Cantor's Proof FAQ
lugi...@gmail.com
The following is Cantor's Proof that there is no 1-1 correspondence
between the natural numbers and the real numbers between 0 and 1, also
called the uncountability of the reals in the interval (0,1):
1. Let A be any ordered list of real numbers between 0 and 1. In
other words, for each natural number x, there exists exactly one real
number A(x) between 0 and 1. Please note that the range of A does not
have to contain all real numbers.
2. Consider the real number f(A) such that the nth digit of f(A) is
equal to 5 if A(n)=4, and is equal to 4 if the A(n) does not equal 4.
3. f(A) cannot be an element of the range of A, since it differs in
the nth digit from the A(n).
4. Therefore, the range of A does not contain every real number. In
other words, A is not a complete list of reals.
5. Since A was arbitrary, there exists no ordered list of real
numbers, and thus the set of real numbers is uncountable.
Too many cranks try to refute this simple, elegant argument in the
following ways:
Q. Step 1 is invalid. There can be no ordered list of arbitrary real
numbers, only lists of specified real numbers. There has to exist some
kind of terminating effective procedure which specifies all the terms
on the list. This is because ordered lists are a type of potential
infinity, not actual infinity, which is impossible/too complicated.
A. Cantor's result pertains to lists of arbitrary real numbers. If
you don't believe in arbitrary real numbers, then why do you even care
to refute this theorem if it pertains to things you don't believe in?
You should just use definable real numbers, and leave stuff about
undefinable real numbers, which are absolutely essential to
mathematics, to the mathematicians who understand enough math to know
that they should embrace infinity.
Q. Step 2 is invalid. f(A) cannot be constructed, since f(A) is
designed to differ in the nth digit from A(n), and f(A) can't differ in
the nth digit from itself!
A. This is based on the misconception that Cantor's proof has to be a
proof by contradiction and must therefore assume for sake of argument
that the range of A must contain all real numbers. If you look at step
1, it explicitly rules out this possibility.
Q. Step 3 is invalid. f(A) can be an element of the range of A
because for m>n for all n,
f(A)=A(m).
A. This is based on the misconception that an infinitely large set
like the set of natural numbers must contain infinitely large members.
If you do believe that N contains infinitely large members, then
consider the set N* of all elements n of N such that n is not
infinitely large. Then N* is still infinitely large, since it has no
largest element, but N* still does not contain infinitely large
members. N is what N* is called in standard terminology.
Q. Step 5 is invalid. Just because A is incomplete, that doesn't mean
all lists of real numbers have to be incomplete. If you add f(A) to A,
you get a complete list of real numbers.
A. No you wouldn't, because you can apply the exact same argument for
A+f(A). In the beginning of the argument, we let A be arbitrary, so
the argument applies to ALL lists.
|-|erc
No need to use arbitrary lists, here's the actual list of all reals.
UTM(n, digit) mod 10 | n = 1,2,3...
When n gets to about 100 digits long, UTM(n) will compute any digit of pi,
it will compute e (for another value of n), just about every integer you can count to
will be computed, it will compute sin(0.5) and cos(sqrt(2)) and a host of other
formula you can put down to pen on paper. Now why are you all so intent
on destroying such a beautiful complete enumeration?
UTM(n, digit) mod 10 | n=1,2,3... computes every possible real number (between 0 and 1)
and maps them from the naturals.
the anti-diagonal real
UTM(digit, digit)+1 mod 10
can be shown not to occur anywhere on the list.
Let the anti-diagonal occur at some row m
UTM(m, digit) mod 10 = UTM(digit, digit)+1 mod 10
when m = digit
UTM(digit, digit) mod 10 = UTM(digit, digit)+1 mod 10
0 = 1 | mod 10
CONTRADICTION
Therefore, the anti-diagonal does not occur on the list.
Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
Notice how I correctly identified the problem after finding a contradiction rather
than assuming larger than infinite sets exist.
Herc
Newberry
Cantor's proof is based on the misconception that in an n x n matrix
the n can be infinite.
Virgil
"|-|erc" <h@r.c> wrote:
What Herc does is to assume that there are only countably many reals,
and use that assumption to prove there are only countably many reals.
In logical circles that is circular and therefore invalid.
|-|erc
> "|-|erc" <h@r.c> wrote:
> > Therefore, the anti-diagonal does not occur on the list.
> > Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
> >
> > Notice how I correctly identified the problem after finding a contradiction
> > rather than assuming larger than infinite sets exist.
> >
> > Herc
>
> What Herc does is to assume that there are only countably many reals,
> and use that assumption to prove there are only countably many reals.
>
> In logical circles that is circular and therefore invalid.
I don't prove there are countably many reals, I merely show that diagonalisation
does not prove there are uncountably many reals.
What does a UTM count to?
Every integer
Every rational
Every maths expression, real, rational, irrational that you can type in your calculator
Every possible sequence of digits (to infinite length) on a real
Really what's left?
Herc
Virgil
"|-|erc" <h@r.c> wrote:
> "Virgil" <vir...@comcast.net> wrote in
> > "|-|erc" <h@r.c> wrote:
> > > Therefore, the anti-diagonal does not occur on the list.
> > > Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
> > >
> > > Notice how I correctly identified the problem after finding a
> > > contradiction
> > > rather than assuming larger than infinite sets exist.
> > >
> > > Herc
> >
> > What Herc does is to assume that there are only countably many reals,
> > and use that assumption to prove there are only countably many reals.
> >
> > In logical circles that is circular and therefore invalid.
>
> I don't prove there are countably many reals, I merely show that
> diagonalisation
> does not prove there are uncountably many reals.
Cantor's first proof suffices.
>
> What does a UTM count to?
> Every integer
> Every rational
> Every maths expression, real, rational, irrational that you can type in
> your calculator
> Every possible sequence of digits (to infinite length) on a real
Where is one of these wondrous UMTs? If you are speaking of Universal
Turing Machines, none exist.
|-|erc
> "|-|erc" <h@r.c> wrote:
>
> > "Virgil" <vir...@comcast.net> wrote in
> > > "|-|erc" <h@r.c> wrote:
> > > > Therefore, the anti-diagonal does not occur on the list.
> > > > Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
> > > >
> > > > Notice how I correctly identified the problem after finding a
> > > > contradiction
> > > > rather than assuming larger than infinite sets exist.
> > > >
> > > > Herc
> > >
> > > What Herc does is to assume that there are only countably many reals,
> > > and use that assumption to prove there are only countably many reals.
> > >
> > > In logical circles that is circular and therefore invalid.
> >
> > I don't prove there are countably many reals, I merely show that
> > diagonalisation
> > does not prove there are uncountably many reals.
>
> Cantor's first proof suffices.
which one is that, UTM(digit, digit)+1 mod 10 doesn't fit, or the box that
contains the numbers of all the boxes that don't contain their own numbers?
> >
> > What does a UTM count to?
> > Every integer
> > Every rational
> > Every maths expression, real, rational, irrational that you can type in
> > your calculator
> > Every possible sequence of digits (to infinite length) on a real
>
> Where is one of these wondrous UMTs? If you are speaking of Universal
> Turing Machines, none exist.
Sure they do, in theory and practice, some of them are consise programs now.
Herc
lugi...@gmail.com
> the anti-diagonal real
> UTM(digit, digit)+1 mod 10
> can be shown not to occur anywhere on the list.
>
> Let the anti-diagonal occur at some row m
> UTM(m, digit) mod 10 = UTM(digit, digit)+1 mod 10
> when m = digit
> UTM(digit, digit) mod 10 = UTM(digit, digit)+1 mod 10
> 0 = 1 | mod 10
> CONTRADICTION
> Therefore, the anti-diagonal does not occur on the list.
> Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
>
know it is improperly specified?
What is your definition of "properly specified real?"
> Notice how I correctly identified the problem after finding a contradiction rather
> than assuming larger than infinite sets exist.
Instead of looking for a specific counterexample to Cantor's result,
examine the proof I provided above which works for an arbitrary lists
of reals. Try to find an error if you can.
>
> Herc
lugi...@gmail.com
lugi...@gmail.com
universal turing machines. They exist, Herc is just using them in his
obviously invalid refutation of Cantor's proof.
Arturo Magidin
I assume you are proposing this and asking for comments?
>The following is Cantor's Proof that there is no 1-1 correspondence
>between the natural numbers and the real numbers between 0 and 1, also
>called the uncountability of the reals in the interval (0,1):
You should add "strictly" before "between 0 and 1".
>1. Let A be any ordered list of real numbers between 0 and 1. In
>other words, for each natural number x, there exists exactly one real
>number A(x) between 0 and 1. Please note that the range of A does not
>have to contain all real numbers.
I would rephrase the clause after "In other words", to avoid
misinterpretation. I would say:
In other words, A is a function whose domain is the natural
numbers, and whose range is contained in (0,1); for each natural
number x, A(x) is a real number between 0 and 1.
>2. Consider the real number f(A) such that the nth digit of f(A) is
>equal to 5 if A(n)=4, and is equal to 4 if the A(n) does not equal 4.
Before doing this, you need to specify that A(x) will be given its
decimal representation, and in the case of dual representations, you
should explicitly pick one; I suggest specifying that you will use a
tail of zeros when there are two of them.
>3. f(A) cannot be an element of the range of A, since it differs in
>the nth digit from the A(n).
First, add that since the decimal representation of f(A) contains only
4s and 5s, it has one and only one decimal representation.
You also need to fix 3 a bit. For example, suppose that A(1) is
.50000000....
Then your definition of f(A) has the first decimal digit equal to
4. However, you cannot conclude from the fact that f(A) has first
decimal digit equal to 4 and A(1) has first decimal digit equal to 5
that A(1) and f(A) are different, because A(1) has a second decimal
representation which ->does<- start with 4. Rather, you need to invoke
the fact that the second digit of f(A) will not be a 9 and so f(A) is
not the "other" decimal representation of A(1).
This should be added ahead of time, when you say that f(A) has only
one decimal representation, and so if it differs from a number x in a
single digit of any decimal representation for x, then it will be
different from x.
>4. Therefore, the range of A does not contain every real number. In
>other words, A is not a complete list of reals.
Replace "real numbers" with "real number strictly between 0 and
1". I mean, we already knew it did not contain 3/2, after all.
--
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
======================================================================
Arturo Magidin
magidin-at-member-ams-org
Bob Kolker
>
> You should add "strictly" before "between 0 and 1".
The open parens indicate the open interval. End points not included.
This is standard mathematical notation. [0,1] is the interval with the
endpoints.
Bob Kolker
Arturo Magidin
Yes, I know.
Since I assumed he was writing it up asking for suggestions, I noted a
few places where he might want to clarify and tighten the
presentation. That was one. He wrote:
The following is Cantor's Proof that there is no 1-1
correspondence between the natural numbers and the real numbers
between 0 and 1, also called the uncountability of the reals in
the interval (0,1)
Now, the proof is by no means restricted to the real numbers in (0,1);
it works equally well if we allow the list to include 0 or 1. So
saying that "it is also called" the uncountability of reals in (0,1)
is a bit of a fudge. Which is why I suggested that, when he first says
what he is doing, he write explicitly "strictly", so that it is clear
that what he says it is also called is indeed the same thing.
Virgil
"|-|erc" <h@r.c> wrote:
> "Virgil" <vir...@comcast.net> wrote ...
> > "|-|erc" <h@r.c> wrote:
> >
> > > "Virgil" <vir...@comcast.net> wrote in
> > > > "|-|erc" <h@r.c> wrote:
> > > > > Therefore, the anti-diagonal does not occur on the list.
> > > > > Therefore, UTM(digit, digit)+1 mod 10 is an improperly specified real.
> > > > >
> > > > > Notice how I correctly identified the problem after finding a
> > > > > contradiction
> > > > > rather than assuming larger than infinite sets exist.
> > > > >
> > > > > Herc
> > > >
> > > > What Herc does is to assume that there are only countably many reals,
> > > > and use that assumption to prove there are only countably many reals.
> > > >
> > > > In logical circles that is circular and therefore invalid.
> > >
> > > I don't prove there are countably many reals, I merely show that
> > > diagonalisation
> > > does not prove there are uncountably many reals.
> >
> > Cantor's first proof suffices.
>
> which one is that
Look it up.
> > > What does a UTM count to?
> > > Every integer
> > > Every rational
> > > Every maths expression, real, rational, irrational that you can type in
> > > your calculator
> > > Every possible sequence of digits (to infinite length) on a real
> >
> > Where is one of these wondrous UMTs? If you are speaking of Universal
> > Turing Machines, none exist.
>
> Sure they do, in theory and practice, some of them are consise programs now.
Is there any UTM currently in existence which is guaranteed to take ANY
finite program, however large, and run it?
What about a program that allows one pass over the contents of the tape
then requires erasing of everything on the tape before proceeding to
wtite something?
Newberry
Step 4. Since f(A) is not in the list and the list contains all the
reals, it clearly follows that f(A) does not exist.
MoeBlee
> Step 4. Since f(A) is not in the list and the list contains all the
> reals, it clearly follows that f(A) does not exist.
Just personally, I'd prefer not to use the notation 'f(A)'. But the
object described in his argument is PROVEN to exist. Moreover, it's
simply a non sequitur to assert that all real numbers are in the list.
MoeBlee.
|-|erc
If you argue UTMs don't exist, then TMs don't exist, and neither does any construction
in mathematics, circles don't exist, pi doesn't exist! Computers can emulate UTMs up to a point.
We can merely define a finite UTM, FUTM which comes with a tolerance rating and
then we make sure we are using the right machine for the right input whenever we are
required to physically run the programs, which for the sake of countability theory may never occur.
Herc
Virgil
"Newberry" <newb...@ureach.com> wrote:
Or that the list does not actually contain all reals.
When a contradiction appears, it may be because any one of the
assumptions leading to is is wrong. In order to conclude a particular
assumtion as the cause, one must demonstrably exclude all others.
|-|erc
> > > Which step of the proof I provided above do you find in error and why?
> >
> > Step 4. Since f(A) is not in the list and the list contains all the
> > reals, it clearly follows that f(A) does not exist.
>
> Or that the list does not actually contain all reals.
>
> When a contradiction appears, it may be because any one of the
> assumptions leading to is is wrong. In order to conclude a particular
> assumtion as the cause, one must demonstrably exclude all others.
Given 2 options, a formula is wrong, or a hyperinfinite set must exist
go with the simpler.
Herc
Newberry
ABSOLUTELY!! Concluding that the list does not contain all the reals
leads to another contradiction: actual infinity. Therefore the
antidiagonal real does not exist any more than r < 3.14 & r > 3.14.
Virgil
"|-|erc" <h@r.c> wrote:
The contradiction arises out of two claimed assumptions:
(1) The list contains all real numbers, and
(2) The number constructed is not on the list.
Since (1) is no part of the original diagonal proof, it is irrelevant.
The original proof merely assumes one has a list of reals, with no other
requirement than that it be a list of reals. There is no original
assumption that it need contains all reals.
Then (2) concludes that the diagonal is not in THAT list, with no
contradiction either stated or implied.
The theorem has established that, given an arbitrary list of reals,
there is a real not in that list.
Those who attempt to require proof by contradiction are apparently
unaware that the original proof did not do so.
|-|erc
That's a misleading way to prove it, because if its arbitrary then it MAY
be complete, so you have to consider that possibility.
STEP 1
GIVEN AN ARBITRARY LIST OF REALS
STEP 2a
THE LIST IS NOT COMPLETE
STEP 3a
THE LIST IS COMPLETE
STEP 4
CONCLUSION
Isn't it easier just to make step 1 assume a complete list of reals and use
proof by contradiction?
Given that the assumption in Cantors proof is a complete list of reals, it
changes the fact that your new construction must fit into the list (as its complete)
and therefore has to be properly encoded, which is where your contradiction
comes from.
Summary:
You can't assume an arbitrary list without assuming a complete list as one option.
You can't asssume a complete list and then ignore that assumption making new
ill defined constructions.
Herc
lugi...@gmail.com
1: "Please note that the range of A does not need to contain all real
numbers." I am just starting with an arbitrary list A of reals,
whether it contains all reals or just some of them. I am then
constructing a real number f(A) that is not on the list. Since A was
arbitrary, for each list, there exists a real number the list does not
contain. I do not need to assume that A contains all reals. So it
does not "clearly follow" that f(A) does not exist. f(A) does exist.
Do you have any other objections about the proof?
Newberry
george
lugi...@gmail.com wrote:
> Cantor's Proof FAQ
> The following is Cantor's Proof that there is no 1-1 correspondence
> between the natural numbers and the real numbers between 0 and 1, also
> called the uncountability of the reals in the interval (0,1):
This is incredibly stupid.
In the first place, you have titled the thread a FAQ, yet there
are no Q's (questions), Frequently Asked or otherwise, in it.
In the second place, Cantor's theorem IS NOT ABOUT
uncountability of the reals. It is about unmappability
of a set onto its powerset. Simply understanding THAT
would ELIMINATE At Least Half of the actually Frequently
Asked Questions.
lugi...@gmail.com
> lugi...@gmail.com wrote:
> > Cantor's Proof FAQ
> > The following is Cantor's Proof that there is no 1-1 correspondence
> > between the natural numbers and the real numbers between 0 and 1, also
> > called the uncountability of the reals in the interval (0,1):
>
> This is incredibly stupid.
> In the first place, you have titled the thread a FAQ, yet there
> are no Q's (questions), Frequently Asked or otherwise, in it.
>
crackpots.
> In the second place, Cantor's theorem IS NOT ABOUT
> uncountability of the reals. It is about unmappability
> of a set onto its powerset. Simply understanding THAT
> would ELIMINATE At Least Half of the actually Frequently
> Asked Questions.
diagonal argument that there are more real numbers than natural
numbers. This is of course provable from Cantor's theorem, but Cantor
proved it separately using his diagonal argument.
Dave Seaman
> lugi...@gmail.com wrote:
>> Cantor's Proof FAQ
>> The following is Cantor's Proof that there is no 1-1 correspondence
>> between the natural numbers and the real numbers between 0 and 1, also
>> called the uncountability of the reals in the interval (0,1):
> This is incredibly stupid.
> In the first place, you have titled the thread a FAQ, yet there
> are no Q's (questions), Frequently Asked or otherwise, in it.
By my count, the OP had 55 lines. The first 19 contained the proof, and
the rest were Q and A.
> In the second place, Cantor's theorem IS NOT ABOUT
> uncountability of the reals. It is about unmappability
> of a set onto its powerset. Simply understanding THAT
> would ELIMINATE At Least Half of the actually Frequently
> Asked Questions.
All of the objections that were addressed in the Q and A could be
rephrased to apply to the proof involving powersets.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
george
lugi...@gmail.com wrote:
> Instead of looking for a specific counterexample to Cantor's result,
> examine the proof I provided above which works for an arbitrary lists
> of reals. Try to find an error if you can.
The "error" is simply that because your original arbitrary
list is infinite, "computing its anti-diagonal" must take an
infinite amount of time, and must therefore never finish,
and therefore your decision to instruct someone to compute
and use the RESULT of such a computation is simply inadmissible:
Itsimply can't be performed in practice.
That this is NOT sufficient to make "anti-diagonal of a denumerably
long list" something "incorrectly specified" is NOT understood by
by the frequent askers of this question.
Randy Poe
george wrote:
> lugi...@gmail.com wrote:
> > Instead of looking for a specific counterexample to Cantor's result,
> > examine the proof I provided above which works for an arbitrary lists
> > of reals. Try to find an error if you can.
>
> The "error" is simply that because your original arbitrary
> list is infinite, "computing its anti-diagonal" must take an
> infinite amount of time, and must therefore never finish,
There is no requirement to compute it.
The question is does it exist. Does it have an n-th digit for
every n.
Do I need to compute the trillion-th digit of pi to conclude
that it has a trillion-th digit?
- Randy
Virgil
"george" <gre...@cs.unc.edu> wrote:
Since for any given number listed, one can show in finite time that the
diagonal is not equal to that number, you objection fails.
george
> george wrote:
> > The "error" is simply that because your original arbitrary
> > list is infinite, "computing its anti-diagonal" must take an
> > infinite amount of time, and must therefore never finish,
>
> There is no requirement to compute it.
Randy Poe wrote:
Of course there is, in "Their" opinion. "They" can always insist that
if it's provably not computable, THAT CONSTITUTES
a proof that it provably doesn't exist. Short of that,
they could insist that they want to be constructive and
that until YOU can compute it, YOU haven't proved that
it must -- OR EVEN CAN -- exist.
>
> The question is does it exist.
Obviously, they say it doesn't.
> Does it have an n-th digit for every n.
Each of those n's is finite.
The overall thing is infinite. Unless you know
that first-order logic is compact, this MIGHT make
a difference. THEY think it makes a difference.
> Do I need to compute the trillion-th digit of pi to conclude
> that it has a trillion-th digit?
Obviously, if Pi doesn't exist, then it doesn't contain any digits
at all, trillionth or otherwise. And if your contention is that Pi
is infinitely long, that for EVERY one of infinitely many n, there
is an nth digit of Pi, AND THAT LEADS TO A CONTRADICTION,
then "they" are perfectly free to pin the "error" on your claim that
every digit of Pi exists. The fact that the contradiction is actually
due to something else entirely will be quite lost on them.
george
There IS NO given number listed.
JEEzus. There is an INFINITE LIST of numbers listed.
> one can show in finite time that the
> diagonal is not equal to that number,
It does NOT follow from that you can anti-diagonalize ANYthing
INfinite with only finite time.
> you objection fails.
But not due to any reason have anything to do with any finite case
(as you wrongly claim here).
T
Randy Poe
george wrote:
> > george wrote:
> > > The "error" is simply that because your original arbitrary
> > > list is infinite, "computing its anti-diagonal" must take an
> > > infinite amount of time, and must therefore never finish,
> >
> > There is no requirement to compute it.
>
>
> Randy Poe wrote:
> Of course there is, in "Their" opinion.
Who are "they". Certainly not anyone advancing Cantor's
proof of the uncountability of reals. That is an existence
proof, not a claim of computability.
> "They" can always insist that
> if it's provably not computable, THAT CONSTITUTES
> a proof that it provably doesn't exist.
No, I've never heard any mathematician claim that the
computable reals are the only ones that exist. In fact,
most reals are known to be uncomputable.
> Short of that,
> they could insist that they want to be constructive and
> that until YOU can compute it, YOU haven't proved that
> it must -- OR EVEN CAN -- exist.
The proof shows that there exists a number which is
different from the n-th number on the list, for every n. It
does that without explicit computation. There is no "they"
insisting on computability. You are arguing with a strawman.
- Randy
lugi...@gmail.com
> george wrote:
> > > george wrote:
> > > > The "error" is simply that because your original arbitrary
> > > > list is infinite, "computing its anti-diagonal" must take an
> > > > infinite amount of time, and must therefore never finish,
> > >
> > > There is no requirement to compute it.
> >
> >
> > Randy Poe wrote:
> > Of course there is, in "Their" opinion.
>
> Who are "they". Certainly not anyone advancing Cantor's
> proof of the uncountability of reals. That is an existence
> proof, not a claim of computability.
>
> > "They" can always insist that
> > if it's provably not computable, THAT CONSTITUTES
> > a proof that it provably doesn't exist.
>
> No, I've never heard any mathematician claim that the
> computable reals are the only ones that exist.
rather stupid philosophy in my opinion, believes that only what can be
"constructed," or computed, actually exist.
lugi...@gmail.com
> > george wrote:
> > > The "error" is simply that because your original arbitrary
> > > list is infinite, "computing its anti-diagonal" must take an
> > > infinite amount of time, and must therefore never finish,
> >
> > There is no requirement to compute it.
>
>
> Randy Poe wrote:
> Of course there is, in "Their" opinion. "They" can always insist that
> if it's provably not computable, THAT CONSTITUTES
> a proof that it provably doesn't exist. Short of that,
> they could insist that they want to be constructive and
> that until YOU can compute it, YOU haven't proved that
> it must -- OR EVEN CAN -- exist.
>
Basically it boils down to this: if someone believes that uncomputable
real numbers do not exist, then the statement that the uncomputable
real numbers are uncountably infinite is vacuously true. For this
reason, constructivists, shouldn't have a problem with Cantor's Proof.
Albrecht
lugi...@gmail.com schrieb:
> george wrote:
> > lugi...@gmail.com wrote:
> > > Cantor's Proof FAQ
> > > The following is Cantor's Proof that there is no 1-1 correspondence
> > > between the natural numbers and the real numbers between 0 and 1, also
> > > called the uncountability of the reals in the interval (0,1):
> >
> > This is incredibly stupid.
> > In the first place, you have titled the thread a FAQ, yet there
> > are no Q's (questions), Frequently Asked or otherwise, in it.
> >
> I do answer questions about it, frequently asked by cranks and
> crackpots.
> > In the second place, Cantor's theorem IS NOT ABOUT
> > uncountability of the reals. It is about unmappability
> > of a set onto its powerset. Simply understanding THAT
> > would ELIMINATE At Least Half of the actually Frequently
> > Asked Questions.
> When did I say this was about Cantor's theorem? This is about Cantor's
> diagonal argument that there are more real numbers than natural
> numbers.
This statement is totally nonsensical and reflects the common
misunderstanding of what is derivable from Cantor's diagonal argument.
There is nothing more than infinitely many and "infinitely many" can't
stand for anything other than "unfinishable", "unrealisable",
"endless", "infinit".
If I must be a crank or crackpot to know this, I really like it to be
one.
>From the math fact that there are algorithms which give convergent
sequences deriving the argument that an arbitrary infinite list of real
numbers could define a new real number is just a logical short-circuit
and shows nothing more than the ill definition of the real numbers.
The unmasking aspect of this thread is, that exactly this people, who
doesn't understand the logical implications of the diagonal argument,
think they are able to explain the critics why they should be wrong.
Best regards
Albrecht S. Storz
Virgil
"Albrecht" <albs...@gmx.de> wrote:
> There is nothing more than infinitely many and "infinitely many" can't
> stand for anything other than "unfinishable", "unrealisable",
> "endless", "infinit".
> If I must be a crank or crackpot to know this, I really like it to be
> one.
>
You appear to have succeeded in this aim beyond your wildest dreams.
MoeBlee
> The unmasking aspect of this thread is, that exactly this people, who
> doesn't understand the logical implications of the diagonal argument,
> think they are able to explain the critics why they should be wrong.
Let us know when you learn the axiom of extensionality, which you are
presently incapable of properly stating. We can take it from there.
MoeBlee
Jesse F. Hughes
> Albrecht wrote:
[...]
> Let us know when you learn the axiom of extensionality, which you are
> presently incapable of properly stating. We can take it from there.
When did he try to state the axiom? My Google skills failed me.
Thanks.
--
"At the Microsoft-sponsored cocktail reception in the Galaxy Ballroom
that evening, Robert Dees urges us 'to network on behalf of the people
of Iraq,'"
-- Naomi Klein reports on Microsoft's efforts to further democracy.
MoeBlee
Jesse F. Hughes wrote:
> "MoeBlee" <jazz...@hotmail.com> writes:
>
> > Albrecht wrote:
>
> [...]
>
> > Let us know when you learn the axiom of extensionality, which you are
> > presently incapable of properly stating. We can take it from there.
>
> When did he try to state the axiom? My Google skills failed me.
I see now that it was Blumschein not Albrecht. My mistake. My apologies
to Albrecht for the misattribution.
Eckard Blumschein post Wed, Dec 6 2006 6:50 am from "Cantor Confusion"
thread:
[begin post]
On 12/5/2006 10:54 PM, Virgil wrote:
> In article <4575A562.60...@et.uni-magdeburg.de>,
> Eckard Blumschein <blumsch...@et.uni-magdeburg.de> wrote:
>> Blissful ignorance of mathematicians does not utter complains if the
>> axiom of (possibly infinite) extensionality claims the existence of a
>> set which has to include all of its elements.
> EB should read what axioms of extensionality actually says before
> pontificating.
Roughly speaking, it just claims that a set is unambiguously determined
by its elements. If i recall correctly A=B<-->(A in B and B in A)
Perhaps the Delphi oracle provided less possibilities of tweaked
interpretation betwixed and between potential and actual infinity.
[end post]
MoeBlee
lugi...@gmail.com
> Obviously, if Pi doesn't exist, then it doesn't contain any digits
> at all, trillionth or otherwise. And if your contention is that Pi
> is infinitely long, that for EVERY one of infinitely many n, there
> is an nth digit of Pi, AND THAT LEADS TO A CONTRADICTION,
> then "they" are perfectly free to pin the "error" on your claim that
> every digit of Pi exists. The fact that the contradiction is actually
> due to something else entirely will be quite lost on them.
contradiction. If you look at my version, you will see that I use no
proof by contradiction.
But even if we work with a proof by contradiction of the uncountability
of the real's like Cantor's original proof was, their objections would
still not be valid. Since the contradiction cannot be derived without
the assumption of countably many real numbers, there must therefore be
some problem with this assumption and not any of the other ones used.
But forget about this proof by contradiction version, since the cranks
find more wrong with that proof than with proofs which are not by
contradiction.
|-|erc
TRUE
> If I must be a crank or crackpot to know this, I really like it to be
> one.
>
> >From the math fact that there are algorithms which give convergent
> sequences deriving the argument that an arbitrary infinite list of real
> numbers could define a new real number is just a logical short-circuit
> and shows nothing more than the ill definition of the real numbers.
Logical short circuit is an excellent description of what the proof shows.
>
> The unmasking aspect of this thread is, that exactly this people, who
> doesn't understand the logical implications of the diagonal argument,
> think they are able to explain the critics why they should be wrong.
>
>
> Best regards
> Albrecht S. Storz
Anyone who jumps to the conclusion, LOOK A NEW R, MUST BE FROM
A LARGER THAN INFINITY SET OF REALS, is looopy.
Herc
Rupert
george wrote:
> > george wrote:
> > > The "error" is simply that because your original arbitrary
> > > list is infinite, "computing its anti-diagonal" must take an
> > > infinite amount of time, and must therefore never finish,
> >
> > There is no requirement to compute it.
>
>
> Randy Poe wrote:
> Of course there is, in "Their" opinion. "They" can always insist that
> if it's provably not computable, THAT CONSTITUTES
> a proof that it provably doesn't exist. Short of that,
> they could insist that they want to be constructive and
> that until YOU can compute it, YOU haven't proved that
> it must -- OR EVEN CAN -- exist.
>
But if you require reals to be computable, you should also require
bijections to be computable. And then the argument still works: there
is no effective enumeration of the set of computable reals.
Gc
lugi...@gmail.com kirjoitti:
> Cantor's Proof FAQ
> The following is Cantor's Proof that there is no 1-1 correspondence
> between the natural numbers and the real numbers between 0 and 1, also
> called the uncountability of the reals in the interval (0,1):
> 1. Let A be any ordered list of real numbers between 0 and 1. In
> other words, for each natural number x, there exists exactly one real
> number A(x) between 0 and 1. Please note that the range of A does not
> have to contain all real numbers.
> 2. Consider the real number f(A) such that the nth digit of f(A) is
> equal to 5 if A(n)=4, and is equal to 4 if the A(n) does not equal 4.
> 3. f(A) cannot be an element of the range of A, since it differs in
> the nth digit from the A(n).
> 4. Therefore, the range of A does not contain every real number. In
> other words, A is not a complete list of reals.
> 5. Since A was arbitrary, there exists no ordered list of real
> numbers, and thus the set of real numbers is uncountable.
Why is the existence of "ordered list of reals" equivalent to that
there is a unique real number for each natural? Is a list always
countable by definition? I can imagine a uncountable ordered list of
all the reals, which I can`t diagonalize, of course, because it is not
a square list.
Virgil
"|-|erc" <h@r.c> wrote:
This pair could serve the classical exemplar of folie a deux"
Virgil
"Gc" <Gcu...@hotmail.com> wrote:
The simpler phrase "list", usually is defined, or at least understood,
to mean a surjection from the set of naturals to the set of objects in
question, with the implied order of the first natural mapped to each
image value. When in doubt, it should be qualified as a countable and
well ordered list.
|-|erc
> > Anyone who jumps to the conclusion, LOOK A NEW R, MUST BE FROM
> > A LARGER THAN INFINITY SET OF REALS, is looopy.
> >
> > Herc
>
> This pair could serve the classical exemplar of folie a deux"
Atleast I can recognise a dodgy proof when I see one.
Now let's play a game, I'll start:
Here is the complete list of all reals between 0 and 1.
UTM(n, digit) mod 10 n=1,2,3...
Now you give me a computable description of any real you think is missing?
Hint: according to your proof you can find extra reals.
Herc
Virgil
"|-|erc" <h@r.c> wrote:
If you have such a list, then my number is:
If your nth number has nth digit 6, mine has 5 otherwise mine has 6.
|-|erc
> "|-|erc" <h@r.c> wrote:
> >
> > Now let's play a game, I'll start:
> >
> > Here is the complete list of all reals between 0 and 1.
> > UTM(n, digit) mod 10 n=1,2,3...
> >
> > Now you give me a computable description of any real you think is missing?
> > Hint: according to your proof you can find extra reals.
> >
> > Herc
>
> If you have such a list, then my number is:
> If your nth number has nth digit 6, mine has 5 otherwise mine has 6.
BZZT! If its computable it must occur somewhere in UTM(n), i.e. there
is some Turing Machine TM-n that computes it.
Let your real be computed by TM-m, then each digit is computed by
UTM(m, digit) mod 10
UTM(m, n) mod 10 = if UTM(n,n)mod10=6 then 5 else 6
But when n=m
UTM(m,m)mod10 = if UTM(m,m)mod10=6 then 5 else 6
If UTM(m,m)mod10 =/= 6 then 5=6 contradiction
If UTM(m,m)mod10 = 6 then 6=5 contradiction
Therefore there is no computable method to make your suggested real.
Actually I think I need a further condition for other UTMs using different
encoding schemes, that none of them will represent the antidiagonal which
I think can be shown since every possible TM will map to another finite
TM in a different UTM scheme. That is, I proved it for UTM(natural, digit),
but I need to prove it for UTMa, UTMb, UTMc... for any possible UTM.
Herc
Ross A. Finlayson
> Do you have any other objections about the proof?
Maybe you might find something deeper in why it's not true.
Consider the antidiagonal argument and why it doesn't work in base two,
three, one (tally marks) or infinity, where the cases of one and
infinity don't lead to real numbers as they are "standardly" defined.
In comparing an alphabet of of four symbols to an alphabet of, say,
two, given one symbol twice as many possibilities exist.
4^1 = 2 * 2^1
Similarly, for ten symbol there are 2^10 many possible values,
4^10 = 2^10 * 2^10
So then there are questions about why if the binary numbers are
sufficient to represent the real numbers, why the antidiagonal argument
_can not be used_ to show the reals in that representation uncountable,
because of dual representation.
(Here Cantor's first or nested intervals is a separate consideration:
well-order the reals.)
The real number, for convenience between zero and one, is represented
as an infinite sequence of these symbols.
4^100 = 2^100 * 2^100
For a sequence of a hundred elements, there are 2^100 more values
representable in base 4 than base 2.
Basically, where as you should know I think the natural/unit
equivalency function is an order preserving bijection between the
naturals and unit interval of reals, i.e. n_i < n_j => EF(n_i) <
EF(n_j), the reals in that image are not standard where the impulse
function is useful, basically I then wonder how I can assuage my
mathematical conscience about the antidiagonal argument, in bases b: 3
< b < oo.
This is where representing a real number as a sequence of post-radix
integral moduli, generally called decimal digits as they are in base
ten, seems a reasonable thing to do, and that while I might never be
able to enumerate all the digits of pi, neither could I enumerate all
the finite counting integers, and I know they all exist.
For any initial segment of the sequence, compared to any other, it is
easy to see the order, the evaluation of the comparator, that they
would have in the natural ordering of the reals. (That the reals
numbers have a natural ordering is not necessarily obvious, accepted,
they have a total linear ordering, except for the vague fugue, and are
said by some to be "projectively well-ordered.") There are cases where
the initial segments could not be distinguishable, in the cases that
they are equal, or, of dual representation.
That's where I'm leading with the notion of varying representations:
that in a higher base, it doesn't necessarily represent more real
numbers, and where it can't, then it must make up somewhere late in the
infinite sequence for in the beginning, as the number of elements in
the sequence diverges, the representation in the higher base has more
possible values for discrete utterances of the symbol.
Consider for example the number, zero. In binary randomly the first
digit is zero or one, at about even odds, and etcetera. So, the
probability of a random 100 digit sequence in base two being zero is 1
/ 2^100. The same value in base 4, of a random 50 digit sequence, 1/
2^100. Now, zero is certainly a regularly occurring number, as nobody
would notice if zero was added to everything many times, but in the
base 4 sequence at 100 digits it is 2^50 times less likely to occur, as
a random value.
(I'm reminded of my recent explanation of a uniform probability
distribution over the naturals, bijecting N^N to R and sampling real
numbers, at uniform random. Why is it not contradicted?)
So, I'm leading towards some notion that either the base two sequence
must combinatorially explode at infinity to have as many sequences as
in base four, or, base four combinatorially implodes, at infinity, to
base two, and even to unary, where an infinite alphabet would only
require one symbol.
The binary symbol in a way has half the precision of the base four
symbol, where precision is generally indicated as the number of
symbols, eg 32-bit precision, four significant (decimal) digits.
They're also smaller.
Skolemize, it's countable. N^G and R are equivalent and defined by
their elements. There's no maximal ordinal to force there being a
model of ZF in ZF, no universe. ZF is the Russell set, there is and is
not, ZF is inconsistent.
Look at how you readily integrate. That's not the area under the curve
for any finite number of rectangles inscribed there, for curvy curves,
and where it's infinitely many, measure is preserved. It _is_ the area
under the curve, the limit _is_ the sum. Newton said so, Leibniz said
so, people today still call it infinitesimal analysis, and there is
iota in the reals. Zeno arrives.
Transfinite cardinals aren't used for things. There are a variety of
mathematical functions that work and use infinitesimals in a way that
is best explained by the naturals bijecting to the unit interval of
reals. They could only do so in a very specific way, in fact the
natural way. The poor things are discriminated against as "not real
functions", but they are quite real.
How do all those who cavalierly, and correctly, integrate and
differentiate with infinitesimals and sums over the infinities of
infinitesimals in units come to terms about that with their
mathematical conscience? It's not necessary, they already do.
You don't get it.
Ross
lugi...@gmail.com
function giving each natural number a unique real number.
Andrew Usher
> lugi...@gmail.com wrote:
>
> > Do you have any other objections about the proof?
>
> Maybe you might find something deeper in why it's not true.
That's your first mistake, Ross!
> (Here Cantor's first or nested intervals is a separate consideration:
> well-order the reals.)
Show me a well ordering of the reals. You can't prove that one exists.
Even theories that do ADMITT that the reals are uncountable.
> Skolemize, it's countable.
Cantor's theorem (and diagonalisation argument) is a theorem of
informal (and second-order) logic where Lowenheim-Skolem doesn't apply,
so you can't use it here.
> How do all those who cavalierly, and correctly, integrate and
> differentiate with infinitesimals and sums over the infinities of
> infinitesimals in units come to terms about that with their
> mathematical conscience? It's not necessary, they already do.
Calculus doesn't require infeinitesimals. That's been known since
Cauchy, if not since Newton!
Andrew Usher
lugi...@gmail.com
Rupert wrote:
> george wrote:
> > > george wrote:
> > > > The "error" is simply that because your original arbitrary
> > > > list is infinite, "computing its anti-diagonal" must take an
> > > > infinite amount of time, and must therefore never finish,
> > >
> > > There is no requirement to compute it.
> >
> >
> > Randy Poe wrote:
> > Of course there is, in "Their" opinion. "They" can always insist that
> > if it's provably not computable, THAT CONSTITUTES
> > a proof that it provably doesn't exist. Short of that,
> > they could insist that they want to be constructive and
> > that until YOU can compute it, YOU haven't proved that
> > it must -- OR EVEN CAN -- exist.
> >
>
> But if you require reals to be computable, you should also require
> bijections to be computable. And then the argument still works: there
> is no effective enumeration of the set of computable reals.
>
Albrecht
lugi...@gmail.com schrieb:
There is also no list of all the things in kosmos.It must be
uncountable many.
Is it really so hard to understand that you are only able to establish
a bijection if both sets or classes are well-orderable or unique
element-to-element-relations are establishable?
If two infinite classes are not bijectable there is no lack of elements
in one of the class since infinite many are enough for all what you
need.
If you put the antidiagonal at the end of the Cantor-list you will get
a new antidiagonal, right. In the moment you get the new, it is at the
end of the new list. Again and again. Who wins? Every list produces a
new antidiagonal, every antidiagonal produces a new list.
Who will decide what happens in the end - since there is no end.
If you switch a lamp infinitely many times on and off, which status may
the lamp have at the end (when the antidiagonal is ready)?
MoeBlee
A 'list' is meant, in this context, to be a function on a countable
ordinal - a finite list being a function on a natural number and a
denumerable list being a function on omega. (Of course, it is trivial
to show that the set of real numbers is not finite, so we need only
consider the denumerable case.) There are functions on uncountable
ordinals, which may be called 'uncountable sequences' or 'uncountable
lists' if you want to call them that. But they are not of issue in the
proof that the set of real numbers is uncountable. To prove that the
set of real numbers is uncountable, since it is trivial to show it is
not finite, it is sufficient to prove that there does not exist a
denumerable list of the set of real numbers. I.e., there is no function
on omega such that the set of real numbers is a subset of the range of
that function.
As to having the set of real numbers as a subset of the range of a
function on an uncountable ordinal, if I'm not mistaken that requires
that we have proven (usually via the axiom of choice) the well ordering
theorem (perhaps someone can verify or qualify this point).
MoeBlee
george
> > This is incredibly stupid.
> > In the first place, you have titled the thread a FAQ, yet there
> > are no Q's (questions), Frequently Asked or otherwise, in it.
>
> By my count, the OP had 55 lines. The first 19 contained the proof, and
> the rest were Q and A.
I blame the evil Google interface. I get a left column and a right
column about unrelated junk. The article is the middle column
only and because it is narrowed from both sides, this took more
than a screenful (more than 30 lines) instead of 19. I just thought it
was over.
> > In the second place, Cantor's theorem IS NOT ABOUT
> > uncountability of the reals. It is about unmappability
> > of a set onto its powerset. Simply understanding THAT
> > would ELIMINATE At Least Half of the actually Frequently
> > Asked Questions.
>
> All of the objections that were addressed in the Q and A could be
> rephrased to apply to the proof involving powersets.
That's hardly obvious. The first objection was to an arbitrary list
of reals. Translated, it would have to be to an arbitrary infinite
set. It is much harder to object to that when THERE IS AN EXPLICIT
AXIOM of ZFC that SAYS THERE IS some arbitrary infinite set.
george
lugi...@gmail.com wrote:
> Q. Step 1 is invalid. There can be no ordered list of arbitrary real
> numbers,
That's not what it says; it says an arbitrary ordered list of real
numbers.
The numbers aren't arbitrary: THE LIST is arbitrary. In other words,
it can be ANY list, INCLUDING the one that you say must be specified by
some rule instead of being arbitrary. This works on EVERY list,
arbitrary
OR NOT.
> only lists of specified real numbers.
IT DOESN'T MATTER: this works ON THEM, TOO.
> There has to exist some
> kind of terminating effective procedure which specifies all the terms
> on the list.
THat's fine. THis proof works if that procedure exists, AND IT ALSO
WORKS IF IT DOESN'T. This JUST DOESN'T MATTER AT ALL.
> A. Cantor's result pertains to lists of arbitrary real numbers. If
> you don't believe in arbitrary real numbers, then why do you even care
> to refute this theorem if it pertains to things you don't believe in?
In the first place, this is still mistaken. The quantifier is over
arbitrary
lists of real numbers, NOT arbitrary real numbers. You may have
arbitrary
real numbers if you like, but if you don't, THAT'S FINE. This works
for
NON-arbitrary real numbers too. The point is simply that it can be
ANY list, that this works for EVERY list. If "arbitrary" scares you
then just replace it with "any and every" everywhere you see it.
> Q. Step 2 is invalid. f(A) cannot be constructed, since f(A) is
> designed to differ in the nth digit from A(n), and f(A) can't differ in
> the nth digit from itself!
This is just stupid. Nobody is alleging that f(A) differs anywhere
from itself. It only has to differ from A(n). For this to lead to a
problem with f(A) differeing from itself would require f(A) to EQUAL
A(n) somewhere. SINCE IT DOESN'T, this is not a contradiction.
> A. This is based on the misconception that Cantor's proof has to be a
> proof by contradiction and must therefore assume for sake of argument
> that the range of A must contain all real numbers.
Well, yes, f(A) would have to equal A(n), for some n, if the range of f
contained every real number. But it doesn't.
> Q. Step 3 is invalid. f(A) can be an element of the range of A
> because for m>n for all n,
> f(A)=A(m).
> A. This is based on the misconception that an infinitely large set
> like the set of natural numbers must contain infinitely large members.
No, it isn't. It's based on failure to understand what "anti-diagonal"
means. Obviously, f(A)does NOT =A(m) because f(A)
differs from A(m) in the mth digit. This has nothing to do with
its differing from A(n) in the nth digit.
> Q. Step 5 is invalid. Just because A is incomplete, that doesn't mean
> all lists of real numbers have to be incomplete.
Well, yes it does; that's just how the universal generalization
inference
rule works. If you don't understand why it works then this explanation
is not likely to help you.
> If you add f(A) to A,
Then you are "changing the problem and cheating", which is not allowed.
> you get a complete list of real numbers.
No, but you do get a refutation of the claim that you can
algorithmically
prove that I can't make a complete list. I mean, all numbers may not
be on the new list, but the one you said COULDN'T be DEFINITELY IS.
So you owe another, or you have failed this proof. So THEY say.
> A. No you wouldn't, because you can apply the exact same argument for
> A+f(A).
And THEY can apply the exact same REBUTTAL of THAT argument.
> In the beginning of the argument, we let A be arbitrary, so
> the argument applies to ALL lists.
In theory. Again, that's just how UG works. But if people don't
understand
that then they just don't understand it. It may help to explain it to
them
synactically: "arbitrary" just means "having a name NOT mentioned
in any of the axioms of the theory" (the prior axiomatic assumptions
CAN'T constrain A if they don't mention it). That is actually harder
for your version because you constrained A to be 1) infinite, 2) a
list,
and 3) made of reals. All these are reasons why "arbitrary" becomes
problematic. You HAVE to do this about a SET A. It doesn't even
MATTER
whether that set is infinite, is a list, or has reals. THEN "they"
have some
hope of knowing that "Arbitrary" *means* ARBITRARY.
Virgil
"Albrecht" <albs...@gmx.de> wrote:
>
> If you put the antidiagonal at the end of the Cantor-list you will get
> a new antidiagonal, right. In the moment you get the new, it is at the
> end of the new list.
But the "antidiagonal" construction, as Cantor defined it is not merely
a number, but a method for finding a number, which simultaneously
defines for every list a number not in that list.
It is a function from the set of lists to the set of reals such that for
each list the value of the function for that list is not listed in that
list.
Dave Seaman
>> >> Cantor's Proof FAQ
>> All of the objections that were addressed in the Q and A could be
>> rephrased to apply to the proof involving powersets.
> That's hardly obvious. The first objection was to an arbitrary list
> of reals. Translated, it would have to be to an arbitrary infinite
> set. It is much harder to object to that when THERE IS AN EXPLICIT
> AXIOM of ZFC that SAYS THERE IS some arbitrary infinite set.
Those raising the objection are not likely to be familiar with the axioms
or to be convinced by them. A "list of reals" is just an infinite set of
ordered pairs, which likewise can be justified by the axioms.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
lugi...@gmail.com
utterly nonsensical: that's because they are. These objections are
frequently asked by cranks who don't really understand enough
mathematics to be taling about Cantor in the first place, let alone
trying to refute him. I'm just trying to clear up their naive
objections.
lugi...@gmail.com
Gc
MoeBlee kirjoitti:
OK. That`s just what I asked, is a list in mathematics defined always
to be countable.
>But they are not of issue in the
> proof that the set of real numbers is uncountable.
Right.
> MoeBlee
MoeBlee
> OK. That`s just what I asked, is a list in mathematics defined always
> to be countable.
I would say not just countable but a function on a natural number or on
omega. So a function on, say, omega+1 is a sequence and it's countable,
but it's not a list, at least as far as I would understand someone
using the word 'list'. I take a sequence to be a function on an ordinal
- so finite ordinals, omega, countable ordinals greater than omega
(greater by the ordinal ordering, not 'greater' by cardinality), and
uncountable ordinals may all serve as domains of sequences. But the
lists are the sequences that have as domain either a natural number or
omega. At least that is the way I would understand the use of the word
'list'.
MoeBlee
MoeBlee
> At least that is the way I would understand the use of the word
> 'list'.
P.S. I find it frustrating when discussions about uncountability get
sidetracked by red herrings in the form of objections regarding lists,
and "grids", and "going back to add another number to the list" and
things like that.
In such simple proofs as are commonly found, what is really at play are
functions. Whether these are called "lists" or whether there is
imagined some infinte "grid" is relevent to picturing the proof but
irrelevent to its validity. For example, to simplify to the bone:
There is no function from omega onto the set of functions from omega
into {0 1}, for if there were such a function f, then the function g on
omega defined by:
g(n)=0 if f(n)(n) = 1 and g(n)=1 if f(n)(n)=0
is a function from omega into {0 1} that is not in the range of f.
And that comes from just first order logic applied to axioms of Z set
theory. There's no need in any step that fills out that proof to use
words such as 'list', 'diagonal', 'grid', etc.
So, one may say that one doesn't accept first order logic or that one
doesn't accept the axioms of Z set theory. But there is no rational
basis - it is indeed being a crank - to deny that there is a sequence
of statements using only first order logic applied to the axioms of Z
set theory that ends with the statement that there does not exist a
function from omega onto the set of functions from omega into {0 1}.
MoeBlee
Virgil
"MoeBlee" <jazz...@hotmail.com> wrote:
I concur. Though a surjection from the set of naturals to an arbitrary
set could also be called a listing, if one allows values to be appear
repeatedly, and then one need not make a special case of a finite list.
At one time there was a commonly understood distinction between
'countable' and 'denumerable', with 'countable' including both the
finite and infinite but 'denumerable' applying only to countably
infinite cases.
That nice distinction seems to have faded away. Pity.
Ross A. Finlayson
>
> So, one may say that one doesn't accept first order logic or that one
> doesn't accept the axioms of Z set theory. But there is no rational
> basis - it is indeed being a crank - to deny that there is a sequence
> of statements using only first order logic applied to the axioms of Z
> set theory that ends with the statement that there does not exist a
> function from omega onto the set of functions from omega into {0 1}.
>
Not if that theory is inconsistent or its axiomatization of infinity is
false, no, that's not the case. If it's inconsistent then statements
and their negations are provable, although there are some questions
about the directness of proof or proofs that don't use the
inconsistency of the system against its direct proofs. Basically in an
inconsistent system that sequence of statements couldn't be said to
have an end, instead it would vacillate. That the axiomatization of
infinity is in a way flawed, or that suppositions about the number
system are incorrectly applied, is somewhat a different matter. For
example, some have there being a point at infinity in the naturals or
reals. So, is there, or not? If there is, is it a necessary feature
or implicit in any construction of those things? Is the
compactification implicit and if so there is an infinite element of the
infinite set? Consider Russell's paradox, there is that thing. People
make good and reasonable, and useful, use of those projections.
Quantify over sets. Via fiat, axiomatization, you can, it's allowed,
yet can't, it's disallowed, build from those a set, the set of them.
It is said there is a bijection between N, in the generic extension
that is N^G, and R. Each is defined by their elements and N^G does not
contain any elements that aren't in N.
Between any two definite (distinct in the mesoscale) irrationals is a
rational. Yet, you would have that between the "uncountably" many
pairs of irrationals there are only "countably" many rationals. You
can't partition them, then what is dx the differential?
Consider bijecting N^N to R, and then sampling R[0,1] at uniform
random, and thus an element of N^N. Inspect the first element of that
permutation, as it were, of the natural integers, is it not at uniform
random?
Cantor's antidiagonal argument is easy to understand. Infinity is less
so.
Transfinite cardinals is finitism, and a false infinitism.
Ross
Gc
MoeBlee kirjoitti:
> In such simple proofs as are commonly found, what is really at play are
> functions. Whether these are called "lists" or whether there is
> imagined some infinte "grid" is relevent to picturing the proof but
> irrelevent to its validity. For example, to simplify to the bone:
>
> There is no function from omega onto the set of functions from omega
> into {0 1}, for if there were such a function f, then the function g on
> omega defined by:
> g(n)=0 if f(n)(n) = 1 and g(n)=1 if f(n)(n)=0
> is a function from omega into {0 1} that is not in the range of f.
Yes, this is the proof in different form.
> And that comes from just first order logic applied to axioms of Z set
> theory. There's no need in any step that fills out that proof to use
> words such as 'list', 'diagonal', 'grid', etc.
No, if they are not defined mathematically.
Ross A. Finlayson
I think there is some consideration that it does, that the
infinitesimals actually still are in the integral calculus, even though
the formalized limit (approximation) is claimed to be that thing.
It seems the invention of the limit is attributed to Gauss, Cauchy and
Weierstrass were among its foremost proponents in establishment. Gauss
did things with infinite series that would fly against the principles
of the fettered infinity of Cauchy and Weierstrass.
Is the definite integral, of a Riemann integrable function, the area
under the curve? It would surely seem to be. The limit gives the
result as if it were the summation of the infinitesimal areas, in a
proscribed way, evaluate the indefinite integral for its upper and
lower bounds and take the difference as the area, the sum of the
differential areas.
For a constant straight line, or square wave step function of sorts,
given finite differences can be used to evaluate the integral, where
that means to determine the area. For the straight line with non-zero
slope, there is no finite number of rectangles that can be placed under
that slope who areas can be summed to give the area under the line. It
is not for any finite number of those regions that the correct answer
results. Yet, in the limit, where the number of differential regions
goes to infinity, the sum of their areas is the correct result. I
realize that pedantically it is denied that the integral is the unique
sum of infinitely many infinitesimal areas, but it is so that that's
what it is. The integral bar is an S, for summation, for a reason.
Consider even the phraseology or notation, "in the limit as the
variable diverges (goes to infinity)", lim_n->oo, it's obviously been
addressed here and elsewhere that it is not generally taught that the
variable ever reaches infinity, yet, it stares right in the face that
it does.
Ah, the antidiagonal argument. It doesn't work in binary, or base
three, or base one (tally marks) or base infinity, where those
unconventional numeric representations of reals of the unit interval
show the successor function f(x) = x +1 leading to the missing element
being the empty set in the set-theoretic powerset result.
Consider well-ordering the reals where then every subset has a least
element. Generally that would map each finite ordinal to a real, then
each countable ordinal to a real, although some have the reals being
non-well-orderable. So, remove the least element from the set and
maintain the same ordering. What happens to the element (inversely)
mapped to omega? If the element mapped to one is now mapped to zero,
is the element mapped to ordinal omega plus one now mapped to omega?
What ordinal then is mapped to the element that just was mapped to
omega? Where omega is a limit ordinal there is no predecessor. What
then of the limit ordinal past that and each limit ordinal less than
the cardinality of the continuum?
There are some other considerations in well-ordering the reals, for
example that it's consistent in regular set theories that the reals are
equivalent to multiple distinct cardinals, violating trichotomy of
cardinals, and thus ordinals. To present one is claimed to have V = L
and those universes don't exist in ZF(C). If the reals are a set and
as well are ordinals they're well-orderable, because there exists for
any set an ordinal that bijects to it, even where ordering-sensitivity
of the set's elements applies.
Skolemize, it's countable. Besides that, no element of ZF is
sufficient to be a model of ZF. What's the class of classes?
Ross
Albrecht
Than there is only one problem left: this sets don't exist.
Merry Christmas
Albrecht S. Storz
Albrecht
There is no more naive objection than saying that there is anything
more than infinite.
lugi...@gmail.com
don't you instead identify what you think is the precise error in the
proof. Which step do you think is in error and why?
Virgil
"Albrecht" <albs...@gmx.de> wrote:
That function does exist in ZFC and NBG, which is enough for us.
If it does not in your world, that is your problem, not ours.
Virgil
"Albrecht" <albs...@gmx.de> wrote:
> There is no more naive objection than saying that there is anything
> more than infinite.
What axioms does Albrecht assume to come to that conclusion?
He cannot conclude it without assuming something.
Ross A. Finlayson
You don't know that. He might be using the axiomless system of natural
deduction. You don't know that.
What's a collection of classes? What's _the_ collection of classes?
What are those of those?
Quantify, your theory is broken.
Merry Christmas. Happy Hanukkah. Happy Chriskwanzannukkahdan day.
Happy Holidays.
Ross
lugi...@gmail.com
> Virgil wrote:
> > In article <1166820604.5...@a3g2000cwd.googlegroups.com>,
> > "Albrecht" <albs...@gmx.de> wrote:
> >
> >
> > > There is no more naive objection than saying that there is anything
> > > more than infinite.
> >
> > What axioms does Albrecht assume to come to that conclusion?
> >
> > He cannot conclude it without assuming something.
>
> You don't know that. He might be using the axiomless system of natural
> deduction. You don't know that.
>
MoeBlee
> MoeBlee wrote:
> >
> > So, one may say that one doesn't accept first order logic or that one
> > doesn't accept the axioms of Z set theory. But there is no rational
> > basis - it is indeed being a crank - to deny that there is a sequence
> > of statements using only first order logic applied to the axioms of Z
> > set theory that ends with the statement that there does not exist a
> > function from omega onto the set of functions from omega into {0 1}.
> >
>
>
> Not if that theory is inconsistent
WRONG. If the theory is inconsistent, then A FORTIORI, there is such a
seqeunce of proof steps as I just mentioned.
> or its axiomatization of infinity is
> false,
The question of whether the proof exists does not depend on the truth
or falsehoof of the axiom of infintity. The proof is a finite object, a
finite sequence of finite sequences of symbols. It exists, and there is
no rational basis to deny that it does.
> If it's inconsistent then statements
> and their negations are provable,
Yes, and so the proof I mentioned would exist a fortiori.
> although there are some questions
> about the directness of proof or proofs that don't use the
> inconsistency of the system against its direct proofs.
There is no rational basis to deny that the proof exists. And it exists
in a system that has no rule of indirect proof but whose only rule is
modus ponens (of course, along with axioms for sentential and predicate
logic).
> Basically in an
> inconsistent system that sequence of statements couldn't be said to
> have an end, instead it would vacillate.
WRONG. That's nonsense.
I'll leave the rest of your usual mumbo jumbo to rot:
MoeBlee
Newberry
I have already shown you where your error is. In Step 4. Since f(A) is
not in the list and the list contains all the reals, it clearly follows
that f(A) does not exist. Why do you keep asking the same question over
and over again?
Jesse F. Hughes
> I have already shown you where your error is. In Step 4. Since f(A) is
> not in the list and the list contains all the reals, it clearly follows
> that f(A) does not exist. Why do you keep asking the same question over
> and over again?
Who said the list contains all the reals? That assumption never
appeared in lugita15's proof.
Read the first "step" again:
1. Let A be any ordered list of real numbers between 0 and 1. In
other words, for each natural number x, there exists exactly one
real number A(x) between 0 and 1. Please note that the range of A
does not have to contain all real numbers.
--
Jesse F. Hughes
"You're ketchup, so I'll put you on meatloaf!"
-- Quincy P. Hughes, age five, tries his hand at insults
Ross A. Finlayson
> Ross A. Finlayson wrote:
> > Virgil wrote:
> > > In article <1166820604.5...@a3g2000cwd.googlegroups.com>,
> > > "Albrecht" <albs...@gmx.de> wrote:
> > >
> > >
> > > > There is no more naive objection than saying that there is anything
> > > > more than infinite.
> > >
> > > What axioms does Albrecht assume to come to that conclusion?
> > >
> > > He cannot conclude it without assuming something.
> >
> > You don't know that. He might be using the axiomless system of natural
> > deduction. You don't know that.
> >
> Then he needs rules of inference. What are the ones he uses.
> > What's a collection of classes? What's _the_ collection of classes?
> > What are those of those?
> >
> > Quantify, your theory is broken.
> >
Hi,
You might consider that it has rules of inference, "logical" axioms,
but not "non-logical" / "proper" axioms.
I've discussed this at length, it's called the null axiom theory. It
has, is, a consistent universe, has that there are no paradoxes, avoids
incompleteness, and is the T.o.E. Have a nice day.
Ross
Ross A. Finlayson
You've misread that. I didn't say the sentence didn't exist, instead
that it's negation would exist in an inconsistent theory. That the
sentence ends is what was disputed. So, no, that's not wrong, and you
can be quite sure I carefully evaluate your statements I address, and
some I don't.
If infinity as stated in, say, ZF is a "false" axiom, compared to some
super/sub-theory where partial or total negation of AoI is a true
axiom, or consequence, not saying there's no infinity but that
regularized infinity is wrong, or only partially/incompletely right,
then there is a rational basis that what actually exists as that
sequence is malformed, so that it does not exist as a true statement,
of true "axioms" of "true" theories, or the true theory.
Where there's a "true" theory, it is "the" true theory.
So, does Zeno ever arrive? Can you make true statements about classes
without quantifying over them?
Ross
Newberry
Jesse F. Hughes wrote:
> "Newberry" <newb...@ureach.com> writes:
>
> > I have already shown you where your error is. In Step 4. Since f(A) is
> > not in the list and the list contains all the reals, it clearly follows
> > that f(A) does not exist. Why do you keep asking the same question over
> > and over again?
>
> Who said the list contains all the reals? That assumption never
> appeared in lugita15's proof.
>
> Read the first "step" again:
>
> 1. Let A be any ordered list of real numbers between 0 and 1. In
> other words, for each natural number x, there exists exactly one
> real number A(x) between 0 and 1. Please note that the range of A
> does not have to contain all real numbers.
It is the other way around. Since the argument fails in case the list
contains all real number it means that the list cannot be arbitrary.
Jesse F. Hughes
> Jesse F. Hughes wrote:
>> "Newberry" <newb...@ureach.com> writes:
>>
>> > I have already shown you where your error is. In Step 4. Since f(A) is
>> > not in the list and the list contains all the reals, it clearly follows
>> > that f(A) does not exist. Why do you keep asking the same question over
>> > and over again?
>>
>> Who said the list contains all the reals? That assumption never
>> appeared in lugita15's proof.
>>
>> Read the first "step" again:
>>
>> 1. Let A be any ordered list of real numbers between 0 and 1. In
>> other words, for each natural number x, there exists exactly one
>> real number A(x) between 0 and 1. Please note that the range of A
>> does not have to contain all real numbers.
>
> It is the other way around. Since the argument fails in case the list
> contains all real number it means that the list cannot be arbitrary.
The argument does not fail. The number f(A) is constructed in the
proof. It is clearly a real number and it is clearly not on the list.
But I'll let others deal with your odd confusion. It doesn't look so
entertaining to me.
--
"It's good for the economy to charge for intellectual property, so
open source software cannot be good, while Microsoft is the most
far-thinking company around and is doing it all for the good of the
public." -- Linus Torvalds paraphrases Microsoft VP Craig Mundie
Newberry
Jesse F. Hughes wrote:
> "Newberry" <newb...@ureach.com> writes:
>
> > Jesse F. Hughes wrote:
> >> "Newberry" <newb...@ureach.com> writes:
> >>
> >> > I have already shown you where your error is. In Step 4. Since f(A) is
> >> > not in the list and the list contains all the reals, it clearly follows
> >> > that f(A) does not exist. Why do you keep asking the same question over
> >> > and over again?
> >>
> >> Who said the list contains all the reals? That assumption never
> >> appeared in lugita15's proof.
> >>
> >> Read the first "step" again:
> >>
> >> 1. Let A be any ordered list of real numbers between 0 and 1. In
> >> other words, for each natural number x, there exists exactly one
> >> real number A(x) between 0 and 1. Please note that the range of A
> >> does not have to contain all real numbers.
> >
> > It is the other way around. Since the argument fails in case the list
> > contains all real number it means that the list cannot be arbitrary.
>
> The argument does not fail. The number f(A) is constructed in the
> proof. It is clearly a real number and it is clearly not on the list.
The number f(A) has not been constructed yet. It is still being
constructed as we speak.
Jesse F. Hughes
>> The argument does not fail. The number f(A) is constructed in the
>> proof. It is clearly a real number and it is clearly not on the list.
>
> The number f(A) has not been constructed yet. It is still being
> constructed as we speak.
Keen! Who's doing it?
Halliburton?
>>
>> But I'll let others deal with your odd confusion. It doesn't look so
>> entertaining to me.
So I was wrong. *This* answer was pretty darned entertaining.
--
"How can people [philosophers] talk like that? Acting as if they're
/glad/ they don't know things! Finding out more and more things they
don't know! It's like children proudly coming to show you a full
potty!" -- Terry Pratchett, /Small Gods/
Virgil
"Newberry" <newb...@ureach.com> wrote:
> Jesse F. Hughes wrote:
> > "Newberry" <newb...@ureach.com> writes:
> >
> > > I have already shown you where your error is. In Step 4. Since f(A) is
> > > not in the list and the list contains all the reals, it clearly follows
> > > that f(A) does not exist. Why do you keep asking the same question over
> > > and over again?
> >
> > Who said the list contains all the reals? That assumption never
> > appeared in lugita15's proof.
> >
> > Read the first "step" again:
> >
> > 1. Let A be any ordered list of real numbers between 0 and 1. In
> > other words, for each natural number x, there exists exactly one
> > real number A(x) between 0 and 1. Please note that the range of A
> > does not have to contain all real numbers.
>
> It is the other way around. Since the argument fails in case the list
> contains all real number it means that the list cannot be arbitrary.
One can NEVER have a list of all real numbers in the first place.
I do not know where you learnt your logic, but you should sue to get
your tuition refunded.
"If (any false statement) then (any statement)" is logically true.
So that when you say "in the case the list contains all real
number [sic]", any conclusion at all may be derived from that false
assumption. But that does not justify the truth of that conclusion.
Virgil
"Newberry" <newb...@ureach.com> wrote:
> Jesse F. Hughes wrote:
> > "Newberry" <newb...@ureach.com> writes:
> >
> > > Jesse F. Hughes wrote:
> > >> "Newberry" <newb...@ureach.com> writes:
> > >>
> > >> > I have already shown you where your error is. In Step 4. Since f(A) is
> > >> > not in the list and the list contains all the reals, it clearly follows
> > >> > that f(A) does not exist. Why do you keep asking the same question over
> > >> > and over again?
> > >>
> > >> Who said the list contains all the reals? That assumption never
> > >> appeared in lugita15's proof.
> > >>
> > >> Read the first "step" again:
> > >>
> > >> 1. Let A be any ordered list of real numbers between 0 and 1. In
> > >> other words, for each natural number x, there exists exactly one
> > >> real number A(x) between 0 and 1. Please note that the range of A
> > >> does not have to contain all real numbers.
> > >
> > > It is the other way around. Since the argument fails in case the list
> > > contains all real number it means that the list cannot be arbitrary.
> >
> > The argument does not fail. The number f(A) is constructed in the
> > proof. It is clearly a real number and it is clearly not on the list.
>
> The number f(A) has not been constructed yet. It is still being
> constructed as we speak.
When one defines a function from the set of lists of real to the reals,
as has been done, then the choice of a list determines the value of the
function.
The "number" f(A) is "constructed" as soon as the rule for constructing
it is complete and a list of reals is chosen.
At least in ZFC or NBG.
Andrew Usher
For the love of God won't you ever try learning something? It doesn't
reach infinity, there is no number called 'INFINITY' that we plug in
for n. It's called a limit. There is a rigorous and intutitive
definition for limit, you know.
You haven't shown what infinitesimals have to do with limits or areas
or even what infinitesimals are in your mathematics.
Without _proof_ it isn't mathematics; it's philosophy at best. Frankly,
your contibutions are worthless as either.
> Consider well-ordering the reals where then every subset has a least
> element. Generally that would map each finite ordinal to a real, then
> each countable ordinal to a real, although some have the reals being
> non-well-orderable. So, remove the least element from the set and
> maintain the same ordering. What happens to the element (inversely)
> mapped to omega? If the element mapped to one is now mapped to zero,
> is the element mapped to ordinal omega plus one now mapped to omega?
> What ordinal then is mapped to the element that just was mapped to
> omega? Where omega is a limit ordinal there is no predecessor. What
> then of the limit ordinal past that and each limit ordinal less than
> the cardinality of the continuum?
It is still mapped to omega. If you remove the first element there are
still omega many elements less than omega. That's precisely BECAUSE it
is a limit ordinal!
Also, you're still assuming AC without saying so, and CH, and making
elementary mistakes about infinity. Do you think before posting at
all???
> There are some other considerations in well-ordering the reals, for
> example that it's consistent in regular set theories that the reals are
> equivalent to multiple distinct cardinals, violating trichotomy of
> cardinals, and thus ordinals. To present one is claimed to have V = L
> and those universes don't exist in ZF(C). If the reals are a set and
> as well are ordinals they're well-orderable, because there exists for
> any set an ordinal that bijects to it, even where ordering-sensitivity
> of the set's elements applies.
If two cardinals are equivalent to R, they are equivalent to one
another and therefore not distinct. Think about what cardinality is:
it's defined by bijection and thus obviously transitive.
> Skolemize, it's countable. Besides that, no element of ZF is
> sufficient to be a model of ZF. What's the class of classes?
Repetition doesn't help, Ross.
Andrew Usher
george
Dave Seaman wrote:
> Those raising the objection are not likely to be familiar with the axioms
Exactly.
> or to be convinced by them.
Oh, PLEASE! That is why the burden of proof
IS ON LUGIT to PHRASE the theorem in the right
language! If you STATE it as a theorem of set theory
then the fact that they are not familiar with the language
becomes quite completely IRrelevant: OBVIOUSLY they
will have to BECOME familiar before they can be ENTITLED
to an opinion. ALL they can say UNTIL then is "I'm sorry,
I don't know what you mean", and that is NOT a frequently
asked question.
> A "list of reals" is just an infinite set of
> ordered pairs,
Ordered pairs OF WHAT?
The whole point is that it MATTERS WHAT a "real" is,
IF you are going to call "Cantor's theorem" and "the theorem
that there are more reals than naturals" the same thing.
People unfamiliar with set theory axioms are NOT going
to perceive "a real" as a subset of N.
But the theorem is IN FACT ABOUT subsets AND NOT
about reals. It is only because reals wind up being so subset-LIKE
that the "reals-version" of the theorem becomes popular.
But the point is simply that that popularity is misguided and
misplaced. Set theory is important. Reals are not.
The theorem is a set theorem. The reals-result is corollary
and ancillary.
Ross A. Finlayson
Repetition doesn't help: go to proper classes to escape perceived
paradoxes of regular set theory, and then the same problems apply to
proper classes. Can you quantify over proper classes to make true
statements about them? Repetition of your non-responses doesn't change
them.
That's not so, about the limit vis-a-vis the sum, and in many cases the
limit being the sum. Newton and Leibni(t)z, for example, called the
inventors of calculus, had it that way, that what we call the limit
actually is the sum. It happens to give perfectly correct answers. I
think I need to re-read that _History of the Calculus and its
Conceptual Development_, where I have read it. Also that Anders Hald
is very good for statistics.
It was and is so that people see the differential as a nilpotent/atomic
infinitesimal, and the integral as their sum. That's why it was and is
called infinitesimal analysis.
Andrew, I see a flaw in the reasoning in assigning the limit to the
sum, for example, the sum of areas under a curve, while denying that
for no finite number of areas is it that sum, and saying it's not so
for some infinite value. The limit is a tool to get that value. For
no finite differential can you evaluate the area under an arbitrary
integrable curve. The limit, as a number, _is_ the sum, they're equal.
Keep in mind: for no finite difference is the approximation the sum.
The limit is the sum, thus it's not a "finite" difference, it's not
zero, because zero's the additive identity, it's an infinitesimal
differential.
Obviously infinity and its meaning, and usage, in mathematics has a
huge corpus of work, and applicable results, about which transfinite
cardinals are irrelevant. Do you know any applications of transfinite
cardinals?
The physical universe as being Ding-an-Sich, which is some philosophy
from Immanuel Kant, does not agree with there not being a universe.
Where does the cumulative hierarchy reside? Is it not ubiquitous?
There's only the null axiom theory.
There are a variety of systems having a point at infinity in the
natural integers, where, the limit of the convergent function is the
sum, as the variable goes to infinity. There are a variety of
counterexamples to standard real analysis.
About the well-ordering of an interval, of the real number line, of the
real numbers, in an onto mapping of each ordinal element less than an
ordinal to an element to that segment, then there is in the domain of
those ordinals each countable limit ordinal. For each, there are only
countably many successor ordinals. Where it takes that one uncountable
ordinal, there's a beginning and an end, and where it doesn't end
there, it never does.
Without AC there isn't trichotomy of cardinals. Where there's
trichotomy of cardinals, AC is an axiom or theorem, because ordinals
are trichotomous and well-ordered and cardinals are ordinals, and sets
are defined by their elements. They're trichotomous in the Archimedean
way. There are a variety of considerations of non-Archimedean
infinities, where for example infinity is less than nothing, a three
symbol proof. There are examples in nature to that effect.
Neither the antidiagonal argument nor nested intervals applies to the
natural/unit equivalency function.
Ross
Dave Seaman
> Dave Seaman wrote:
>> Those raising the objection are not likely to be familiar with the axioms
> Exactly.
>> or to be convinced by them.
> Oh, PLEASE! That is why the burden of proof
> IS ON LUGIT to PHRASE the theorem in the right
> language! If you STATE it as a theorem of set theory
> then the fact that they are not familiar with the language
> becomes quite completely IRrelevant: OBVIOUSLY they
> will have to BECOME familiar before they can be ENTITLED
> to an opinion. ALL they can say UNTIL then is "I'm sorry,
> I don't know what you mean", and that is NOT a frequently
> asked question.
Look, I didn't come here to defend Lugit. I merely pointed out that your
original claims in this thread were completely unfounded, a fact that you
have admitted. Since then you have been engaged in a desparate attempt
to save face, as if you had not completely misread the OP. I am not
interested in playing your silly game.
>> A "list of reals" is just an infinite set of
>> ordered pairs,
> Ordered pairs OF WHAT?
Here's a hint. Everything in ZFC is a set. If you can't figure it out
beyond that, then get someone else to explain it to you.
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Ross A. Finlayson
> Here's a hint. Everything in ZFC is a set. If you can't figure it out
> beyond that, then get someone else to explain it to you.
>
With all the recent discussion of "sets of all sets" and "universes",
and so on, it seems like there's a difference between "for each" and
"for every", and even between "for any" and "for every". I think it
has to do with the transfer principle, that there is the universal
quantifier with the extra feature that the collection of elements has
the same property as the element.
That is to say, there is no "everything" in ZF, there's no totality, no
universe. Where X = {x, x=x}, X is not a set, defined by its elements,
because then it would contain itself, be irregular and contradict at
least one of ZF's axioms, its existence would be contradictory to ZF.
The ability to talk about "everything" in ZF makes ZF inconsistent.
Where each thing in ZF is a set, collect them, as it is said can be
done in ZF, and that is X as above and contradicts ZF.
The set of ordinals would be an ordinal, the set of sets a set, these
are called Burali-Forti's and Cantor's "paradoxes." Thus, there is a
rational basis that there is no set of ordinals nor sets in ZF, for it
would be "the" set. The universe of ZF is the Russell set, thus it's
irregular, and regularity/foundation is a false axiom.
Is there work already done about refining the universal quantifier into
separately forany, forevery, and foreach? I've addressed that notion
before.
Ross
george
> On 23 Dec 2006 08:34:24 -0800, george wrote:
>
> > Dave Seaman wrote:
> >> Those raising the objection are not likely to be familiar with the axioms
>
> > Exactly.
>
> >> or to be convinced by them.
Right.
Dave Seaman wrote:
> Look, I didn't come here to defend Lugit.
SHut up. THat is exactly what you are doing.
> I merely pointed out that your
> original claims in this thread were completely unfounded,
Now you are just lying.
Pointing out that there were some questions in the OP is NOT
the ONLY thing you have done here. You have done a lot mroe
than MERELY that. More to the point, I made a lot MORE
claims than MERELY that there were no questions in the OP,
and you CERTAINLY have NEVER pointed out that ANY of the OTHER
claims were completely unfounded. So, now, I am just pointing out
that your insulting over-generalizations are completely unfounded.
> a fact that you
> have admitted.
What UTTER bullshit. I admitted that I overlooked the questions but
that OBVIOUSLY HAS NO bearing on anything ELSE I might've said about
why the OP was misguided or why stating the theorem in set theory
would obviate FAQs.
> Since then you have been engaged in a desparate attempt
> to save face,
Oh, go fuck yourself.
> as if you had not completely misread the OP.
I OBVIOUSLY did not "completely misread" the OP.
I quoted and responded to the parts I did read. Just because
I didn't read the questions does not invalidate anything I may have
said about what I did read. Your allegations here that one oversight
might've tainted all replies "completely" is not worthy of any LOGICAL
context.
> I am not
> interested in playing your silly game.
But you are playing a game, a most assholic one at that,
AS OPPOSED to trying to say anything about the proof
or associated pedagogy. And it is indeed a silly game.
And you suck at it.
Rupert
george wrote:
> > On 23 Dec 2006 08:34:24 -0800, george wrote:
> >
> > > Dave Seaman wrote:
> > >> Those raising the objection are not likely to be familiar with the axioms
> >
> > > Exactly.
> >
> > >> or to be convinced by them.
>
> Right.
>
> Dave Seaman wrote:
> > Look, I didn't come here to defend Lugit.
>
> SHut up. THat is exactly what you are doing.
>
Has it ever occurred to you that maybe you should do something about
the fact that you're such an unpleasant person? Most other people here,
when they disagree with someone, are capable of expressing themselves
politely. Didn't your parents ever teach you the value of that?
MoeBlee
Newberry wrote:
> I have already shown you where your error is. In Step 4. Since f(A) is
> not in the list and the list contains all the reals, it clearly follows
> that f(A) does not exist.
That's not valid first order logic you're using. To make your argument,
you're glossing over the actual sentential logic in the proof. You're
confused as to some very basic sentential logic here. You should seek
tutoring.
MoeBlee
MoeBlee
> It is the other way around. Since the argument fails in case the list
> contains all real number it means that the list cannot be arbitrary.
You are in dire need of tutoring in basic sentential logic.
MoeBlee
george
lugi...@gmail.com wrote:
> I addressed the constructivist objection you pointed out in my FAQ.
If you want to CALL that "addressing". Objective
observers would say you totally whiffed.
> Basically it boils down to this: if someone believes that uncomputable
> real numbers do not exist,
OK, they believe that.
> then the statement that the uncomputable
> real numbers are uncountably infinite is vacuously true.
IT IS NOT, dumbass. If uncomputable real numbers do not
exist then the set of uncomputable real numbers IS EMPTY,
and the EMPTY set IS COUNTABLE. BY ZERO or by ITSELF
if the empty set is representing zero.
> For this
> reason, constructivists, shouldn't have a problem with Cantor's Proof.
For this reason, you're an idiot.
Jeezus.
Has it never occurred to you that smart people really do have trouble
seeing things from stupid people's perspective from time to time?
"Vacuously true" that something is UNCOUNTABLY INFINITE?
You really should be completely ashamed.
george
Rupert wrote:
> But if you require reals to be computable, you should also require
> bijections to be computable. And then the argument still works: there
> is no effective enumeration of the set of computable reals.
This is COMPLETELY missing the point.
"Computability" arises ONLY in an INfinitary context.
The theorem PROPERLY phrased does not have ANYthing
to say ABOUT infinity because it is ALSO TRUE OF *FINITE*
sets! NONE of those considerations CAN be relevant!
Rupert
george wrote:
> Rupert wrote:
> > But if you require reals to be computable, you should also require
> > bijections to be computable. And then the argument still works: there
> > is no effective enumeration of the set of computable reals.
>
> This is COMPLETELY missing the point.
> "Computability" arises ONLY in an INfinitary context.
Why do you think that, exactly? The theorem I stated can be proved in
Primitive Recursive Arithmetic.
> The theorem PROPERLY phrased does not have ANYthing
> to say ABOUT infinity because it is ALSO TRUE OF *FINITE*
> sets!
Interesting argument. Sounds like a load of bollocks to me, to be
perfectly frank.
Albrecht
lugi...@gmail.com schrieb:
> > There is no more naive objection than saying that there is anything
> > more than infinite.
> >
> Instead of just talking about stuff like "more than infinite," why
> don't you instead identify what you think is the precise error in the
> proof. Which step do you think is in error and why?
The problem lies not so much in the steps you have listed. You use some
implications which are not so much ensured and leading me to the
feeling that your argumentation bases on the platonic view that some
math objects in some sense exists more true than other.
One of this implicated assumptions is the consideration that real
numbers build up a set with the definition that any arbitrary infinite
sequence of ciphers build up a real number. But which sense may have an
arbitrary infinite sequence of ciphers if you can't have one of them
definite if it isn't defined by a law of it's construction? It means
that you imply the existence of unconstructable numbers in your basic
assumptions.
There are two possibilities: All real numbers are constructable. Than
they are countable.
Or: There are unconstructable real numbers too. Than, if you choose an
arbitrary real number the possibility that this number is
unconstructable is 1 and the possibility that this number is
constructable is zero.
Who decides?
And you seem to accept the existence of infinite sets without use of
the axiom of infinity. Maybe you think that the use of ZF or ZFC is
common. But your argumenation has a diction of absolute (platonic)
thruth. You seem to think that Cantor's diagonal argument is senseful
without an elaborated and cleared set theory like e.g. ZFC.
It seems to be natural to you that infinite sets as the set of natural
numbers, the set of real numbers, the set of rational numbers, ...,
exists. But in the sense of actual infinite sets they exists only if
you state the axiom of infinity. It's an axiom. One can choose other.
So, why do you want to discuss your argumentation without stating all
assumptions you use? Why do you think one could only criticise your
listed steps?
Virgil
"Albrecht" <albs...@gmx.de> wrote:
> lugi...@gmail.com schrieb:
>
> The problem lies not so much in the steps you have listed. You use some
> implications which are not so much ensured and leading me to the
> feeling that your argumentation bases on the platonic view that some
> math objects in some sense exists more true than other.
> One of this implicated assumptions is the consideration that real
> numbers build up a set with the definition that any arbitrary infinite
> sequence of ciphers build up a real number.
If by "cyphers" you mean numbers, no one claims anything of the sort.
There are certain infinite sequences of "cyphers", i.e., rational
numbers, which "converge" in the Cauchy sense, and such sequences do
produce real number limits.
> But which sense may have an
> arbitrary infinite sequence of ciphers if you can't have one of them
> definite if it isn't defined by a law of it's construction? It means
> that you imply the existence of unconstructable numbers in your basic
> assumptions.
> There are two possibilities: All real numbers are constructable. Than
> they are countable.
> Or: There are unconstructable real numbers too. Than, if you choose an
> arbitrary real number the possibility that this number is
> unconstructable is 1 and the possibility that this number is
> constructable is zero.
Except that the possibility of "choosing" any unconstructable number
must be zero, or it would become constructable by that choice.