listing the mistakes of Cantor and his diagonal argument Chapt28 Summary, Review and Reminders #1210 Correcting Math 3rd ed

[Archimedes Plutonium]

Sometimes in mathematics it is grueling work to making clear what it
is you understand. You know something and understand it, but have
still a lot of roughness and patches of fog. Some of it is because
there is still a lot to learn about a topic even though we know it and
understand it. Perhaps it is that we understand it in large degree,
but still have smaller degrees of learning more.

In the case of Cantor's diagonal argument for larger infinities, I
know he was wrong and mistaken and can prove it. But I can still make
improvements on clarity.

What prompted me into writing this post is a admonition by LWalk
yesterday who said words to the effect "make it as clear as possible
and let others then judge whether they get enough information to be
persuaded by the argument". And so I still have room for improvement
on making the case against Cantor more clear.

Plus, also, I see a thread going in sci.math asking whether any peer
reviewed math journal raises the issue of the falsity of Cantor's
diagonal argument.

So let me improve on clarity of why Cantor was wrong. And let me start
with the Probability theory of counting of a true false test with a
number of questions. It is called the Fundamental Counting Principle
so that a true false test of 2 questions has 4 possible cases TT, TF,
FT, and FF. For 3 questions it is 2^3 = 8 for 4 questions it is 2^4 =
16.

Now Cantor and every other mathematician since Cantor would not argue
the idea that the Reals are all-possible-digit-arrangements. And we
focus only on the Reals between 0 and 1 such as the Real 0.11100000..
or the Real 0.01010101..

And Cantor and all subsequent mathematicians would not argue against
the idea of making the Reals from 0 to 1 in binary base rather than in
decimal base representation. So that the list of all Reals from 0 to 1
has only the two digits of either 0 or 1.

Now, to make the proof that Cantor was wrong, the easiest way possible
is to show an example and then to expand on that example.

So the example is All Reals of two digits and two place values
so that we have .00 and .01 and .10 and .11 as the total universe of 2
digits to 2 place value. Now of course Cantor did not restrict himself
to just 2 digits and 2 place value, but we can expand that. And if we
collect a missing Real in 2 digits and 2 place value, we can then
agree that Cantor was true, but if we cannot collect a Real not on the
list in 2 digits and 2 place value, we have to begin to worry that
Cantor's proof was a fake and we have to see how Cantor may have been
lead astray.

So for 2 digits in 2 place value the Fundamental Counting Principle
tells us there are 2^2 = 4 total possible cases or, in our problem 4
Reals in all possible digit arrangement and they are

00
10
01
11

where I have deleted the decimal point and well do so in the rest of
this post, for we all know they are Reals between 0 and 1.

Now, immediately we can see a problem for which Cantor never addressed
and which Cantor likely failed and made a fake proof.
The problem is that I have listed all the Reals for 2 digits and 2
place value but I cannot arrange them in a geometry to where the
diagonal performs a Cantor Diagonal to manufacture a new Real not on
the list.

Let me show you:

I can take two of those Reals and perform a Cantor diagonal, and say I
take 00 and 01

I make a Cantor list as such:

00
01

And the diagonal is 45 degrees cutting into 0 in the top row
and the 1 in the bottom row.

Now according to Cantor's proof argument in the 1800s, he altered the
0 to be 1 in the top row and then he altered the 1 to be 0 in the
bottom row and thus Cantor manufactured the number 10.

Now is it a new and missing Real, this 10? Why yes of course when you
list only 00 and 01 where you can perform a geometry diagonal.

But that would be a menacing flaw of Cantor's proof, because he cannot
list all the Reals of 2 digits to 2 place value and have a working
diagonal that alters every Real on the list.

There is no geometrical means of listing all the 4 Reals that belong
to 2 digits to 2 place value

Could Cantor make some geometrical list of

00
01
10
11

and have a diagonal or something else that alters all those 4 Reals
and produces a new missing Real.

It is late at night here and I have not gone to bed yet, but had a
shower and I for the life of me see no way possible of making a
geometrical arrangement that alters every one of those 4 Reals and
manufactures a new Real.

As shown above, if we deleted 10 and 11 and work with only
00 and 01 we manufacture the Cantor method we manufacture the Real 10
and it is different from 00 and 01 but it is not different from any of
the 4 Reals that should have been listed.

So here we begin to see the huge major flaw of Cantor and why his
proof is not true but a fake.

Cantor assumed he could list all the Reals and apply a diagonal on all
those Reals. In the case of just all the Reals of 2 digits to 2 place
value, no geometry exists that can form a list where the diagonal
works on all those Reals. In the case of 2 digits to 3 place values we
have 2^3 =8 total Reals possible represented as this:

000
001
010
100
110
101
011
111

Now we ask Cantor to take that list of all possible Reals with 2
digits and 3 place value and to come up with some geometry
configuration of where he can apply his diagonal and produce a Real
number that is different from those 8 listed Reals. And the answer is
impossible.

Cantor can take three of them and do a diagonal

000
010
111

and by doing the diagonal, Cantor will have manufactured the number
100 and that number is not on the list, but it is on the list of all
Reals and it is not on the list of the diagonal because the geometry
forbids it to be on the list.

So here we begin to see the huge internal flaw of Cantor's argument.
Cantor made a large list with alot of dots indicating large numbers
like this:

0000000000000000000000000......
0101010101010111010101010......
0000000000011000000000000......
0101010101110101010101010......
0000000000001110000000000......
0101010101010101011101010......
0111000000000000000000000......
0101011111010101010101010......
.
.
.
.
.

In other words Cantor made his list look large and full of
dots to indicate it was even larger than what it looks and he
did this because when you then draw that diagonal and presume
you have all the Reals on that list that the diagonal will deliver you
a new missing number.

But in fact, as the case of 2 digits 2 place value or 2 digits 3 place
value or 2 digits, infinity place value shows us by math induction,
that in every such case example, the only way to manufacture a missing
Real is because the diagonal cannot accommodate all the Reals that had
to be listed.

So the flaw of Cantor diagonal is that he applied geometry and
diagonal when such an application was never warranted.

Now maybe some place value such as perhaps 2 digits to the 100 place
value will allow a geometry with a 45 degree diagonal to cover all the
Reals that have to be covered takes place? But the answer to that is
negative, because as the place value gets larger, the number of Reals
produced gets exponentially larger.

So, the major flaw of Cantor is that he could never list all the Reals
and cut them by a diagonal to produce a new missing Real.

Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud 
are galaxies

[Archimedes Plutonium]

On Oct 29, 7:38 am, Archimedes Plutonium

So let me list the three major flaws of Cantor in his diagonal
argument:

(1) He assumed you can list all the Reals and such a list would be a
square in geometry with a 45 degree diagonal cutting through that
list. That was a pretty grotesque assumption, if I may say so. Because
in all the smaller cases the Reals never form a square configuration
for a diagonal to penetrate each Real.

(2) Cantor never used the concept of all-possible-digit-arrangements.
If he had known of that concept and applied it, he would have
instantly recognized that no geometrical array will manufacture a new
missing Real. Because if you think you have manufactured a missing
Real, then you have not listed all possible digit arrangements. It is
the same as asking where the missing possibility is of a true false
test of 2 questions when you have TT, TF, FT, FF and expecting
something new and missing. Those 4 outcomes are all the possible
outcomes and no more. So if Cantor had known of the concept All
Possible Digit Arrangements and realized it was the set of all Reals,
that the alteration of any of them only yields a Real in the list
already.

(3) Cantor tried to marry geometry to a problem of manufacturing Reals
not listed. Trouble is that the only such marriage is where you delete
Reals by intention and then you can form a square with a 45 degree
diagonal and produce a Real that you intentionally deleted.

Now in overall perspective, the Cantor alleged proof was a eye-
delusion. Do you remember the eye delusions of which is larger:

<---->


>----<

and to our surprise we find out our eyes were deluded in an optical
delusion and the two figures are equal in distance.

On page 20 of Jacobs book MATHEMATICS A Human Endeavor lists many
optical math illusions.

In the case of Cantor's diagonal argument, will in the future be
listed as another optical illusion of producing missing Reals when
none can be produced.

[Zuhair]

On Oct 29, 3:38 pm, Archimedes Plutonium

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

> whole entire Universe is just one big atom
> where dots of the electron-dot-cloud 
are galaxies

That is certainly not a proof against Cantor, not even a plan of a
proof.

First mistake of yours is that you think that at actual infinite level
matters must copy what happens at finite level, well this not the
case.

Take for example the set {1,2,3} can you have a bijection from it to a
proper subset of it? the answer is NO, and so it is for all finite
sets, but take the infinite set N of all natural numbers {1,2,3,...},
here we can have a bijection with a proper subset of it.

Again I'll come back to your analogy:

If we take n-place digits of two values 0,1 as you wrote, then the
question that present itself is the following:

Can we place all possible binary n-place digits with elements of the
set {1,2,...,n} in one-once correspondence?

let's try when n=2,

For example we may have

01
10

Simply apply the diagonal and we'll have 11 missing from the above.

We can have

00
11

Simply apply the diagonal and we'll have 10 missing from the above.
Actually this will apply to any couple of two place binary digits, so
from this we conclude that it is not possible to place all two place
binary digits with all elements of the set {1,2}!

Moving from here one can easily see that this do generalize over all n-
place binary digits, if we list n many
of them, then we'll get the geometrical arrangement necessary to carry
on the diagonal argument and the diagonal will always be an n-place
binary digit that is missing from the n sized list.

So to state this: we CANNOT place all possible binary n-place digits
with elements of the set {1,2,...,n} in one-once correspondence! This
is when n is finite natural.

The fact of the matter is because infinity showed different features
from what is occurring at finite level (like having a bijection to a
proper subset), then Cantor was testing whether the above diagonal
feature applicable to the finite world would still persist at infinite
level, so he took Aleph_0 place binary digits, and tried to see if
when listing Aleph_0 of them would that escape the diagonal argument
occurring at finite level. The answer was no. Still the argument is
applicable and so the case remains as it with the finite world i.e. we
cannot have all possible Aleph_0 place binary digits placed in one-one
correspondence with all elements of Aleph_0. So in this sense the
Reals are larger in amount than the naturals.

You are simply not capturing the intended argument of Cantor's. Cantor
never said that we can list all possible Aleph_0 place binary digits
with Aleph_0 in one-one correspondence, he said IF we assume that was
possible, then this leads to a contradiction (as it is the case with
finite sets), so the conclusion is that it is not possible, this is an
argument by contradiction "aka:Reductio ad absurdum".


Zuhair

[MoeBlee]

On Oct 29, 8:18 am, Archimedes Plutonium

That's a bizarre malconception of the diagonal argument. That the the
word 'diagonal' is used to name the agument does not entail that
there's use of a literal geometric angle in the proof. And there's no
'deleting' in the proof.

The proof is that given any enumeration f of denumerable binary
sequences, there is a denumerable binary sequence not in the range of
f. Specifically, consider the denumerable binary sequence g defined by
g(n)=0 if f(n)(n)=1 and g(n)=1 if f(n)(n)=0. g is not in the range of
f.

No mention of the word 'diagonal' is needed in the actual proof, let
alone a literal 45 degree line!

> Do you remember the eye delusions of which is larger:
>
> <---->
>
> >----<

They look to be the same length to me.

> and to our surprise we find out our eyes were deluded in an optical
> delusion and the two figures are equal in distance.

I experienced no such illusion (let alone DElusion!). The two figures
looked to be of equal distance to me

MoeBlee

[Archimedes Plutonium]

Oh really, so that your opinions account for the truth and not someone
demonstrating the truth. So that when Cantor makes a diagram and
people vote for it, while AP makes a different diagram, and people
marshal opinion against is the truth reached.

I should have expected a flood of philosophers come charging in.

But let me show you how mistaken Cantor is and how mistaken you are.

Let me start first with a Optical ILLusion of Jacobs 1970 book
Mathematics a Human Endeavor where he has a large number of optical
illusions such as this one:

_____________
_____________
_____________
_____________
_____________
_____________


And the illusion in Jacobs book of that is "it looks like a rectangle,
but in fact it is a square.

What Cantor got away with in truth of mathematics is that he built a
geometry model with a diagonal and he just listed enough Reals but not
all the Reals he knew he had to list. He listed just enough to be able
to make a square and then make a diagonal.

So what Cantor failed in Logic is that he imposes upon the Reals a
geometry that the Reals could never accommodate and that when
alters his selected few Reals that form the square when the Reals form
a rectangle that Cantor then manufactures a Real that he never
included because he could not include since he forced only so many
Reals that form a Square when the Reals form a rectangle.

Now I looked up some dates and it appears that Cantor had his diagonal
argument around the 1890s and that Dedekind had his Dedekind Cut
argument around 1890s also. That would mean that neither Cantor nor
Dedekind ever had the notion or idea of
All Possible Digit Arrangements. If either mathematician had ever
had the inkling of All Possible Digit Arrangements as the Reals then
both would have been preempted of their folly.

But let me straighten out Zuhair.

So if we make the Universe Space of all the Reals with 2 digits to 2
place value we have this rectangle:

00
01
10
11

If we make the Universe Space of all Reals with 2 digits to 3 place
value we have this rectangle:

000

001

010
100
110
101

011
111

Those are All Possible Digit Arrangements

Now a Cantor or a Zuhair coming in and wanting to make a "fake proof"
would say the set of all Reals when listed are

000
001
010

and proceed to make the diagonal and altering the digit on the
diagonal to manufacture 111 which is not on the list. Well of course
it is not on the list because Cantor and Zuhair are not interested in
fully listing all the Reals which would form a Rectangle and never a
Square, and so Cantor and Zuhair like the optical illusions of Jacob's
book disregard the All Possible Digit Arrangements that the Reals
truly are and form a rectangle where a diagonal is impossible to
encompass all the Reals into a act of alteration.

Now perhaps Zuhair, being more of a philosopher than a mathematician
deems it well to patch up Cantor's magic trick of deleting Reals he
should not have deleted, just to shape the Reals into a square at
infinity when the Reals always formed a rectangle at infinity.

So that Zuhair on 2 digits to 2 place values want the further magic
trick of this:

00--
01--
10--
11--

So now Zuhair would have accomplished the magic trick of optical
illusion that Cantor foisted upon the world with his 1890s published
fake proof. Now a diagonal can be drawn because the Reals were all
included but those are not numbers of mathematics but filler space of
dash marks.

So when Zuhair draws the diagonal through to 0 and then 1 then a dash
mark then another dash mark, what does he alter the dash mark into
becoming?

What Cantor assumed was that he would fill up the dash mark with
zeroes and then alter them with a 1 by the diagonal.

So the error, the flaw, the fakery of Cantor and all his followers of
the next century was that Cantor never understood that Reals are All
Possible Digit Arrangements and that geometrically they never form a
square to apply a Cantor diagonal. If you leave out many of the Reals
you do form a square and you pick up what you deleted.

Archimedes Plutonium

http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom

where dots of the electron dot cloud are galaxies

[Archimedes Plutonium]

On Oct 29, 3:40 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Oct 29, 10:18 am, Zuhair <zaljo...@gmail.com> wrote:

(snipped to save space)

Alright, I reached what I call Critical Mass Clarity. It is a term for
psychology and not for the sciences. It is a term that means when in
the pursuit of scientific truth, there comes a moment in which you
have gathered enough information to be able to see the problem and its
solution with clarity. Not to say there is an end to learning new
information about the problem and its solution, but rather, that
enough data, information is there to know the truth has been reached.

The failure of Cantor and Zuhair over the diagonal argument involves
these concepts:

(1) when a rectangle is formed versus a square and the diagonal
involved

(2) Reals as All Possible Digit Arrangements

(3) The Fundamental Counting Principle of Probability theory such as a
true-false test questions are counted by 2^2 then 2^3 then 2^4
etc etc.

Those three concepts are critical and here is the argument laid bare
and why and how Cantor and Zuhair failed.

The argument all reduces down to the Reals between 0 and 1 and even
Cantor acknowledges that. Also, a further reduction to simplicity is
that the Reals are represented in binary base of its two digits of 0
and 1. These make the argument easier and can easily be ramped up to
include the more general proof that Cantor had a fake proof.

Proof that Cantor diagonal argument is a fake proof:
(a)Reals are All Possible Digit Arrangements so that if given
0.0000000000... and any of those digits are altered into being a 1
rather than a 0 digit in a specific place-value, that it is still a
Real number. So that All Possible Digit Arrangements is the set of All
Reals.

(b) All Possible Digit Arrangements as equal to the set of All Reals
has a arithmetic measure in the form of the Fundamental Counting
Principle with its arithmetic of 2^2 then 2^3 then 2^4 etc etc

(c) the arithmetic of 2^2, 2^3, 2^4, . . is a rectangle forming
arithmetic as can been seen by these diagrams of all the Reals with 2
digits to 2 place value and all the Reals with 2 digits to 3 place
value:

00
01
10
11

and this:

000
001


010

100

110
101


011
111

(d) so when listing Reals, Reals always form rectangles since all the
Reals of a specific place value have a 2^2,2^3,2^4,. . . measure.

(e) Since any listing of Reals that is not a rectangle cannot have a
diagonal that includes all the Reals. Which means that whenever anyone
like Cantor or Zuhair list the Reals as a square with a diagonal, they
could only do that by deleting over half of the Reals that they were
demanded to list. And that their diagonal alteration manufactures a
Real which they had deleted beforehand in order to corral their
remaining Reals into a square figure.

QED

I reached a point of Critical Mass Clarity and I learned something
spectacularly new which I never before realized about the set of All
Reals. The set of all Reals forms a rectangle and never a square,
unless you want to discard over 1/2 of the Reals and then form a
square with the remaining depleted set.

[Archimedes Plutonium]

So the argument is really going to end up as those who believe in
Cantor must believe that the Reals between 0 and 1 always forms a
perfect square for which a diagonal can be drawn and execute a
alteration of each Real.

The people who believe Cantor was full of beans and offered a fakery
would point to the idea that the Reals, all of them between 0 and 1
forms a geometrical rectangle for which no diagonal can be formed that
allows a operation of alteration. And the evidence, the strong
evidence is 2 digits (binary base) in place values follows a
arithmetic measure of 2^2, 2^3, 2^4, etc etc, the Fundamental
Principle of Counting in Probability theory.

So as you go to infinity, you go to it in the Reals by 2^2, 2^3, 2^4,
etc etc and that is a rectangle. It is this pattern:

00
01
10
11
and this:
000
001
010
100
110
101
011
111

So in Cantor's fakery, he just ignores the fact that the Reals never
form a perfect square, unless you are willing to delete over half of
the Reals that should not be deleted and then you manufacture a new
Real, only because the new Real is one of the many that you deleted to
allow a square to form. In the case of 2 digits to 3 place value
Cantor kept just

000
001
010

and deleted all the others of these:

100
110
101
011
111

And then when you formed the diagonal and did the alteration he
manufactured the Real 111, but he thought he manufactured a Real from
a assumed list of All the Reals. Too bad for him, he deleted 5 of the
Reals from a set of 8 Reals in order for him to form a perfect square.

So, the people who believe in Cantor as a proof of mathematics, are
going to be terribly let down and depressed, because they never really
examined the trickery of the argument, as if they were conned by a
optical illusion.

But now, it took me several years to push the fakery of Cantor's proof
to this point of clarity. For me, back in early 1990, I had the
instinct that Cantor was wrong and that his alleged proof was a
fakery. It would take me until 2012 to make this crystal clear of how
large a fakery it was. But back in 1991 thereabouts, the concept of
All Possible Digit Arrangements came to me and if you have a highly
logical mind, just a single concept of All Possible Digit Arrangements
is enough to not only convince you that Cantor was wrong to prove it
in just three sentences.

Mind you, I have looked through the literature of mathematics and from
what I see, I am the first mathematician to have discovered and
enunciated this idea or concept of the Reals as All Possible Digit
Arrangements.

So how does that concept alone debunk Cantor's diagonal?

Easy.

If you form all the Reals between 0 and 1 as All Possible Digit
Arrangements. And no matter whether those Reals form a rectangle or a
square, regardless, and you then include the diagonal and do the
alteration. And then you manufacture what you think is a new Real
missing from the list. Contradiction, in that All Possible Digit
Arrangements negates any claim that you manufactured a new Real not
already present. So that in the early 1990s, I had the instinct that
Cantor was wrong, and the moment I discovered All Possible Digit
Arrangements, would serve as a proof that Cantor was wrong. But as
LWalk spoke of clarity, most people would not see that All Possible
Digit Arrangements is enough. Most people have so little logic, even
professors of mathematics, that they need the full deal and scope of
squares, rectangles, place values, Fundamental Counting Principle.

To a person with strong logic, they need only an instinct and then a
good concept.
Now all of this is going to sound like bragging but I am not writing
for my generation but for the future generations to learn a very
important idea. The idea is that most mathematicians are very poor in
logic. And that within a century, there maybe just one or two
mathematicians having logical abilities and all the rest are parrots
of mathematics. And those parrots cause more damage to mathematics
than good by filling up math with fakery proofs and concepts. And then
it takes that logical mathematician to remove the trash of the
parrots. So what sounds like bragging and gloating is rather, my
warning and teaching for the future generations of mathematicians. And
the true logical mathematician the few and far that come, their
abilities in mathematics or science is not a genetic skill nor a
environment skill but is a skill given them by superdeterminism. The
best mathematicians were not genetic or nature or nurture, but were
blessed by the gods of the Atom Totality. There are no genes for math
genius nor is there a environment in which to nurture a math genius.
It comes purely from "out of this world, from the nucleus of the Atom
Totality via superdeterminism."

Did you ever wonder how a new born bird knows where to fly to migrate
to home? Did you ever wonder how some can multiply in their head large
numbers or memorize pi to thousands of digits. It is not because of
genes nor nurture, but because every thought, every idea, every action
is superdetermined by the Atom Totality. I guess it is getting close
to November 7 and starting my preaching early for the holiday of
Plutonium Day.

Archimedes Plutonium
http://www.iw.net/~a_plutonium/

[Archimedes Plutonium]

Now I probably started a topic a few days ago I should not have
started. I am finding it increasingly more difficult to remember where
I left off on a topic of the past. My memory is no longer the sharp
memory I had when I was younger and my memory is in decline. So
sometimes I am going to embark on a topic for which I had resolved
earlier, but not remembering how I resolved it. So I have to keep an
eye out and alert for topics I embark on when I should not.

A few days ago I embarked on the idea that the Rationals are all
possible digit arrangements of the integers via division.

Now I should not have embarked on that because with the finite to
infinity borderline being 10^603 means that the smallest nonzero
number is 10^-603 and it serves as a successor function axiom in the
same manner as the Successor function serves in the Peano axioms for
the Natural Numbers.

So that starting with 0 and 1 in Peano axioms we derive all the
Naturals as 0, 1, 2, 3, 4, ...

And starting with 0 and 1*10^-603 and continually adding by 10^-603 we
end up with all the Reals between 0 and 10^603.

So we can arrive at all the numbers from 0 to 10^603 in two manners.
We can start with 0 and 10^-603 and successively add 10^-603 and we
end at 10^603 and end up with 10^1206 numbers in full.

Or, we can do a all-possible-digit-arrangements of the numbers between
the integers from 0 to 10^603 with digits of no more than
603 digits rightward of the decimal point.

So now, why should I have never embarked upon the division definition
of Rationals?

Well, if you can derive all the Reals from 0 to 10^603 as successions
of 10^-603, then all of those numbers are Reals, but they are also
Rationals. Every number of those 10^1206 numbers are no longer able to
be classified as either rational or irrational or algebraic or
transcendental.

When mathematics recognizes the borderline of finite with infinity as
10^603, then mathematics loses the distinction between Rational and
Irrational and Algebraic and Transcendental. When mathematics keeps a
illogical and fuzzy notion of infinity with no borderline, then
mathematics invents these fake-definitions of numbers that are
irrational and transcendental. Both the numbers pi and e, when
infinity starts at 10^603, both pi and e are rational numbers as are
all numbers of mathematics.

So, sorry I made that detour and I should monitor my memory better.

[Zuhair]

On Oct 29, 11:40 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Oct 29, 10:18 am, Zuhair <zaljo...@gmail.com> wrote:

Dear plutonium,

I think it is better if we take matters step by step.

It is helpful to try to know what the question is, before starting to
solve it.

The question is can we have one-one correspondence between the set R
of all Reals and the set N of all naturals?

This question is also equivalently stated as

Can the set of all Aleph_0 (the cardinality of N) sized {0,1} digit
sequences have a one-one correspondence with the set N?

Before we answer this question, lets ask it about the finite world,
and see what happens there, and whether it could be paralleled at the
infinite level, the question would be:

Can the set of all n sized {0,1} digit sequences have a one-one
correspondence with the set {1,2,3,...,n} where n is a natural number?

Now to solve this question we actually use the 'diagonal' argument of
Cantor's!

We say for a proof by negation lets ***ASSUME*** that ALL n sized
{0,1} digit sequences have a one-one correspondence with the set
{1,2,3,...,n}, so it is helpful to make a list that "follow this
ASSUMPTION":

1 -> d1_1, d1_2, d1_3,..., d1_n
2 -> d2_1, d2_2, d2_3,..., d2_n
.
.
.
n -> dn_1, dn_2, dn_3,..., dn_n

where each variable di_j in the above list either take value 0 or 1.

Again the above square list is just the list portraying the ASSUMPTION
made above (the list above is made to portray the assumption given
above and not to copy the Reality of listing all binary n long
sequences, you need to hammer that in your mind).

Now make diagonal alternations i.e. change the value of each di_i, so
for example if d3_3 had the value 0 then change it to 1. Now denote
each changed digit by d'i_i.

So we'll have a new diagonal that is

d'1_1, d'2_2, ...., d'n_n

which is not in the above list.

So the conclusion is that AT FINITE level we cannot have a one-one
correspondence between the set of all n sized {0,1} digit sequences
and the set {1,2,...,n}.

The diagonal argument at FINITE level is a TRUE argument. NOBODY
disputes that!

Now the question is Can we carry on this argument into infinite
levels. The answer is YES. You can follow the exact argument even on
infinite listing since identification of the position of the altered
digit in the diagonal is always recognizable (because it is finite)
and infinity wont affect that.

So first we make the ASSUMPTION that there is such a correspondence
and check the consequence of this Assumption:

To see that in a quasi-illustrative manner:

1 -> d1_1, d1_2, d1_3, ....
2-> d2_1, d2_2, d3_3,....
.
.
.

The same occurs as with the **assumption** made at finite level, we'll
have a square list, but this is just the consequence of the assumption
made before, it is not a statement of the Reality of listing all the
reals.

now change the value of each di_i digit in the above list to obtain
d'i_i digit, and the sequence

d'1_1, d'2_2,.....

is a new sequence that is not present in the list.

So here the same exact argument given to show what is happening at
FINITE level, still works at INFINITE level.

So the answer is we cannot place all Aleph_0 size binary sequences
with all naturals!

YOU are confused, you are not doing the correct matching, you are
simply confusing the assumptive for the Real, there is where your
confusion is!

Let me show you what I mean:

The REALITY of listing ALL n sized binary valued sequences where n is
finite is that except for the case where n=1 we'll always have a
RECTANGULAR list!!!

Of course that is YOUR statement, which is correct, no doubt, it is a
TRUE statement that NOBODY denies, and you showed the examples of two
place and three place binary sequence complete listings.

Those rectangular lists you've made are portrayals of the REALITY of
how the list of ALL possible n sized binary sequences look for a
particular n. They are not lists of the ASSUMPTION that all possible n
sized binary sequences are in one-one correspondence with an n sized
set, for a particular n.

So there are two kinds of lists, the REALITY lists, and the ASSUMPTIVE
lists.

For example: The REALITY list of all two place binary digit sequences
is

00
01
10
11

A rectangular list!!

But an assumptive list of having a one-one correspondence between all
two placed binary digit sequences and the set {1,2} would be written
in variable notation listing like that

d1_1, d2_2
d2_1, d2_2

where each di_j either equal 0 or 1.

This assumptive listing is SQUARE shaped, and it reflects the
assumption made in the argument, it is not the Reality list.

In other words what Cantor's argument is proving is that the list of
all Reals do not make a square! in other words the list of all Reals
is rectangular as it is the case for the list of all n sized binary
sequences for n>1.

In nutshell so that you know where your error is

The list that portrays the reality of listing of all n sized binary
sequences is ALWAYS rectangular as far as n>1, whether n is finite or
not!!! Cantor's diagonal argument proves that!

The list that portrays the assumptive argument that there is a one-one
correspondence between all n-placed binary digit sequences and an n
sized set would of course be a square list, but it is shown by
Cantor's diagonal argument that this list is not possible even when n
is infinite!

With Cantor's argument I'm seeing a consistent form of thinking
between Reality listing and assumptive listing, while in YOURS there
is a confusion, you object to the assumptive listing being a square
list because the reality listing is not forming a square, you are
simply not being able to follow the logic of Argumentation by
Contradiction, you are confusing the result for the means to reach at
it, you simply confused the assumptive for the real!!! And one would
not expect the TRUTH of the matter to be envisaged by a confused
endeavor.

Zuhair

[Archimedes Plutonium]

I wish Dik Winter was still around now, for somewhere in the early
2000s he was telling me factual data of the original Cantor diagonal
written in German. My memory is sketchy but I seem to recall that
Cantor's argument was unlike the above described by Zuhair.

Zuhair's argument follows on the lines of what appears in Wikipedia
under Cantor diagonal with the 1-1 correspondence and stating:

"Let T be a set consisting of all infinite sequences of 0s and 1s."

But now here again, both the Cantor and the Zuhair argument will fail.

But let me just say a few words about a different issue. In your above
critique Zuhair, I feel you have no right to even mention finiteness
or infiniteness in any sentence until and unless you specify the
border between finite and infinity. I made a project of
discovery and I place infinity at 10^603 and pi*10^603.

So what that gives me is the fact that all the Reals between 0 and 1
are in a 1-1 correspondence with all the finite Naturals:


1 to 1*10^-603
2 to 2*10^-603
3 to 3*10^-603
.
.
.
((10^603)-2) to ((1)- (2*10^-603))
((10^603)-1) to ((1)- (1*10^-603))

And for a 1-1 correspondence of Naturals with All the Reals which are
all the numbers from 0 to 10^603 involves the Naturals out to 10^1206.

But that is a story for another time.

The error of the above scheme of diagonal with sequences is the same
error as Cantor with Reals between 0 and 1.

It is fine to show a big list that fills the page. But there is no
guarantee that those sequences forms a perfect square.

Zuhair, the only guarantee is that 2 digits to 2 place value and
increasing place value never form a perfect square. They all form
rectangles.

And so, Cantor failed to guarantee his list of all Reals in 0 to 1
forms a perfect square or whether he had to delete many Reals to make
a perfect square. With the Zuhair list of all sequences, what
guarantee does Zuhair have that such a list is a perfect square or a
rectangle?

The only guarantee we do have is the place value forms rectangles
since it is from the Fundamental Counting Principle 2^2, 2^3, 2^4, etc
etc

A microscopic Cantor diagonal argument is 2 digits to 2 place value:

00
01
10
11

Cantor and Zuhair would delete the 10 and the 11 because they spoil
the forming of a square to make a diagonal. And with the diagonal on

00
01

you manufacture 10 in the alteration. And so Cantor and Zuhair would
claim they found a new Real not on the list of All Reals, when in fact
they deleted 10 and 11 just so they could form a square.

Whether Cantor uses All Reals or Zuhair uses All Sequences, both
have no guarantee that they form a perfect square for the diagonal to
work properly.

The Place-Value due to Counting Principle dictates that they form
Rectangles, not squares. So that dooms Cantor and Zuhair.

Zuhair, three questions.

(1) You obviously reject the logic of All Possible Digit Arrangements.
That if we define the Reals as such, and then no matter what the Reals
form, whether a square or rectangle, that the moment you alter those
Reals, the manufactured Real already exists within the list. Why do
you not buy that?

(2) How did Cantor define the Reals, keeping in mind he was in the
1870s to 1890s. The definition of Reals was not really all that good
in the 1800s.

(3) Why is it that you never seem to ask yourself for a borderline
between finite and infinity, such as 10^603? Do you think that
finiteness is on a different track than infinity and that they never
approach one another. In order for Germany and France to exist as
countries, means there must exist a borderline between them and you
would fail as a geographer if you never state the borderline. And you
should fail as a mathematician if you never
insist on a borderline crossing between finite and infinity. So the
question is why have you never considered the borderline of finite to
infinite, Zuhair?

[MoeBlee]

On Oct 30, 7:03 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:

> > In other words what Cantor's argument is proving is that the list of
> > all Reals do not make a square!

AGAIN, there's no need to prove anything makes or does not make a
"square".

Here, AGAIN, is the argument:

Claim: If S is a set of denumerable binary seqeuences, and f is an
enumeration of S, then there is a denumerable binary sequence not in
the range of f.

Proof: Consider the denumerable binary sequence g defined by g(n)=0 if

f(n)(n)=1 and g(n)=1 if f(n)(n)=0. g is not in the range of
f.

> (2) How did Cantor define the Reals, keeping in mind he was in the
> 1870s to 1890s.

If I'm not mistaken, he took reals to be equivalence classes of Cauchy
sequences.

MoeBlee

[Archimedes Plutonium]

On Oct 30, 3:48 am, Zuhair <zaljo...@gmail.com> wrote:

> On Oct 29, 11:40 pm, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > On Oct 29, 10:18 am, Zuhair <zaljo...@gmail.com> wrote:
>
> Dear plutonium,
>
> I think it is better if we take matters step by step.
>

Hi Zuhair, I hope you are up for your blunders. I caught a full night
worth of sleep and am able now to show you your blunder in full and
that of Wikipedia.

Let me please skip to the heart of your blunder:

>
> We say for a proof by negation lets ***ASSUME*** that ALL n sized
> {0,1} digit sequences have a one-one correspondence with the set
> {1,2,3,...,n}, so it is helpful to make a list that "follow this
> ASSUMPTION":
>
> 1 -> d1_1, d1_2, d1_3,..., d1_n
> 2 -> d2_1, d2_2, d2_3,..., d2_n
> .
> .
> .
> n -> dn_1, dn_2, dn_3,..., dn_n
>
> where each variable di_j in the above list either take value 0 or 1.
>
> Again the above square list is just the list portraying the ASSUMPTION
> made above (the list above is made to portray the assumption given
> above and not to copy the Reality of listing all binary n long
> sequences, you need to hammer that in your mind).
>

Now Wikipedia on "Cantor diagonal argument" write Zuhair's above as
this:

s1 = 0000000
s2 = 1111111
s3 = 0101010
s4 = 1010101
s5 = 1101011
s6 = 0011011
s7 = 1000100

And Wikipedia's alteration after the diagonal is

s0 = 1011101

So, a simple question Zuhair, since you make the same blunder as the
Wikipedia authors make.

Can the Cantor diagonal system work if the diagonal is straight up or
down at a 90 degree angle rather than a square with a exact 45 degree
angle.

For in yours, Zuhair, if a 90 degree diagonal works for Cantor, yours
would be

d1_1
d2_1
.
.
.
dn_1

with alterations of a 90 degree diagonal.

Wikipedia's would be alterations on
0
1
0
1
1
0
1

which would be a S0 of 1010010 and not a S0 of 1011101

So the question I am asking you Zuhair, since I cannot ask a author of
the Wikipedia page which shows the same blunder.

Is how crucial is it for Cantor Diagonal to have a precise 45 degree
diagonal in order to make anything work correctly in the Cantor
method? Can we have diagonals at 90 degrees or 89 degrees or
say 43.52 degrees.

So, since the answer, you must agree upon eventually, is that the
Cantor diagonal method can only work if the listing is a perfect
square, never a rectangle.

And since you, nor anyone else making lists of numbers is certain that
your list is a perfect square and the diagonal is exactly 45 degrees.
That you have no method and no proof of anything other than your
imagination run amok.

Now here is another proof that the Cantor method is horribly flawed.

Suppose I asked you to pick up the alleged new Real you created in
your diagonal and add it as a member to your already existing list.
And ditto the same for Wikipedia where they obtained

1011101

So they threw back that number onto the list and now they did another
diagonal.

I think you know where this is leading to. For if you were correct and
Wikipedia was correct, that your utter disregard for the fact that the
Cantor method works only on perfect-squares and not rectangles and
only with a exact 45 degree diagonal, that when you throw back your
new manufactured number to be also altered by the diagonal that you
run into a heap of big trouble because that 45 degree diagonal does
not intercept the new number for the listing is no longer a square
that you conveniently picked out.

So Zuhair, please try to answer these questions:

(1) Is it essential that the Cantor method uses only perfect squares?

(2) Does the diagonal have to be exactly 45 degrees and no other angle
can work in the Cantor diagonal such as a straight up and down angle
of 90 degree diagonal on the first member of each row in the list?

If your answer is "yes" to both of these, then your blunder and
Wikipedia's blunder and Cantor's blunder is that none of you can
guarantee a list of Reals in the interval 0 to 1 forms a perfect-
square unless you cherry-pick a perfect square.

[MoeBlee]

On Oct 30, 12:56 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:

> Is how crucial is it for Cantor Diagonal to have a precise 45 degree
> diagonal in order to make anything work correctly in the Cantor
> method? Can we have diagonals at 90 degrees or 89 degrees or
> say 43.52 degrees.

You have to be stupid or dishonest to impose on the proof a literal
geometric interpretation. The proof does not involve ANGLES or
DEGREES.

Again, though you continue to ignore:

Claim: If S is a set of denumerable binary seqeuences, and f is an
enumeration of S, then there is a denumerable binary sequence not in
the range of f.

Proof: Consider the denumerable binary sequence g defined by g(n)=0
if
f(n)(n)=1 and g(n)=1 if f(n)(n)=0. g is not in the range of
f.

No mention of diagonals, angles, or degrees.

MoeBlee

[Zuhair]

On Oct 30, 3:03 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Oct 30, 3:48 am, Zuhair <zaljo...@gmail.com> wrote:

[snip]

>
> A microscopic Cantor diagonal argument is 2 digits to 2 place value:
>
> 00
> 01
> 10
> 11
>
> Cantor and Zuhair would delete the 10 and the 11 because they spoil
> the forming of a square to make a diagonal. And with the diagonal on
>
> 00
> 01
>
> you manufacture 10 in the alteration. And so Cantor and Zuhair would
> claim they found a new Real not on the list of All Reals,

No this is not the case, try to read fully what I wrote to you about
assumptive and real listing so that you know what Cantor and I mean by
our argumentation, and so that you know where your confusion lies.

Neither Cantor nor I are making short listing or deleted listing or
even saying that the list of all reals is a square list nor anything
like that, all of those claims are your own misinterpretation of what
is said.

YOU simply do not understand the logical form of "Argument by
contradiction", that's your basic problem.

In argumentation by contradiction we suppose that we are confronted
with a situation that can only have one of two solutions lets say it
is either B or not B.

Suppose I think the true solution is B, now in argumentation by
negation we first suppose the negation of what we think it is true, in
this case we start by supposing that "not B" is true, Now you need to
understand that this is just an initial supposition, it is not what we
believe as the final truth. Now we proceed and show that supposing
"not B" is true will lead to a contradiction! so we go from there and
conclude that "not B" is not the solution, so the real solution to the
problem must be: B.

In the case of the Reals and naturals, we saw that at all finite cases
(finite defined in the standard manner after Tarski) the list of all n
sized binary sequences is Rectangular in shape as you said and we
agree to that, so based on this observation we think that the real
situation with the Reals is also the same, i.e. we think that the
reality of the matter is that the list of all countably infinite sized
binary sequences (this list has the same size as that of the list of
all reals) is a RECTANGULAR list!!! Now in order to prove that we use
argument by negation. This goes in the following manner:

(1) Either the list of all countably infinite sized binary sequences
is:

a) Rectangular (i.e.Reals more than naturals)
or
b) Square (i.e. Reals as many as naturals)

(2) Assume that b) is true, then ***according to that assumption*** we
would have the following list of all
countably infinite sized binary sequences:

d1_1, d1_2, d1_3, ...
d2_1, d2_2, d2_3,....
d3_1, d3_2, d3_3,....
d4_1, d4_2, d4_3,....
.
.
.

where each di_j takes the value 0 or the value 1

Now this list is clearly what you call as square list, now simply
change the value of each di_i in the list (for all
i=1,2,3,...........) to obtain d'i_i with a different value from di_i
value, for all i values, then we'll construct the diagonal string

d'1_1, d'2_2, d'3_3,..........

Now this string is different from all the others in the list above
because every string of those is of the form

di_1, di_2, di_3,....., di_i, di_i+1,......

where i=1 or 2 or 3 or 4 or ......

Now each of those strings will have a di_i in it that is different
from the value of d'i_i in the diagonal string, that's why the
diagonal string is not in any of the listed strings.
Now this contradicts (2) which is the assumption that the listing is
complete.

So assumption (2) have lead to a contradiction, so we conclude that
assumption (2) is not possibly true. In other words we conclude that
the square list is never a complete list of all countably infinite
sized binary sequences!

So the true solution is (a) i.e. the list is Rectangular! In other
words the conclusion is that there are more Reals than naturals!!!
because the opposite leads to a contradiction as I showed above.

Can you get what I'm saying here. You need to follow the line of
argumentation of "Argument by contradiction".

Zuhair

[MoeBlee]

On Oct 30, 1:24 pm, Zuhair <zaljo...@gmail.com> wrote:
> You need to follow the line of
> argumentation of "Argument by contradiction".

Why even bother with that? The argument is as simple as this:

Claim: If S is a set of denumerable binary seqeuences, and f is an
enumeration of S, then there is a denumerable binary sequence not in
the range of f.

Proof: Consider the denumerable binary sequence g defined by g(n)=0 if
f(n)(n)=1 and g(n)=1 if f(n)(n)=0. g is not in the range of
f.

MoeBlee

[MoeBlee]

On Oct 30, 1:24 pm, Zuhair <zaljo...@gmail.com> wrote:

> try to read fully what I wrote to you about
> assumptive and real listing so that you know what Cantor and I mean by
> our argumentation, and so that you know where your confusion lies.

You might as well be asking a chimpanzee to please be careful to use
the correct zip code when addressing a letter.

MoeBlee

[Zuhair]

On Oct 30, 8:56 pm, Archimedes Plutonium

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

> whole entire Universe is just one big atom
> 
where dots of the electron dot cloud are galaxies

The square analogy is just an illustration, it is not really meant to
be as such, as Moe Blee mentioned to you.

Anyhow what we mean by a square list is simply having the number (i.e.
cardinality) of digits in each sequence being equal to the number of
sequences in the list

Just an example take the following list:

01
11

Now this list has two sequences and each sequence has two digits, so
the number of sequences in the list is equal to the number of digits
in each sequence so by definition it is a square list.

Generally for the finite case we right this as:

d1_1, d1_2,..., d1_n
d2_1, d2_2,..., d2_n
.
.
.
dn_1, dn_2,..., dn_n

where n is any finite natural number. and each di_j takes the value 0
or the value 1. You see we have n many digits in each sequence and
also we have n many sequences in the list, so it is a square list by
definition.

Now the diagonal of the above list is defined as the following
sequence:

d1_1, d2_2, d3_3,...., dn_n

Now in the infinite case we represent a square list in the following
manner:

d1_1, d1_2, d1_3, ...
d2_1, d2_2, d2_3, ...

d3_1, d3_2, d3_3, ...
.
.
.
.

This is called a square list because by definition the number of
digits in each sequence is countably infinite, i.e. it is specifically
Aleph_0, and also the number of sequences in the list is also Aleph_0,
so by definition it is called a square list.

You need to know that the infinite have different properties from the
finite, so for example lets delet the first sequence of the above
list, so we'll obtain the following sequence

d2_1, d2_2, d2_3, ...
d3_1, d3_2, d3_3, ...
.
.
.
.

This is also a square list because it is still the case that we have
the same number digits in each sequence being equal to the number of
sequence in the list both are Aleph_0.

The same thing applies if I for example add a sequence of the original
list that is also countably infinite sized sequence, still the
resulting list will be a square list.

Now the definition of the diagonal also has a strict definition: the
diagonal of each square list is the sequence of the first digit in the
first sequence, followed by the second digit int he second sequence,
etc...

So the diagonal of the original list above would be

d1_1, d2_2, d3_3,....

While the diagonal of the following list

d2_1, d2_2, d2_3, ...
d3_1, d3_2, d3_3, ...
.
.
.
.

Is actually: d2_1, d3_2, d4_3,...,di_i-1,.....

Hope that helps!

Zuhair

[Zuhair]

This is too short for AP to grasp.

Zuhair

[MoeBlee]

On Oct 30, 2:25 pm, Zuhair <zaljo...@gmail.com> wrote:

> This is too short for AP to grasp.

Right, like there is a right length that he would ever understand.

MoeBlee

[Archimedes Plutonium]

Your first major error, for you call it a finite natural number
and then later talk about finite versus infinite, yet you never told
us what is the borderline between finite and infinite so as to know
whether say 10^1000 is a finite number or an infinite number.

If you were in geography and never able to draw the borderline between
France and Germany you would fail geography. If you are in mathematics
and use the concept finite and infinity and never tell where the
border crossing is between finite and infinity, you would be a failure
of mathematics also.

Alright, good good for you define the diagonal such that it need not
be a list that is a perfect square and a diagonal that need not be a
45 degree angle. With that definition and the Cantor method, we can
now prove that the Natural Numbers have no 1-1 correspondence with the
Natural Numbers. That given any list of Natural Numbers, the Cantor
method yields a Natural Number not on that list of all Natural
Numbers. So that the Natural Numbers, using the Cantor method is a set
that is of a larger infinity than the original set of Natural Numbers
we start with.

> So the diagonal of the original list above would be
>
> d1_1, d2_2, d3_3,....
>
> While the diagonal of the following list
>
> d2_1, d2_2, d2_3, ...
> d3_1, d3_2, d3_3, ...
> .
> .
> .
> .
>
> Is actually: d2_1, d3_2, d4_3,...,di_i-1,.....
>
> Hope that helps!
>
> Zuhair

So in the above, Zuhair claims that the diagonal need not be a 45
degree angle and that the listed set of all Reals between 0 and 1 need
not be a perfect square.

So applying that to the list of All Natural Numbers and using the
Cantor method we prove that the Natural Numbers are a larger infinity
than the Natural Numbers we start with.

Put the Natural Numbers in binary base representation and list them as
this:

0
01
011
0010
.
.
.

Now I just give one possible listing as in the flavor of Cantor
diagonal method but I am sure others can configure a better listing of
all the Natural Numbers in a Cantor style listing.

Now we apply the Cantor Diagonal as prescribed by Zuhair and we can
even just do a 90 degree downward diagonal because the definition of
the diagonal as Zuhair wrote above is so flexible that we can run the
diagonal as he says, a step across then a step down, step across, then
step down, or, we can just go completely down, down, down and do the
alterations.

Either way or any other way using the Cantor diagonal method we end up
with the absurd conclusion that the infinity of the Natural Numbers is
larger than the infinity of the Natural Numbers.

And the reason it works is because the disconnect in Cantor's proof,
the disconnect in Zuhair's proof, the disconnect in Wikipedia's proof
is they claim to list ALL the Reals in interval 0 to 1, or ALL the
Natural Numbers. The new number they manufacture
is merely the reflection of the fact that adding one to the largest
number in a list yields a new number not in the list.

So that instead of going through the process of Cantor diagonal to
fetch a new number not in the list, why not just pick out the largest
Real listed and add something to it to manufacture a new number not in
the list. Why bother to Cantor diagonal the Natural Numbers, just pick
out the largest in the list and add 1 and thus claim no list of All
Natural Numbers is the infinity of the Natural Numbers.

Of course, the reader will realize that when you have a fake idea in
mathematics, you produce more fakery such as the Natural Numbers are a
larger infinity than the Natural Numbers.

Now, the burden so far in this conversation between Zuhair and myself,
the burden has been on me to show Zuhair where his mistakes are. But
now, let me shift the burden on Zuhair and refusal to play game is
admission of error. I played Zuhair's game, now he should play mine.

I know Cantor's diagonal is fakery because it lacks the understanding
of one important concept-- All Possible Digit Arrangements and that
concept is integral to Probability theory in that of the Fundamental
Counting Principle that the total possibilities of 2 digits in 2 place
value is 2^2 and in the 3 place value is 2^3 and in the 4 place value
is 2^4, etc etc.

So the error of Cantor is that All Reals in the interval 0 to 1 is all
the Reals of All Possible Digit Arrangements of place values from 0 to
1.

THEREFOR: no matter what you list or how you list it or whether the
list is a square or rectangle or whether the diagonal is a Zuhair
diagonal or some other diagonal. The fact that the list is All REALS
of All Possible Digit Arrangements, forces the alteration of any Real
on that list, to have a identical representative already in the list.
If you argue that you have manufactured some alien Real not in the
list, then you have violated the fact that the list is All Possible
Digit Arrangements.

That is the proof that the Reals from 0 to 1 have the same identical
infinity as the Natural Numbers from 0 to infinity and that infinity
comes in one and only one type of measure.

So, now, Zuhair does not believe the proof I have just given using All
Possible Digit Arrangements.

The burden is on Zuhair to point out what error or what mistake I
made, if any, by proving that the Reals are the same infinity as the
Natural Numbers.

The burden is on Zuhair to show that All Possible Digit Arrangements
of Reals from 0 to 1 preempts any goofy Cantor diagonals to
manufacture some foreign, alien Real not already on the list.

Here is the PROOF in essence:

List all the Reals between 0 and 1 as All Possible Digit Arrangements.
Alter any one of those or combinations of those, and you never
manufacture a new Real since the list is All Possible Digit
Arrangements.

Zuhair seems unable to comprehend how All Possible Digit Arrangements
preempts a Cantor diagonal.

So, what is wrong with that, Zuhair?

Archimedes Plutonium

[MoeBlee]

On Oct 30, 3:35 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> you never told
> us what is the borderline between finite and infinite

There is no greatest natural number to serve as such a "border".

A set is finite iff there there is a well ordering R of the set such
that the converse of R is also a well ordering of the set.

A set is infinite iff the set is not finite.

Those are exact definitions. There's no need to invoke a notion of
"border".

> so as to know
> whether say 10^1000 is a finite number or an infinite number.

It's finite. It's trivial to prove that 10^1000 is a natural number.

Indeed, take any two natural numbers and apply an arithmetic operation
to them (such as exponentiation) and the result is still a natural

number.

> If you were in geography and never able to draw the borderline between
> France and Germany you would fail geography.

This is not geography.

(By the way, I don't think being able to draw a borderline (which is a
curvy and very complicated line) between France and Germany is
required for a passing grade in geography.)

MoeBlee

[Patricia Shanahan]

I think maybe some of the confusion may be due to thinking about finite
sequences, and assuming g cannot be longer than the longest f(n), so I
would modify this to pick e.g. 1 if f(n) has less than n digits.

Patricia

[MoeBlee]

[MoeBlee]

My previous post was submitted by mistake; there's nothing but quote
in it.

This is the post I meant to make:

On Oct 30, 4:59 pm, Patricia Shanahan <p...@acm.org> wrote:

> I think maybe some of the confusion may be due to thinking about finite
> sequences, and assuming g cannot be longer than the longest f(n), so I
> would modify this to pick e.g. 1 if f(n) has less than n digits.

There are no finite sequences involved in my proof. S is a set of
denumerable (i.e., countably infinite) binary sequences.

There's no reason to consider an S that has objects other thann
denumerable binary sequences. There is a 1-1 correspondence between
the real interval [0 1] and the set of denumerable binary sequences.
Thus, by showing that the set of denumerable binary sequences is not
enumerable, we show also that the interval [0 1] is not enumerable,
thus the set of real numbers is not enumerable.

MoeBlee

[Archimedes Plutonium]

Now while I wait for Zuhair to point out the errors, if any in my
proof that the Reals are countable, since the Reals are All Possible
Digit Arrangements and that infinity comes in only one type-- a
countable infinity.

Proof: List all the Reals of interval 0 to 1. Define Reals as All
Possible Digit Arrangements. Do any type or sort of Cantor Diagonal,
any and all. Because no Cantor Diagonal will manufacture a number that
is not already on the list. The reason being is that if you understand
the meaning of All Possible Digit Arrangements, that there does not
exist any Real absent from the list. And by making alterations, can
never create a Real which is absent.

QED

So now, while I wait for Zuhair to correct any mistakes of that proof,
let me do some more correcting of Zuhair's Cantor explanation, and
some correcting of Wikipedia's Cantor diagonal and some correcting of
Cantor's argument.

Zuhair does not yet realize, although I alluded to it in the last post
by saying if the Cantor method were true, then it will prove that the
Naturals are a uncountable infinity.

So let me just tease out that flaw some more.

Notice that Zuhair and Cantor use binary base representation and
restrict their alleged proof to the interval 0 to 1. And they also
delete the decimal point which is no cause for alarm since it is
generalizable.

But what Cantor never noticed, nor Zuhair ever notice, or the authors
of the Wikipedia entry on the Cantor diagonal, is that every
conclusion drawn for the Reals between 0 and 1, is just as apt of a
conclusion for all the Natural Numbers. In other words, Cantor and
Zuhair and Wikipedia believed they had proven that the list of all
Reals in interval 0 to 1 cannot be listed and that those Reals must
then be Uncountable Infinite. What they never realized was that the
very same conclusion is drawn for the Natural Numbers, in that the
Cantor diagonal proves the list of all Natural Numbers is an
Uncountable Infinite list because the mechanism (the trickery and
deceit) of the diagonal says the very same thing for the Natural
Numbers as it says for the Reals in interval 0 to 1. We just do not
include the decimal point, and we look at the Naturals in binary base
representation.

So when Cantor or Zuhair or Wikipedia listed the Reals in interval 0
to 1 and did the diagonal mechanism and then concluded that Reals are
an Uncountable Infinity, they also concluded, although they never saw
it coming, that the Natural Numbers are an Uncountable Infinity.

Their mistake, their gross colossal mistake is two mistakes. They
never stated a borderline crossing between what is a finite number and
what is a infinite number. And secondly, they never dawned in their
minds that the Reals are All Possible Digit Arrangements.

Now we can also define the Naturals as All Possible Digit
Arrangements.

And let me make a first stab at this definition:

Consider all the Naturals from 0 to 100 as the two place value
Naturals. So, we have this list:

00
01
02
03
.
.
.
98
99
100

Now that list contains 101 Natural Numbers including 0.

The point I am making is that given any two place value Natural, for
example 34 or 87 or 16. That the list contains all the Naturals of two
place value.

What Zuhair or Cantor or Wikipedia tried to do is run a Cantor
diagonal and concluded they found a Natural Number of YX which is a
Natural Number and belongs somewhere between 0 and 100 but is not
listed. You see, Zuhair, Cantor, Wikipedia never understood that the
concept of All Possible Digit Arrangements preempts the foolishness of
a Cantor diagonal. And the other flaw, which is a flaw even larger
than comprehending All Possible Digit Arrangements is the flaw that
you can have finite and infinity without ever crossing a borderline
between them. If France and Germany had no borderline between them,
then one can say that Dresden is in France and that Paris is in
Germany. Likewise, Zuhair or Cantor can say that 10^899 is finite
whereas I would say it is infinite since the borderline is 10^603.

When mathematics has a borderline between finite and infinity, then
all of the Cantor double-speak disappears and evaporates into the fake
dust that it is. But until then, people like Zuhair or Wikipedia need
our help to rescue him/them out of their fakery abyss.

[MoeBlee]

On Oct 30, 6:09 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:

> my
> proof that the Reals are countable, since the Reals are All Possible
> Digit Arrangements and that infinity comes in only one type-- a
> countable infinity.

In other words, the set of reals is countable because all sets are
countable.

Brilliant.

MoeBlee

[Archimedes Plutonium]

Alright I am in a sort of snag with my physics work of m_s, quantum
spin, but I invite this interruption with mathematics, because that
snag is brewing. You see, in a snag, it is best to take a break from
it and during the break, something may give. But of course I will
return to it. In the meantime I am having fun exploiting the Cantor
diagonal fakery.

Now I am waiting and hopeful that Zuhair is going to critique whether
my proof is correct or wrong for him. If he does not, then he admits I
am correct. If he does apply what he believes is my mistakes, well,
then I must improve my proof.

Here is my proof that I asked Zuhair to correct, since he does not buy
it.

Why Cantor is wrong:

Proof: List all the Reals of interval 0 to 1. Define Reals as All
Possible Digit Arrangements. Do any type or sort of Cantor Diagonal,

any and all such as square list with 45 degree diagonal or any other
configuration. Because no Cantor Diagonal will manufacture a number

that 
is not already on the list. The reason being is that if you
understand 
the meaning of All Possible Digit Arrangements, that there
does not 
exist any Real absent from the list. And by making
alterations, can 
never create a Real which is absent.
QED

Now to give an example is all the numbers of 2 digits to 2 place-value
by the Fundamental Counting Principle is 2^2 = 4 total possibilities

And thus we have a total list of
00
01
10
11

Now Cantor would go in and remove or delete from that list two of
those entries because he cannot form a diagonal that covers all 4.
And Cantor would form

00
01

and yield 10 and claim it was not on the original list, but only
because he deleted or forgot to include it.

Or, Cantor could say, make a diagonal that is 90 degree angle and
include all 4 entries:

00
01
10
11

And now the downward diagonal encompasses 0011 forming 1100
and then Cantor would say that he manufactured a new number not on the
list of all the possible 2 digits to 2 place value. So did Cantor
cheat because 1100 is truly not on the list, nor is the number 0011,
which both are 2 digits to 4 place value.

Zuhair and the believers of Cantor would say that Cantor did not
cheat. But I would say he did cheat, and for a good reason.

In the Cantor alleged proof, there is a moment in the proof where he
has to state "this list is the list of All the Reals in interval
0 to 1" And if the list is a list of All Possible Digit Arrangements,
then there is no new number which is different from all the numbers
already present.

So the difference between believers of Cantor and myself, is that they
believe the Cantor method can override All Possible Digits. Whereas I
believe, if a list is All Possible Digits Arrangements, that the
Cantor method is preempted.

Now I showed in prior posts that if the Cantor method is valid, then
it also proves the Natural Numbers are a uncountable infinity just as
the Reals are a uncountable infinity, because when we use the Reals in
the interval 0 to 1, they are interchangeable with the Natural Numbers
to base 2 binary system. So whatever conclusion the Reals begot from
diagonal method is the same conclusion that the Natural Numbers would
be cloaked in. Now that is a hilarious outcome to think that there is
no 1-1 correspondence of the Counting Numbers to themselves, because
Cantor diagonal manufactures Counting Numbers that exist in the set of
Counting Numbers and that we never knew were there, sort of like a
imaginary whole number that resides between 2 and 3 and we never knew
it before we did the Cantor diagonal on the Natural Numbers.

And we can see the folly of the Cantor Diagonal on the decimal system
numbers by looking at all the counting numbers from 0 to 100

00
01
02

.
.
98
99
100

Now that is a All Possible Digit Arrangement of Natural Numbers for 2
place value.

If you give me the ten digits, 0,1,2,3,4,5,6,7,8,9 and required me to
arrange them in all possible arrangements to 2 place value
then that list above is such a list. There is no other arrangement of
those 10 digits to 2 place value.

Now in the Cantor diagonal of Reals in interval 0 to 1 we drop the
decimal point and we list them as such:

0
00
001
1
01
02
.
.
.

Now what I am doing is playing around, for what I am striving to show
next is that there is a 1-1 correspondence of the Reals in interval 0
to 1 with the Natural Numbers.

Now maybe I should avoid decimal base and cling to binary base which
is easier.

So the binary base is

00
01
10
11
100
101
110
111
1000
etc etc


Now I can include 0s to the leftward in binary base and not upset the
number it is, such as 3 being 11 and is 011. Now that helps to
correspond to the Real of .011 in the interval 0 to 1.

So what I am doing is configuring a 1-1 correspondence of all the
Reals in interval 0 to 1 and all the Natural Numbers (in binary base).

So in other words, where Cantor tried to make lists with diagonals to
show that Reals were a different infinity than Naturals, what his
lists turn out doing is showing there is a 1-1 correspondence between
Reals and Naturals and all because of the concept of All Possible
Digit Arrangements.

Now I want to also add a remark about Probability theory, for the
concept of All Possible Digit Arrangements comes from Probability
theory such as Fundamental Counting Principle. Also Probability theory
has its basis on the Reals in the interval 0 to 1 where 0 is never
happening and where 1 is probability certainty, and all other
probabilities in between 0 and 1. The comment is that if Cantor is
correct, it makes a upheaval of Probability theory, because there are
probabilities of what we would have to call-- transcendental-
probability, for there would be numbers that a sequence or counting
numbers would not be able to reach. So if Cantor were correct, then
Probability theory takes a wound or hit.

So let us see if Zuhair can correct the proof above. And let me see if
I can rig the binary Naturals to have a 1-1 correspondence with the
Reals in interval 0 to 1. I already showed a 1-1 correspondence but I
had to invoke the borderline of infinity at 10^603 to apply the 1-1
correspondence. Here I maybe able to fetch the 1-1 correspondence
without invoking the infinity borderline.

[Archimedes Plutonium]

On Oct 31, 12:24 am, Archimedes Plutonium

Alright, I think this is going to work, but I am going to sleep on it
before declaring victory for there maybe a iron in the fire that
prevents it from working.

The idea is to use the Cantor diagonal listing of Reals in the
interval 0 to 1. Where the Reals are listed in binary base and decimal
point removed.

Now the Naturals are this in binary base

0= 0
1= 1
2= 10
3= 11
4= 100
5= 101
6= 110
7= 111
etc etc

Now the Reals in binary in interval 0 to 1 can be made as such:

0
1
01
11
001
101
011
111
etc etc

Where the Real corresponds to the reverse image of the Natural.

Now I think I captured all the Reals in that interval with a
corresponding Natural. And will sleep on it to see if true by morning.

Now if true it is a remarkable construction because it does not invoke
infinity border of 10^603 which then forces the smallest postive
nonzero Real to be 10^-603 and where all the Reals from 0
to 10^603 are incremental adding of 1*10^-603. So in that invocation
of infinity all the Reals between 0 and 1 are in a 1-1 correspondence
with all the Integers from 0 to 10^603.

What is special about the above if true, is there is no invoking of
infinity number and the 1-1 correspondence grows out of the fact of
the concept of All Possible Digit Arrangements for Reals and All
Possible Digit Arrangements for the Natural Numbers.

I hope it is true. For it would shut down much of the noise of the
Cantor fake diagonal method.

[Archimedes Plutonium]

On Oct 31, 1:11 am, Archimedes Plutonium

Now I have not gone to bed yet and slept on it. But in these waning
hours of the night here, I am feeling confident that I found a 1-1
Correspondence of the Naturals to the Reals. It involves the above
scheme where the correspondence is a reverse image:

1 <--> .1
2 <--> .01
3 <--> .11
4 <--> .001
etc etc

Of course I converted the binary Natural to decimal base. I did not
convert the binary Real in interval 0 to 1.

I may have a problem of what Natural corresponds to the integer 1, or
I could just leave it out as the open set (0,1) which is no blemish on
the argument.

The remarkable aspect of the above is that it does not invoke the
borderline of infinity to make the correspondence. When the borderline
10^603 is invoked the correspondence is a breeze to make. But the
above is free of invoking infinity and for more than a century, people
have looked for a 1-1 Correspondence of Reals with Naturals without
invoking infinity and now one is found.

Question immediately comes to fore. What allows the correspondence to
be made? The answer is actually quite simple. We focus on the All
Possible Digit Arrangements of Reals and the the All Possible Digit
Arrangements of Naturals and that concept is so powerful in organizing
that the 1-1 Correspondence is there for the picking. With All
Possible Digit Arrangements, there is no breakage in the pattern of
forming one Natural and then the next succeeding Natural. It is more
difficult to impose a succession on the Reals, but the All Possible
Digit Arrangement imposes a succession on the Reals by reversal of the
Natural

> Now I think I captured all the Reals in that interval with a
> corresponding Natural. And will sleep on it to see if true by morning.
>
> Now if true it is a remarkable construction because it does not invoke
> infinity border of 10^603 which then forces the smallest postive
> nonzero Real to be 10^-603 and where all the Reals from 0
> to 10^603 are incremental adding of 1*10^-603. So in that invocation
> of infinity all the Reals between 0 and 1 are in a 1-1 correspondence
> with all the Integers from 0 to 10^603.
>
> What is special about the above if true, is there is no invoking of
> infinity number and the 1-1 correspondence grows out of the fact of
> the concept of All Possible Digit Arrangements for Reals and All
> Possible Digit Arrangements for the Natural Numbers.
>
> I hope it is true. For it would shut down much of the noise of the
> Cantor fake diagonal method.
>

I am still going to sleep on it, for maybe there is something I
missed.

[Archimedes Plutonium]

Alright, I pretty much destroyed Cantor's diagonal method and his
uncountable infinity. But let me give some respect to Cantor for a
change. He did do a true proof of a 1-1 Correspondence with Naturals
to Rationals and if you remember his method he listed the Rationals in
a large rectangle or square array and starting in the upper left
corner, kept snaking back and forth in a diagonal pattern to link a
Natural to a Rational.

So having shown a 1-1 Correspondence tonight of the Naturals with
Reals in interval 0 to 1, I need to generalize to showing a 1-1
Correspondence of the Naturals to all the Reals.

So I look for a means similar to the correspondence set up with a
snaking back and forth diagonal. Now we know the Reals in interval 0
to 1 is the same cardinal number of Reals in the interval 1 to 2 or 2
to 3 etc etc.

But tonight I want to sleep and think about whether there is a cleaner
way of a 1-1 Correspondence of all the Reals with the Naturals. If
not, then the best is the way of doing the Rationals linked with
Naturals.

And I am happy to give credit to Cantor for something that he did
which is true in mathematics, and not leave on a negative note or
tone.

[Zuhair]

On Oct 31, 8:24 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:

>
> Why Cantor is wrong:
>
> Proof: List all the Reals of interval 0 to 1. Define Reals as All
> Possible Digit Arrangements. Do any type or sort of Cantor Diagonal,
> any and all such as square list with 45 degree diagonal or any other
> configuration. Because no Cantor Diagonal will manufacture a number
> that 
is not already on the list. The reason being is that if you
> understand 
the meaning of All Possible Digit Arrangements, that there
> does not 
exist any Real absent from the list. And by making
> alterations, can 
never create a Real which is absent.
> QED
>

The question is whether the list of all possible digit arrangements is
countable or not? that's the question. And that's what you are not
addressing.

Of course when you have a complete list of all Reals (or equivalently
of all possible digit arrangements) then you cannot ever produce (by
any method) a Real (or a digit arrangement) that do not belong to it,
that's clear. That's not the issue, the issue is Can we have a
bijection from the set of all naturals to the set of all Reals (or of
all digit possible arrangements) ? What Cantor showed is that ***IF***
you say that there is such a bijection then by using the diagonal
method a real (or a digit arrangement) will be constructed that is not
in the range of that bijection BUT the range of that bijection is the
set of ALL reals (or possible digit arrangements) by definition of
that bijection, A flagrant contradiction!!!!! Thus the conclusion is
that: There cannot be a bijective function from the set of all
naturals to the set of ALL reals (or of all possible digit
arrangements).

Neither Cantor nor I did say that we can apply the diagonal method to
the true Set of ALL reals (or of all possible digit arrangements) and
get a new real (or a digit arrangement) that is not in that set, it is
clear that this is contradictory, it is crystal clear that such a
thing CANNOT be done. Cantor and I applied the diagonal method to a
set that is ASSUMED to be of ALL reals and assumed to be of the same
size as that of the naturals. Cantor applied the diagonal argument on
the range of the bijection from the set of all naturals to that of all
reals. And it is clear that ***IF*** such a bijection could ever
exist, then we CAN apply the diagonal method on its RANGE to derive a
new real that is not in that Range, it is clear that such a thing can
be done, but as I said by then the range of such a bijection would
always be INCOMPLETE and thus it would never represent the set of ALL
reals (or of all possible digit arrangements), so this prove that our
assumption that there can be such a bijection is a FALSE assumption.

And also to address your question the diagonal method of Cantor do not
prove that the naturals are more than the naturals as you said.
However it does prove that there are more sequences of naturals than
naturals, and so you cannot have a bijective function from the set of
all naturals to the set of all sequences of naturals, this is
something that the diagonal of Cantor's prove, of course.

Zuhair

[Frederick Williams]

Archimedes Plutonium wrote:

>
> And I am happy to give credit to Cantor for something that he did
> which is true in mathematics, and not leave on a negative note or
> tone.

Archy, you're a true gentleman. I don't usually read your posts because
I don't have your intellect, but it's heart-warming to see a person of
your standing giving credit to others.

--
I have seen elephants paint very competently... but not in Cumbria.

[MoeBlee]

On Oct 31, 5:41 am, Zuhair <zaljo...@gmail.com> wrote:

> Cantor and I applied the diagonal method to a
> set that is ASSUMED to be of ALL reals and assumed to be of the same
> size as that of the naturals. Cantor applied the diagonal argument on
> the range of the bijection from the set of all naturals to that of all
> reals.

In what paper did Cantor write that?

You can yourself prove the theorem that way, but I don't know where
Cantor himself did it that way.

Rather, as far as I know, Cantor did it basically the way I suggested:

Let f be an enumeration of any set of denumerable binary sequences.
Let g be the denumerable binary sequence defined: g(n)=0 iff f(n)
(n)=1.
So g is not in the range of f.
So f is not an enumeration of all the denumerable binary sequence.

There's no mention of an assumption that f is an enumeration of ALL
the denumerable binary sequences; and no assumption of a bijection.

MoeBlee

[Zuhair]

I mean Cantor's argument can be equivalently interpreted as such and
such.....

Zuhair

[Zuhair]

On Oct 31, 1:41 pm, Zuhair <zaljo...@gmail.com> wrote:
> On Oct 31, 8:24 am, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
>

Let me put matters in a more clear context for you using matters that
you like.

Lets take some set S of possible digit arrangements, let's agree that
this is a set of possible Aleph_0 sized sequences of 0 , 1. Where of
course Aleph_0 is the cardinality of the set N of ALL natural numbers
defined in the standard manner. I need you to concentrate here S is
just a non specific set in which ALL of its elements are Aleph_0 sized
sequences of 0,1, IT IS NOT necessarily the set of ALL possible digit
arrangements. So the only thing we know about S is that it is non
empty and that it can only contain elements that are countably
infinite sized binary digit sequences, so S might be the set of ALL
possible digit arrangements or it might be some set of some but not
all possible digit arrangements.

Now the question is: For any set S so defined, Suppose we have a
bijection from N to S, then could S be the set of ALL possible digit
arrangements?

Notice that we don't know whether S could be the set of all possible
digit arrangements or not, all what we know is that S is non empty and
that any element of S is countably infinite binary digit sequence, OK.

Now it can be proved that any such S that have a bijection to N would
be missing some possible digit arrangement, the proof is established
by Cantor's diagonal argument, as you know the diagonal so constructed
after Cantor's will provably be countably infinite sized binary digit
sequence that is not in the Range of the bijection from N to S. From
this we conclude that any such S set would be incomplete. And thus we
conclude that if S is ANY set of possible digit arrangements, and if S
is bijective to N, THEN S is incomplete, i.e. S do not contain ALL
possible digit arrangements. That's what Cantor's argument mounts to.

Zuhair

[MoeBlee]

On Oct 31, 12:25 pm, Zuhair <zaljo...@gmail.com> wrote:

> I mean Cantor's argument can be equivalently interpreted as such and
> such.....

Sure. Fair enough.

MoeBlee

[JRStern]

On Wed, 31 Oct 2012 03:41:15 -0700 (PDT), Zuhair <zalj...@gmail.com>
wrote:

>On Oct 31, 8:24 am, Archimedes Plutonium
><plutonium.archime...@gmail.com> wrote:
>
>>
>> Why Cantor is wrong:
>>
>> Proof: List all the Reals of interval 0 to 1. Define Reals as All
>> Possible Digit Arrangements. Do any type or sort of Cantor Diagonal,
>> any and all such as square list with 45 degree diagonal or any other
>> configuration. Because no Cantor Diagonal will manufacture a number

>> that ?is not already on the list. The reason being is that if you
>> understand ?the meaning of All Possible Digit Arrangements, that there
>> does not ?exist any Real absent from the list. And by making
>> alterations, can ?never create a Real which is absent.

>> QED
>>
>
>The question is whether the list of all possible digit arrangements is
>countable or not? that's the question. And that's what you are not
>addressing.

That certainly seems to be a question.

>Of course when you have a complete list of all Reals (or equivalently
>of all possible digit arrangements) then you cannot ever produce (by
>any method) a Real (or a digit arrangement) that do not belong to it,
>that's clear.

Is that clear? I'd sure like anything in this to be clear.

> That's not the issue,

Whoa there, maybe it *is* the issue, or again, at least *an* issue.

> the issue is Can we have a
>bijection from the set of all naturals to the set of all Reals (or of
>all digit possible arrangements) ? What Cantor showed is that ***IF***
>you say that there is such a bijection then by using the diagonal
>method a real (or a digit arrangement) will be constructed that is not
>in the range of that bijection BUT the range of that bijection is the
>set of ALL reals (or possible digit arrangements) by definition of
>that bijection, A flagrant contradiction!!!!! Thus the conclusion is
>that: There cannot be a bijective function from the set of all
>naturals to the set of ALL reals (or of all possible digit
>arrangements).

But we've already lost our clarity, what are we mapping to? If it's a
complete list, then I thought it was "clear" that the diagonal doesn't
work. If it's not a complete list because, um, just whyever you think
that might be, then we're not REALLY even testing the thesis, are we?

The point of contention then is whether this paper representation does
or does not represent an uncountable set. But hey, we write aleph
characters and claim they DO represent uncountable sets, but there
seems to be something wrong, with this matrix? Just how is that
determined?

J.

[Virgil]

In article <jfs298pmsdt57p85k...@4ax.com>,

JRStern <JRS...@foobar.invalid> wrote:

> >The question is whether the list of all possible digit arrangements is
> >countable or not? that's the question. And that's what you are not
> >addressing.
>
> That certainly seems to be a question.

A more appropriate question is whether any such thing as

"the list of all possible digit arrangements"

even exists.

Particularly since such things have been proven not to exist
--

[JRStern]

Well again, I'm confused by such a claim.

Is it because any possible "arrangement" is by definition enumerable,
and we are claiming to "enumerate" a nondenumerable set?

It doesn't seem the kind of thing amenable to proof, though it might
be to definition.

J.

[Archimedes Plutonium]

On Oct 31, 5:41 am, Zuhair <zaljo...@gmail.com> wrote:

Thanks, you did address an error on my part and will have to fix it.

I have to answer Patricia's concern, which in fact will amend Zuhair's
concern also.

However, Zuhair is still wrong on the question of the overall Cantor
method, because, the Naturals are an uncountable infinity, since the
method is still useable on just the Naturals alone and where a
"missing Natural is obtained".

But let me answer Patricia's question for that will solve Zuhair's.

Archimedes Plutonium

http://www.iw.net/~a_plutonium/

[MoeBlee]

On Oct 31, 1:54 pm, JRStern <JRSt...@foobar.invalid> wrote:
> If it's a
> complete list, then I thought it was "clear" that the diagonal doesn't
> work.

Assume there exists a list of all the real numbers. Take the anti-
diagonal of the list to a get real number that is not on the list.
Then we have a contradiction with the assumption that all the real
numbers are on the list. Therefore, the intital assumption (that there
exists a list of all the real numbers) has entailed a contradiction.
Therefore we infer the negation of the assumption. I.e. we infer that
there does not exist a list of all the real numbers.

MoeBlee

[Zuhair]

On Oct 31, 9:54 pm, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 03:41:15 -0700 (PDT), Zuhair <zaljo...@gmail.com>

> J.

I understand your concern, sometimes the wording is indeed confusing.
I already rephrased this wording in my latest response to AP.

Lets be a little bit more clear.

Lets identify a real with a countably infinite binary digit sequence.

Now we define the predicate "is a set of reals" in the following
manner:

S is a set of reals <-> (for all y. y in S -> y is a real).

So S may be empty, or S might contain some but not all reals, or S
might contain all reals in which case it will be called the set of all
reals.

Now our question can be modified to the following:

Suppose that S is a set of reals and Suppose that there exist a
bijection from N to S, then can S be the set of ALL reals?

of course N above refers to the set of all naturals.

Cantor's diagonal argument answer this question to the negative!

The proof is that for any S that is a set of reals and that has a
bijection from N to it, then we can define a diagonal on the range of
that bijection (which is S of course) and then perform the needed
alternations on that diagonal to derive a countably infinite binary
digit sequence (i.e. a real) that is provably missing from the range
of the bijective function from N to S (which is S) And this proof
generalizes to every such bijective function from N to S.

From this we conclude that any set S of reals where there exist a
bijection from N to S, then there must exist some real that is not an
element of S, so S is always an incomplete set of reals. So the answer
is any such set S cannot ever be the set of all reals.

Zuhair

[Archimedes Plutonium]

On Oct 31, 10:25 am, Patricia Shanahan <p...@acm.org> wrote:
> On 10/31/2012 12:52 AM, Archimedes Plutonium wrote:
> ...

>
> > 1 <--> .1
> > 2 <--> .01
> > 3 <--> .11
> > 4 <--> .001
> > etc etc
>

> ...
>
> Which natural, in this scheme, corresponds to 1/3? Which one corresponds
> to pi?
>
> Patricia

Alright, I slept on this last night and this morning I still feel
triumphant over it. The question Patricia asks is important, not only
in this context but in the context of Cantor's diagonal method
overall.

I said it was a remarkable proof above, because I did not need the
borderline of infinity to enter the picture. But to answer Zuhair, I
need to answer Patricia's with that infinity borderline.

Now a few posts back I used the word "viniculum" (spelling) for those
three dots that indicate infinity, but the correct word to use is
"ellipsis" for those three dots that indicate infinity.

1 <--> .1
10 <--> .01
11 <--> .11
100 <--> .001

Now I converted the above back to its original where the Naturals are
binary base and the Reals in interval 0 to 1 are in binary base. And
the claim I am making is that this is a 1-1 Correspondence of all the
Reals in the open interval (0,1) with the Natural Numbers.

Now if I invoke the borderline of infinity to be 10^603, then
mathematics as a science of precision stops at 10^603 where it starts
to fall apart as you go beyond 10^603, because our definitions of
being prime, being irrational, being transcendental no longer hold up
once you trespass further beyond 10^603. Physics in fact has no need
or use or "has anything physical" beyond 10^603. It is only human
imagination that thinks mathematics is still mathematics beyond 10^603
and of course humans also believe in witches, ghosts and fire
breathing dragons.

Patricia, when I invoke the infinity borderline 10^603, then I can
drop all ellipsis in numbers and also, the smallest nonzero Real is
10^-603. Since mathematics falls apart beyond 10^603, it falls apart
between 0 to 10^-603. So the numbers of mathematics have tiny holes
and gaps all the along a line stretching from 0 to 10^603.

So that we can list all the Reals from 0 to 10^603 as a successor
function of adding 10^-603. So all the Reals are:

0
1*10^-603
2*10^-603
3*10^-603
.
.
10^603 subtract 10^-603
10^603

So, Patricia, you can see that the above set of All the Reals is
countable by Naturals of 10^1206 (Naturals are not broken apart yet).
And the Reals between 0 and 1 are countable by the Natural of 10^603
for there are exactly 10^603 Reals in that interval.

Now I could do the same illustration for Patricia and for Zuhair if we
say that infinity borderline is the number 10 instead of 10^603.
In that restricted Universe of mathematics where math starts to fall
apart after 10 and the only precision math we can use are the Naturals
0 to 10, then the smallest nonzero Real is 1/10 or 0.1. There are no
Reals between 0 and 0.1 and all the Reals in the interval 0 to 1 are
the ten Reals .1, .2, .3, .4, .5, .6, .7, .8, .9.

They are countable and no Cantor Diagonal Argument is able to exist
when mathematics is precise about the border between finite and
infinity.

So now, let me answer Patricia's question when infinity borderline is
not invoked. When we ignore what mathematics is all about-- precision,
precision precision. When we ignore that our first duty was to specify
the border of what we have to call the end of finite and the start of
infinity.

When I ignore the borderline or when Patricia ignores the borderline
or when Cantor or Zuhair ignore the borderline we all use the
ellipsis. We are lazy and we are ignoring of precision so we use
ellipsis.

So in answer to Patricia, the Real of 0.3333... is a infinity of 3s
where we ignore the borderline of finite into infinity. The Natural
Number of 33333... when we ignore the borderline is a Natural Number
because we fail to specify the borderline and chose to ignore the
responsibility of stating the borderline. If we had specified 10 to be
infinity borderline then the Real 0.33333.... is the Real 0.3 and the
Natural Number that is the 1-1 correspondence is the number 3.

So, where Cantor, where Zuhair, where Patricia, where all
mathematicians who fail the first responsibility of mathematics in
precision of specifying the borderline between finite and infinity and
chose to use an ellipsis to cover up their failure of precision, then
the Natural Number that corresponds to the Real 0.33333..........

is the Natural Number

3333333.........

Archimedes Plutonium

http://www.iw.net/~a_plutonium/

[MoeBlee]

On Oct 31, 2:17 pm, JRStern <JRSt...@foobar.invalid> wrote:

> Is it because any possible "arrangement" is by definition enumerable,
> and we are claiming to "enumerate" a nondenumerable set?
>
> It doesn't seem the kind of thing amenable to proof, though it might
> be to definition.

A list is an enumeration. An enumeration is a function whose domain is
a natural number or is the set of natural numbers. An enumeration is
an enumeration of the range of the enumeration.

Now, take any infinite enumeration F of a set whose members are only
real numbers. Apply the diagonal method to get a real number that is
not in the range of F. So F is not an enumeration of the set of ALL
real numbers.

MoeBlee

[Zuhair]

On Oct 31, 10:30 pm, Archimedes Plutonium

No you cannot use the diagonal method to obtain a missing natural. You
CAN use the diagonal method to obtain a missing "sequence of naturals"
from a denumerable list of sequences of naturals.

In other words every denumerable list of sequences of naturals is
incomplete, i.e. there will always be a missing sequence from such
list! Cantor's diagonal proves that!

Zuhair

[MoeBlee]

On Oct 31, 2:30 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> the Naturals are an uncountable infinity,

That's nonsense.

The set of natural numbers is not uncountable.

Df: S is countable iff (there is a bijection between S and the set of
natural number or there is a bijection between S and some natural
number)

So the set of natural numbers is countable since there is a bijection
between the set of natural numbers and the set of natural numbers.

Df: S is uncountable iff S is not countable.

So, since S is countable, we have that S is not uncountable.

MoeBlee

[Archimedes Plutonium]

On Oct 31, 1:54 pm, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 03:41:15 -0700 (PDT), Zuhair <zaljo...@gmail.com>

Hello J Stern, I see you started a thread in sci.math on a question of
a peer reviewed article over the validity of Cantor's Diagonal Method.

I just want to drop an opinion to that thread and its inferences.

I want to say that the Mathematics Journals are the problem, a major
problem of mathematics and not a venue of a solution. It was the Math
Journal system of the past that created this mess of the Cantor
diagonal. Because once you publish a piece of math that is fakery into
the math journals, that it has a way of sticking around longer and
confusing mathematicians than if Cantor's method had never been
published.

So the question of publishing in math journals the invalidity of
Cantor's method has to be questioned because it was the Math Journals
that caused the harm in the first place.

So what I am saying in this post is that the correction of Cantor's
method was caused by the Usenet, sci.math and that the problem was the
Math Journals and their practice of dishing out fake math.

So what I am advocating is that we ditch the Math Journal System of
mathematics and let all of new math publication be handled by free
speech purviews such as Sci.Math or Sci.Logic.

If you look at Math Journals, their prime concern is not true
mathematics, but their prime concern is a career ladder for pay
promotion by colleges. Much of that which is published in math
journals is for the intent of promoting some math professor to more
money at his college.

So, J. Stern, I recommend that we ditch the entire math journal system
and we set up Sci.Math as the world venue of **new math**.

Do the same for all the sciences.

There is no more need of the old ways, the ways of censoring, of
delaying publication, the ways of filling science with fakery such as
Cantor's diagonal. Over in Physics, there is not a science journal
that is going to publish the fact that Saturn's Ring is solid body
rotation and that throws gravity and General Relativity out the window
into the trash pile.

So we make Sci. newsgroups the venue of now and the venue of the
future. Science reporting needs freedom of speech and the old journal
system were anything but freedom of speech.

[Frederick Williams]

Archimedes Plutonium wrote:

>
> Now a few posts back I used the word "viniculum" (spelling) for those

vinculum

> three dots that indicate infinity, but the correct word to use is
> "ellipsis" for those three dots that indicate infinity.

[Patricia Shanahan]

On 10/31/2012 1:05 PM, Archimedes Plutonium wrote:
...

> Now if I invoke the borderline of infinity to be 10^603, then
> mathematics as a science of precision stops at 10^603 where it starts
> to fall apart as you go beyond 10^603, because our definitions of
> being prime, being irrational, being transcendental no longer hold up
> once you trespass further beyond 10^603. Physics in fact has no need
> or use or "has anything physical" beyond 10^603. It is only human
> imagination that thinks mathematics is still mathematics beyond 10^603
> and of course humans also believe in witches, ghosts and fire
> breathing dragons.

...

If you are only proving things about rationals that have terminating
binary expansions, why do you claim to be able to say anything about the
reals?

Patricia

[Lord Androcles, Zeroth Earl of Medway]

"Patricia Shanahan" <pa...@acm.org> wrote in message news:9POdncuNsJHiCwzN...@earthlink.com...

============================================

In a word, autism.

Autism is a disorder of neural development characterized by impaired social interaction and communication, and by restricted and repetitive behavior. The diagnostic criteria require that symptoms become apparent before a child is three years old.Autism affects information processing in the brain by altering how nerve cells and their synapses connect and organize; how this occurs is not well understood.

 http://en.wikipedia.org/wiki/Autism

 

Never mind limits, if dy/dx = [f(x+h) – f(x)] / h then by definition h is the smallest

indivisible value that is greater than zero, for if j = h/2 then dy/dx = [f(x+j)-f(x)]/j

is a better approximation to dy/dx than [f(x+h) – f(x)] / h.

Plutonium is debating the value of h which he says is 1/10^603 because he cannot

multiply it by 2 or merely add a least significant digit, and along with his witches,

dragons and ghosts, demonstrates his impaired social interaction and communication.

It is other people that believe in witches, ghosts and fire-breathing dragons, Plutonium

believes h has a numerical value rather than a concept.

 

-- This message is brought to you from the keyboard of
Lord Androcles, Zeroth Earl of Medway

 

 

 

[Archimedes Plutonium]

Alright, let us go back a bit before that question of yours.

You pride yourself in being a mathematician (I am guessing).
And a mathematician goes around with a tool box full of math tools.
The box carries axioms, theorems, definitions, methods, etc.

One thing that should not be in the tool box is a theorem that says
1 = 2, or a theorem that says Euclidean triangles sum to greater than
180 degrees because these are contradictory-tools.

Now some may carry these contradictory tools yet never realize it and
they pull them out of their toolbox on occasion and use them, not
knowing or realizing that they have ceased doing mathematics because
they are using a tool which is not mathematics.

Now a tool in every mathematician's tool box should have a definition
of finite number and infinite number. So that when ever Patricia uses
the word "infinity or infinite or finite" that Patricia has a
definition tool of finite compared to infinite in her toolbox.

So that if Patricia meets another mathematician who asks Patricia
whether the number 10^700 is finite or infinite number, Patricia can
sneek into her toolbox and answer the question.

Now Patricia would instinctively know that finite moves or transitions
into infinity. Something that Zuhair denies, because anyone who is not
looking for a borderline has to deny that finite transitions into
infinity. But I think Patricia is smart enough to not deny that finite
moves into infinite. It is just that she never bothered to think that
such implies a borderline and one must find that borderline in order
to be a "mathematician".

And then, in that toolbox, if you realize and understand that finite
moves into infinite, that there must be a borderline between them and
it has to be specified whenever one is doing mathematics where finite
and infinite is involved.

So, apparently Patricia has never looked hard enough inside her
toolbox of mathematics to answer where is the borderline, so that she
can give a mathematical-answer as to whether 10^700 is a finite number
or an answer just from her opinion box, that it looks finite because
she never made a real effort to see where the borderline rests.

The only people who can ask a question about finite and infinite are
those that pinpoint the borderline between finite and infinite and
then they can make those silly questions go away.

The pseudosphere area equals the respective sphere when infinity
border is 10^603 and is precisely equal and precisely 1/2 the volume
when it reaches 10^603.

The Riemann Hypothesis is false because it is false at 10^603.

The Euler formulas on Regular Polyhedra are true in the first instance
when pi reaches 603 digits rightward of the decimal point.
Before it reaches 10^603, the Euler formulas are false.

So, in answer to Patricia. Please do not ask me silly non math
questions when you yourself never takes the time to pinpoint a
borderline between finite and infinity. Your mathematics is nonmath
until you specify the borderline.

Archimedes Plutonium
http://www.iw.net/~a_plutonium

[JRStern]

On Wed, 31 Oct 2012 13:02:47 -0700 (PDT), Zuhair <zalj...@gmail.com>

OK.

>Now we define the predicate "is a set of reals" in the following
>manner:
>
>S is a set of reals <-> (for all y. y in S -> y is a real).

OK.

>So S may be empty, or S might contain some but not all reals, or S
>might contain all reals in which case it will be called the set of all
>reals.

OK.

>Now our question can be modified to the following:
>
>Suppose that S is a set of reals and Suppose that there exist a
>bijection from N to S, then can S be the set of ALL reals?

Let's move slowly here for my benefit.

When we make the diagonal argument, is that a claim that there is a
bijection from N to S?

>of course N above refers to the set of all naturals.
>
>Cantor's diagonal argument answer this question to the negative!
>
>The proof is that for any S that is a set of reals and that has a
>bijection from N to it, then we can define a diagonal on the range of
>that bijection (which is S of course) and then perform the needed
>alternations on that diagonal to derive a countably infinite binary
>digit sequence (i.e. a real) that is provably missing from the range
>of the bijective function from N to S (which is S) And this proof
>generalizes to every such bijective function from N to S.

OK, again, slowly.

So the claim is that for some set S that may not be a complete set,
the diagonal argument may work. And then by induction we find it true
of all sets and the complete set.

But I yet remain unconvinced, along the same lines as the local
"cranks". What exactly is this set S for which it is claimed that the
diagonal argument works? What is the set, and what is the
representation, and which one, set or representation, exactly are we
talking about at any moment?

Typically we use some small but still unenumerable subset of S, like
the segment from 0 to 1. For that matter, it can be any real segment
of the tiniest dimension, right, like from 1.0^E-23 to what, 1.0^E-23
+ C, where C = 1.0E-23456789. It's just inconvenient to draw that on
paper. For that matter it doesn't even have to be a continuous
segment, as the argument goes, but this is a bit difficult, as we're
going to assume that the noncontinuous set still has the same
cardinality as if it were a segment. We're trying to avoid a certain
difficulty, which I can't see really succeeds. We're trying to define
just what *is* the difference between one line and another in our
*representation*. If we give ourselves the freedom of a noncontiguous
set, we can try to whistle past this. Otherwise, we're basically
introducing an infinitesimal argument, which I guess is not generally
considered respectable. There is no successor function in the reals,
right? So I can't see what the argument is, that the diagonal is
supposed to be drawn on. As soon as you draw it, you appear to accept
the enumerability of the set yet then you don't, for if this were
merely indicative of the *actual* segment then it would be "clear"
that the diagonal argument would fail. So no paper and ink
representation is exaustive, but then we all KNEW that. So what is
really being argued?

So the entire argument comes down to definitional, and the problem
isn't so much that the real are unenumerable, because you can't really
(fully) enumerate even a single real to infinite countable digits, nor
even an integer of infinite size.* It only proves the convention
about representation, but the convention was there from the start.
You've proven your convention. It may be a convention of an actual
mathematical truth, but it's a circular argument at best.**

--

So when the cranks, plus or minus me, want to argue that it is
incomprehensible how any real *could* be left out of a full
enumeration, even that is a misleading claim because the real (sic)
"problem" is that we can't even represent even a single rational to
infinite degree except by a set of conventions - which we happily pull
out from under your feet, just when you thought you had something
going on it.


Josh


*Nor even a large integer, except by our base 10 conventions rather
than a countable number of strokes! For that matter, not even a
*small* integer, except by our convention of base 10 numerals, etc.

**But isn't that all you expect to see in a mathematical proof,
especially an argument by contradiction, you ask. Well, yes and no.
My discomfort here is in swapping the interpretation of the
enumeration from line to line. When the argument is rephrased in pure
mapping terms, it is at least harder to pin down, at least if you want
to take issue with it then, you are probably forced to constructivist
terminology. I'm trying to do much the same thing here not with
constructivist arguments but with some kind of symbolic/formalist
argument, which may be intertranslatable to constructivist or not, I'm
not entirely certain. It's kind of to figure that out, that I
wandered in here.

Again, my argument is with some pretty specific details of the
diagonal argument, and not with whether reals are countable or not. I
accept that they are not. If anything, I just want to be that much
clearer on when and how they are not, as I want to make certain
arguments within finite (but potentially large) domains, where I can
be a little more certain that my symbols aren't running away with
things!

[MoeBlee]

On Oct 31, 5:24 pm, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 13:02:47 -0700 (PDT), Zuhair <zaljo...@gmail.com>

No need to bring in 'bijection' here.

>
> Let's move slowly here for my benefit.
>
> When we make the diagonal argument, is that a claim that there is a
> bijection from N to S?

No need to bring 'bijection' in here.

> >of course N above refers to the set of all naturals.
>
> >Cantor's diagonal argument answer this question to the negative!
>
> >The proof is that for any S that is a set of reals and that has a
> >bijection from N to it, then we can define a diagonal on the range of
> >that bijection (which is S of course) and then perform the needed
> >alternations on that diagonal to derive a countably infinite binary
> >digit sequence (i.e. a real) that is provably missing from the range
> >of the bijective function from N to S (which is S) And this proof
> >generalizes to every such bijective  function from N to S.
>
> OK, again, slowly.
>
> So the claim is that for some set S that may not be a complete set,
> the diagonal argument may work.

No, no, no.

The proof is that for ANY enumeration of ANY set or reals, there is a
real not in the range of the enumeration.

> And then by induction we find it true
> of all sets and the complete set.

There's no induction involved.

/

Please throw out the extra confusions you've having. Just take it one
step at a time:

(I'm going to say 'real number' here where, actually it should be
'denumerable binary sequence', so allow for that bit of informality
here.)

(1) Let N be the set of all and only the naturual numbers. Let R be
the set of all and only the real numbers.

(2) We claim that there is no function from N onto R. And we prove
that as follows:

(3) Let f be any function from N onto any set S whose members are only
real numbers.

(4) The antidiagonal of f is a real number not in S.

(5) Therefore, since the the antidiagonal is not in the range of f, we
have that f is not onto R.

(6) Since f and S are ARBITRARY, there is no surjection from N onto R.

Now, if there is a step there that you don't see as inescapably
logical, then please say what step it is and what you think is not
inescabably logical about it.

MoeBlee

[JRStern]

On Wed, 31 Oct 2012 15:39:36 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>>> >Of course when you have a complete list of all Reals (or equivalently
>>> >of all possible digit arrangements) then you cannot ever produce (by
>>> >any method) a Real (or a digit arrangement) that do not belong to it,
>>> >that's clear.
>>>
>>> Is that clear?  I'd sure like anything in this to be clear.
>>>
>>> > That's not the issue,
>>>
>>> Whoa there, maybe it *is* the issue, or again, at least *an* issue.

...

>> >Lets be a little bit more clear.
>>
>> >Lets identify a real with a countably infinite binary digit sequence.
>>
>> OK.
>>
>> >Now we define the predicate "is a set of reals" in the following
>> >manner:
>>
>> >S is a set of reals <-> (for all y. y in S -> y is a real).
>>
>> OK.
>>
>> >So S may be empty, or S might contain some but not all reals, or S
>> >might contain all reals in which case it will be called the set of all
>> >reals.
>>
>> OK.
>>
>> >Now our question can be modified to the following:
>>
>> >Suppose that S is a set of reals and Suppose that there exist a
>> >bijection from N to S, then can S be the set of ALL reals?
>
>No need to bring in 'bijection' here.

OK. Surjection.

>> Let's move slowly here for my benefit.
>>
>> When we make the diagonal argument, is that a claim that there is a
>> bijection from N to S?
>
>No need to bring 'bijection' in here.

OK.

>> >of course N above refers to the set of all naturals.
>>
>> >Cantor's diagonal argument answer this question to the negative!
>>
>> >The proof is that for any S that is a set of reals and that has a
>> >bijection from N to it, then we can define a diagonal on the range of
>> >that bijection (which is S of course) and then perform the needed
>> >alternations on that diagonal to derive a countably infinite binary
>> >digit sequence (i.e. a real) that is provably missing from the range
>> >of the bijective function from N to S (which is S) And this proof
>> >generalizes to every such bijective  function from N to S.
>>
>> OK, again, slowly.
>>
>> So the claim is that for some set S that may not be a complete set,
>> the diagonal argument may work.
>
>No, no, no.

So, if S *is* the complete set of reals, we can still find a number,
by this process or any other, that is not in the complete set of
reals?

I thought we heard it was "clear" that if S were the complete set of
reals, this could not be the case.


J.

[MoeBlee]

On Oct 31, 6:47 pm, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 15:39:36 -0700 (PDT), MoeBlee <modem...@gmail.com>

If you want to put it that way.

If S is the set of all real numbers and we find another real number
not in S, then that is a contradiction, so we infer that S is not the
set of all real numbers.

Remember basic logic: Assume P, show a contradiction, so infer not-P.

And we could do it that way. Assume there is an enumeration of the set
of all real numbers. Derive a contradiction, Then infer that there is
not an enumeration of the set of all real numbers.

But that's not how I actually did it. Again, what I did:

Let S be any set of real numbers (whether the set of all real numbers
or a proper subset of the set of all real numbers, I don't care,
rather just let S be any set that has only real numbers as members).

Let f be a function from N onto S.

Then the antidiagonal is a real number not in the range of f.

So there is a real number not in S.

So there is no function from N onto the set of real numbers.

Done.

MoeBlee

[MoeBlee]

Again, I sugges taking a moment to clear your mind of the confusions.
For a moment, throw away what you've been thinking about this.

Just consider this and get back to me on it:

(I'm going to say 'real number' here where, actually it should be
'denumerable binary sequence', so allow for that bit of informality
here.)


(1) Let N be the set of all and only the naturual numbers. Let R be
the set of all and only the real numbers.


(2) We claim that there is no function from N onto R. And we prove
that as follows:


(3) Let f be any function from N onto any set S whose members are
only
real numbers.


(4) The antidiagonal of f is a real number not in S.


(5) Therefore, since the the antidiagonal is not in the range of f,
we
have that f is not onto R.

(6) Since f and S are ARBITRARY, there is no function from N onto R.

[Jesse F. Hughes]

I'm not sure whether you're being purposely difficult, but the claim is
simple enough:

No function f:N -> R is a surjection.

If you insist on complicating it as you have, by adding a pointless
subset S of R, then we can state it thus:

Let S c R. If there exists a surjection f:N -> S, then S != R.

Proof:

Suppose S c R and let f:N -> S be a surjection. Construct, in the
usual manner, an element r e R such that r is not in image(f).
Conclude r is not in S (since image(f) = S).

I've no idea why you think that this awkward restatement is more
interesting than the usual theorem, but there you have it.

--
Jesse F. Hughes
-- A lesson in meta-honesty --
Baba: Thanks for being honest.
Quincy (age 7): I won't be honest next time. And that's more honesty.

[MoeBlee]

On Oct 31, 7:08 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:

> the claim is
> simple enough:
>
>   No function f:N -> R is a surjection.
>
> If you insist on complicating it as you have, by adding a pointless
> subset S of R,

The point is to allow, just for the hell of it, comparison with
Cantor's own argument that does not take the form of assuming, toward
a contradicdion, that there is a function from N onto R.

Let f be a function from N onto S.

The antidiagonal is a real not in S.
Therefore, S is not R.
Since S and f are arbitrary, there is no function from N onto R.

As opposed to:

Suppose f is a function from N onto R.
The antidiagonal is a real not in the range of f.
Contradiction.
Since f is arbitrary, there is no function from N onto R.

Both are fine, except, if I'm not mistaken (maybe I am?), the first is
closer to Cantor's own argument. So, when, for whatever reason, we
might like to hew close to the historical argument, the first is
better.

Moreover, the poster was confused by the argument that assumes, toward
a contradiction, that there is a function from N onto R. So I
suggested that we use a different argument. Though, I would say that
the poster should later go back to get rid of his confusion about
arguing toward a contradiction.

MoeBlee

[JRStern]

On Wed, 31 Oct 2012 17:02:49 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>Again, I sugges taking a moment to clear your mind of the confusions.
>For a moment, throw away what you've been thinking about this.

I will table several issues for the moment.

>Just consider this and get back to me on it:
>
>(I'm going to say 'real number' here where, actually it should be
>'denumerable binary sequence', so allow for that bit of informality
>here.)
>
>
>(1) Let N be the set of all and only the naturual numbers. Let R be
>the set of all and only the real numbers.
>
>
>(2) We claim that there is no function from N onto R. And we prove
>that as follows:
>
>
>(3) Let f be any function from N onto any set S whose members are
>only real numbers.
>
>
>(4) The antidiagonal of f is a real number not in S.
>
>
>(5) Therefore, since the the antidiagonal is not in the range of f,
>we have that f is not onto R.

(I'm uncertain as to hopping between R and S here)

--


I'm (further) confused about exactly what the claim is here.

Is it:

A. That we REALLY HAVE found a number not in S?

or

B. That we have a CLAIM to have found a number not in S, but since S
is of the cardinality of R (?) this is a contradiction, so we have
proven that SOMETHING upstream is rotten.

J.

[Virgil]

In article <9vt298hdqjuj17imr...@4ax.com>,

JRStern <JRS...@foobar.invalid> wrote:

> On Wed, 31 Oct 2012 13:05:19 -0600, Virgil <vir...@ligriv.com> wrote:
>
> >In article <jfs298pmsdt57p85k...@4ax.com>,
> > JRStern <JRS...@foobar.invalid> wrote:
> >
> >> >The question is whether the list of all possible digit arrangements is
> >> >countable or not? that's the question. And that's what you are not
> >> >addressing.
> >>
> >> That certainly seems to be a question.
> >
> >A more appropriate question is whether any such thing as
> > "the list of all possible digit arrangements"
> >even exists.
> >
> >Particularly since such things have been proven not to exist
>
> Well again, I'm confused by such a claim.

My claim is that such things as "the list of 'all' possible digit
arrangements" have been proven not really to be ALL of them.

>
> Is it because any possible "arrangement" is by definition enumerable,
> and we are claiming to "enumerate" a nondenumerable set?

It is because any such a list of ALL arrangements can be shown not to
contain ALL of them.

>
> It doesn't seem the kind of thing amenable to proof, though it might
> be to definition.

It was proved by Cantor! That is what his diagonal proof was all about!
--

[Virgil]

In article <dvd398h2oiud3uubp...@4ax.com>,

The Cantor diagonal argument applies only to LISTS, which a "complete
set of reals" is not and cannot be.
--

[Virgil]

In article
<f468a561-63f9-4569...@r5g2000yqo.googlegroups.com>,

MoeBlee <mode...@gmail.com> wrote:

> On Oct 31, 7:08 pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
>
> > the claim is
> > simple enough:
> >
> >   No function f:N -> R is a surjection.
> >
> > If you insist on complicating it as you have, by adding a pointless
> > subset S of R,
>
> The point is to allow, just for the hell of it, comparison with
> Cantor's own argument that does not take the form of assuming, toward
> a contradicdion, that there is a function from N onto R.
>
> Let f be a function from N onto S.
> The antidiagonal is a real not in S.
> Therefore, S is not R.
> Since S and f are arbitrary, there is no function from N onto R.

Cantor's own proof does not make any such assumption, but only shows
that that is a legitimate conclusion. The only assumption the Cantor
argument requires is existence of an injection from N to 2^N

>
> As opposed to:
>
> Suppose f is a function from N onto R.
> The antidiagonal is a real not in the range of f.
> Contradiction.

The Cantor form of argument only assumes a function from N to 2^N which
he easily shows not to be surjective.

> Since f is arbitrary, there is no function from N onto R.
>
> Both are fine, except, if I'm not mistaken (maybe I am?), the first is
> closer to Cantor's own argument.

Cantor's own original argument was based on any function from N to the
set of all functions from N to {w,m}, i.e., sequences of letters from
{m,w}.

>
> Moreover, the poster was confused by the argument that assumes, toward
> a contradiction, that there is a function from N onto R. So I
> suggested that we use a different argument. Though, I would say that
> the poster should later go back to get rid of his confusion about
> arguing toward a contradiction.
>
> MoeBlee

--

[Archimedes Plutonium]

On Oct 31, 5:23 pm, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Oct 31, 4:22 pm, Patricia Shanahan <p...@acm.org> wrote:
>

(snipped)

>
> > If you are only proving things about rationals that have terminating
> > binary expansions, why do you claim to be able to say anything about the
> > reals?
>
> > Patricia

A more direct answer from me is that the Reals and Naturals have two
things in common. They both are All Possible Digit Arrangements.

Since they share that characteristic the Cantor Diagonal Method is
doomed to failure.

The way to see that is your previous question of how do I correspond
1/3 with a Natural in this 1-1 correspondence:

1 <--> .1
10 <--> .01
11 <--> .11
100 <--> .001

etc etc.

My reply was that

3 to .3
33 to .33
333 to .333
3333 to .3333

Your retort above says you do not like that and that my construction
is not good enough for you.

For which I had to interject with the other characteristic of never a
precision defining of finite with infinity by a borderline.

Actually, you know the ellipsis, which unfortunately in the history of
mathematics was given the symbol . . . to mean "to infinity" but a
wiser denotation to mean "to infinity" would have been a triple
question mark like this 1/3 = 0.333??? So that could have alerted the
smarter people in mathematics that they needed to apply more logic to
that part of mathematics. Because three dots lulls people into
thinking everything is fine, when everything is sunk under water.

So, Patricia, was not happy with my reply that

as the Naturals expand in 3s:

3
33
333
3333
33333

that they cover the expansion of 3s in the Real 1/3
.3
.33
.333
.3333
.33333

Somehow, Patricia was not happy with the idea that the expansion of
3s as a Counting Number seemed far less wholesome than the expansion
of 3s in 0.333333.... and that the ellipsis dots in 1/3 was so warmly
and pleasingly healthy, but that the expansion of 3s
in Natural Numbers was somehow a disease or cancer as the whole
numbers grew larger.

My reply to Patricia was that her mind was not conducive enough to see
that I had given her a proper answer.

That as she was expanding the 3s in
0.3333333333...

I was expanding the 3s equally in the Natural Numbers

33333333.....

as fast as Patricia was including another 3 in the Real I was
including another 3 in the Whole Number 33333....

But then all of the mathematics community had become a religious
mathematics where if one is told that all Natural Numbers are finite
then no-one is allowed to write 33333....

They are allowed to write 0.3333.... in religious-math because no math
scripture says all Reals are finite, where the Real 1 is finite but
the Real 1/3 is finite but the Real 0.33333.... is a hybrid between
finite and infinite since its 3s are never-ending.

So, I wanted to cut the chase, and tell Patricia the solution up
front.

The solution is that there are two issues involved.

(1) All Possible Digit Arrangements and that is why there is a 1-1
Correspondence between Naturals and Reals

(2) Finite transitions into Infinite and that means a borderline and
when that borderline is specified, that causes a 1-1 correspondence
between Naturals and Reals.

Patricia has always used the concept of finite and infinite and I
suspect deep down knows there must be a borderline between them but
has never given it any thought. If she said that the borderline is
10^603, which I have claimed, then she would admit that all the Reals
between 0 and 1 amount to 10^603 Reals and thus are a 1-1
Correspondence.

And where every integer of this form:
3
33
333
3333
33333

corresponds to every Real of this form
.3
.33
.333
.3333

And Patricia would not have asked that question, because my math is up
to par, but Patricia's math is flagging since she uses finite and
infinite concept regularly but has never constructed a definition of
either to tell her if a number is a finite number or a infinite
number. But she is not alone, for the entire rest of the mathematics
community is in the same boat.

[JRStern]

On Wed, 31 Oct 2012 22:55:06 -0600, Virgil <vir...@ligriv.com> wrote:

>> I thought we heard it was "clear" that if S were the complete set of
>> reals, this could not be the case.
>
>The Cantor diagonal argument applies only to LISTS, which a "complete
>set of reals" is not and cannot be.

Interesting.

I don't want to fragment this, can you work that out with MoeBlee and
his story?


J.

[Archimedes Plutonium]

I raised this issue some years back. I raised it because there are at
least three instances where this comes up. The most important case is
the finite versus infinite issue and that affects the entire whole of
mathematics. But there are two other cases involving borderlines that
have turned mathematicians into circus clowns. These two other cases
are the Kepler Packing Problem and the 4 Color Mapping Problem.

Now if it were a single case of the finite versus infinite, I would
not bring this subject up for it would be a isolated case history
study. But because it occurred in two other famous math problems there
is more than meets the eye as to why borderlines make mathematicians
into fools. In a post years ago I gave a speculation answer that in
the roots and soul of humanity we want freedom and wide open freedom
and that we are biased against the world having boundaries as if being
imprisoned and that subconsciously shows up in our mathematics that we
fight any imprisonment sort of thing.

So let me list those three cases:

(1) Borderline between finite versus infinite. We all know that finite
moves or transitions into infinite. That implies there must exist a
borderline so that if the borderline is 10^603 then all numbers larger
are infinite numbers and it implies that no nonzero numbers are
smaller than 10^-603.

(2) The second case is the Kepler Packing Problem and it was not a
problem hard to prove, but a problem that was snagged up in
interpretation of boundaries and borders. What is the densest packing
of uniform small spheres in space. The snag was that mathematicians
did not allow the proof to be a volume of a container that holds the
sphere. So they demanded that you remove any thought of container,
while you kept the concept of density. Well, let me tell you, the
concept of density cannot exist with the concept of container. That is
why no-one has ever proven the Kepler Packing Problem because it is
not even a logical-formulated problem. It is illogic, as asking to
prove that "Up is high".

(3) And worst of all, is the 4 Color Mapping. Here they do a "One Flew
Over the Cuckoo's Nest" of mathematics. Here, they ask you to take a
country and with a sharp razor or knife cut away the border of every
country so that you cannot color it and color only interiors. This, to
me, is like asking if you removed the sides of a triangle or rectangle
would you even have a figure? If you removed the skin of a human,
would the body disintegrate? But leave it to ivory towered
mathematicians to gum up a problem or conjecture. The mathematics
community currently think they solved the problem, but it is a
Computer modeling alleged proof and in my opinion is not solved. If
they kept the borderlines, then the proof of the 4 Color Mapping is
easily tackled in a half hour with the Jordan Curve Theorem.

So, what is this disease by mathematicians whenever borderlines are in
the problem? Is it due to some deep seated prejudice of limitations?
Does mathematics seek freedom and not truth?

It is a disease and perhaps some of the other sciences can put the
mathematics community under the psychoanalysis table.

[Archimedes Plutonium]

Through the months and years I kept using a analogy to Geography as to
the blindness of the mathematics community on the issue of a

borderline between finite and infinity.

I suspect not a single person in all of the mathematics community
would disagree that finite moves or transitions into infinite. That
both are on the same track or stream or river and that as we go from a
region of finite we finally enter a region of infinite. That the two
concepts are not disjoint and on different tracks. But that belief
then entails a crisp and sharp conclusion. It entails there must exist
a borderline crossing. So that on one side of the border are finite
numbers and on the other side are the infinite numbers.

Yet, surprisingly, every one in the math community would agree with
the above, but no-one wants to research where the exact border is
located. I did at least a year's worth of research and concluded the
borderline was 10^603 or pi*10^603. The number 10^603 is the first
time that the digits in pi have 3 zero digits in a row allowing for
the Euler formulas on Regular Polyhedra to be true. And 10^603 is the
first time that the pseudosphere surface area equals exactly the
respective sphere area. There are other confirmations that 10^603 is
the borderline. But what I want to discuss is the analogy between
geography and the lack of logic by
mathematicians for a borderline.

Now if Geography were mathematics and where geographers behaved like
mathematicians concerning borderlines. Then all the countries on Earth
would not have borderlines. All the countries would be like numbers.
So that starting with Europe, that UK is 1 and France 2 and Germany 3
etc etc and move to the Americas where USA is say 11 on the list and
Canada 12 and moving over to Asia etc etc. So all the countries are a
number. But all the countries have no borderline, no borders.

Now we have a question arise, and the question is whether Paris is in
the USA because there are no borders between countries? One may say
that since there is a ocean between them that Paris is not in the USA.
Likewise, one can ask if the number 333333.... is in the set of
Counting Numbers since mathematics refuses to have a borderline
between finite and infinite.

Now one may ask if Paris is in Russia because there are no
borderlines, and likewise one may ask if 99999.... is a finite number
since there is no borderline.

Now one may ask a perfectly normal question whether the world is all
Chinese since the world geography has no borderlines that the answer
must be yes since no borderline stops us from saying that Middle East
and Africa are all Chinese. If we had a borderline between countries
we would not be asking these questions but since we do not we ask
them. Likewise, we would ask if the infinity of the Reals is larger
than the infinity of the Naturals since current mathematics does not
see fit to explore the borderline between finite and infinite.

In other words, the message I am driving home is that the science of
geography has a lot of down to Earth commonsense, common logic. The
science of mathematics, is feebly spurting along in a intellectual
desert as they continue to ignore that finite transitions into
infinite and that demands a borderline crossing.

[Virgil]

In article <n94498l8sd24c9q7k...@4ax.com>,

I suspect he will agree that what I just said is valid.
--

[Zuhair]

On Nov 1, 1:24 am, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 13:02:47 -0700 (PDT), Zuhair <zaljo...@gmail.com>

No you didn't get the idea.

The claim of the diagonal argument is about EVERY set of Reals that
have a bijection to the set N of all natural numbers.

(although as Moe Blee said we don't need the strong claim of having a
bijection, but yet I use it here for clarity).

So EVERY set S of reals that have a bijection to the set of all
natural numbers, then this set S will be subject to the diagonal
argument of Cantor's. And the result will be that there would exist a
real that is not in S.

So the diagonal argument of Cantor works on all sets of reals that
have bijections to the set of all naturals. (of course the argument
can be extended to include higher cases but this is not the subject
now)

Now suppose that S is a set of reals and also suppose that there is NO
bijection between S and N, so one may ask, would such set S be subject
to the diagonal argument of Cantor, in other words can we construct a
diagonal on such set S to define a real that is not in S in an EXACTLY
similar manner to how we did it for any set of reals that is bijective
to N. The answer is NO! because the diagonal argument of Cantor
depends on using the bijective function from N to S in order to
construct a missing real in S, so if there is no such a bijection from
N to S then we cannot apply that argument. So if you say that S is a
set of reals and that there do not exist any bijection between S and
N, then the Diagonal argument of Cantor cannot touch S.

Now we know that the set R of ALL reals is by definition a set of
reals. Now the question is would R be a set of reals that is bijective
to N or R would be a set of reals that is not bijective to N? That's
the question that the diagonal argument of Cantor's addresses

To answer this question, we first affirm that R is either a set of
reals that is bijective to N or is a set of reals that is not
bijective to N, it cannot be both and it cannot be neither, so if we
prove that one of the above two conditions is impossible then this
entails that the other condition is true.
So from here we proceed to prove that one of the conditions above is
impossible by proving that if we assume it being true then we'd derive
a contradiction, and after we establish that, then this would as I
said above entail that the other condition is true. This is called
Argument by Contradiction:

We proceed in the following manner

lets say that R is a set of reals that is bijective to N.
But by then R would be subject to the Diagonal argument of Cantor
which states that any set S of reals that is bijective to N would be
missing a real, so R would be missing a real which is against the
definition of R. (remember R is defined as the set of ALL reals).
A CONTRADICTION.

So we conclude that R cannot be a set of reals that have a bijection
to N, because this leads to the above contradiction. Accordingly the
other possibility is the True one.

So R must be a set of reals that do not have a bijection to N.

More explicitly the Set of All reals cannot be place in one-one
correspondence with the Set of All naturals.

QED

I hope this is clear enough.

Zuhair

[Zuhair]

On Nov 1, 8:46 am, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> On Oct 31, 5:23 pm, Archimedes Plutonium
>
>

> That as she was expanding the 3s in
> 0.3333333333...
>
> I was expanding the 3s equally in the Natural Numbers
>
> 33333333.....
>

Matters really boil down to the following:

You thought that just by reversing the order of the string of digits
after the decimal point in any real then this will yield a natural
number.

Patricia pointed out that 1/3 which is a real having the decimal
expansion

0.333.....

If we reverse that then we get

333....

Which is NOT a natural number.

Now instead of you admitting that your argument was erroneous, what
you did is to defend your bad argument by another even worse claim
that of 333... being a natural number, which is CLEARLY false, and you
know that, a natural number can ONLY be represented by a FINITE string
of digits, while 333... is an infinite string of digits, so by
definition it is NOT a natural number.

The reals differ from the naturals in that they can be represented by
an INFINITE string of digits, but the naturals cannot.

Just admit that Patricia Knocked you out.

Zuhair

[Zuhair]

On Nov 1, 10:04 am, Archimedes Plutonium

> Archimedes Plutoniumhttp://www.iw.net/~a_plutonium

> whole entire Universe is just one big atom
> 
where dots of the electron dot cloud are galaxies

Of course there is a boarderline between the finite and the infinite.
It is Omega

If n < Omega then n is finite

If n >= Omega then n is infinite

Omega is the cardinality of the set of all naturals.

All these issues has been defined clearly, and modern set theory
clearly shows all of that. You just need to read.

Your approach of regarding some Large finite as an infinite is not
correct. You are thinking of an upper bound on finities beyond which
what you call infinities starts. While in ZFC we do have a lower bound
on infinities below which the finities starts. Anyhow the idea of an
upper bound on naturals might be true but your phrasing of it is not
correct. If you believe that the truth is that there is an upper bound
on naturals, which indeed might be the case, then this is generally
called an ultrafinitist context and Cantor's argument altogether is
not applicable to it. Not only that even if you believe in a POTENTIAL
infinite, which is a stronger claim than ultrafinistism, then Cantor's
argument will not thrive in such media. Cantor's argument only thrives
on the assumption of ACTUAL infinity, which is running against
Aristotle so to say. If you refuse the assumption of existence of the
Actual infinite, then there is no need to go and involve yourself with
particulars of the diagonal argument, since the diagonal argument
already assumes that.

Zuhair

[Archimedes Plutonium]

On Nov 1, 4:28 am, Zuhair <zaljo...@gmail.com> wrote:
> On Nov 1, 8:46 am, Archimedes Plutonium
>
> <plutonium.archime...@gmail.com> wrote:
> > On Oct 31, 5:23 pm, Archimedes Plutonium
>
> > That as she was expanding the 3s in
> > 0.3333333333...
>
> > I was expanding the 3s equally in the Natural Numbers
>
> > 33333333.....
>
> Matters really boil down to the following:
>
> You thought that just by reversing the order of the string of digits
> after the decimal point in any real then this will yield a natural
> number.
>

Hi Zuhair, you are reading my posts too fast and losing comprehension
of what I am saying.

You have the above sentence backwards. I start with the Naturals since
they have order whereas the Reals have no order (both in binary base)

1 <--> .1
10 <--> .01
11 <--> .11
100 <--> .001

101 <--> .101
110 <--> .011
etc etc

Now as I move along with every Natural and then reverse it before a
decimal point, as you and Patricia and JStern can see, I capture every
Real.

Now the reason I can say I capture every Real is because the Naturals
are All Possible Digit Arrangements of 0 and 1 to any place value, as
well as the Reals are All Possible Digit Arrangements of 0 and 1 to
any place value.

With All Possible Digit Arrangements means there is no hole in either
set.

Now the only objection is from Patricia, who wants to create a Real in
an instant of time rather than equal bijections in a steady time of
walking out the Naturals. Patricia wants me to instantly state the
Natural for the Real 1/3.

And I explained her error, although the error needs more discussion.

As I said to her, as we walk out the 3s in the Naturals:

3
33
333
3333
33333

we correspondingly biject to the Reals
.3
.33
.333
.3333


But if you wanted me to immediately biject

.333333....
with those ellipsis symbols I would say stop, for you have now gone
nonmath because you never defined what "gone to infinity" means and
you should provide a borderline of where finite stops and infinity
starts.

So I respond with the Natural Number

333333..... since Patricia does not want to wait as we walk down digit
by digit the bijection of the Naturals with the Reals.

So if Patricia had said, infinity border is 10^1000, then the
bijection is simple and easy for there are only exactly 10^1000 Reals
between 0 and 1, and there are exactly 10^2000 Reals between
0 and 10^1000.

So the errors that Cantor made was that he never had the concept of
All Possible Digit Arrangements, for then the Naturals are instantly
bijected into the Reals since whenever any two sets that go to
infinity (borderline or no borderline) that share All Possible Digit
Arrangements are bijectable. The only reason that Cantor pulled out a
alleged missing Real is because his list never listed all the Reals
that he should have listed.

Just as the list of all 10 digits to 2 place value is contained in the
first 100 numbers and no matter what sort of diagonal setup you
contrive, the diagonal and alteration are not going to give you a
number not already present. Any combinations of 10 digits to 2 place
value is present already in the numbers from 0 to 100.

And the second flaw of Cantor diagonal is that finite transitions into
infinity, which means there must be a borderline. And whatever that
borderline is creates a smallest nonzero Real for which all the other
Reals are incremental adding of that smallest Real. I believe the
borderline is 10^603 so the smallest nonzero Real is 10^-603 and that
means automatically the Reals are a bijection with the Naturals. So
the whole entire Cantor episode is washed up and washed away worse
than Hurricane Sandy.

I hope you were not a victim of Hurricane Sandy and on battery power
to send out your post.

> Patricia pointed out that 1/3 which is a real having the decimal
> expansion
>
> 0.333.....
>
> If we reverse that then we get
>
> 333....
>
> Which is NOT a natural number.
>

No, you are not following the argument. The argument is that we start
with the Naturals since they have order and we walk along the line of
all the Naturals starting with 0.

So, as we build up the Naturals walking down the line we
correspondingly build up the Reals. And I know, and you know, and
Patricia knows that as we walk down that line we eventually pick up
every Real and bijected to every Natural because both sets are All
Possible Digit Arrangements.

The only hardship is when you or Patricia is too tired and want to try
to throw a monkey wrench into the picture by demanding what is
0.333333.... or say 0.666666...... with those ellipsis, for neither
you or Patricia state what the borderline is for finite with infinity.

> Now instead of you admitting that your argument was erroneous, what
> you did is to defend your bad argument by another even worse claim
> that of 333... being a natural number, which is CLEARLY false, and you
> know that, a natural number can ONLY be represented by a FINITE string
> of digits, while 333... is an infinite string of digits, so by
> definition it is NOT a natural number.
>
> The reals differ from the naturals in that they can be represented by
> an INFINITE string of digits, but the naturals cannot.
>
> Just admit that Patricia Knocked you out.
>
> Zuhair

Wrong again.

So I need to see what you think of this Claim of Logic:

Claim of Logic:
Finite versus Infinity Structure:
Finite is a condition in which we consider a quantity is bounded and
comes to an end. Infinity is a condition in which we think a quantity
has no end. And we consider that finite is moving or transitioning
into infinite. That entails a borderline. A borderline that separates
finite from infinite.

So, Zuhair, do you agree, accept the above or do you not agree and why
not?

[Archimedes Plutonium]

On Nov 1, 4:46 am, Zuhair <zaljo...@gmail.com> wrote:
> On Nov 1, 10:04 am, Archimedes Plutonium
>

Hi again, I did not read this post before writing my prior post, for I
asked you the question of whether you believed in a borderline, and
you answered it in this post.

>
> Of course there is a boarderline between the finite and the infinite.
> It is Omega
>

Alright, I am going to squabble and fight with you on that.

You do agree there is a borderline.

Now what I am going to say is that if finite are numbers, then the
Infinity borderline must also be a number. Finite is a number
representing the amount of quantity, so that we have say 2000 objects.
But now Omega is not a number is it.

> If n < Omega then n is finite
>
> If n >= Omega then n is infinite
>
> Omega is the cardinality of the set of all naturals.
>
> All these issues has been defined clearly, and modern set theory
> clearly shows all of that. You just need to read.
>

Depends on what is read, for if you read propaganda, then you end up
with Omega as borderline.

> Your approach of regarding some Large finite as an infinite is not
> correct. You are thinking of an upper bound on finities beyond which
> what you call infinities starts. While in ZFC we do have a lower bound
> on infinities below which the finities starts. Anyhow the idea of an
> upper bound on naturals might be true but your phrasing of it is not
> correct. If you believe that the truth is that there is an upper bound
> on naturals, which indeed might be the case, then this is generally
> called an ultrafinitist context and Cantor's argument altogether is
> not applicable to it. Not only that even if you believe in a POTENTIAL
> infinite, which is a stronger claim than ultrafinistism, then Cantor's
> argument will not thrive in such media. Cantor's argument only thrives
> on the assumption of ACTUAL infinity, which is running against
> Aristotle so to say. If you refuse the assumption of existence of the
> Actual infinite, then there is no need to go and involve yourself with
> particulars of the diagonal argument, since the diagonal argument
> already assumes that.
>
> Zuhair

Zuhair, you never bothered to think deeper into this subject of
borderline.

Your entire above goes down the drain, if you consider that the
Pseudosphere surface area is equal the respective sphere area at
pi*10^603 and the volume is exactly 1/2 the respective sphere.

In other words, the pseudosphere goes to infinity and at 10^603 it
obeys surface area and volume, but if you extend the pseudosphere
beyond 10^603, to say 10^604 it loses its equal area and 1/2 volume.

The Euler formula for Regular Polyhedra is not obeyed until it reaches
10^603 since all regular polyhedra and regular polygons are evenly
divisible by 120 = 5!

The Riemann Hypothesis falls apart at 10^603.

To name just three items in math that depend on a borderline of finite
versus infinite.

Your problem, Zuhair is that you read too many books that has polluted
your mind.

But I will think about what logic demands the borderline to be a
actual number rather than a ghost-number Omega.

The first problem with Omega is again, what is the last finite
number.

So if Omega were truly the borderline number, Zuhair, you should be
able to state the last finite number and you cannot so Omega is not
the borderline.

Archimedes Plutonium

[LudovicoVan]

"Archimedes Plutonium" <plutonium....@gmail.com> wrote in message
news:ef6821d3-2b7d-4216...@v3g2000yqb.googlegroups.com...

> But I will think about what logic demands the borderline to be a
> actual number rather than a ghost-number Omega.

Maybe relevant , maybe not: anyway extended multi-precision numbers (which
are *effective* implementations of "big" numbers) already have this feature,
that is they belong to finite sets of the form (considering the example of
unsigned integers) { 0, 1, ..., Max, Overflow, Infinity }, with the obvious
ordering, where Infinity is a constant taken to be the result of division by
zero, Overflow is another constant taken to be the result of any operation
whose result is greater than Max, and of course Max would be the "border
line".

Differently from your approach, though, Max is a parameter of the system,
not an absolutely fixed value, i.e. it is an expression of the capabilities
of the specific implementation: for instance, on 32-bit unsigned integers
the greatest Max could be is 2^32-3 (to leave room for the special values),
but with 64-bit unsigned integers it could be up to 2^64-3, etc.

-LV

[JRStern]

On Thu, 1 Nov 2012 02:10:38 -0700 (PDT), Zuhair <zalj...@gmail.com>
wrote:

>> >I understand your concern, sometimes the wording is indeed confusing.
>> >I already rephrased this wording in my latest response to AP.
>>
>> >Lets be a little bit more clear.
>>
>> >Lets identify a real with a countably infinite binary digit sequence.
>>
>> OK.
>>
>> >Now we define the predicate "is a set of reals" in the following
>> >manner:
>>
>> >S is a set of reals <-> (for all y. y in S -> y is a real).
>>
>> OK.
>>
>> >So S may be empty, or S might contain some but not all reals, or S
>> >might contain all reals in which case it will be called the set of all
>> >reals.
>>
>> OK.
>>
>> >Now our question can be modified to the following:
>>
>> >Suppose that S is a set of reals and Suppose that there exist a
>> >bijection from N to S, then can S be the set of ALL reals?
>>
>> Let's move slowly here for my benefit.
>>
>> When we make the diagonal argument, is that a claim that there is a
>> bijection from N to S?

I would like to move this question again.

Also to clarify, to others' comments, whether we really need a
bijection, or whether it is what - injection, surjection? Being
careful (if it is not bijection) to state from which to which.

Thanks.


Josh

[MoeBlee]

On Oct 31, 8:22 pm, JRStern <JRSt...@foobar.invalid> wrote:
> (I'm uncertain as to hopping between R and S here)

S is just the range of f.

If S confuses you, then I'll say it without S (Jesse, per his
suggestion, will appreciate that too):

First, review definitions:

f is a function from X into Y
<->
f is a function & dom(f)=X & range(f) is a subset of Y

f is a function from X onto Y
<->
f is a function & dom(f)=X and range(f)=Y

Claim: There is no function from N onto R

Proof:
(1) Let f be an arbitrary function from N into R. We will show that f
is not onto R.
(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
range(f), so f is not onto R.
(3) Since f was arbitrary, we may conclude that there is no function
from N onto R.
Done.

Please which, if any, step you don't understand.

> we REALLY HAVE found a number not in S?

Yes.

> B.  That we have a CLAIM to have found a number not in S, but since S
> is of the cardinality of R (?) this is a contradiction, so we have
> proven that SOMETHING upstream is rotten.

I said NOTHING about "cardinality" so why even bring it in?

Again, the argument:

Claim: There is no function from N onto R

Proof:
(1) Let f be an arbitrary function from N into R. We need to show that

f is not onto R.

(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
range(f), so f is not onto R.
(3) Since f was arbitrary, we may conclude that there is no function
from N onto R.
Done.

Please say what, if anything, in those steps you don't understand.

MoeBlee

[MoeBlee]

On Nov 1, 1:06 am, JRStern <JRSt...@foobar.invalid> wrote:
> On Wed, 31 Oct 2012 22:55:06 -0600, Virgil <vir...@ligriv.com> wrote:

> >The Cantor diagonal argument applies only to LISTS, which a "complete
> >set of reals" is not and cannot be.

> I don't want to fragment this, can you work that out with MoeBlee and
> his story?

I'm not the expert, so we don't have to go through me for each
assertion.

But, yes, what Virgil says is good enough.

In mathematics 'list' and 'enumeration' mean the same thing. And one
definition of "f is an enumeration of X" is "f is a function from N
onto X" (or, in the finite case, a function from a member of N onto X;
but here we're concerned with the infinite case since the set of real
numbers is infinite).

What the diagonal argument proves is that there is no list of R. There
is no enumeration of the set of real numbers.

MoeBlee

[MoeBlee]

On Nov 1, 10:18 am, JRStern <JRSt...@foobar.invalid> wrote:

> Also to clarify, to others' comments, whether we really need a
> bijection, or whether it is what - injection, surjection? Being
> careful (if it is not bijection) to state from which to which.

You're confusing yourself unnecessarily. Here is the argument in stark
simplicity:

Definitions:

f is a function from X into Y
<->
f is a function & dom(f)=X & range(f) is a subset of Y

f is a function from X onto Y
<->
f is a function & dom(f)=X and range(f)=Y

Claim: There is no function from N onto R

Proof:
(1) Let f be an arbitrary function from N into R. We will show that f
is not onto R.
(2) The antidiagonal of f is in R, but the antidiagonal of f is not
in
range(f), so f is not onto R.
(3) Since f was arbitrary, we may conclude that there is no function
from N onto R.
Done.

In the above there's no mention of "cardinality", "injection",
"surjection", or "bijection". Put that all aside. Clear your mind.
Just think through the above argument. What exactly, if anything, do
you not understand about it, just as it is writtten and not dragging
in extraneous confusions regarding terminology that is not even in the
argument above.

MoeBlee

[JRStern]

On Thu, 1 Nov 2012 09:46:07 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>On Oct 31, 8:22 pm, JRStern <JRSt...@foobar.invalid> wrote:
>> (I'm uncertain as to hopping between R and S here)
>
>S is just the range of f.
>
>If S confuses you, then I'll say it without S (Jesse, per his
>suggestion, will appreciate that too):
>
>First, review definitions:
>
>f is a function from X into Y
><->
>f is a function & dom(f)=X & range(f) is a subset of Y
>
>f is a function from X onto Y
><->
>f is a function & dom(f)=X and range(f)=Y
>
>Claim: There is no function from N onto R
>
>Proof:
>(1) Let f be an arbitrary function from N into R. We will show that f
>is not onto R.
>(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
>range(f), so f is not onto R.

Does a function have a diagonal?

The diagonal is the range of f, right?

I'm mildy curious what kind of function it is that pulls n'th digits
out of its arguments, but OK.

>(3) Since f was arbitrary, we may conclude that there is no function
>from N onto R.

But f was not arbitrary, it was highly specific, and makes not the
least attempt to be onto.

>Done.
>
>Please which, if any, step you don't understand.
>
>> we REALLY HAVE found a number not in S?
>
>Yes.

OK (that is, that clarifies it), but then this is a very different
argument from the one that Zuhair is making in some other branches
here.

>> B.  That we have a CLAIM to have found a number not in S, but since S
>> is of the cardinality of R (?) this is a contradiction, so we have
>> proven that SOMETHING upstream is rotten.
>
>I said NOTHING about "cardinality" so why even bring it in?

Because the ACTUAL diagonal argument is NOT about N to R, it is about
some kind of illustrated, indicated set, the exact nature of which
cannot just be blithely assumed, it must be very carefully specified.

Is that set finite, countably infinite, or uncountable?

You cannot just hop over that detail and make a claim that it does not
matter and try to baffle the conversation by asserting the conclusion.

If you assume it finite, you will not find it infinite. That is not
arguing to a contradiction, it is arguing back to an assumption.


J.

[MoeBlee]

On Nov 1, 12:07 pm, JRStern <JRSt...@foobar.invalid> wrote:

> On Thu, 1 Nov 2012 09:46:07 -0700 (PDT), MoeBlee <modem...@gmail.com> wrote:

> >(1) Let f be an arbitrary function from N into R. We will show that f
> >is not onto R.
> >(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
> >range(f), so f is not onto R.
>
> Does a function have a diagonal?

Certain kinds of functions do. And we don't even need the word
'diagonal'. Where I say 'diagonal' and 'anti-diagonal' I could instead
just give the explicit specification.

First, we have reals represented as denumerable binary sequences.

The function f is from N into the set of denumerable binary sequences.
So for each n in N, we have that f(n) is itself a function from N into
{0 1}.

The "diagonal of f" is the function d defined by d(n)=f(n)(n).

Now take the function g from N into {0 1} defined by g(n)=0 if d(n)=1
and g(n)=1 if d(n)=0. So g is the "anti-diagonal of f".

So we don't need the words "diagonal" and "anti-diagonal".

> The diagonal is the range of f, right?

No. The diagonal is as I described above.

> >(3) Since f was arbitrary, we may conclude that there is no function
> >from N onto R.
>
> But f was not arbitrary, it was highly specific, and makes not the
> least attempt to be onto.

No, f is an arbitrary function from N into R. We don't assume that f
is onto R and we don't assume that f is not onto R.

If you don't understand that step in the proof then you don't
understand basic mathematical argumentation (codified by the rule of
universal generalization).

> Because the ACTUAL diagonal argument is NOT about N to R, it is about
> some kind of illustrated, indicated set, the exact nature of which
> cannot just be blithely assumed, it must be very carefully specified.
>
> Is that set finite, countably infinite, or uncountable?
>
> You cannot just hop over that detail and make a claim that it does not
> matter and try to baffle the conversation by asserting the conclusion.
>
> If you assume it finite, you will not find it infinite.  That is not
> arguing to a contradiction, it is arguing back to an assumption.

Now I see that you are not familiar with basic mathematical
argumentation.

I'll explain by an analogy (this analogy is not an argument that my
method is correct, but rather the analogy is given just so that you
can better picture the basic logic).

In a mathematical argument, I might say "Let T be an arbitrary
triangle". I don't say what kind of triangle - whehter equilateral,
having area greater than a certain number, lying in a certain quadrant
of the bi-coordinate plane - nothing except that T is a triangle. Then
I prove that T, simpy by virture of being a triangle, has certain
properties. Then I am allowed to infer that ALL triangles have those
properties, since T was an arbitrary triangle and in my argument I
assumed nothing about T except that it is a triangle.

Here the argument is "Let f be an arbitrary function from N into R".
All I assume about f is that it is a function from N into R. Then I
prove that, by virtue of being a function from N into R, we have that
f has the property of not being onto R. And, since I assumed nothing
about f except that it is function from N into R, I can conclude that
ANY function from N into R has the property of not being onto R, i.e.,

that there is no function from N onto R.

If you don't grasp this basic kind of mathematical reasoning, then you
need to read more mathematics and/or study a bit of symbolic logic
that codifies such principles in precise form.

MoeBlee

[Shmuel Metz]

In <jfs298pmsdt57p85k...@4ax.com>, on 10/31/2012
at 11:54 AM, JRStern <JRS...@foobar.invalid> said:

>Is that clear?

It's clear that archiepu is begging the question when he writes
"Proof: List all the Reals of interval 0 to 1." Keep that and the
definition of complete in mind while rereading Zuhair's article and
his statement may be clearer.

>But we've already lost our clarity, what are we mapping to?

The set of all reals, which is *not* a list.

>If it's a complete list, then I thought it was "clear" that the
>diagonal doesn't work.

No, it's clear that the antidiagonal procedure does work, showing that
no list of reals can be complete.

>The point of contention then is whether this paper representation
>does or does not represent an uncountable set.

What paper representation. The proof involves the set of all reals and
a list of reals, and shows that the list cannot be complete.

>But hey, we write aleph
>characters and claim they DO represent uncountable sets,

No.

--
Shmuel (Seymour J.) Metz, SysProg and JOAT <http://patriot.net/~shmuel>

Unsolicited bulk E-mail subject to legal action. I reserve the
right to publicly post or ridicule any abusive E-mail. Reply to
domain Patriot dot net user shmuel+news to contact me. Do not
reply to spam...@library.lspace.org

[Shmuel Metz]

In <rs6398h2n6ma5jcmu...@4ax.com>, on 10/31/2012

at 03:24 PM, JRStern <JRS...@foobar.invalid> said:

>When we make the diagonal argument, is that a claim that there is a
>bijection from N to S?

No.

>So the claim is that for some set S that may not be a complete set,
>the diagonal argument may work.

May? It does work.

>And then by induction

No induction is required.

>What exactly is this set S for which it is claimed that the
>diagonal argument works?

It works for any set S of reals, and is, in fact, a red herring.

>Typically we use some small but still unenumerable subset of S,
>like the segment from 0 to 1.

The argument doesn't depend on the set S.

>as we're going to assume that the noncontinuous set still has the
>same cardinality as if it were a segment.

No.

>We're trying to define just what *is* the difference between one
>line and another in our *representation*.

No. We're trying to show that any list of reals is incomplete; only
this and nothing more.

>Otherwise, we're basically introducing an infinitesimal argument,

No.

>There is no successor function in the reals, right?

Nor do we need one.

>So I can't see what the argument is, that the diagonal is supposed
>to be drawn on.

It isn't supposed to be drawn.

>So what is really being argued?

Given a sequence s:N->R, there is a real that is not in the sequence,
i.e. (\exist r \in R) ¬(\exist i \in N)s(i)=r.

>So the entire argument comes down to definitional, and the problem
>isn't so much that the real are unenumerable, because you can't
>really (fully) enumerate even a single real to infinite countable
>digits,

You're conflating unrelated issues. The Cantor antidiagonal procedure
is about the inability to enumerate all of the reals and has nothing
to do with enumerating the digits in a positional representation of a
real. Further, the term enumeration has nothing to do with making
marks on paper.

>circular

No. Look at the argument as presented and don't try to hang unrelated
baggage on it.

[Shmuel Metz]

In <virgil-8C4A3C....@bignews.usenetmonster.com>, on
10/31/2012

at 10:14 PM, Virgil <vir...@ligriv.com> said:

>My claim is that such things as "the list of 'all' possible digit
>arrangements" have been proven not really to be ALL of them.

What has been proven is that there is no such list.

>It is because any such a list of ALL arrangements can be shown not
>to contain ALL of them.

What has been proven is that there is no such list.

>It was proved by Cantor!

FSVO it.

[Shmuel Metz]

In <virgil-23824D....@bignews.usenetmonster.com>, on
10/31/2012
at 11:05 PM, Virgil <vir...@ligriv.com> said:

>Cantor's own proof

Which one?

[JRStern]

On Thu, 1 Nov 2012 10:31:22 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>On Nov 1, 12:07 pm, JRStern <JRSt...@foobar.invalid> wrote:
>> On Thu, 1 Nov 2012 09:46:07 -0700 (PDT), MoeBlee <modem...@gmail.com> wrote:
>
>> >(1) Let f be an arbitrary function from N into R. We will show that f
>> >is not onto R.
>> >(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
>> >range(f), so f is not onto R.
>>
>> Does a function have a diagonal?
>
>Certain kinds of functions do. And we don't even need the word
>'diagonal'. Where I say 'diagonal' and 'anti-diagonal' I could instead
>just give the explicit specification.
>
>First, we have reals represented as denumerable binary sequences.
>
>The function f is from N into the set of denumerable binary sequences.
>So for each n in N, we have that f(n) is itself a function from N into
>{0 1}.

Then I cannot any longer recognize this as being about the diagonal
argument. It is in some way shape or form about N and R, as is the
diagonal argument. But it is a different story. I do not see that it
includes the diagonal argument.

>The "diagonal of f" is the function d defined by d(n)=f(n)(n).
>
>Now take the function g from N into {0 1} defined by g(n)=0 if d(n)=1
>and g(n)=1 if d(n)=0. So g is the "anti-diagonal of f".
>
>So we don't need the words "diagonal" and "anti-diagonal".
>
>> The diagonal is the range of f, right?
>
>No. The diagonal is as I described above.

No. The diagonal argument I am asking about is from some
underspecified set S, and the nature of that set is the crux of the
issue, it now appears to me.

(not to foreclose other arguments, btw)

>> >(3) Since f was arbitrary, we may conclude that there is no function
>> >from N onto R.
>>
>> But f was not arbitrary, it was highly specific, and makes not the
>> least attempt to be onto.
>
>No, f is an arbitrary function from N into R. We don't assume that f
>is onto R and we don't assume that f is not onto R.

Then you lose nothing by choosing a specific f and talking about it,
and much of my question is about the specifics of a function, I'm not
sure I have anything to say about any general case. I fully grant
that I expect valid arguments can be made about the general case, and
about N to R. My question are about the specifics of a diagonal
argument.

>If you don't understand that step in the proof then you don't
>understand basic mathematical argumentation (codified by the rule of
>universal generalization).

You have lost the topic.

>> Because the ACTUAL diagonal argument is NOT about N to R, it is about
>> some kind of illustrated, indicated set, the exact nature of which
>> cannot just be blithely assumed, it must be very carefully specified.
>>
>> Is that set finite, countably infinite, or uncountable?
>>
>> You cannot just hop over that detail and make a claim that it does not
>> matter and try to baffle the conversation by asserting the conclusion.
>>
>> If you assume it finite, you will not find it infinite.  That is not
>> arguing to a contradiction, it is arguing back to an assumption.
>
>Now I see that you are not familiar with basic mathematical
>argumentation.

No, you have not grasped the question, and are telling a story that is
not related.

J.

[MoeBlee]

On Nov 1, 12:56 pm, JRStern <JRSt...@foobar.invalid> wrote:
> On Thu, 1 Nov 2012 10:31:22 -0700 (PDT), MoeBlee <modem...@gmail.com>

> wrote:
>
> >On Nov 1, 12:07 pm, JRStern <JRSt...@foobar.invalid> wrote:
> >> On Thu, 1 Nov 2012 09:46:07 -0700 (PDT), MoeBlee <modem...@gmail.com> wrote:
>
> >> >(1) Let f be an arbitrary function from N into R. We will show that f
> >> >is not onto R.
> >> >(2) The antidiagonal of f is in R, but the antidiagonal of f is not in
> >> >range(f), so f is not onto R.
>
> >> Does a function have a diagonal?
>
> >Certain kinds of functions do. And we don't even need the word
> >'diagonal'. Where I say 'diagonal' and 'anti-diagonal' I could instead
> >just give the explicit specification.
>
> >First, we have reals represented as denumerable binary sequences.
>
> >The function f is from N into the set of denumerable binary sequences.
> >So for each n in N, we have that f(n) is itself a function from N into
> >{0 1}.
>
> Then I cannot any longer recognize this as being about the diagonal
> argument.

It is the diagonal argument, pretty much just as Cantor gave it.

"diagonal argument" is a nickname for a kind of argument in
mathematics. And most specifically, Cantor's diagonal argument
regarding the reals. I've given you the diagonal argument.

> It is in some way shape or form about N and R, as is the
> diagonal argument.  But it is a different story.  I do not see that it
> includes the diagonal argument.

It couldn't be more exactly the diagonal argument.

d is the diagonal. g is the anti-diagonal. The words 'diagonal' and
'anti-diagonal' are not themselves needed for the proof. Again,
"diagonal argument" is a nickname for a kind of approach in
mathematical proof. The argument I gave is a diagonal argument. The
argument I gave is virtually exactly Cantor's argument except I used
more modern notation.

> >The "diagonal of f" is the function d defined by d(n)=f(n)(n).
>
> >Now take the function g from N into {0 1} defined by g(n)=0 if d(n)=1
> >and g(n)=1 if d(n)=0. So g is the "anti-diagonal of f".
>
> >So we don't need the words "diagonal" and "anti-diagonal".
>
> >> The diagonal is the range of f, right?
>
> >No. The diagonal is as I described above.
>
> No.  The diagonal argument I am asking about is from some
> underspecified set S, and the nature of that set is the crux of the
> issue, it now appears to me.

I'm giving you the classic diagonal argument. Whether I use 'S' or
not, doesn't matter. S is just the range of f.

Use either my earlier proof that uses 'S' or my more streamlined
version now. They're essentially the same. The use of 'S' is
inessential. I used it originally only as I thought it might help you
to better visulize the range of f.

> >> >(3) Since f was arbitrary, we may conclude that there is no function
> >> >from N onto R.
>
> >> But f was not arbitrary, it was highly specific, and makes not the
> >> least attempt to be onto.
>
> >No, f is an arbitrary function from N into R. We don't assume that f
> >is onto R and we don't assume that f is not onto R.
>
> Then you lose nothing by choosing a specific f and talking about it,

No, that is EXACTLY the opposite of how such a proof should work.

We want to prove that ANY function from N into R has a certain
property (namely the property of not being onto R). So we must NOT
choose a specific f but rather we consider an ARBITRARY f that is a

function from N into R.

'f' is just a variable. We are saying, "Let the variable 'f'
temporarily stand for a function from N into R, and we assume nothing
else about f - only that it is a function from N into R."

> and much of my question is about the specifics of a function, I'm not
> sure I have anything to say about any general case.  I fully grant
> that I expect valid arguments can be made about the general case, and
> about N to R.  My question are about the specifics of a diagonal
> argument.
>
> >If you don't understand that step in the proof then you don't
> >understand basic mathematical argumentation (codified by the rule of
> >universal generalization).
>
> You have lost the topic.

No, the discussion with you has led to the point that you don't
understand the most basic form of the diagonal argument, which begins
with letting f be an ARBITRARY function from N into R. If you don't
understand such basic mathematical reasoning, then you cannot
understand the diagonal argument.

MoeBlee

[MoeBlee]

JRStern:

I'm giving you the benefit of the doubt that you're not trolling -
setting up incorrect objections just for the sake of getting people to
waste their time composing replies to you.

But if you are not trolling, then it's an acute fact that you don't
understand the most basic thing about mathematical argumentation that
is the very foundation and starting point of the diagonal arument, as
I explained in my previous post.

If you do not come to grips with understanding the method used in
taking f to be an arbitrary function from N into R, then there is no
point in trying to help you undersand the diagonal argument.

I gave you my recommendation: Familiarize yourself with mathematical
reasoning by reading some more mathematics, or look into a book on
symbolic logic, which will explain even the exact codification of
basic mathematical reasonikng. If you would look at mathematical
proofs in just about any field of mathematics, you will see at work
the very basic method of letting a variable (such as 'f') stand for an
arbitrary object and not a specific one.

MoeBlee

[Archimedes Plutonium]

On Nov 1, 7:09 am, "LudovicoVan" <ju...@diegidio.name> wrote:
> "Archimedes Plutonium" <plutonium.archime...@gmail.com> wrote in message

Thanks, that is relevant for it tells me what I am entering has had
some thought already given to the topic matter.

I did not know that Zuhair and the rest of the math community that
believes in ZFC system even had a thought or inkling of a borderline,
for Zuhair now says that ZFC had the borderline concept and that they
said the borderline was Omega, the cardinality of the Naturals.

So that using LV's above, then Zuhair and ZFC mathematics would have
the borderline as this, as per LV's notation:

{ 0, 1, ..., Omega, Omega + 1, Omega + 2 }

Now the problem with that is it does not work for it breaks the rules
of logic, since we cannot climb down once given Omega. It breaks the
rules of logic because it is not a borderline. A borderline has to
have the ability to move backwards as it has the ability to move
forwards.

So in other words, the borderline of finite with infinity must be of
the same class of numbers as 0,1,2,... albeit a large number so that
we can move backwards and then be able to say, this is the last finite
number and the next number starts infinite numbers.

Now in my analogy to Geography borderlines. If we cannot move
backwards as well as move forwards then we do not have a borderline.
If we start in Paris France and move in a straight line on the globe
we have to be able to move forward and reach a point on the map that
is the last and largest point that is still France and not Germany.
For if we cannot move backwards at the borderline and say, this is the
last and final point of France and the next point is the first point
of Germany, then we have no borderline between France and Germany, but
just some goofy mathematics bandied about by goofy mathematicians who
do not mind breaking the rules of logic.

So thanks, I needed to know that some people have been working on
these ideas and that it is not totally all new.

[JRStern]

On Thu, 1 Nov 2012 11:17:30 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>> Then I cannot any longer recognize this as being about the diagonal
>> argument.
>
>It is the diagonal argument, pretty much just as Cantor gave it.

...

>I'm giving you the classic diagonal argument. Whether I use 'S' or
>not, doesn't matter. S is just the range of f.

What is S, that is, the domain of f, in Cantor's argument, that
appears to be the question, before asking what we can say about the
range of f or draw any conclusions from or about it.

J.

[Archimedes Plutonium]

On Nov 1, 10:18 am, JRStern <JRSt...@foobar.invalid> wrote:

> On Thu, 1 Nov 2012 02:10:38 -0700 (PDT), Zuhair <zaljo...@gmail.com>

It is a bijection.

The mistake of Cantor, of Zuhair and his fellow ZFC is that they hide
behind a ellipsis. Here is their failure outlined:

We start with Naturals and we list them and reverse them to obtain the
bijection into the Reals. Since we use the ordering of the Naturals it
may look as though it is not a bijection:

1 <--> .1

10 <--> .01
11 <--> .11
100 <--> .001
101 <--> .101
110 <--> .011
etc etc

Now every Natural has a Real to correspond to it and the reason that
is true is because both sets are All Possible Digit Arrangements and
neither set has a hole in it.

Now what Zuhair and ZFC followers are trapped and deluded by, is that
they do not want to wait as the above covers all the Naturals and thus
all the Reals, so they jump far far ahead. They jump to a number like
0.333333.... with its ellipsis.

If they had waited and done the link-up
3 to .3
33 to .33
333 to .333
etc etc

If they had waited to do each link up they would not have raised the
complaint of 0.33333.... and what is the Natural.

But they are impatient and demand what is the Natural for that
ellipsis Real of 0.3333..... and here is where Zuhair, Cantor, and ZFC
followers do their cheating. They allow 0.3333.... as a legitimate
Real number and never define the borderline of Finite into Infinity,
but when I say that 33333.... is the Natural that corresponds to
0.33333.... they raise their illegitimate red flags saying that the
Peano axioms calls all Naturals as finite entities and that you are
not allowed under threat of imprisonement or a firing squad of
attaching an ellipsis to a Natural.

So, Cantor and Zuhair and ZFC followers can be vague or false about
the borderline of finite with infinite and attach ellipsis to any and
all Reals, whenever they want, but when a Logical and Correct thinking
person attaches the ellipsis to a Natural of 33333.... to follow the
0.33333..... they all freak out.

So let us say that Zuhair, Cantor and ZFC followers had a gram of
logical intelligence and applied it. They would then say Infinity
border is 10^603. That immediately removes the ellipsis in all
mathematics and replaces it with 603 digits to the rightwards of every
decimal point.

Thus, when Zuhair, Cantor or ZFC followers ask for the Natural that
corresponds to the Real 0.3333..33 with its 603 digits of 3s, the
Natural Number is 3333..33 which is 603 digits of 3s.

So Zuhair, Cantor and ZFC followers cheated and continue to cheat, for
they conveniently hide behind the ellipsis when it comes to Reals and
refuse to specify the borderline, but when it comes to Naturals, they
penalize you for using the ellipsis since they had Peano state in an
axiom that no Natural has an ellipsis.

The problem is all solved when ZFC followers acknowledge that a
borderline exists and must be a number so that there is a last and
final finite Natural such as 10^603. And what that does is peg the
smallest nonzero Real as being 1*10^-603.

Now, to finally answer JStern about whether it is a bijection.

It is indeed a bijection because I can start with the Naturals since
they are in a ordering already.

But also, I can start with the Reals in the interval 0 to 1 since they
are in a ordering and very much so.

That ordering is

0, then 1*10^-603, then 2*10^-603, then 3*10^-603

And the bijection using the ordering of the Reals would be this:

0 <==> 0
1*10^-603 <==> 10^603
2*10^-603 <==> 10^603 -1
etc etc

So the entire fault, the fakery of Cantor of Zuhair of ZFC followers
is that they fail to recognize All Possible Digit Arrangements
preempts any missing number whether in the Reals or in the Naturals
and thus you cannot produce a contradiction, coupled onto the fact
that they fail to apply a borderline between finite and infinite and
use the cheating excuse that Peano axioms disallow the ellipsis for
Naturals. Pretty musty, dirty and stinky was ZFC.

This story reminds me of George King of England during the time of Ben
Franklin and his experiments in electricity. Where Franklin discovered
some truths about electricity and the King of England was so enraged
that he made a royal decree against the truths of the laws of
electricity. Of course, no human laws can go against the natural laws
of physics. But Cantor and ZFC followers impose the same silly human
laws by decree that they never specify a borderline and they penalize
anyone who attaches an ellipsis to Naturals, yet they freely attach
ellipsis to Reals. So Cantor and ZFC followers are not mathematicians
but cheaters.

[Jesse F. Hughes]

JRStern <JRS...@foobar.invalid> writes:

> But we've already lost our clarity, what are we mapping to? If it's a

> complete list, then I thought it was "clear" that the diagonal doesn't

> work. If it's not a complete list because, um, just whyever you think
> that might be, then we're not REALLY even testing the thesis, are we?
>

> The point of contention then is whether this paper representation does

> or does not represent an uncountable set. But hey, we write aleph
> characters and claim they DO represent uncountable sets, but there
> seems to be something wrong, with this matrix? Just how is that
> determined?

With all due respect, you seem to be taking the ramblings of Archimedes
Plutonium seriously. This is probably not a good idea.

Of course, each argument stands or falls on its merits, and so we
shouldn't reject mathematical arguments just because their proponents
also happen to claim, say, that the universe is a giant plutonium atom.

That said, I daresay you're going to have problems finding any clearly
stated mathematical argument in AP's scratchings. There must surely be
a better source to defend than he.

--
Jesse F. Hughes
"Maybe I screwed up on one of my assumptions [...]. Otherwise, um,
it's very easy to factor, and things are about to get really, really
weird." -- James S. Harris

[Archimedes Plutonium]

Sorry that post is #1234 and corrected it on the original

[Virgil]

In article <5092878c$14$fuzhry+tra$mr2...@news.patriot.net>,

Shmuel (Seymour J.) Metz <spam...@library.lspace.org.invalid> wrote:

> In <virgil-23824D....@bignews.usenetmonster.com>, on
> 10/31/2012
> at 11:05 PM, Virgil <vir...@ligriv.com> said:
>
> >Cantor's own proof
>
> Which one?

The "anti-diagonal" one. Cantor said that given any sequence of binary
sequences (in his argument the individual binary sequences were strings
of m's and w's) necessarily failed to contain all possible such
sequences, by reason of showing that there was always at least one
binary sequence that any original sequence of such binary sequences did
not contain..
--

[JRStern]

On Thu, 1 Nov 2012 11:24:30 -0700 (PDT), MoeBlee <mode...@gmail.com>
wrote:

>JRStern:
>
>I'm giving you the benefit of the doubt that you're not trolling -
>setting up incorrect objections just for the sake of getting people to
>waste their time composing replies to you.

I appreciate your efforts, however if they are just repeating the text
and brushing aside the question, they are of limited effectiveness,
and insults do not comprise either pedagogy or explanation.

I understand the limits of Usenet in general and it's a rare day that
either side represents well, but all we can do is try.

J.

[JRStern]

On Thu, 01 Nov 2012 15:39:52 -0400, "Jesse F. Hughes"
<je...@phiwumbda.org> wrote:

>JRStern <JRS...@foobar.invalid> writes:
>
>> But we've already lost our clarity, what are we mapping to? If it's a
>> complete list, then I thought it was "clear" that the diagonal doesn't
>> work. If it's not a complete list because, um, just whyever you think
>> that might be, then we're not REALLY even testing the thesis, are we?
>>
>> The point of contention then is whether this paper representation does
>> or does not represent an uncountable set. But hey, we write aleph
>> characters and claim they DO represent uncountable sets, but there
>> seems to be something wrong, with this matrix? Just how is that
>> determined?
>
>With all due respect, you seem to be taking the ramblings of Archimedes
>Plutonium seriously. This is probably not a good idea.

There just might be some kernel of truth in the ramblings, I certainly
can't support everything I see there.

>Of course, each argument stands or falls on its merits, and so we
>shouldn't reject mathematical arguments just because their proponents
>also happen to claim, say, that the universe is a giant plutonium atom.
>
>That said, I daresay you're going to have problems finding any clearly
>stated mathematical argument in AP's scratchings. There must surely be
>a better source to defend than he.

Thanks to the posters here, I believe I have located the point of the
current issue as a question about exactly what the domain is supposed
to be for the diagonal function, and whether that constitutes assuming
the conclusion that is then supposed to follow from it.


J.

[Archimedes Plutonium]

>
> 0 <==> 0
> 1*10^-603 <==> 10^603
> 2*10^-603 <==> 10^603 -1
> etc etc
>

Sorry, silly me, a better bijection is this:

0 <==> 0
1*10^-603 <==> 1

2*10^-603 <==> 2
.
.
0.999..99 with 603 9s <==> 9999..99 with 603 9s
1 <==> 10^603

Now that allows a closed set of the Reals in 0 to 1.

And if you can discern, the battle that constantly wages in
mathematics as to whether 0.9999.... is 1 or not 1 is finally
laid to rest, because when mathematics gets off its lazy behind and
specifies the borderline of finite with infinite, it removes the
silliness of the ellipsis and with that, the silliness of 0.9999... as
1.

As I said earlier in a post, it is a shame the history of mathematics
never used three question marks for something like
1/3 = 0.33333??? rather than 0.33333.... for the question mark may
have alerted those in mathematics to define with precision the
borderline of finite with infinite much earlier than having to wait
until 2012.

[Virgil]

In article <i1d598here1tdrcoc...@4ax.com>,

JRStern <JRS...@foobar.invalid> wrote:

> >The function f is from N into the set of denumerable binary sequences.
> >So for each n in N, we have that f(n) is itself a function from N into
> >{0 1}.
>
> Then I cannot any longer recognize this as being about the diagonal
> argument. It is in some way shape or form about N and R, as is the
> diagonal argument. But it is a different story. I do not see that it
> includes the diagonal argument.

The original Cantor diagonal argument sowed that there was no surjection
from n the any set of all infinite binary sequences.

There are then proofs, though not as simple, that any such set of all
binary sequences can be bijected with the set of all reals.

Which allows one to conclude that there is no surjection from N to R.

Those who are ignorant of how to biject such a set of all binary
sequences with the set of reals may well be confused by their ignorance.

Like you apparently are!
--

[Jesse F. Hughes]

JRStern <JRS...@foobar.invalid> writes:

> On Thu, 01 Nov 2012 15:39:52 -0400, "Jesse F. Hughes"
> <je...@phiwumbda.org> wrote:
>
>>JRStern <JRS...@foobar.invalid> writes:
>>
>>> But we've already lost our clarity, what are we mapping to? If it's a
>>> complete list, then I thought it was "clear" that the diagonal doesn't
>>> work. If it's not a complete list because, um, just whyever you think
>>> that might be, then we're not REALLY even testing the thesis, are we?
>>>
>>> The point of contention then is whether this paper representation does
>>> or does not represent an uncountable set. But hey, we write aleph
>>> characters and claim they DO represent uncountable sets, but there
>>> seems to be something wrong, with this matrix? Just how is that
>>> determined?
>>
>>With all due respect, you seem to be taking the ramblings of Archimedes
>>Plutonium seriously. This is probably not a good idea.
>
> There just might be some kernel of truth in the ramblings, I certainly
> can't support everything I see there.

Yeah, well, good luck with that.

But, I can understand your initial desire -- a request for legitimate,
published counter-arguments regarding the validity of Cantor's theorem.
When you start trying to make sense of AP's silliness, you've fallen far
below that original mark.

>
>>Of course, each argument stands or falls on its merits, and so we
>>shouldn't reject mathematical arguments just because their proponents
>>also happen to claim, say, that the universe is a giant plutonium atom.
>>
>>That said, I daresay you're going to have problems finding any clearly
>>stated mathematical argument in AP's scratchings. There must surely be
>>a better source to defend than he.
>
> Thanks to the posters here, I believe I have located the point of the
> current issue as a question about exactly what the domain is supposed
> to be for the diagonal function, and whether that constitutes assuming
> the conclusion that is then supposed to follow from it.

Don't keep us hanging. Do you believe that you have found that the
theorem is invalid or not?

--
Jesse F. Hughes
"The way that she did what she did
when she did what she did to me
made me think of you." --- Delbert McClinton

[Jesse F. Hughes]

"Jesse F. Hughes" <je...@phiwumbda.org> writes:

> JRStern <JRS...@foobar.invalid> writes:
>
>> On Thu, 01 Nov 2012 15:39:52 -0400, "Jesse F. Hughes"
>> <je...@phiwumbda.org> wrote:
>>
>>>JRStern <JRS...@foobar.invalid> writes:
>>>
>>>> But we've already lost our clarity, what are we mapping to? If it's a
>>>> complete list, then I thought it was "clear" that the diagonal doesn't
>>>> work. If it's not a complete list because, um, just whyever you think
>>>> that might be, then we're not REALLY even testing the thesis, are we?
>>>>
>>>> The point of contention then is whether this paper representation does
>>>> or does not represent an uncountable set. But hey, we write aleph
>>>> characters and claim they DO represent uncountable sets, but there
>>>> seems to be something wrong, with this matrix? Just how is that
>>>> determined?
>>>
>>>With all due respect, you seem to be taking the ramblings of Archimedes
>>>Plutonium seriously. This is probably not a good idea.
>>
>> There just might be some kernel of truth in the ramblings, I certainly
>> can't support everything I see there.
>
> Yeah, well, good luck with that.

Just to be clear, you should realize that AP's meanderings include the
claim that 10^603 is infinite. Literally.

>> Thanks to the posters here, I believe I have located the point of the
>> current issue as a question about exactly what the domain is supposed
>> to be for the diagonal function, and whether that constitutes assuming
>> the conclusion that is then supposed to follow from it.
>
> Don't keep us hanging. Do you believe that you have found that the
> theorem is invalid or not?
--

"There are people [...] who think it's socially acceptable to level
accusations of mental illness in insulting exchanges to make
points[...] [They] are rather sick [them]selves, and in reality, are
sociopathic." --- James Harris, evidently a self-described sociopath

[Virgil]

In article <bi45985i991ui1pi1...@4ax.com>,

JRStern <JRS...@foobar.invalid> wrote:

> On Thu, 1 Nov 2012 02:10:38 -0700 (PDT), Zuhair <zalj...@gmail.com>
> wrote:
>
>
> >> >I understand your concern, sometimes the wording is indeed confusing.
> >> >I already rephrased this wording in my latest response to AP.
> >>
> >> >Lets be a little bit more clear.
> >>
> >> >Lets identify a real with a countably infinite binary digit sequence.
> >>
> >> OK.
> >>
> >> >Now we define the predicate "is a set of reals" in the following
> >> >manner:
> >>
> >> >S is a set of reals <-> (for all y. y in S -> y is a real).
> >>
> >> OK.
> >>
> >> >So S may be empty, or S might contain some but not all reals, or S
> >> >might contain all reals in which case it will be called the set of all
> >> >reals.
> >>
> >> OK.
> >>
> >> >Now our question can be modified to the following:
> >>
> >> >Suppose that S is a set of reals and Suppose that there exist a
> >> >bijection from N to S, then can S be the set of ALL reals?
> >>
> >> Let's move slowly here for my benefit.
> >>
> >> When we make the diagonal argument, is that a claim that there is a
> >> bijection from N to S?
>
> I would like to move this question again.

Your even asking that question shows your lack of comprehension of what
is being discussed.

>
> Also to clarify, to others' comments, whether we really need a
> bijection, or whether it is what - injection, surjection? Being
> careful (if it is not bijection) to state from which to which.
>

Cantor postulated having two sets, N and S, N being the set of naturals
and S being a set of all binary sequences, e.g., all mappings from N to
some two element set such as {m,w}, and some unspecified function, f,
from N to S.

By his 'diagonal' argument, Cantor showed that no such function, f,
could be surjective.

Later others modified his argument in a variety of ways to show also
that no mapping from N to R could be surjective.

The issues that you are trying to bring up all miss the essentials, and
are irrelevancies.
--

[netzweltler]

On 1 Nov., 20:40, Archimedes Plutonium

<plutonium.archime...@gmail.com> wrote:
> We start with Naturals and we list them and reverse them to obtain the
> bijection into the Reals. Since we use the ordering of the Naturals it
> may look as though it is not a bijection:
>
> 1 <--> .1
> 
10 <--> .01
> 11 <--> .11
> 100 <--> .001
> 101 <--> .101
> 110 <--> .011
> etc etc
>
> Now every Natural has a Real to correspond to it and the reason that
> is true is because both sets are All Possible Digit Arrangements and
> neither set has a hole in it.

The naturals correspond to the reals which end up in an infinite
number of 0's, right?

1 <--> .10000...
10 <--> .01000...
11 <--> .11000...
...

which naturals do represent the binary complement of the reals above?

? <--> .01111...
? <--> .10111...
? <--> .00111...
...

which natural does represent .010101(01)...?

--
netzweltler