then apply it infinites.
# 0000...........(n-1)0
##0000...........(n-2)0
### 0000.........(n-3)0
... .
... .
................. .
... .
(n)#######..........(n-n)0
In reality the last row has no zeros so it is full of #, n is a finite
natural, so the last vertical column is not all zeros.
For example let n=4 , the sequare will look like that.
#000
##00
###0
####
In reality if you read the chapter "Paradoxical Results" in My article
"The Infinite Calculus"(or Preliminary Infinite Calculus ) present in
my web site: http://zaljohar.tripod.com . you can see that I've
depended on such imagination to count the number of # in that figure
and the number of # is {(n^2)/2} +{n/2), Apply it for n=4 above =
[16/2]+2= 10 .
Now let's calculute when n=Omega=w
w of # can be symbolized in a unitary manner as ####.......
it means that the first # has 1-1 relationship wich number 0, the
second # has 1-1 relationship with the second natural (number 1),
etc....
so we can say that w is the number of all natural number HORIZONTALLY
PLACED IN ACENDING SEQUENCE one after the other ( and this is not
equivalent to the number of all natural numbers which is something
uncountable by using the infinite,see my illustration of that in my
article "the comparison between cantor's transfintie math and the
unitary infinite math"- topic on reflexiveness in my web site ).
Now back to our problem.
The number of # when n=w is simply= [(w^2)/2]+ w/2
Alot of the contradiction you are encountering is because you believe
that just placing the naturals one after the other horizontall you can
have a set that contains all the natural numbers which is impossible.
In order to let you forget this idea , let us construct Q like below
135..........
024.......... =Q
012.......... =N
See that Q has 2-1 correspondance with N , and so Q contains double the
amount of natural numbers contained in N.
And in order to make a reply to Tony who always think that their is a
finite number bigger than an infinite set of finite numbers this is a
good example above, more clearly if we say that Q is the union of the
two disjount sets Q1 and Q2 where Q1 contains the first w or natural
numbers
ie Q1=N(starting with zero) , while Q2 contains the second w set of
natural numbers ( of coarse in ascending order) , then min Q2= lim Q1
Now if we say that the size of any natural number set beginning from
zero
0,1,2,3,....,n is always n+1 and if we symbolize it as S
then S(0,1,2,3,4,.....,n)=n+1 and it is a finite number.
Then: min Q2= S(0,1,2,3,......)=lim N
All of what I wanted to say is that their is a great deal of confusion
between the concepts of Infinity and the Concept of totality. Alot of
people think that the set of all natural numbers is a kind of a set
that can have a 1-1 correspondance with a certain infinite set and this
is impossible. In reality the number of all natural numbers is not
countable and it may be the kind of a number I used to refer to as The
inconsistent number( for details about that see Sci.math-Topic:The
meaning of 0/0?
Best
Zuhai
Peter Webb
not defined by its largest element. You have to form an explicit bijection
to prove equal cardinality; that's what cardinality is defined to mean.
That the largest element of a set is a useless way to define something even
resembling cardinality, consider the set of integers {..-3, -2, -1, 0, 1, 2
..}. Now consider the those less than or equal to 1 {..-3, -2, -1, 0, 1}. B
y your measure, this has "cardinality" of 1, so 1 = infinity.
The countable set of Rationals x such that 0<x<1 has 0.9, 0.99, 0.999 etc
but not 1.000, and so has a bounded value but no maximum element in the set.
Whaddya going to do here?
Keep on trying.
albs...@gmx.de
>
> I don't know were is the problem. May be I am not the first one who
> advocated aproaching the infinite by unary approach , but I don't know
> another one who advocated that .
> I made a lot of viusal trainning on using the unitary system in
> infinity and I find it a very useful way in understanding the infinite
> and I have some experience with that, yet I don't see any problem in
> the figure you just said, in reality I made such a figure of yours four
> years ago and I saw no problems then , let me review some of what you
> say.
>
> You said: The first vertical row of #s could not
> > exceed the biggest vertical row of Os.
>
> Why it couldn't exceed?
It could not exceed, because it's the advise to build the considererd
set. The question is not, if it can exceed, the question is, what
happens to the set if it not exceed.
> if you want the reality it exceeds it by
> one.because the last vertical column contains # at its end and it is
> not full of zeros as you imagined .
i don't image it. i want it. and i see it. because it is.
Regards
AS
zuhair
equivalent to the infinite number.
Since I believe in a natural infinite number, therefore their is no
contradiction on my side.
In my views the smalles natural infinite number is the one that comes
after all finite natural numbers horizontally placed. so at the end
your rectangle will have w by w+1 dimentions , were w means the first
natural infinite number which is equivalent to the number of finite
natural numbers horizontally palced in ascending manner from zero.
Best
Zuhair
David R Tribble
> [...]
> Since there is no biggest number and since there is no infinite number,
> the size of the set of numbers in form of sets of #s is undefined as
> the biggest natural number is undefined.
>
> But the sequence of the sets of # fullfill the peano axiomes. So this
> set must be infinite.
>
> The cardinality of a set is not able to be infinite and "not defined"
> at the same time.
> This is the contradiction.
I don't see the contradiction. The size of the set is "not defined"
to be the same as any natural number, and the set size is obviously
infinite. This is no contradiction, since no natural number is
infinite.
The thing that is "not defined" is the largest natural, which obviously
does not exist. But the set size is infinite, and is nicely defined
by an infinite cardinal.
You seem to be mixing the two concepts of "natural" and "cardinal"
numbers to create a supposed contradiction, but that does not work.
ste...@nomail.com
> ste...@nomail.com wrote:
>> albs...@gmx.de wrote:
>>
>> > But there is a slight difference. Since there is no infinite natural in
>> > form of a set of Os and since after every set of #s there should be a
>> > O, the size of the set of the naturals as sets of #s could not extend
>> > the "biggest" number of the naturals in form of sets of Os.
>> > Since there is no biggest number and since there is no infinite number,
>> > the size of the set of numbers in form of sets of #s is undefined as
>> > the biggest natural number is undefined.
>>
>> Whoever said the size of a set has anything to do with the "biggest"
>> element?
> You are unable to recognize the primitivest form of bijection?
That is a meaningless sentence. Bijections make no
mention of "biggest".
>>
>> > But the sequence of the sets of # fullfill the peano axiomes. So this
>> > set must be infinite.
>>
>> > The cardinality of a set is not able to be infinite and "not defined"
>> > at the same time.
>>
>> > This is the contradiction.
>>
>> No, the contradiction is assuming that cardinality has
>> anything to do with the "biggest" element. Cardinality
>> is not defined in terms of the largest element. It
>> is defined in terms of bijections. Your post says
>> nothing about bijections, so it says nothing about
>> cardinality.
>>
>> Stephen
> I did not use the word "bijection"? This must be really the flaw of my
> proof.
If you want to talk about cardinality, which is defined in
terms of bijections, then you should be talking about
bijections. You instead are talking about "biggest" elements
which have nothing to do with cardinality. You claimed
that the cardinality was undefined, because there is
no biggest element. That is just simply and obviously
wrong.
Stephen
ste...@nomail.com
What does it mean for a line to be infinite in your
infinite triangle? Presumably the line has one end point
somewhere. Where is the other end point? At infinity?
An infinite line does not end. It does not have another
end point.
You are German. Translate the following sentence into German:
The line ends at infinity.
<snip>
> The lenghts of the sides of the sequence of this triangles follows the
> peano axiomes. Consequently there are infinitely many steps if a is
> infinite. So the sum of the x-components of c is infinite. But the sum
> of the y-components of c isn't infinite. It is undefined. A really
> logical concept.
Infinite geometric figures really do not make much sense.
You are just assuming conclusions about them without
any sort of proof.
Stephen
albs...@gmx.de
You are not able to understand that there is no difference between
numerals and sets. My sketches shows this exactly.
Cantor proofs his wrong conclusion with the same mix of potential
infinity and actual infinity. But there is no bijection between this
two concepts. The antidiagonal is an unicorn.
There is no stringend concept about infinity. And there is no aleph_1,
aleph_2, ... or any other infinity.
Regards
AS
Tony Orlow
> David R Tribble wrote:
> > Albrecht S. Storz wrote:
> > > [...]
> > > Since there is no biggest number and since there is no infinite number,
> > > the size of the set of numbers in form of sets of #s is undefined as
> > > the biggest natural number is undefined.
> > >
> > > But the sequence of the sets of # fullfill the peano axiomes. So this
> > > set must be infinite.
> > >
> > > The cardinality of a set is not able to be infinite and "not defined"
> > > at the same time.
> > > This is the contradiction.
> >
> > I don't see the contradiction. The size of the set is "not defined"
> > to be the same as any natural number, and the set size is obviously
> > infinite. This is no contradiction, since no natural number is
> > infinite.
> >
> > The thing that is "not defined" is the largest natural, which obviously
> > does not exist. But the set size is infinite, and is nicely defined
> > by an infinite cardinal.
> >
> > You seem to be mixing the two concepts of "natural" and "cardinal"
> > numbers to create a supposed contradiction, but that does not work.
>
>
> You are not able to understand that there is no difference between
> numerals and sets. My sketches shows this exactly.
measure of a set of units, and the only common feature that all sets share is
size. or number of these units. I wouldn't sat a quantity IS set, but rather a
salient FEATURE of a set. Particular sets have other features and properties,
depending on the properties of the elements.
> Cantor proofs his wrong conclusion with the same mix of potential
> infinity and actual infinity. But there is no bijection between this
> two concepts. The antidiagonal is an unicorn.
The antidiagonal serves to prove that there are more digital strings than
whatever maximum length string you choose. The list of digital numbers is
longer than it is wide. So, given an infinite number of digits, you have a
larger infinite number of strings. That's all.
> There is no stringend concept about infinity. And there is no aleph_1,
> aleph_2, ... or any other infinity.
Here we disagree. How many points, how many real numbers, are on the number
line betweeen 0 and 1?
>
> Regards
>
> AS
>
>
--
Smiles,
Tony
Tony Orlow
ordered set with distinct endpoints. It is valid to speak of what happens at
infinity.
>
> You are German. Translate the following sentence into German:
> The line ends at infinity.
>
> <snip>
>
> > The lenghts of the sides of the sequence of this triangles follows the
> > peano axiomes. Consequently there are infinitely many steps if a is
> > infinite. So the sum of the x-components of c is infinite. But the sum
> > of the y-components of c isn't infinite. It is undefined. A really
> > logical concept.
>
> Infinite geometric figures really do not make much sense.
> You are just assuming conclusions about them without
> any sort of proof.
diagram that whatever length/value you allow the unary string elements to
assume, that is also equal to the number of strings in the set. It's a very
clear visual argument, the likes of which should be more prevalent in
mathematics. It illsutrates exactly what I've been trying to get through.
>
> Stephen
>
--
Smiles,
Tony
Tony Orlow
> albs...@gmx.de wrote:
> > ste...@nomail.com wrote:
> >> albs...@gmx.de wrote:
> >>
> >> > But there is a slight difference. Since there is no infinite natural in
> >> > form of a set of Os and since after every set of #s there should be a
> >> > O, the size of the set of the naturals as sets of #s could not extend
> >> > the "biggest" number of the naturals in form of sets of Os.
> >> > Since there is no biggest number and since there is no infinite number,
> >> > the size of the set of numbers in form of sets of #s is undefined as
> >> > the biggest natural number is undefined.
> >>
> >> Whoever said the size of a set has anything to do with the "biggest"
> >> element?
>
>
> > You are unable to recognize the primitivest form of bijection?
>
> That is a meaningless sentence. Bijections make no
> mention of "biggest".
great. He is making a bijection between the lengths of the strings of 0's and
#'s, the one representing the values of the elements, and the other repreenting
the count of the elements. The values of both are equal. That's the point.
> >>
> >> > But the sequence of the sets of # fullfill the peano axiomes. So this
> >> > set must be infinite.
> >>
> >> > The cardinality of a set is not able to be infinite and "not defined"
> >> > at the same time.
> >>
> >> > This is the contradiction.
> >>
> >> No, the contradiction is assuming that cardinality has
> >> anything to do with the "biggest" element. Cardinality
> >> is not defined in terms of the largest element. It
> >> is defined in terms of bijections. Your post says
> >> nothing about bijections, so it says nothing about
> >> cardinality.
> >>
> >> Stephen
>
>
> > I did not use the word "bijection"? This must be really the flaw of my
> > proof.
>
> If you want to talk about cardinality, which is defined in
> terms of bijections, then you should be talking about
> bijections. You instead are talking about "biggest" elements
> which have nothing to do with cardinality. You claimed
> that the cardinality was undefined, because there is
> no biggest element. That is just simply and obviously
> wrong.
he is perfectly justified in saying that if one is not defined, then neither is
the other.
>
> Stephen
>
--
Smiles,
Tony
Tony Orlow
1, which will have all 0's in the Nth column. He could have started with 1 and
gone through N for N elements, and had N-1 #'s in the first row, and N-1 0's on
the last. So, his point is valid in either case.
>
> In reality if you read the chapter "Paradoxical Results" in My article
> "The Infinite Calculus"(or Preliminary Infinite Calculus ) present in
> my web site: http://zaljohar.tripod.com . you can see that I've
> depended on such imagination to count the number of # in that figure
world of programming. :D
>
> and the number of # is {(n^2)/2} +{n/2), Apply it for n=4 above =
> [16/2]+2= 10 .
>
> Now let's calculute when n=Omega=w
>
> w of # can be symbolized in a unitary manner as ####.......
> it means that the first # has 1-1 relationship wich number 0, the
> second # has 1-1 relationship with the second natural (number 1),
> etc....
>
> so we can say that w is the number of all natural number HORIZONTALLY
> PLACED IN ACENDING SEQUENCE one after the other ( and this is not
> equivalent to the number of all natural numbers which is something
> uncountable by using the infinite,see my illustration of that in my
> article "the comparison between cantor's transfintie math and the
> unitary infinite math"- topic on reflexiveness in my web site ).
>
> Now back to our problem.
>
> The number of # when n=w is simply= [(w^2)/2]+ w/2
(Martin's version). Fancy that. We all got the same answer for the sum of the
naturals! Is this a monkey-typewriter-shakespeare phenomenon?
>
> Alot of the contradiction you are encountering is because you believe
> that just placing the naturals one after the other horizontall you can
> have a set that contains all the natural numbers which is impossible.
>
> In order to let you forget this idea , let us construct Q like below
>
> 135..........
> 024.......... =Q
>
> 012.......... =N
>
> See that Q has 2-1 correspondance with N , and so Q contains double the
> amount of natural numbers contained in N.
equating the sizes of those sets with the set of all naturals, I think that is
a mistake.
>
> And in order to make a reply to Tony who always think that their is a
> finite number bigger than an infinite set of finite numbers this is a
> good example above, more clearly if we say that Q is the union of the
> two disjount sets Q1 and Q2 where Q1 contains the first w or natural
> numbers
> ie Q1=N(starting with zero) , while Q2 contains the second w set of
> natural numbers ( of coarse in ascending order) , then min Q2= lim Q1
I was saying either. Perhaps you could rephrase this?
>
> Now if we say that the size of any natural number set beginning from
> zero
> 0,1,2,3,....,n is always n+1 and if we symbolize it as S
> then S(0,1,2,3,4,.....,n)=n+1 and it is a finite number.
smallest "ordinal" greater than n, which for all finites, is the first limit
ordinal, which is infinite. It should be considered simply n+1 always, so if n
is finite, then so is the size of the set. This discontinuity doesn't belong
here.
>
> Then: min Q2= S(0,1,2,3,......)=lim N
>
> All of what I wanted to say is that their is a great deal of confusion
> between the concepts of Infinity and the Concept of totality. Alot of
> people think that the set of all natural numbers is a kind of a set
> that can have a 1-1 correspondance with a certain infinite set and this
> is impossible. In reality the number of all natural numbers is not
> countable and it may be the kind of a number I used to refer to as The
> inconsistent number( for details about that see Sci.math-Topic:The
> meaning of 0/0?
of this set, it can still serve as a unit of measure for other infinite sets,
given the identity relationship between element count and element value. One
can compare sets over the infinite real range formulaically. While you do not
get particular numbers for such set sizes, what you do get is a rich means of
comparison for infinite sets.
>
> Best
> Zuhai
>
>
--
Smiles,
Tony
William Hughes
albs...@gmx.de wrote:
<snip>
>
> If we accept the uncountability as a form of infinity, this leads to
> the paradoxon that the natural numbers are not countable.
No, the natural numbers are countable precisely because
they do count themselves. The fact that there is no
natural number that repsresents this "count" is not a paradox
because the "count" is defined in terms of bijections. [You
may not like the use of the terms "count" and "countable"
because you think they should imply something different. So
be it. However, you cannot say "you are using a term which
I think should mean somthing different, so you must mean
not what you mean but what I mean"]
- William Hughes
Tony Orlow
> Tony Orlow wrote:
> > albs...@gmx.de said:
> > >
> > > (Hint: The sketches would not looks good in proportional fond)
> > >
> > > Let's start with a representation of the natural numbers in unitary
> > > (1-adic) system as follows:
> > >
> > >
> > >
> > > O O O O O O O O O ...
> > > O O O O O O O O ...
> > > O O O O O O O ...
> > > O O O O O O ...
> > > O O O O O ...
> > > O O O O ...
> > > O O O ...
> > > O O ...
> > > O ...
> > > .
> > > .
> > > .
> > >
> > >
> > > 1 2 3 4 5 6 7 8 9 ...
> > >
> > >
> > > Each vertical row shows a natural number. Horizontally, it is the
> > > sequence of the natural numbers. Since the Os, the elements, are local
> > > distinguished from each other, we can also look at the rows as sets. A
> > > set may contain the coordinates of the elements as their representation
> good
> > graphic illustration of the equality between element value and element co=
> unt
> > for the natural numbers. It would seem very hard to argue that the array =
> with
> > its diagonal is somehow longer than it is wide, using this unary notation=
> . I
> > believe that you have constructed a representation of a set which is
> wed to
> > become infinite in both directions. Good job! Danke!
> > >
> > >
> > > Best regards
> > >
> > > Albrecht S. Storz, Germany
> > >
> > >
> >
> > --
> > Smiles,
> >
> > Tony
>
>
> The idea results from the understanding, that every number is a set. A
> number is the unchanged aspect of a simultanity of endlessly many
> objects which only and absolutly only has common aspects in the number
> of their elements.
> That's the only, or one possible definition of a number.
counting operations, those measurements are only pssobile given discrete units
of measurement, and amount to the equivalent of a size of a set of units.
>
> The similarity of my sketches with the usual representation of the
> Cantor diagonal argument is not an accidently happend effect.
Using base 1 kind of destroys the argument.
>
> We could interpret the struktures in the sketches alternatively. In one
> sense, the lines or columns are natural numbers, in the form of the
> 1-adic system or in the other sense as sets which completeness follows
> the peano axiomes.
>
> Numbers are sets. A set without number isn't a set. Since there is no
> infinite numeral, there is no set with an infinite number of elements.
the same in decimal as 0.3333...? Isn't that an infinite numeral? Is there any
reason I cannot have an infinite number of digits to the elft of the digital
point, like ...3333.0? Okay, this can get ambiguous, so can't I declare a digit
at some infinite position in common with other numbers and compare them? Think
of the points in (0,1] as going from 0.000...001 through 1.000...000. Now
multiply by N=1:000...000.0. We get 1.000...000 through 1:000...000.0, for a
complete bijection between the infinite set of reals in (0,1] and the set of
natural numbers, finite and infinite. Again, isn't the set of points in (0,1]
infinite, or is there some finite number of points in that set?
>
> But if someone don't like this interpretation, he may think, that
> infinity is something as "undefined". Infinity, endlessly,
> uncountability means the same aspect in different "dimensiones".
> Endlessly is the aspect of infinity in space and time, uncountability
> the aspcet of infinity of sets of discret things.
from space or time in that respect?
>
> If we accept the uncountability as a form of infinity, this leads to
> the paradoxon that the natural numbers are not countable. That's
> paradox since the natural numbers count themself.
>
> The most mathematics shurely say, that the word "uncountability" is
> just a word. It's accidentally another word for infinity. Since
> infinity is defined. Infinity is that, what could be biject to a part
> of itself.
> With infinity you can do very interesting things.
> You can find two of them: potential infinity and actual infinity.
> Perhaps three? Undefined?
infinite, since they are all restricted to finite numbers of iterations, and
thus elements. Uncountable infinities are actually infinite, but seem countable
in a variety of ways. The whole countability criterion is misguided, IMHO.
>
> Infinity is just a "facon de parler". In this sense it's the strongest
> tool of math.
>
> I know, the reactions of the most other posters will be the usual one.
> You know them. A dream can be stronger than every argument. So I thank
>
>
> AS
>
>
--
Smiles,
Tony
Tony Orlow
For every natural value represented by 0's in the diagram there is an equal
count represeted by #'s. This is the identity relationship between count and
value that I've been talking about. Think of it as the limit of a square as the
side goes to oo. Your objection is just another form of "No Largest Finite!! No
Diagonal Corner!!! (jingle jangle)" Oh, nice wind chime!!
>
> <snip>
>
> > It says everything about the size of the set of naturals compared with the
> > values in the set. Can the square be wider than it is tall? No? Well, then,
> > what does that say to you? (probably nothing, of course)
>
> As there is no square in the first place, it is irrelevant
> if a square can be wider than it is tall.
>
> Stephen
>
--
Smiles,
Tony
David Kastrup
There is no "diagonal" for something that has only one corner.
> For every natural value represented by 0's in the diagram there is
> an equal count represeted by #'s.
It does not make sense to talk about "an equal count" for things that
don't end.
> This is the identity relationship between count and value that I've
> been talking about. Think of it as the limit of a square as the side
> goes to oo. Your objection is just another form of "No Largest
> Finite!! No Diagonal Corner!!! (jingle jangle)" Oh, nice wind
> chime!!
Well, too bad that you insist on making the same mistake all over
again. Small wonder you get your nose rubbed into it all over again.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
David R Tribble
>> [...]
>> Since there is no biggest number and since there is no infinite number,
>> the size of the set of numbers in form of sets of #s is undefined as
>> the biggest natural number is undefined.
>>
>> But the sequence of the sets of # fullfill the peano axiomes. So this
>> set must be infinite.
>>
>> The cardinality of a set is not able to be infinite and "not defined"
>> at the same time.
>> This is the contradiction.
>
David R Tribble wrote:
>> I don't see the contradiction. The size of the set is "not defined"
>> to be the same as any natural number, and the set size is obviously
>> infinite. This is no contradiction, since no natural number is
>> infinite.
>>
>> The thing that is "not defined" is the largest natural, which obviously
>> does not exist. But the set size is infinite, and is nicely defined
>> by an infinite cardinal.
>>
>> You seem to be mixing the two concepts of "natural" and "cardinal"
>> numbers to create a supposed contradiction, but that does not work.
>
Albrecht S. Storz wrote:
> You are not able to understand that there is no difference between
> numerals and sets.
I have no problem seeing the correspondence between natural numbers
and von Neumann sets. But neither of these are the same as
cardinalities, which are not numbers, but measures (sizes) of sets.
> My sketches shows this exactly.
> Cantor proofs his wrong conclusion with the same mix of potential
> infinity and actual infinity. But there is no bijection between this
> two concepts. The antidiagonal is an unicorn.
> There is no stringend concept about infinity. And there is no aleph_1,
> aleph_2, ... or any other infinity.
For that to be true, there must be a bijection between an infinite
set (any infinite set) and its powerset. Bitte, show us a bijection
between N and P(N).
zuhair
Please read my simplified article I wrote latelly to fully understand
my views about the infinite.Read this at:
http://zaljohar.tripod.com/infinite.txt
I will be more than happy to discuss it with you , or with any open
minded mathematician in that group.
Zuhair
Tony Orlow
like about it? It is valid.
--
Smiles,
Tony
David R Tribble
>> Cantor proofs his wrong conclusion with the same mix of potential
>> infinity and actual infinity. But there is no bijection between this
>> two concepts. The antidiagonal is an unicorn.
>> There is no stringend concept about infinity. And there is no aleph_1,
>> aleph_2, ... or any other infinity.
>
David R Tribble said:
>> For that to be true, there must be a bijection between an infinite
>> set (any infinite set) and its powerset. Bitte, show us a bijection
>> between N and P(N).
>
Tony Orlow wrote:
> I already showed you the bijection between binary *N and P(*N).
> What didn't you like about it? It is valid.
No, you showed a mapping between *N and R, which is equivalent
to a mapping between *N and P(N). That's easy.
But you have not provided a mapping between any set and its powerset,
infinite or otherwise.
David R Tribble
>> The number of # when n=w is simply= [(w^2)/2]+ w/2
>
Tony Orlow wrote:
> Gee, that looks amazingly similar to (N^2+N)/2 (my version) or N(N+1)/2
> (Martin's version). Fancy that. We all got the same answer for the sum of the
> naturals! Is this a monkey-typewriter-shakespeare phenomenon?
Okay, so the triangular number T(n) = n(n+1)/2, which is certainly
true for any finite n. So what? Just because someone agrees with
you doesn't make you right. It's not in the least bit surprising
that all of you got the same answer by making the same incorrect
assumptions. Fancy that.
None of you three (or four, depending on the counting) have proven
that infinite naturals must exist, and certainly none of you has
proven that arithmetic with infinite numbers works the same as
finite arithmetic.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> David R Tribble said:
>
> > For that to be true, there must be a bijection between an infinite
> > set (any infinite set) and its powerset. Bitte, show us a
> > bijection between N and P(N).
> >
> >
> I already showed you the bijection between binary *N and P(*N). What
> didn't you like about it? It is valid.
There are at least two things wrong with it.
(1) It is not valid by any standard mathematics or logic (only in
TOmatics which is irrelevant to both standard logic and mathematics).
(2) Unless *N is bijectable with N, no such bijection on N* is relevant
to bijections on N.
albs...@gmx.de
You can not see, that in my demonstration there is no distinguishing
betwéen sets, naturals, cardinalities. You can not see, that this,
what holds for natural numbers must also holds for cardinalities.
>
>
> > My sketches shows this exactly.
> > Cantor proofs his wrong conclusion with the same mix of potential
> > infinity and actual infinity. But there is no bijection between this
> > two concepts. The antidiagonal is an unicorn.
> > There is no stringend concept about infinity. And there is no aleph_1,
> > aleph_2, ... or any other infinity.
>
> For that to be true, there must be a bijection between an infinite
> set (any infinite set) and its powerset. Bitte, show us a bijection
> between N and P(N).
At first, you should show, that bijection means something to
notwellordered infinite sets.
Bijection is a clear concept on finite sets, it also works on
wellordered infinite sets of the same infinite concept.
Aber: Show me a bijection between two infinite sets with the same
cardinality, where one of the sets is still not wellorderable.
Than I will show you a bijection between N and P(N) or N and R or P(N)
and P(P(N)) or what you want.
But this is not the issue of this thread. You are free to critisize my,
in the startposting demonstrated, argumentation.
Regards
AS
zuhair
It's not in the least bit surprising
that all of you got the same answer by making the same incorrect
assumptions. Fancy that.
-----------
Well although we reached into the same results, but their is some
difference
to me their is no number which can discribe the multiplicity of all
natural numbers, but their are numbers which can discrible the sum of a
specifically defined infinite sets of natural numbers.
Examples:
The sum of all natural numbers in N defined as N ={ 1,2,3,4,........}
is:
1+2+3+4+............. = [(w^2)/2]+ w/2
But the sum of all natural numbers in Q defined as a set of natural
numbers which has 2-1 correspondance with setN above :
246............
135............ =Q
Now the sum of numbers in Q is
2 4 6
+ + + + + +.................. = [((2w)^2)/2]+ w = 2 w^2 + w
1 3 5
While the sum of all natural numbers for U where U has 3-1
correspondance with set N. is
[((3w)^2)/2]+ 3w/2
In general the sum of all natural numbers in set X where X has a -1
correspondance with setN is:
[((aw)^2)/2]+ aw/2
Best
Zuhair
Tony Orlow
> Albrecht S. Storz wrote:
> >> Cantor proofs his wrong conclusion with the same mix of potential
> >> infinity and actual infinity. But there is no bijection between this
> >> two concepts. The antidiagonal is an unicorn.
> >> There is no stringend concept about infinity. And there is no aleph_1,
> >> aleph_2, ... or any other infinity.
> >
>
> David R Tribble said:
> >> For that to be true, there must be a bijection between an infinite
> >> set (any infinite set) and its powerset. Bitte, show us a bijection
> >> between N and P(N).
> >
>
> Tony Orlow wrote:
> > I already showed you the bijection between binary *N and P(*N).
> > What didn't you like about it? It is valid.
>
> No, you showed a mapping between *N and R, which is equivalent
> to a mapping between *N and P(N). That's easy.
No, it was specifically a bijection between two sets of infinite binary strings
representing, on the one hand, the whole numbers in *N starting from 0, both
finite and infinite, in normal binary format, and on the other hand, the
specification of each subset of whole numbers in *N, where each bit which, in
the binary number, represents 2^n denotes membership of n in the subset. This
is a bijection between the whole numbers in *N and P(*N), using an intermediate
bijection with a common set of infinite binary strings. This has nothing to do
with the reals. Sorry.
>
> But you have not provided a mapping between any set and its powerset,
> infinite or otherwise.
Have too.
>
>
--
Smiles,
Tony
Tony Orlow
> zuhair said:
> >> The number of # when n=w is simply= [(w^2)/2]+ w/2
> >
>
> Tony Orlow wrote:
> > Gee, that looks amazingly similar to (N^2+N)/2 (my version) or N(N+1)/2
> > (Martin's version). Fancy that. We all got the same answer for the sum of the
> > naturals! Is this a monkey-typewriter-shakespeare phenomenon?
>
> Okay, so the triangular number T(n) = n(n+1)/2, which is certainly
> true for any finite n. So what? Just because someone agrees with
> you doesn't make you right. It's not in the least bit surprising
> that all of you got the same answer by making the same incorrect
> assumptions. Fancy that.
maintains the same constant relationships based on the definition of the
geometrical pattern? What is your assumption? That it goes the way of the
vase's balls, or that you can say whatever you fancy about it?
>
> None of you three (or four, depending on the counting) have proven
> that infinite naturals must exist, and certainly none of you has
> proven that arithmetic with infinite numbers works the same as
> finite arithmetic.
some oats as well.
>
>
--
Smiles,
Tony
Tony Orlow
> In article <MPG.1dbdd9753...@newsstand.cit.cornell.edu>,
> Tony Orlow <ae...@cornell.edu> wrote:
>
> > David R Tribble said:
> >
> > > For that to be true, there must be a bijection between an infinite
> > > set (any infinite set) and its powerset. Bitte, show us a
> > > bijection between N and P(N).
> > >
> > >
> > I already showed you the bijection between binary *N and P(*N). What
> > didn't you like about it? It is valid.
>
> There are at least two things wrong with it.
>
> (1) It is not valid by any standard mathematics or logic (only in
> TOmatics which is irrelevant to both standard logic and mathematics).
>
> (2) Unless *N is bijectable with N, no such bijection on N* is relevant
> to bijections on N.
between the elements, can't you? Where do you define the end of that
relationship? That's irrelevant anyway. The point was making a bijection
between a set and its power set. Again, Virgil misses the point.
>
--
Smiles,
Tony
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> > But you have not provided a mapping between any set and its powerset,
> > infinite or otherwise.
> Have too.
TO's delusions about what can represent what are not valid outside the
twilight zone of TOmatics.
TO simultaneously wants to represent each member of *N as (1) an
infinite binary string, and (2) a one bit in one digit of an infinite
binary string.
Only in the twilight zone of TOmatics!
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> David R Tribble said:
> > zuhair said:
> > >> The number of # when n=w is simply= [(w^2)/2]+ w/2
> > >
> >
> > Tony Orlow wrote:
> > > Gee, that looks amazingly similar to (N^2+N)/2 (my version) or N(N+1)/2
> > > (Martin's version). Fancy that. We all got the same answer for the sum of
> > > the
> > > naturals! Is this a monkey-typewriter-shakespeare phenomenon?
> >
> > Okay, so the triangular number T(n) = n(n+1)/2, which is certainly
> > true for any finite n. So what? Just because someone agrees with
> > you doesn't make you right. It's not in the least bit surprising
> > that all of you got the same answer by making the same incorrect
> > assumptions. Fancy that.
> What incorrect assumptions?
The first incorrect assumption you all made was that any of you have any
abilities with either mathematics or logic at all.
The second is that appending assumptions which are contradictory to the
axiom systems they are being appended to improves those axiom systems.
> That the unending geometrical pattern continues and maintains the
> same constant relationships based on the definition of the
> geometrical pattern? What is your assumption? That it goes the way of
> the vase's balls, or that you can say whatever you fancy about it?
One of our assumptions is that no ball remains in a vase after having
been removed from it. TO obviously rejects this asumption.
> >
> > None of you three (or four, depending on the counting) have proven
> > that infinite naturals must exist, and certainly none of you has
> > proven that arithmetic with infinite numbers works the same as
> > finite arithmetic.
> You can't make a horse drink.
The end of the horse that we are addressing does not drink.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> Virgil said:
> > In article <MPG.1dbdd9753...@newsstand.cit.cornell.edu>,
> > Tony Orlow <ae...@cornell.edu> wrote:
> >
> > > David R Tribble said:
> > >
> > > > For that to be true, there must be a bijection between an infinite
> > > > set (any infinite set) and its powerset. Bitte, show us a
> > > > bijection between N and P(N).
> > > >
> > > >
> > > I already showed you the bijection between binary *N and P(*N). What
> > > didn't you like about it? It is valid.
> >
> > There are at least two things wrong with it.
> >
> > (1) It is not valid by any standard mathematics or logic (only in
> > TOmatics which is irrelevant to both standard logic and mathematics).
> Why, exactly. It's not a fly you can wave away.
In standard mathematics, it has been proven that no bijection can exist
between any set and its power set.
So that if TO says that he has a bijection between some set and its
power set, then, at least outside of the twilight zone of TOmatics,
either that "set" is not a set (possibly a proper class), or the "power
set" is not actually its power set or the "bijection" is not actually a
bijection, or several of these.
> >
> > (2) Unless *N is bijectable with N, no such bijection on N* is relevant
> > to bijections on N.
> Accrding to you, it should be.
Not by me, it isn't. For me N is the Dedekind infinite set of finite
naturals representable by finite strings of digits in any natural base.
By any reasonable standard, *most* members of TO's *N are not members of
N.
You can certainly define a 1-1 correspondence
> between the elements, can't you?
Between elements of what set and elements of what other set?
Sometimes bijections do not exist, as, for example, between the empty
set and any non-empty set.
Between any standard version of N and TO's *N no bijection exists.
> Where do you define the end of that
> relationship? That's irrelevant anyway. The point was making a bijection
> between a set and its power set. Again, Virgil misses the point.
TO missed the point that, outside of the twilight zone of TOmatics, no
bijection between any set and its power set is possible.
What happens in that twilight zone of TOmatics is irrelevant to
mathematics, as no proper mathematics can ever be done in a place where
both every statement and its negation are theorems.
ste...@nomail.com
>> infinite or otherwise.
> Have too.
No you have not Tony. The proof that there does not exist
a bijection between a set and its power set is quite short.
Let f be a function from S to P(S).
Define the set w as follows:
w= { x : x in S and x is not in f(x) }
Clearly w is a subset of S, and so w is an element of P(S).
We now show that w is not in the image of f. That is,
there does not exist a y such that f(y)=w.
Suppose such a y exists. If such a y exists, it must
either be an element of w, or not.
If y is an element of w, then y is in f(y), which means
it is not an element of w.
If y is not an element of w, then y is not in f(y), which
means it is an element of w.
These are both contradictions. So y cannot be an element
of w, and it cannot not be an element of w. So y
cannot exist.
So there is at least one element in P(S) which is
not in the image of f, so f is not an onto function,
and it is not a bijection.
What is wrong with this proof in your opinion?
Stephen
Virgil
albs...@gmx.de wrote:
> David R Tribble wrote:
> At first, you should show, that bijection means something to
> notwellordered infinite sets.
>
> Bijection is a clear concept on finite sets, it also works on
> wellordered infinite sets of the same infinite concept.
> Aber: Show me a bijection between two infinite sets with the same
> cardinality, where one of the sets is still not wellorderable.
If either of two sets is well-orderable and there is a bijection between
the sets, then that bijection induces well-order-ability on the other.
So that you cannot have a bijection between two sets when one is
well-orderable and the other is not.
David R Tribble
>> I already showed you the bijection between binary *N and P(*N).
>> What didn't you like about it? It is valid.
>
David R Tribble said:
>> No, you showed a mapping between *N and R, which is equivalent
>> to a mapping between *N and P(N). That's easy.
>
Tony Orlow wrote:
> No, it was specifically a bijection between two sets of infinite binary
> strings representing, on the one hand, the whole numbers in *N starting
> from 0, both finite and infinite, in normal binary format, and on the other
> hand, the specification of each subset of whole numbers in *N, where each
> bit which, in the binary number, represents 2^n denotes membership of n in
> the subset. This is a bijection between the whole numbers in *N and P(*N),
> using an intermediate bijection with a common set of infinite binary strings.
But that's an incomplete mapping, because there are not enough infinite
binary strings in *N to enumerate all of the subsets of *N. Try it,
if you don't believe me.
David R Tribble
>> Cantor proofs his wrong conclusion with the same mix of potential
>> infinity and actual infinity. But there is no bijection between this
>> two concepts. The antidiagonal is an unicorn.
>> There is no stringend concept about infinity. And there is no aleph_1,
>> aleph_2, ... or any other infinity.
>
David R Tribble wrote:
>> For that to be true, there must be a bijection between an infinite
>> set (any infinite set) and its powerset. Bitte, show us a bijection
>> between N and P(N).
>
Albrecht S. Storz wrote:
> At first, you should show, that bijection means something to
> notwellordered infinite sets.
>
> Bijection is a clear concept on finite sets, it also works on
> wellordered infinite sets of the same infinite concept.
> Aber: Show me a bijection between two infinite sets with the same
> cardinality, where one of the sets is still not wellorderable.
> Than I will show you a bijection between N and P(N) or N and R or P(N)
> and P(P(N)) or what you want.
I see. I'm supposed to show you a proof before you can show me your
proof. Okay, I give up, you win, so your proof must be correct.
Come on, now. It's up to you to prove your own claim, especially
when it contradicts established mathematics. I know you cannot show
a bijection between N and P(N).
P.S.
Let
D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,...
This is the i-th binary digit of natural n
L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i
This is the number of binary digits of natural n, or ceil(log2(n))
Let
M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1
This reverses the binary digits of n.
Then M(n) is a mapping N -> N, from all n in N to M(n) in N,
but the set of all M(n) is not a well-ordered set.
Happy?
Tony Orlow
> Tony Orlow <ae...@cornell.edu> wrote:
> > David R Tribble said:
> >>
> >> But you have not provided a mapping between any set and its powerset,
> >> infinite or otherwise.
> > Have too.
>
> No you have not Tony.
Have, too!
> The proof that there does not exist
> a bijection between a set and its power set is quite short.
Then it shouldn't take too much looking to see where it goes wrong....
>
> Let f be a function from S to P(S).
Our proposed mapping bijection....
>
> Define the set w as follows:
>
> w= { x : x in S and x is not in f(x) }
So, w is the set of all elements which are not members of the subsets which
they map to through f(x)....
>
> Clearly w is a subset of S, and so w is an element of P(S).
Clearly...but properly?
>
> We now show that w is not in the image of f. That is,
> there does not exist a y such that f(y)=w.
So, there can be no y such that it maps to the subset of all elements which do
not map to subsets containing themselves? We'll see.....
>
> Suppose such a y exists. If such a y exists, it must
> either be an element of w, or not.
One or the other. I'll accept the excluded middle....
>
> If y is an element of w, then y is in f(y), which means
> it is not an element of w.
If y is in w, this means y is a member of the set of elements which map to
subsets which do not contain themselves. This means that y is in S but not in f
(y). So, indeed, it IS a member of w. There is no reason why both x and y
cannot map to subsets which do not contain themselves.
If y is a member of w, then all you can say is that y does not map to any
subset which contains itself. Y does not map to w. Neither does x.
>
> If y is not an element of w, then y is not in f(y), which
> means it is an element of w.
If y is not a member of w, then y maps to a subset which contains y as a
member, and y IS in f(y). Is this possible? We'll see what you think.....
>
> These are both contradictions. So y cannot be an element
> of w, and it cannot not be an element of w. So y
> cannot exist.
Neither of those possibilities causes a contradiction. You are getting confused
with your double negatives. If w is the set of all elements which do not map to
subsets containing themselves, then being a member of w means simply that y
does not map to a subset containing y, which is perfectly possible. Not being a
member of w means that an element maps to a subset which DOES contain itself.
Where is the contradiction?
>
> So there is at least one element in P(S) which is
> not in the image of f, so f is not an onto function,
> and it is not a bijection.
Sorry, not so.
>
> What is wrong with this proof in your opinion?
You confused yourself with double negatives. If I misinterpreted any of what
you said, please clarify. On the face of it, you ain't got no proof.
Let's put this in terms of the bijection between *N and P(*N). Given the common
binary string representation of the binary naturals and the subset
specifications, and starting from 0, let's see what's in w. 0 maps to a string
of all 0's, representing the null set. 1 maps to the set containing only zero.
2 maps to the set containing only 1. 3 maps to the subset containing 0 and 1. 4
maps to the subset containign only 2. f(5)={0,2}, f(6)={1,2}, f(7)={0,1,2}, f
(8)={3}. Do you see a pattern here? Do you see ANY elements of *N which map to
subsets of *N which contain themselves? There aren't any. Your proof doesn't
apply. there is no contradiction in it. *N bijects with P(*N) through the set
of infinite binary strings.
Now, w seems to be the entire set S, since there is no element which is a
member of the subset it denotes as a binary string. So what element maps to w?
It would appear to be an infinite string of 1's, one for every element. You
might also think this infinite string of 1's represents the largest element of
*N, in which case you might think that this is the one number that is not in w.
But, be careful! If you have a string of N 1's for the N elements of S, then in
binary this string of N digits equals 2^N-1, and you have 2^N subsets, from 0
through this number. So, despite the fact that you can create the bijection,
the two sets are clearly different sizes, and bijection is shown NOT to
indicate equivalence between sets.
Give me a Q! Q!!! Give me an E! E!!! Give me a D! D!!! Whaddya got? QED!!!!!
>
> Stephen
>
--
Smiles,
Tony
Tony Orlow
that I cannot construct a bijection between the two on an element-by-element
basis which continues infinitely through the set of binary strings? Where does
it stop? You aren't beginning to see that determining the infinite value range
is crucial to this problem after all, are you? What do you want me to try,
anyway, and infinite mapping, element-by-element? A bijection's a bijection,
right? These sets are obviously of the same cardinality, since I have a
bijection which carries to infinity, right? For every subset there is a unique
natural and for every natural there is a unique subset. If you disagree, please
state which of either has no corresponding element in the other.
Please try to frame your objection in a more operative manner. How do you know
there aren't a large enough infinity of bits for the power set vs the values in
the set?
--
Smiles,
Tony
Tony Orlow
to have an infinite number of finite naturals, so you would have infinite bit
strings in P(N) and only finite bit strings in N. But, of course, this is an
artificial and incorrect interpretation of the situation, as I have repeatedly
tried to convey.
>
>
> P.S.
>
> Let
> D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,...
> This is the i-th binary digit of natural n
> L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i
> This is the number of binary digits of natural n, or ceil(log2(n))
> Let
> M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1
> This reverses the binary digits of n.
>
> Then M(n) is a mapping N -> N, from all n in N to M(n) in N,
> but the set of all M(n) is not a well-ordered set.
> Happy?
ordered? Doesn't it have a first element, which is all 0's, and a successor to
every string in the set? Hmmm..... What is your point?
>
>
--
Smiles,
Tony
ste...@nomail.com
> Have, too!
If y is in w, then y is in f(y), because w=f(y). Remember,
we are assuming there exists a y such that f(y)=w. However
w is defined such that y can only be in w, if y is not in f(y).
> If y is a member of w, then all you can say is that y does not map to any
> subset which contains itself. Y does not map to w. Neither does x.
What is x? If y is a member of w, then y is in f(y), and
by the definition of w, w is not a member of w.
>>
>> If y is not an element of w, then y is not in f(y), which
>> means it is an element of w.
> If y is not a member of w, then y maps to a subset which contains y as a
> member, and y IS in f(y). Is this possible? We'll see what you think.....
>>
>> These are both contradictions. So y cannot be an element
>> of w, and it cannot not be an element of w. So y
>> cannot exist.
> Neither of those possibilities causes a contradiction. You are getting confused
> with your double negatives. If w is the set of all elements which do not map to
> subsets containing themselves, then being a member of w means simply that y
> does not map to a subset containing y, which is perfectly possible. Not being a
> member of w means that an element maps to a subset which DOES contain itself.
> Where is the contradiction?
The contradiction is that if y is in w, then it is not in w,
and if y is not in w, then it is in w. That is a pretty
obvious contradiction.
You apparently failed to grasp the very important fact that w=f(y).
So once again, is y in w?
If y is in w, then y is in f(y), because w=f(y).
If y is in f(y), then y is not in w, because w is defined
as containing the elements x in S such that x is not in f(x).
w cannot contain y, because y is in f(y).
If y is not in w, then y is not in f(y), because w=f(y).
If y is not in f(y), then y is in w, because w is defined
as containing the elements x in S such that x is not in f(x).
w must contain y, because y is not in f(y).
>>
>> So there is at least one element in P(S) which is
>> not in the image of f, so f is not an onto function,
>> and it is not a bijection.
> Sorry, not so.
Yes. The fact that you cannot understand a simple
proof does not make the proof invalid.
>>
>> What is wrong with this proof in your opinion?
> You confused yourself with double negatives. If I misinterpreted any of what
> you said, please clarify.
You apparently misintrepted all of it.
Do you understand that we are assuming that w=f(y)?
Do you understand that y is in w if and only if y is not in f(y)?
Do you understand that
y is in w if and only if y is not in w
is a contradiction?
Stephen
David R Tribble
>> But you have not provided a mapping between any set and its powerset,
>> infinite or otherwise.
>
Tony Orlow wrote:
>> Have too.
Stephen said:
>> No you have not Tony.
>
Tony Orlow wrote:
> Have, too!
>
Stephen said:
>> The proof that there does not exist
>> a bijection between a set and its power set is quite short.
>> Define the set w as follows:
>> W = { x : x in S and x is not in f(x) }
>>
>> Clearly W is a subset of S, and so w is an element of P(S).
>> We now show that W is not in the image of f. That is,
>> there does not exist a y such that f(y)=w.
Here's a more visual example of that proof...
Let
B(n,i) = floor(n/2^i) mod 2, for all n = 0,1,2,3,...
So B(n,i) is the i-th binary digit of natural n.
Let
F(n) = { i where B(n,i+1) > 0 for all i = 0,1,2,3,...}
for all n = 0,1,2,3,...
So F(n) is the set of all naturals representing non-zero binary digits
of natural n.
Example:
n = 69
n = 1000101 (binary)
n = 1x2^0 + 0x2^1 + 1x2^2 + 0x2^3 + 0x2^4 + 0x2^5 + 1x2^6
B(69,0) = 1
B(69,2) = 1
B(69,6) = 1
so
F(69) = {0,2,6}
Thus we have a mapping function that generates a subset F(n) of N
for any natural n in N.
Now we list the subsets F(n) for all n:
F(0) = {}
F(1) = {0}
F(2) = {1}
F(3) = {0,1}
F(4) = {2}
F(5) = {0,2}
F(6) = {1,2}
F(7) = {0,1,2}
F(8) = {3}
F(9) = {0,3}
...
Clearly the F(n) sets are all the finite subsets of N, with each
subset corresponding to a unique finite natural n in N (and vice
versa). It is also clear that since every F(n) is subset of N, it
is also a member of P(N), the powerset of N.
Now, applying Stephen's (Cantor's) set W, we let
W = { n : n in N and n is not in F(n) }
In other words, W is the set of all naturals n in N where n is not
itself a member of subset F(n). From the list of F(n)'s above, it is
obvious that none of the F(n) subsets contains the n that generated
it, i.e., for every given n, F(n) does not contain n. (This is
because for any n, the largest member of F(n) is ceil(log2(n)), which
is always less than n, so n cannot itself be a member of F(n).)
This means that set W is simply N itself, since none of the members
in N appear in their corresponding F(n) subset; thus every member n
of N is a member of W, so W = N.
The final step of the proof is:
there does not exist a y in N such that F(y) = W.
Assume that y exists and that F(y) = W. Is y in W?
If y is a member of W (and N), then y is a member of F(y), because
we just assumed that W = F(y). But the definition of W means that
y cannot be in F(y).
Contradiction.
If y is not a member of W (or N), then y is a not a member of F(y)
either, because W = F(y). But the definition of W means that
since y is not a member of F(y), y must be a member of W.
Contradiction.
So either way, y cannot both be a member and not be a member of
F(y) = W = N. So there does not exist a y in N that maps to N
itself.
Which means that the P(N), the powerset of N, has members that
cannot be mapped by any member of N itself, which means that P(N)
has more members than N.
(This proof also works for infinite sets with infinite naturals,
such as *N and P(*N).)
David R Tribble
>> I already showed you the bijection between binary *N and P(*N).
>> What didn't you like about it? It is valid.
>
David R Tribble said:
>> No, you showed a mapping between *N and R, which is equivalent
>> to a mapping between *N and P(N). That's easy.
>
Tony Orlow wrote:
>> No, it was specifically a bijection between two sets of infinite binary
>> strings representing, on the one hand, the whole numbers in *N starting
>> from 0, both finite and infinite, in normal binary format, and on the other
>> hand, the specification of each subset of whole numbers in *N, where each
>> bit which, in the binary number, represents 2^n denotes membership of n in
>> the subset. This is a bijection between the whole numbers in *N and P(*N),
>> using an intermediate bijection with a common set of infinite binary
>> strings.
>
David R Tribble said:
>> But that's an incomplete mapping, because there are not enough infinite
>> binary strings in *N to enumerate all of the subsets of *N. Try it,
>> if you don't believe me.
>
Tony Orlow wrote:
> Not enough infinite binary strings? How many in *N and in P(*N)?
> Are you saying that I cannot construct a bijection between the two on an
> element-by-element basis which continues infinitely through the set of
> binary strings?
That's exactly what I'm saying.
card(*N) = c, but card(P(*N)) = 2^c, and c < 2^c.
> Where does it stop? You aren't beginning to see that determining the
> infinite value range is crucial to this problem after all, are you?
If you think that approach will work, then show us how.
By the way, what is the "range" of a powerset, whose members are
sets themselves?
> What do you want me to try, anyway, and infinite mapping,
> element-by-element? A bijection's a bijection, right?
Yes, that would be nice. Please show us your bijection.
> These sets are obviously of the same cardinality, since I have a
> bijection which carries to infinity, right?
No, their sizes are different cardinalities, which happen to be
different infinities. Both sets are infinite, but one set is
larger than the other.
> For every subset there is a unique natural and for every natural there is
> a unique subset. If you disagree, please state which of either has no
> corresponding element in the other.
Like I said, there are not enough naturals to map to every subset
of the naturals.
This is true whether you've got infinite naturals or not; there are
not enough members in N to enumerate all the subsets in N, and
likewise there are not enough members in *N to enumerate all the
subsets in *N.
> Please try to frame your objection in a more operative manner. How do you
> know there aren't a large enough infinity of bits for the power set vs the
> values in the set?
Because I can prove it (and it's a very old proof). A powerset of
a nonempty set contains more elements that the set. Can you prove
otherwise?
David R Tribble
>> At first, you should show, that bijection means something to
>> notwellordered infinite sets.
>
David R Tribble said:
>> Let
>> D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,...
>> This is the i-th binary digit of natural n
>> L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i
>> This is the number of binary digits of natural n, or ceil(log2(n))
>> Let
>> M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1
>> This reverses the binary digits of n.
>>
>> Then M(n) is a mapping N -> N, from all n in N to M(n) in N,
>> but the set of all M(n) is not a well-ordered set.
>> Happy?
>
Tony Orlow wrote:
> Is that supposed to make me happy?
It was supposed to make Albrecht happy, since I was responding to him.
Slight correction:
M(n) = sum{i=0 to L(n)} 2^L(n)2^(L(n)-i+1) where D(n,i) = 1
David Kastrup
"David R Tribble" <da...@tribble.com> writes:
> Because I can prove it (and it's a very old proof). A powerset of a
> nonempty set contains more elements that the set.
Drop the "nonempty", it is not necessary. P({}) = {{}} contains more
elements than {}, and the old proof works even for that case.
--
David Kastrup, Kriemhildstr. 15, 44793 Bochum
albs...@gmx.de
Sad!
First of all you don't argue on my claim of the thread.
Second, your above argueing is not clear to me, since both sets are
well-ordered. But it's nice, so I give this:
N
{1},{2},{3}, ...
N/{1},N/{2},N/{3}, ...
{1,2},{1,3},{2,3},{1,4},...
N/{1,2}, ...
...
Now count in diagonal sequence. You may think of Cantor's first
diagonal proof.
Which subset of N is not included?
Regards
AS
David R Tribble
>> Because I can prove it (and it's a very old proof). A powerset of a
>> nonempty set contains more elements that the set.
>
David Kastrup wrote:
> Drop the "nonempty", it is not necessary. P({}) = {{}} contains more
> elements than {}, and the old proof works even for that case.
Yep, I forgot about that one!
ste...@nomail.com
Is this supposed to be a list? My reading
of this is that your list is:
N,
{1},
{2},
{3},
...
N/{1},
Right here we have a problem. What is the element
before N/{1} in your "list"? There is no end to the
list
{1}, {2}, {3}, ...
so you cannot put N/{1} after the end of that list.
Remember each element in a list must be indexed
by a natural number.
> Now count in diagonal sequence. You may think of Cantor's first
> diagonal proof.
What diagonal?
> Which subset of N is not included?
N/{1} for one. What is the index of N/{1} in your
"list" above? The index of N is 0. The index of {1}
is 1. The index of {2} is 2. What is the index
of N/{1}?
Also your "enumeration" above only includes finite
sets, or sets whose complement with respect to N is finite.
That excludes an awful lot of sets. In fact it excludes
"most" of them. For example, where is the set of primes going
to show up?
Stephen
David R Tribble
>> Come on, now. It's up to you to prove your own claim, especially
>> when it contradicts established mathematics. I know you cannot show
>> a bijection between N and P(N).
>>
>
Albrecht S. Storz wrote:
> Second, your above argueing is not clear to me, since both sets are
> well-ordered. But it's nice, so I give this:
>
> N
> {1},{2},{3}, ...
> N/{1},N/{2},N/{3}, ...
> {1,2},{1,3},{2,3},{1,4},...
> N/{1,2}, ...
> ...
>
> Now count in diagonal sequence. You may think of Cantor's first
> diagonal proof.
>
> Which subset of N is not included?
I don't know what you mean by "count in diagonal sequence".
Let's try a simpler approach...
Let S0 = {}, the empty set.
Let S1 = {{0}, {1}, {2}, {3}, ...},
so it is the set of all singleton subsets of N.
Let S2 = {{0,1}, {0,2}, {1,2}, {2,3}, ...},
so it is the set of all two-member subsets of N.
And so on, defining S3, S4, S5, etc., so that Sn(N) is the set of
all n-member subsets of N.
Now let U = S0 u S1 u S2 u S3 u ...
It seems to me that N is as large as S1, since for every member n
in N, there is a member {n} in S1, right? So shouldn't U be even
larger than N? And isn't U just P(N)?
[I realize this might not be completely correct as per standard set
theory; but I'd like to see Albrecht's response.]
William Hughes
Sigh. Try {2,4,6,8,...} Or any other infinite subset
that is not the complement of a finite set.
It is common for people to note that the finite subsets of N
are countable and incorrectly claim that the subsets of N are
countable. Adding in the complements of the finite subsets
does not change things very much.
Those who do not study anti-cantor cranks are doomed to
repeat their idiocies.
- William Hughes
P.S. No TO, going to TO-infinite rows is not going to help us.
The problem is that for any TO-finite row, the subsets
listed will have an bounded finite number of elements, or be
the complement of a subset with a finite number of elements.
Yes, you can claim (without a shred of motivation) that
when you get to TO-infinite rows the subsets will suddenly
have an unbounded number of elements, but all this tells us
is that a bijection from the TO-naturals to P(N) exists.
What we need is a bijection from the finite TO-naturals
to P(N).
albs...@gmx.de
>
> Because I can prove it (and it's a very old proof). A powerset of
> a nonempty set contains more elements that the set. Can you prove
> otherwise?
This argument is stupid. Is there any magic in the powerfunction? A
hidden megabooster for transcendental overflow? What is the very
special aspect of the powerfunction to be so magic?
Why should all operations with transfinite numbers lead to results with
the same "level" of infinity, but only powerfunction beams up to the
next level?
Is not true: a^2 = a*a? 2a = a+a?
Is the powerfunction something other than a very shortcut for multiple
additions?
Which amount you are able to reach with powerfunction which is
unreachable by succesor operation?
I'm very sensible about this because this argument is found in very
much books although it's total meaningless.
(Weak minds might be impressed by the big numbers which are easily
produced by powerfunction.)
What in finity holds may not (or do not) hold in infinity.
Regards
AS
albs...@gmx.de
That's right.
So biject two non-well-orderable sets.
Regards
AS
William Hughes
albst...@gmx.de wrote:
> David R Tribble wrote:
>
> >
> > Because I can prove it (and it's a very old proof). A powerset of
> > a nonempty set contains more elements that the set. Can you prove
> > otherwise?
>
> This argument is stupid. Is there any magic in the powerfunction? A
> hidden megabooster for transcendental overflow? What is the very
> special aspect of the powerfunction to be so magic?
> Why should all operations with transfinite numbers lead to results with
> the same "level" of infinity, but only powerfunction beams up to the
> next level?
> Is not true: a^2 = a*a? 2a = a+a?
Yes, but the powerfunction does not look like a^2 but 2^a.
> Is the powerfunction something other than a very shortcut for multiple
> additions?
Yes. You cannot represent 2^x as multiple additions.
> Which amount you are able to reach with powerfunction which is
> unreachable by succesor operation?
Infinity for one. You cannot get from a finite quantity to
an infinite quantity by using the successor operation (unless
like TO you are willing to wave a circular magic wand and apply
the successor operation an infinite number of times).
>
> I'm very sensible about this because this argument is found in very
> much books although it's total meaningless.
>
I suspect that you mean "sensitive" not "sensible".
> (Weak minds might be impressed by the big numbers which are easily
> produced by powerfunction.)
Strong minds are impressed with the fact that there is no
bijection between X and P(X).
>
>
> What in finity holds may not (or do not) hold in infinity.
Words to live by. Start by noting that a finite set has a
"number of elements" that can be described by a natural number
while an infinite set (e.g. the set of natural numbers) does
not have a "number of elements" that can be described by a
natural number. However, some things are true for both
finite and infinite sets. e.g. the fact that there is no
bijection between X and P(X).
-William Hughes
Daryl McCullough
>N
>{1},{2},{3}, ...
>N/{1},N/{2},N/{3}, ...
>{1,2},{1,3},{2,3},{1,4},...
>N/{1,2}, ...
>...
>
>Now count in diagonal sequence. You may think of Cantor's first
>diagonal proof.
>
>Which subset of N is not included?
I'm not sure exactly what your list is supposed to mean.
I think you're trying to say that
row 2 = all sets with 1 element
row 4 = all sets with 2 elements
row 6 = all sets with 3 elements
etc.
row 1 = the set containing all elements
row 3 = the sets missing 1 element
row 5 = the sets missing 2 elements
etc.
If that's what you mean, then you have left
out the set of all even numbers. You've left
out the set of all odd numbers. You've left
out the set of all prime numbers. You've left
out the set of all powers of two.
You've left out a whole bunch of sets.
--
Daryl McCullough
Ithaca, NY
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> ste...@nomail.com said:
What x has f(x) = w? If there is no such x, then f is not a surjection.
> >
> > So there is at least one element in P(S) which is not in the image
> > of f, so f is not an onto function, and it is not a bijection.
> Sorry, not so.
Sorry, but it is only "not so" in TOmatics, it is so everywhere else.
> >
> > What is wrong with this proof in your opinion?
> You confused yourself with double negatives. If I misinterpreted any
> of what you said, please clarify. On the face of it, you ain't got no
> proof.
>
> Let's put this in terms of the bijection between *N and P(*N). Given
> the common binary string representation of the binary naturals and
> the subset specifications, and starting from 0, let's see what's in
> w.
TO assumes what is not in evidence here, that there is a bijection
between *N and P(*N) through his non-existent double-ended endless
sequences. Having assumed it, TO has no difficulty in proving it follows
from its assumption. Psuedo-proof deleted.
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> David R Tribble said:
> > Tony Orlow wrote:
> > >> I already showed you the bijection between binary *N and P(*N).
> > >> What didn't you like about it? It is valid.
> > >
> >
> > David R Tribble said:
> > >> No, you showed a mapping between *N and R, which is equivalent
> > >> to a mapping between *N and P(N). That's easy.
> > >
> >
> > Tony Orlow wrote:
> > > No, it was specifically a bijection between two sets of infinite binary
> > > strings representing, on the one hand, the whole numbers in *N starting
> > > from 0, both finite and infinite, in normal binary format, and on the
> > > other
> > > hand, the specification of each subset of whole numbers in *N, where each
> > > bit which, in the binary number, represents 2^n denotes membership of n
> > > in
> > > the subset. This is a bijection between the whole numbers in *N and
> > > P(*N),
> > > using an intermediate bijection with a common set of infinite binary
> > > strings.
> >
> > But that's an incomplete mapping, because there are not enough infinite
> > binary strings in *N to enumerate all of the subsets of *N. Try it,
> > if you don't believe me.
> >
> >
> Not enough infinite binary strings?
Precisely.
albs...@gmx.de
> David R Tribble wrote:
> >> Come on, now. It's up to you to prove your own claim, especially
> >> when it contradicts established mathematics. I know you cannot show
> >> a bijection between N and P(N).
> >>
> >> [Example snipped]
> >
>
> Albrecht S. Storz wrote:
> > Second, your above argueing is not clear to me, since both sets are
> > well-ordered. But it's nice, so I give this:
> >
> > N
> > {1},{2},{3}, ...
> > N/{1},N/{2},N/{3}, ...
> > {1,2},{1,3},{2,3},{1,4},...
> > N/{1,2}, ...
> > ...
> >
> > Now count in diagonal sequence. You may think of Cantor's first
> > diagonal proof.
> >
> > Which subset of N is not included?
>
> I don't know what you mean by "count in diagonal sequence".
It's a usual visualisation of Cantors first diagonal proof. You can go
through diagonally e.g.
N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ...
It's a good way to understand, that the most subsets of the powerset
are this, which are infinit and and also their counter parts. And at
the end, are unconstructable.
But this is far away from the intend of this thread.
Exists the powerset? Exists the sum of all naturals?
The definition of infinity is wrong. I had shown an easy understandable
argument.
Regards
AS
Virgil
Tony Orlow <ae...@cornell.edu> wrote:
> David R Tribble said:
> > Come on, now. It's up to you to prove your own claim, especially
> > when it contradicts established mathematics. I know you cannot
> > show a bijection between N and P(N).
> The only reason you cannot show a bijection between those two is that
> you claim to have an infinite number of finite naturals, so you would
> have infinite bit strings in P(N) and only finite bit strings in N.
> But, of course, this is an artificial and incorrect interpretation of
> the situation, as I have repeatedly tried to convey.
But that peculiar situation only holds in the wild woolly world of
TOmatics where everything is both provable and disprovable, and
definitely does not hold in the standard world where the negaation of
any provable statement is necessarily unprovable.
ste...@nomail.com
> David R Tribble wrote:
>> David R Tribble wrote:
>> >> Come on, now. It's up to you to prove your own claim, especially
>> >> when it contradicts established mathematics. I know you cannot show
>> >> a bijection between N and P(N).
>> >>
>> >> [Example snipped]
>> >
>>
>> Albrecht S. Storz wrote:
>> > Second, your above argueing is not clear to me, since both sets are
>> > well-ordered. But it's nice, so I give this:
>> >
>> > N
>> > {1},{2},{3}, ...
>> > N/{1},N/{2},N/{3}, ...
>> > {1,2},{1,3},{2,3},{1,4},...
>> > N/{1,2}, ...
>> > ...
>> >
>> > Now count in diagonal sequence. You may think of Cantor's first
>> > diagonal proof.
>> >
>> > Which subset of N is not included?
>>
>> I don't know what you mean by "count in diagonal sequence".
> It's a usual visualisation of Cantors first diagonal proof. You can go
> through diagonally e.g.
> N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ...
Let us restate that as a function from N to your above list:
1 -> N
2 -> {1}
3 -> {2}
4 -> N/{1}
5 -> {1,2}
6 -> {3}
7 -> {4}
8 -> N/{3}
9 -> {1,3},
...
So if we consider
w = { x : x is not in f(x) }
we get
w = { 2, 3, 5, 6, 7, 9, ....
Where does that appear in your list?
> It's a good way to understand, that the most subsets of the powerset
> are this,
No. Most of the elements of the power set are not this.
This is a countable subset of an uncountable set.
> which are infinit and and also their counter parts. And at
> the end, are unconstructable.
At the end? The end of what? The end of the unending list?
There is no end to an unending list. That is what unending
means. Unendlich!!!
> But this is far away from the intend of this thread.
> Exists the powerset? Exists the sum of all naturals?
> The definition of infinity is wrong. I had shown an easy understandable
> argument.
You showed that you do not understand the definition
of cardinality.
Stephen
Virgil
albs...@gmx.de wrote:
> David R Tribble wrote:
>
> >
> > Because I can prove it (and it's a very old proof). A powerset of
> > a nonempty set contains more elements that the set. Can you prove
> > otherwise?
>
> This argument is stupid. Is there any magic in the powerfunction?
"Proofs" are not stupid until they can be refuted. The proof that for an
arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone.
Virgil
albs...@gmx.de wrote:
With or without an axiom of choice?
Virgil
albs...@gmx.de wrote:
> It's a usual visualisation of Cantors first diagonal proof. You can go
> through diagonally e.g.
>
> N, {1}, {2}, N/{1}, {1,2}, N/{2}, {3}, {4}, N/{3}, {1,3}, ...
This version of sequencing subsets of the set of naturals omits most
subsets of the set of naturals. it includes only those which ase finite
or whose relative compliments are finite, but most subsets are both
infinite and with infinite compliments.
imagin...@despammed.com
ste...@nomail.com wrote:
> albs...@gmx.de wrote:
<snip: my goodness this stuff goes on and on...>
> > First of all you don't argue on my claim of the thread.
> > Second, your above argueing is not clear to me, since both sets are
> > well-ordered. But it's nice, so I give this:
>
> > N
> > {1},{2},{3}, ...
> > N/{1},N/{2},N/{3}, ...
> > {1,2},{1,3},{2,3},{1,4},...
> > N/{1,2}, ...
> > ...
>
> Is this supposed to be a list? My reading
> of this is that your list is:
>
> N,
> {1},
> {2},
> {3},
> ...
> N/{1},
>
> Right here we have a problem. What is the element
> before N/{1} in your "list"? ...
>
> > Now count in diagonal sequence. You may think of Cantor's first
> > diagonal proof.
>
> What diagonal?
Come on, come on! He means the zigzag diagonal, as in the standard
demonstration that the rationals _are_ countable. This isn't "Cantor's
first diagonal proof", but if you're going to argue with cranks you
must expect them to be pretty muddled about things.
Anyway, it's obvious that *if* the OP shows a "list of lists" that
include all the subsets that is enough. In practice, of course he's
given the standard crank non-list.
Brian Chandler
http://imaginatorium.org
albs...@gmx.de
I see, you are the real checker. You knows it all. You are famous. You
are apodictic. All the authors who speak of the first diagonal proof of
Cantor are wrong.
> but if you're going to argue with cranks you
> must expect them to be pretty muddled about things.
>
> Anyway, it's obvious that *if* the OP shows a "list of lists" that
> include all the subsets that is enough. In practice, of course he's
> given the standard crank non-list.
>
> Brian Chandler
> http://imaginatorium.org
You and many of the other checkers are not able to discuss my starting
argument. You are only able to respond to the usual wrong arguments you
know. And I think, your answers are memorized because you are unable to
think your own thoughts.
Regards
AS
imagin...@despammed.com
> imagin...@despammed.com wrote:
> > ste...@nomail.com wrote:
> > > albs...@gmx.de wrote:
> > <snip: my goodness this stuff goes on and on...>
> > > > Now count in diagonal sequence. You may think of Cantor's first
> > > > diagonal proof.
> > >
> > > What diagonal?
> >
> > Come on, come on! He means the zigzag diagonal, as in the standard
> > demonstration that the rationals _are_ countable. This isn't "Cantor's
> > first diagonal proof",
>
> I see, you are the real checker. You knows it all. You are famous. You
> are apodictic. All the authors who speak of the first diagonal proof of
> Cantor are wrong.
Remind me: "Cantor's first diagonal proof" shows what?
Look, you're a crank, and you're muddled. We expect that. I'm trying to
*help* you by pointing out that even if what you are trying to say is
not quite right (the term "count in diagonal sequence" is not
standard), at least we can see what you mean.
> You and many of the other checkers are not able to discuss my starting
> argument. You are only able to respond to the usual wrong arguments you
> know. And I think, your answers are memorized because you are unable to
> think your own thoughts.
Fool. OK, here's a thought of mine - I see you claim to have a list
(well, actually a list of lists, but it comes to the same thing) of all
subsets of the naturals. In this list, which comes first: the set of
even numbers, or the set of prime numbers, and why?
Brian Chandler
http://imaginatorium.org
albs...@gmx.de
William Hughes wrote:
> albst...@gmx.de wrote:
> > David R Tribble wrote:
> >
> > >
> > > Because I can prove it (and it's a very old proof). A powerset of
> > > a nonempty set contains more elements that the set. Can you prove
> > > otherwise?
> >
> > This argument is stupid. Is there any magic in the powerfunction? A
> > hidden megabooster for transcendental overflow? What is the very
> > special aspect of the powerfunction to be so magic?
> > Why should all operations with transfinite numbers lead to results with
> > the same "level" of infinity, but only powerfunction beams up to the
> > next level?
> > Is not true: a^2 = a*a? 2a = a+a?
>
> Yes, but the powerfunction does not look like a^2 but 2^a.
>
> > Is the powerfunction something other than a very shortcut for multiple
> > additions?
>
> Yes. You cannot represent 2^x as multiple additions.
Since we talk about x e {1,2,3,4,5,...}, why not?
>
> > Which amount you are able to reach with powerfunction which is
> > unreachable by succesor operation?
>
> Infinity for one. You cannot get from a finite quantity to
> an infinite quantity by using the successor operation (unless
> like TO you are willing to wave a circular magic wand and apply
> the successor operation an infinite number of times).
But with 2^n you will reach infinity? Than you also will reach it with
1+1+1+1+...
>
> >
> > I'm very sensible about this because this argument is found in very
> > much books although it's total meaningless.
> >
>
> I suspect that you mean "sensitive" not "sensible".
Of course.
>
>
> > (Weak minds might be impressed by the big numbers which are easily
> > produced by powerfunction.)
>
> Strong minds are impressed with the fact that there is no
> bijection between X and P(X).
>
> >
> >
> > What in finity holds may not (or do not) hold in infinity.
>
> Words to live by. Start by noting that a finite set has a
> "number of elements" that can be described by a natural number
> while an infinite set (e.g. the set of natural numbers) does
> not have a "number of elements" that can be described by a
> natural number. However, some things are true for both
> finite and infinite sets. e.g. the fact that there is no
> bijection between X and P(X).
>
> -William Hughes
No.
Regards
AS
albs...@gmx.de
I think it is more interesting for the most people without AC, since AC
is not widely loved by the mathematics.
But you may show both if you want.
Regards
AS
albs...@gmx.de
Even if you think that the powersets of finite and infinite sets have
both a greater cardinality than their starting sets, you would not
really think it depends on the same cause in both cases.
You must proof it independently for finite and for infinite sets. In
this sense the argument is stupid.
Regards
AS
David Kastrup
Uh, no. The proof depends merely on the fact that some value has to
be either a member of a set, or not. And if some value is in one set,
bur not another, then those two sets are different.
That's all. Finiteness or infiniteness does not even play into it.
The proof just constructs a set which differs by the membership of at
least one particular value with every target set in the assumedly
complete mapping of set to powerset.
albs...@gmx.de
David R Tribble wrote:
> Albrecht S. Storz wrote:
> >> Since there is no biggest number and since there is no infinite number,
> >> the size of the set of numbers in form of sets of #s is undefined as
> >> the biggest natural number is undefined.
> >>
> >> But the sequence of the sets of # fullfill the peano axiomes. So this
> >> set must be infinite.
> >>
> >> The cardinality of a set is not able to be infinite and "not defined"
> >> at the same time.
> >> This is the contradiction.
> >
>
> David R Tribble wrote:
> >> I don't see the contradiction. The size of the set is "not defined"
> >> to be the same as any natural number, and the set size is obviously
> >> infinite. This is no contradiction, since no natural number is
> >> infinite.
> >>
> >> The thing that is "not defined" is the largest natural, which obviously
> >> does not exist. But the set size is infinite, and is nicely defined
> >> by an infinite cardinal.
> >>
> >> You seem to be mixing the two concepts of "natural" and "cardinal"
> >> numbers to create a supposed contradiction, but that does not work.
> >
>
> Albrecht S. Storz wrote:
> > You are not able to understand that there is no difference between
> > numerals and sets.
>
> I have no problem seeing the correspondence between natural numbers
> and von Neumann sets. But neither of these are the same as
> cardinalities, which are not numbers, but measures (sizes) of sets.
Natural numbers are sets. Why be so delicate about this? It's not only
a correspondence between them. It's identity. Only a fool is unable to
see that fact if the numbers are shown in unitary (1-adic) system.
And now the cardinality. The definition of cardinality bases on sets.
The existence of infinite sets bases on definition. The meaning of the
cardinality of an infinite set is unknown.
Coincidently natural numbers and cardinalities are undistinguishable in
finity. Cardinality is just a artificial concept with no sens and
meaning.
Define the existence of unicorns and be happy.
Regards
AS
>
>
> > My sketches shows this exactly.
> > Cantor proofs his wrong conclusion with the same mix of potential
> > infinity and actual infinity. But there is no bijection between this
> > two concepts. The antidiagonal is an unicorn.
> > There is no stringend concept about infinity. And there is no aleph_1,
> > aleph_2, ... or any other infinity.
>
> For that to be true, there must be a bijection between an infinite
> set (any infinite set) and its powerset. Bitte, show us a bijection
> between N and P(N).
albs...@gmx.de
David Kastrup wrote:
> Tony Orlow <ae...@cornell.edu> writes:
>
> > ste...@nomail.com said:
> >> Tony Orlow <ae...@cornell.edu> wrote:
> >> > ste...@nomail.com said:
> >> >>
> >> >> > But there is a slight difference. Since there is no infinite
> >> >> > natural in form of a set of Os and since after every set of #s
> >> >> > there should be a O, the size of the set of the naturals as
> >> >> > sets of #s could not extend the "biggest" number of the
> >> >> > naturals in form of sets of Os. Since there is no biggest
> >> >> > number and since there is no infinite number, the size of the
> >> >> > set of numbers in form of sets of #s is undefined as the
> >> >> > biggest natural number is undefined.
> >> >>
> >> >> "biggest" element?
> >> > Stephen, did you even look at the diagrams he presented? Do you
> >> > not see that the width of the square and the height are the
> >> > same. Do you not see that the width is the count of naturals and
> >> > the height is the value? The picture said so, that's who.
> >>
> >> What square? The sides of a square are line segments. The four
> >> corners of the square are defined by the ends of those line
> >> segments. If your lines extend indefinitely, then there is no
> >> square.
> >>
> >> This is not a square:
> >> +-----------.....
> >> |
> >> |
> >> .
> >> .
> >> .
> >>
> >> A square has four corners. This only has one "corner".
> >> Remember, infinite lines do not end. Not even "at infinity".
> > (sigh) As Albrecht said, the square is defined by the diagonal at 45
> > degrees.
>
> There is no "diagonal" for something that has only one corner.
>
> > For every natural value represented by 0's in the diagram there is
> > an equal count represeted by #'s.
>
> It does not make sense to talk about "an equal count" for things that
> don't end.
>
> > This is the identity relationship between count and value that I've
> > been talking about. Think of it as the limit of a square as the side
> > goes to oo. Your objection is just another form of "No Largest
> > Finite!! No Diagonal Corner!!! (jingle jangle)" Oh, nice wind
> > chime!!
>
> Well, too bad that you insist on making the same mistake all over
> again. Small wonder you get your nose rubbed into it all over again.
>
> --
> David Kastrup, Kriemhildstr. 15, 44793 Bochum
You are not able to respond to my concept. You prefer to correct all
over again the same mistakes (if you are shure to accord with the
majority in this aspect). It's the usual dishonest of the dogmatic
people. Or are you anxious to disgrace yourself?
If you have two straight lines, suptending at point Zero, what is the
rectangular distance from a point in infinity laying on the one
straight line to the other?
Regards
AS
albs...@gmx.de
William Hughes wrote:
> albs...@gmx.de wrote:
>
> <snip>
>
> >
> > If we accept the uncountability as a form of infinity, this leads to
> > the paradoxon that the natural numbers are not countable.
>
> No, the natural numbers are countable precisely because
> they do count themselves.
This is exact my argument: there are uncountable many natural numbers
(nothing other means infinity many) but the natural numbers are shurely
countable since they count themself.
You are not able to recognise a paradoxon if you see it.
> The fact that there is no
> natural number that repsresents this "count" is not a paradox
> because the "count" is defined in terms of bijections. [You
> may not like the use of the terms "count" and "countable"
> because you think they should imply something different. So
> be it. However, you cannot say "you are using a term which
> I think should mean somthing different, so you must mean
> not what you mean but what I mean"]
>
> - William Hughes
Regards
AS
David Kastrup
> You are not able to respond to my concept. You prefer to correct all
> over again the same mistakes (if you are shure to accord with the
> majority in this aspect).
They don't go away by ignoring them.
David Kastrup
> William Hughes wrote:
>> albs...@gmx.de wrote:
>>
>> <snip>
>>
>> >
>> > If we accept the uncountability as a form of infinity, this leads to
>> > the paradoxon that the natural numbers are not countable.
>>
>> No, the natural numbers are countable precisely because
>> they do count themselves.
>
> This is exact my argument: there are uncountable many natural numbers
> (nothing other means infinity many)
Whining does not make it so. "countable" has a precise definition in
mathematics, and sets that can be placed into bijection with natural
numbers are both infinite and countable.
> but the natural numbers are shurely countable since they count
> themself. You are not able to recognise a paradoxon if you see it.
There is no paradoxon. You just have to read the definition of
"countable" instead of making up your own and then complaining that it
does not correspond to the established one.
albs...@gmx.de
I don't know what you are talking about. The proof for finite sets
needs just a complete induction. This will not hold for infinity I
think.
Regards
AS
William Hughes
albs...@gmx.de wrote:
> William Hughes wrote:
> > albs...@gmx.de wrote:
> >
> > <snip>
> >
> > >
> > > If we accept the uncountability as a form of infinity, this leads to
> > > the paradoxon that the natural numbers are not countable.
> >
> > No, the natural numbers are countable precisely because
> > they do count themselves.
>
> This is exact my argument: there are uncountable many natural numbers
> (nothing other means infinity many)
This is silly. For a set to be infinite
does not mean that it is uncountable (both "infinite"
and "countable have precise defintions, and they do not
contradict one another).
A set is infinite if there is a bijection between
the set and a proper subset of itself.
A set is countable if there is a bijection between
the set and a subset of the natural numbers.
Note that using these definitions the set of natural numbers
is both infinite and countable.
<snip>
Try reareading this next bit
David Kastrup
> I don't know what you are talking about. The proof for finite sets
> needs just a complete induction.
Nonsense.
> This will not hold for infinity I think.
If it needed complete induction in any manner, then only countable
infinities could be covered. But where do you see complete induction
in the following?
Given a set A and a presumed complete mapping f(.) from A to P(A),
consider the set X={B in A|B not in f(A)}. Now X =/= f(C) for all C
in A, since C in A <=> C not in f(A), and so X is not covered by the
mapping f(.).
This is not complete induction by any means. It is not related to the
naturals at all, nor to finiteness or infiniteness.
Daryl McCullough
>I don't know what you are talking about. The proof for finite sets
>needs just a complete induction. This will not hold for infinity I
>think.
There is no induction involved in the finite case, and the
infinite case is *exactly* the same proof as the finite case.
Here it is once again:
Let A be any set whatsoever, finite or infinite, it doesn't matter.
Let f be any function from A to P(A).
Let w = { x in A | x is not an element of f(x) }.
Let x = any set in A.
Let u = f(x). We prove that u is not equal to w.
By definition of w, we have x in w <-> x is not an element of f(x).
So x in w <-> x is not an element of u. That means that there are
two cases: Case 1: x in w, and x is not in u. In that case, u cannot
equal w. Case 2: x is not in w, and x is in u. In that case, u cannot
equal w.
So what we have proved is that forall x, w is not equal to f(x). So
w is not in the image of f. So f is not a bijection between A and P(A).
There's no induction. There's no assumption that A is finite.
William Hughes
Actually you should learn the definition of cardinality before
making statements about it. This might stop you uttering such
idiocy as "The meaning of the cardinality of an infinite set
is unknown."
> Coincidently natural numbers and cardinalities are undistinguishable in
> finity.
They are very similar, but they are not quite "undistinguishable".
A natural number is a set, a cardinality is an equivalence class.
>Cardinality is just a artificial concept with no sens and
> meaning.
Again, learn the definition of cardinality before criticizing it,
> Define the existence of unicorns and be happy.
After you learn the definition of cardinality, please explain
how it is possible for a set not to have a cardinality.
- William Hughes
William Hughes
Perhaps you are thinking about proving that n<2^n for all natural
numbers n? This can be done by induction, and indeed this proof
doen not extend to infinite numbers. However, this is not the
proof usually used to show that there is no bijection between a
set and its powerset. The standard proof does not use induction,
nor does it require finiteness.
-William Hughes
>
>
> Regards
> AS
Tony Orlow
> >> >> infinite or otherwise.
> >> > Have too.
> >>
> >> No you have not Tony.
>
> > Have, too!
>
> >> The proof that there does not exist
> >> a bijection between a set and its power set is quite short.
> > Then it shouldn't take too much looking to see where it goes wrong....
> >>
> >> Let f be a function from S to P(S).
> > Our proposed mapping bijection....
> >>
> >> Define the set w as follows:
> >>
> >> w= { x : x in S and x is not in f(x) }
> > So, w is the set of all elements which are not members of the subsets which
> > they map to through f(x)....
> >>
> >> Clearly w is a subset of S, and so w is an element of P(S).
> > Clearly...but properly?
>
>
> >>
> >> We now show that w is not in the image of f. That is,
> >> there does not exist a y such that f(y)=w.
> > So, there can be no y such that it maps to the subset of all elements which do
> > not map to subsets containing themselves? We'll see.....
> >>
> >> Suppose such a y exists. If such a y exists, it must
> >> either be an element of w, or not.
> > One or the other. I'll accept the excluded middle....
>
> >>
> >> If y is an element of w, then y is in f(y), which means
> >> it is not an element of w.
> > If y is in w, this means y is a member of the set of elements which map to
> > subsets which do not contain themselves. This means that y is in S but not in f
> > (y). So, indeed, it IS a member of w. There is no reason why both x and y
> > cannot map to subsets which do not contain themselves.
>
> If y is in w, then y is in f(y), because w=f(y). Remember,
> we are assuming there exists a y such that f(y)=w. However
> w is defined such that y can only be in w, if y is not in f(y).
Okay, I hit this one at the end of the day, and got confused halfway through it
and forgot that you're assuming y is mapped to w. Sorry about that. It is clear
that no element in the set maps to a subset that contains itself, as I
illustrated below. If f(y)=w, then y can't be in w, but then that means y IS in
f(y), which means it's in w. Got it.
That certainly causes a bit of a contradiction, based on the largest-finite
kind of argument, since you are noting that whatever subset you choose, it
never contains the natural that maps to it, and the question remains what
number maps to the set containing just that natural. You are assuming some
completed w, where some identifiable y is the number that maps to it. But that
number has to be larger than any of the elements in w. So, you draw a
contradiction from the assumption that y is in N and also maps to N. Clearly,
the power set is LARGER than the set.
However, given the definition of bijections, it is easy to map any well-ordered
infinite set to its power set through the binary representation, such that each
element's position starting at element 0, represented as a binary natural, also
represents a subset of the naturals, where the bit in the 2^n position denotes
membership of the nth element. You may have a discrepancy in the values of the
naturals and the values in the subsets, but you have a complete bijection
nonetheless. Discrepancies in value don't bother you elsewhere. Why is this
bijection so different?
>
> > If y is a member of w, then all you can say is that y does not map to any
> > subset which contains itself. Y does not map to w. Neither does x.
>
> What is x? If y is a member of w, then y is in f(y), and
> by the definition of w, w is not a member of w.
Yes, there is no element y in any given set S which maps to S. That element
would be element 2^|S|-1, outside the scope of S.
>
> >>
> >> If y is not an element of w, then y is not in f(y), which
> >> means it is an element of w.
> > If y is not a member of w, then y maps to a subset which contains y as a
> > member, and y IS in f(y). Is this possible? We'll see what you think.....
>
> >>
> >> These are both contradictions. So y cannot be an element
> >> of w, and it cannot not be an element of w. So y
> >> cannot exist.
> > Neither of those possibilities causes a contradiction. You are getting confused
> > with your double negatives. If w is the set of all elements which do not map to
> > subsets containing themselves, then being a member of w means simply that y
> > does not map to a subset containing y, which is perfectly possible. Not being a
> > member of w means that an element maps to a subset which DOES contain itself.
> > Where is the contradiction?
>
> The contradiction is that if y is in w, then it is not in w,
> and if y is not in w, then it is in w. That is a pretty
> obvious contradiction.
Well, y is in w, as are all the elements. y does not map to w. For any given
set, there is no element in the set which maps this way to the set itself. And
yet, when you have infinite sets such as this, can't I just map element 2^S-1
to subset S?
>
> You apparently failed to grasp the very important fact that w=f(y).
> So once again, is y in w?
Not if w=f(y). If w=f(y), then y is not in S, and therefore not in w.
>
> If y is in w, then y is in f(y), because w=f(y).
> If y is in f(y), then y is not in w, because w is defined
> as containing the elements x in S such that x is not in f(x).
> w cannot contain y, because y is in f(y).
>
> If y is not in w, then y is not in f(y), because w=f(y).
> If y is not in f(y), then y is in w, because w is defined
> as containing the elements x in S such that x is not in f(x).
> w must contain y, because y is not in f(y).
Got it. y is not in S, and therefore not in w.
>
> >>
> >> So there is at least one element in P(S) which is
> >> not in the image of f, so f is not an onto function,
> >> and it is not a bijection.
> > Sorry, not so.
>
> Yes. The fact that you cannot understand a simple
> proof does not make the proof invalid.
I got a little confused, but you still haven't proven anything like the
impossibility of a bijection with the power set. In other cases bijections are
performed without regard to such discrepancies.
>
> >>
> >> What is wrong with this proof in your opinion?
> > You confused yourself with double negatives. If I misinterpreted any of what
> > you said, please clarify.
>
> You apparently misintrepted all of it.
not entirely.
>
> Do you understand that we are assuming that w=f(y)?
No, I forgot that in figuring out what we were talking about specifically. I
don;t see anything in it proving bijection impossible. If w=f(y) then y is not
in S, assuming some limit to S. But, S goes on forever and ever, and we can
always borrow for our bijection, having sets containing elements with at most
the log2 of the value of the element that maps to it. As long as there's a 1-1
correspondence, what's the problem?
> Do you understand that y is in w if and only if y is not in f(y)?
Yes.
> Do you understand that
> y is in w if and only if y is not in w
> is a contradiction?
If y is not in S, then y is not in w.
>
> Stephen
>
--
Smiles,
Tony
ste...@nomail.com
No, that is not clear at all. It is entirely possible that an
element maps to a set that contains itself. However no element
maps to w. You still do not get it.
Here is a simple mapping from N to P(N).
1 -> {1}
2 -> {1,2}
3 -> {1,2,3}
4 -> {1,2,3,4}
...
Every element is contained in the subset that it is mapped
to. For this mapping, w={}, and no element is mapped to {}.
> That certainly causes a bit of a contradiction, based on the largest-finite
> kind of argument, since you are noting that whatever subset you choose, it
> never contains the natural that maps to it,
No. Try again.
What is element "2^S-1"? S is a set. Element "2^S-1" means
nothing to me.
>>
>> You apparently failed to grasp the very important fact that w=f(y).
>> So once again, is y in w?
> Not if w=f(y). If w=f(y), then y is not in S, and therefore not in w.
If y is not in S, then it is not part of the bijection.
f is function from S to P(S). It makes no sense to plug
in a value that is not in S into f.
>>
>> If y is in w, then y is in f(y), because w=f(y).
>> If y is in f(y), then y is not in w, because w is defined
>> as containing the elements x in S such that x is not in f(x).
>> w cannot contain y, because y is in f(y).
>>
>> If y is not in w, then y is not in f(y), because w=f(y).
>> If y is not in f(y), then y is in w, because w is defined
>> as containing the elements x in S such that x is not in f(x).
>> w must contain y, because y is not in f(y).
> Got it. y is not in S, and therefore not in w.
No. You still do not get it.
>>
>> >>
>> >> So there is at least one element in P(S) which is
>> >> not in the image of f, so f is not an onto function,
>> >> and it is not a bijection.
>> > Sorry, not so.
>>
>> Yes. The fact that you cannot understand a simple
>> proof does not make the proof invalid.
> I got a little confused, but you still haven't proven anything like the
> impossibility of a bijection with the power set. In other cases bijections are
> performed without regard to such discrepancies.
There is no bijection between a set and its powerset.
It does not matter what the set is, it does not matter
if it is finite, or infinite.
>>
>> >>
>> >> What is wrong with this proof in your opinion?
>> > You confused yourself with double negatives. If I misinterpreted any of what
>> > you said, please clarify.
>>
>> You apparently misintrepted all of it.
> not entirely.
>>
>> Do you understand that we are assuming that w=f(y)?
> No, I forgot that in figuring out what we were talking about specifically. I
> don;t see anything in it proving bijection impossible. If w=f(y) then y is not
> in S, assuming some limit to S. But, S goes on forever and ever, and we can
> always borrow for our bijection, having sets containing elements with at most
> the log2 of the value of the element that maps to it. As long as there's a 1-1
> correspondence, what's the problem?
If nothing in S maps to w, and w is in P(S), then f is not a bijection.
It is that simple. Nothing in S maps to w. If you think otherwise,
tell us what element of S is mapped to w.
You still do not understand the proof at all. It is a relatively
simple proof.
Here are some examples that might help you out. Although
you will likely snip them as "tedious nonsense."
First, look at the finite case. Let S={a,b,c}.
Lets look at
f(a) = { a }
f(b) = { a, c }
f(c) = { a, b, c }
w = { x : x not in f(x)}, so w={b}. {b} is not in the image of f.
We can try again.
f(a) = { a }
f(b) = { b }
f(c) = { a, b, c }
Now w={}, which is not in the range of f.
How about
f(a) = { }
f(b) = { a }
f(c) = { a, b }
Now w={a,b,c}. We can keep playing this game, but w will
never be in the image of f.
Of course in the finite case it is obvious that there
cannot be a bijection from S to P(S) because P(S) has
2^|S| elements, and when |S| is finite it is obvious that
2^|S| > |S|.
However the above proof makes no mention or use
of the set being finite, and it applies equally well
to finite and infinite sets.
Lets look at the natural numbers and the "bijection"
albstorz proposed:
1 -> N
2 -> {}
3 -> {1}
4 -> N\{1}
5 -> {2}
6 -> N\{2}
7 -> {1,2}
8 -> N\{1,2}
9 -> {3}
10 -> N\{3}
11 -> {1,3}
12 -> N\{1,3}
13 -> {4}
14 -> N\{4}
15 -> {1,4}
16 -> N\{1,4}
17 -> {2,3}
18 -> N\{2,3}
19 -> {5}
20 -> N\{5}
...
In this case
w={ 2,3,5,7,9,11,13,15,17,19 .... }
Where does this show up in teh above list? If you
claim it shows up in position y, then is y in w or not?
Stephen
albs...@gmx.de
William Hughes wrote:
>
> > Coincidently natural numbers and cardinalities are undistinguishable in
> > finity.
>
> They are very similar, but they are not quite "undistinguishable".
> A natural number is a set, a cardinality is an equivalence class.
You make me hopefull. Some experts make "äääh", "hömm" and
"üüüh" if I said "A natural number is a set." One sees a
correspondence between natural numbers and von Neumann sets after all.
You are free to say "A natural number is a set." without "äääh-",
"hömm-" and "üüüh-" comments. Be lucky. You are right.
And natural numbers don't behave in any other way than sets. So, if
there is an infinite set there is an infinite number. If there is no
infinite number there is no infinite set. And vic versa.
Regards
AS
Tony Orlow
> Tony Orlow wrote:
> >> I already showed you the bijection between binary *N and P(*N).
> >> What didn't you like about it? It is valid.
> >
>
> David R Tribble said:
> >> No, you showed a mapping between *N and R, which is equivalent
> >> to a mapping between *N and P(N). That's easy.
> >
>
> Tony Orlow wrote:
> >> No, it was specifically a bijection between two sets of infinite binary
> >> strings representing, on the one hand, the whole numbers in *N starting
> >> from 0, both finite and infinite, in normal binary format, and on the other
> >> hand, the specification of each subset of whole numbers in *N, where each
> >> bit which, in the binary number, represents 2^n denotes membership of n in
> >> the subset. This is a bijection between the whole numbers in *N and P(*N),
> >> using an intermediate bijection with a common set of infinite binary
> >> strings.
> >
>
> David R Tribble said:
> >> But that's an incomplete mapping, because there are not enough infinite
> >> binary strings in *N to enumerate all of the subsets of *N. Try it,
> >> if you don't believe me.
> >
>
> Tony Orlow wrote:
> > Not enough infinite binary strings? How many in *N and in P(*N)?
> > Are you saying that I cannot construct a bijection between the two on an
> > element-by-element basis which continues infinitely through the set of
> > binary strings?
>
> That's exactly what I'm saying.
>
> card(*N) = c, but card(P(*N)) = 2^c, and c < 2^c.
At what point does the bijection break down? It certainly works for all finite
cases, does it not? What makes you think it falls apart at some point? For
which element is there not a corresponding subset? For which subset is there no
corresponding element? Name either one of these, as a counterexample, or
explain why you think the bijection fails.
>
>
> > Where does it stop? You aren't beginning to see that determining the
> > infinite value range is crucial to this problem after all, are you?
>
> If you think that approach will work, then show us how.
>
> By the way, what is the "range" of a powerset, whose members are
> sets themselves?
Sets don't have an inherent measure beyond raw size, and the power set is not
well ordered as far as subset size goes, so it doesn't make sense to talk about
a value range on the power set exactly. However, one might think of the binary
mapping of the power set to the set, and say that the power set of an ordered
set has an order based on that mapping, taking each bitstring as a natural. In
this case, you might be tempted to say, if the set has N elements, the power
set has a range of 2^N-1.
>
>
> > What do you want me to try, anyway, and infinite mapping,
> > element-by-element? A bijection's a bijection, right?
>
> Yes, that would be nice. Please show us your bijection.
f(0) = ...000 = {}
f(1) = ...001 = {0}
f(2) = ...010 = {1}
f(3) = ...011 = {0,1}
f(4) = ...100 = {2}
f(5) = ...101 = {0,2}
f(6) = ...110 = {1,2}
f(7) = ...111 = {0,1,2}
etc. Any questions?
>
>
> > These sets are obviously of the same cardinality, since I have a
> > bijection which carries to infinity, right?
>
> No, their sizes are different cardinalities, which happen to be
> different infinities. Both sets are infinite, but one set is
> larger than the other.
But I have constructed a bijection between the two using an intermediate binary
representation. What is the specific rule I have broken concerning the
construction of bijections. If I haven't broken any such rule, then is it true
that a bijection between two sets means that have the same size, or even
cardinality?
>
>
> > For every subset there is a unique natural and for every natural there is
> > a unique subset. If you disagree, please state which of either has no
> > corresponding element in the other.
>
> Like I said, there are not enough naturals to map to every subset
> of the naturals.
Which subset, specifically, is left unmapped? If you claim there is one, then
surely you can name it?
>
> This is true whether you've got infinite naturals or not; there are
> not enough members in N to enumerate all the subsets in N, and
> likewise there are not enough members in *N to enumerate all the
> subsets in *N.
But, as the keepers of that standard say, what holds for the finite case does
not necessarily hold for the infinite case. Once you have infinite sets,
neither one ends, and there is always a natural for any subset, and always a
subset for any natural. Isn't this the way the standard treatment of bijection
goes for infinite sets?
>
>
> > Please try to frame your objection in a more operative manner. How do you
> > know there aren't a large enough infinity of bits for the power set vs the
> > values in the set?
>
> Because I can prove it (and it's a very old proof). A powerset of
> a nonempty set contains more elements that the set. Can you prove
> otherwise?
No, I fully agree with that conclusion. However, there is nothing that
precludes a bijection between any ordered infinite set and its power set,
despite the different sizes. My point is that bijection alone does not mean
equal size, as this example shows well. Two infinite sets may have a bijection
while one contains some finite number of elements more than the other, some
finite multiple of the other's number of elements, or some formulaic relation
that shows one to be infinitely greater than the other, like N^2. Such
bijections are taken to imply equal size, or at least cardinality, and yet, a
bijection CAN be constructed between any ordered infinite set and its power
set, which contradicts the power set rule.
>
>
--
Smiles,
Tony
William Hughes
You have made exactly this mistake before. Yes every number
is a set. No, not every set is a number. For example
{peach, apple, plum, fiddle} is a set but not a number.
Just because you have a set does not mean you have a number.
So yes, there is an infinite set. But this does not mean
that this set is a number. Indeed, no infinite set is a number.
- William Hughes
P.S Actually it is not true that natural numbers must be sets, but
they can be. As you insist on using a model in which the natural
numbers are sets, I am playing along to be polite.
Randy Poe
What is "the" bijection? The proof that card(S) < card(P(S))
shows that no bijection exists. That proof has been given a
few times in this thread and I see you're trying to get through
it. Whether you believe the proof or not yet, you do realize
that the proposition is "Let f(S) be any mapping from S to
P(S). Then f can't be a bijection." Right?
If the proof is correct (as it is), then ALL bijections break
down. The proof consists of showing that no matter what
mapping you choose, there's an unmapped element of P(S).
> It certainly works for all finite cases, does it not?
Absolutely not. Let S = {1,2,3}.
Then P(S) has 8 elements:
{},
{1}, {2}, {3},
{1,2}, {2,3}, {1,3},
{1,2,3}
There is no bijection between the 3 elements of S and the
8 elements of P(S).
> What makes you think it falls apart at some point?
It falls apart for EVERY set.
> For which element is there not a corresponding subset?
It breaks apart in the other direction: there is (at least
one) subset for which there is no corresponding element.
First you have to define your mapping rule. Given any mapping
rule, a subset can be identified for which there is no
corresponding element.
- Randy
imagin...@despammed.com
<showing that there cannot be a bijection from a set to its power
set...>
Yes, that sounds fairly typical. You are about to show that eleven
million (whatever, I've forgotten the numbers already, and I made them
up anyway) mathematicians have been getting it wrong for 100 years with
an 8-line proof. But you hit it at the end of the day and got confused.
> and forgot that you're assuming y is mapped to w. Sorry about that. It is clear
> that no element in the set maps to a subset that contains itself, as I
> illustrated below. If f(y)=w, then y can't be in w, but then that means y IS in
> f(y), which means it's in w. Got it.
>
> That certainly causes a bit of a contradiction, based on the largest-finite
> kind of argument, since you are noting that whatever subset you choose, it
> never contains the natural that maps to it, and the question remains what
> number maps to the set containing just that natural. You are assuming some
> completed w, where some identifiable y is the number that maps to it. But that
> number has to be larger than any of the elements in w. So, you draw a
> contradiction from the assumption that y is in N and also maps to N. Clearly,
> the power set is LARGER than the set.
Well, that was a jumble, wasn't it? What is a "completed" w? Normal set
theory talks about sets, and a set contains its members. It contains
all of its members, all the time, never contains anything else, never
becomes tired, "unidentifiable", or "tenuous", just sits there
containing all the elements it contains. (Notice that this immediately
means that normal set theory can't accommodate things like sets that
contain different collections of all of something, one of the direct
consequences of the "infinite numbers are just the same as finite
numbers only bigger" crank line.)
Notice also how you are again unable to consider abstract sets, and
keep mumbling about "numbers". Why on earth should the y be "larger"
than any element in w? We've proved that y cannot exist at all, but
there is no a priori reason any notion of "largeness" is involved. If
we were considering the set of all finite simple groups or the set of
all Platonic solids, there would be no "larger", but the proof applies
exactly the same.
> However, given the definition of bijections, it is easy to map any well-ordered
> infinite set to its power set ....
Blah blah blah. This one line after a proof that no such bijection
exists. Well, it's no wonder you can't do mathematics.
Brian Chandler
http://imaginatorium.org
Tony Orlow
> > > >> Cantor proofs his wrong conclusion with the same mix of potential
> > > >> infinity and actual infinity. But there is no bijection between this
> > > >> two concepts. The antidiagonal is an unicorn.
> > > >> There is no stringend concept about infinity. And there is no aleph_1,
> > > >> aleph_2, ... or any other infinity.
> > > >
> > >
> > > David R Tribble wrote:
> > > >> For that to be true, there must be a bijection between an infinite
> > > >> set (any infinite set) and its powerset. Bitte, show us a bijection
> > > >> between N and P(N).
> > > >
> > >
> > > > At first, you should show, that bijection means something to
> > > > notwellordered infinite sets.
> > > >
> > > > Bijection is a clear concept on finite sets, it also works on
> > > > wellordered infinite sets of the same infinite concept.
> > > > Aber: Show me a bijection between two infinite sets with the same
> > > > cardinality, where one of the sets is still not wellorderable.
> > > > and P(P(N)) or what you want.
> > >
> > > I see. I'm supposed to show you a proof before you can show me your
> > > proof. Okay, I give up, you win, so your proof must be correct.
> > >
> > > Come on, now. It's up to you to prove your own claim, especially
> > > when it contradicts established mathematics. I know you cannot show
> > > a bijection between N and P(N).
> > >
> > >
> > >
> > > Let
> > > D(n,i) = floor(n/2^i) mod 2, for all i=0,1,2,3,...
> > > This is the i-th binary digit of natural n
> > > L(n) = i where D(n,i) = 1 and D(n,j) = 0 for all j > i
> > > This is the number of binary digits of natural n, or ceil(log2(n))
> > > Let
> > > M(n) = sum{i=0 to L(n)} 1/2^(i+1) where D(n,i) = 1
> > > This reverses the binary digits of n.
> > >
> > > Then M(n) is a mapping N -> N, from all n in N to M(n) in N,
> > > but the set of all M(n) is not a well-ordered set.
> > > Happy?
> >
> >
> > Sad!
> > First of all you don't argue on my claim of the thread.
> > Second, your above argueing is not clear to me, since both sets are
> > well-ordered. But it's nice, so I give this:
> >
> > N
> > {1},{2},{3}, ...
> > N/{1},N/{2},N/{3}, ...
> > {1,2},{1,3},{2,3},{1,4},...
> > N/{1,2}, ...
> > ...
> >
> > diagonal proof.
> >
>
> Sigh. Try {2,4,6,8,...} Or any other infinite subset
> that is not the complement of a finite set.
>
> It is common for people to note that the finite subsets of N
> are countable and incorrectly claim that the subsets of N are
> countable. Adding in the complements of the finite subsets
> does not change things very much.
>
> Those who do not study anti-cantor cranks are doomed to
> repeat their idiocies.
>
> - William Hughes
>
> P.S. No TO, going to TO-infinite rows is not going to help us.
> The problem is that for any TO-finite row, the subsets
> listed will have an bounded finite number of elements, or be
> the complement of a subset with a finite number of elements.
> Yes, you can claim (without a shred of motivation) that
> when you get to TO-infinite rows the subsets will suddenly
> have an unbounded number of elements, but all this tells us
> is that a bijection from the TO-naturals to P(N) exists.
> What we need is a bijection from the finite TO-naturals
> to P(N).
>
>
Look, it's certainly not my position that the pwoer set is the same sie as the
set. It's clearly not. I just see a bijection between them, which only bolsters
my argument that bijection alone is not sufficient to equate the sizes of two
sets.
When it comes to the evens (let's start with 0), the value 0:010.......1010101
represents such a subset, and is essentially binary N/3.
--
Smiles,
Tony
Tony Orlow
> David R Tribble wrote:
>
> >
> > Because I can prove it (and it's a very old proof). A powerset of
> > a nonempty set contains more elements that the set. Can you prove
> > otherwise?
>
> hidden megabooster for transcendental overflow? What is the very
> special aspect of the powerfunction to be so magic?
combinations of elements is larger than the set of single elements, especially
when that power set essentially includes the set, in the form of the singleton
subsets?
> Why should all operations with transfinite numbers lead to results with
> the same "level" of infinity, but only powerfunction beams up to the
> next level?
It shouldn't. I think the set of squares shpuld be considered to have a size
which is the square root of the set of naturals. At least the standard theory
recognizes that the power set is larger than the set. Unfortunately, that's
about as deep as the standard distinctions go.
> Is not true: a^2 = a*a? 2a = a+a?
> Is the powerfunction something other than a very shortcut for multiple
> additions?
> Which amount you are able to reach with powerfunction which is
> unreachable by succesor operation?
As far as bijections go, none. Bijections alone are insufficient with infinite
sets. Certainly, you don't disagree that any set's power set is larger than
itself? Or, perhaps you do.
>
> I'm very sensible about this because this argument is found in very
> much books although it's total meaningless.
>
> (Weak minds might be impressed by the big numbers which are easily
> produced by powerfunction.)
It's the rice on the chessboard and the wealth of the king. :)
>
>
> What in finity holds may not (or do not) hold in infinity.
Or, it may very well hold for all cases, finite and infinite.
>
> Regards
>
> AS
>
>
--
Smiles,
Tony
Tony Orlow
>
> albst...@gmx.de wrote:
> > David R Tribble wrote:
> >
> > >
> > > Because I can prove it (and it's a very old proof). A powerset of
> > > a nonempty set contains more elements that the set. Can you prove
> > > otherwise?
> >
> > This argument is stupid. Is there any magic in the powerfunction? A
> > hidden megabooster for transcendental overflow? What is the very
> > special aspect of the powerfunction to be so magic?
> > the same "level" of infinity, but only powerfunction beams up to the
> > next level?
>
>
> > additions?
>
>
> > unreachable by succesor operation?
>
> an infinite quantity by using the successor operation (unless
> like TO you are willing to wave a circular magic wand and apply
> the successor operation an infinite number of times).
>
> >
> > much books although it's total meaningless.
> >
>
>
>
> > produced by powerfunction.)
>
> bijection between X and P(X).
>
> >
> >
> > What in finity holds may not (or do not) hold in infinity.
>
> "number of elements" that can be described by a natural number
> while an infinite set (e.g. the set of natural numbers) does
> not have a "number of elements" that can be described by a
> natural number. However, some things are true for both
> bijection between X and P(X).
Did you have an objection to the bijection between *N and P(*N)? What did I do
wrong there, bijection-wise?
>
> -William Hughes
>
>
--
Smiles,
Tony
Tony Orlow
> In article <MPG.1dbf2de3...@newsstand.cit.cornell.edu>,
> Tony Orlow <ae...@cornell.edu> wrote:
>
> > David R Tribble said:
> > > Tony Orlow wrote:
> > > >> I already showed you the bijection between binary *N and P(*N).
> > > >> What didn't you like about it? It is valid.
> > > >
> > >
> > > David R Tribble said:
> > > >> No, you showed a mapping between *N and R, which is equivalent
> > > >> to a mapping between *N and P(N). That's easy.
> > > >
> > >
> > > Tony Orlow wrote:
> > > > No, it was specifically a bijection between two sets of infinite binary
> > > > strings representing, on the one hand, the whole numbers in *N starting
> > > > from 0, both finite and infinite, in normal binary format, and on the
> > > > other
> > > > hand, the specification of each subset of whole numbers in *N, where each
> > > > bit which, in the binary number, represents 2^n denotes membership of n
> > > > in
> > > > the subset. This is a bijection between the whole numbers in *N and
> > > > P(*N),
> > > > using an intermediate bijection with a common set of infinite binary
> > > > strings.
> > >
> > > binary strings in *N to enumerate all of the subsets of *N. Try it,
> > > if you don't believe me.
> > >
> > >
>
>
So, you need more than an infinite amount? How many more? Do you need more than
an infinite amount to list all the integral multiples of 1/10?
--
Smiles,
Tony
Tony Orlow
>
> > David R Tribble wrote:
> >
> > >
> > > Because I can prove it (and it's a very old proof). A powerset of
> > > a nonempty set contains more elements that the set. Can you prove
> > > otherwise?
> >
> > This argument is stupid. Is there any magic in the powerfunction?
>
> arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone.
>
measly counterexample?
--
Smiles,
Tony
Tony Orlow
>
> imagin...@despammed.com wrote:
> > ste...@nomail.com wrote:
> > > albs...@gmx.de wrote:
> >
> > <snip: my goodness this stuff goes on and on...>
> >
> > > > First of all you don't argue on my claim of the thread.
> > > > Second, your above argueing is not clear to me, since both sets are
> > > > well-ordered. But it's nice, so I give this:
> > >
> > > > N
> > > > {1},{2},{3}, ...
> > > > N/{1},N/{2},N/{3}, ...
> > > > {1,2},{1,3},{2,3},{1,4},...
> > > > N/{1,2}, ...
> > > > ...
> > >
> > > of this is that your list is:
> > >
> > > N,
> > > {1},
> > > {2},
> > > {3},
> > > ...
> > > N/{1},
> > >
> > > before N/{1} in your "list"? ...
> >
> > >
> > > > Now count in diagonal sequence. You may think of Cantor's first
> > > > diagonal proof.
> > >
> >
> > Come on, come on! He means the zigzag diagonal, as in the standard
> > demonstration that the rationals _are_ countable. This isn't "Cantor's
> > first diagonal proof",
>
> I see, you are the real checker. You knows it all. You are famous. You
> are apodictic. All the authors who speak of the first diagonal proof of
> Cantor are wrong.
>
>
> > but if you're going to argue with cranks you
> > must expect them to be pretty muddled about things.
> >
> > Anyway, it's obvious that *if* the OP shows a "list of lists" that
> > include all the subsets that is enough. In practice, of course he's
> > given the standard crank non-list.
> >
> > Brian Chandler
> > http://imaginatorium.org
>
> You and many of the other checkers are not able to discuss my starting
> argument. You are only able to respond to the usual wrong arguments you
> know. And I think, your answers are memorized because you are unable to
> think your own thoughts.
>
> Regards
> AS
>
>
Albrecht, much as I appreciate some of your ideas, I am missing this one a
little. If you are trying to create an enumeration, then you must be doing what
Brian suggests, with the zigzag diagonal like for the enumeration of the
rationals. However, you only have one set on the top line and the bottom, which
you don't show, but which would have the null set. There is not really a zigzag
diagonal covering the set listed this way, as far as I can tell, because it's
not really a rectangular list. I highly recommend the binary natural ordering.
--
Smiles,
Tony
Tony Orlow
>
> Virgil wrote:
> > In article <1129684276.2...@g47g2000cwa.googlegroups.com>,
> > albs...@gmx.de wrote:
> >
> > > David R Tribble wrote:
> > >
> > > >
> > > > Because I can prove it (and it's a very old proof). A powerset of
> > > > a nonempty set contains more elements that the set. Can you prove
> > > > otherwise?
> > >
> > > This argument is stupid. Is there any magic in the powerfunction?
> >
> > "Proofs" are not stupid until they can be refuted. The proof that for an
> > arbitrary set S, Card(S) < Card(P(S)) has not been refuted by anyone.
>
>
> both a greater cardinality than their starting sets, you would not
> really think it depends on the same cause in both cases.
>
> You must proof it independently for finite and for infinite sets. In
> this sense the argument is stupid.
>
>
> AS
>
>
Albrecht, do you accept the axiom of induction? If so, it is easily provable
inductively that the power set of a set of size n has size 2^n, and since this
is an equality property, it holds for the infinite case. The power set of an
infinite set is infinite, but a larger infinity than the set.
--
Smiles,
Tony