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Guest · Posted: Thu Jul 31, 2008 9:41 pm Post subject: Re: the problem with Cantor

On 31 Jul, 22:26, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 31, 1:06 pm, ju...@diegidio.name wrote:

On 31 Jul, 21:02, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 31, 12:58 pm, ju...@diegidio.name wrote:

I guess the argument goes like: "IF there is such a list, THEN no
sequence can ever escape it."

Let>s assume we have such a list. Then we can apply a suitable
diagonalisation function and get a sequence that differs from the
first sequence in the first place, from the second sequence in the
second place, and so on.

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.

"the last sequence", "the last place", and "the limit sequence". All
undefined nonsense.

It is not undefined nonsense:

"That is to say that, at the limit, any diagonal sequence is still a
sequence in the list, namely the limit sequence."

I am not trying to be ambiguous: what is unclear?

Please READ what I wrote:

We give you our EXPLICIT axioms, rules, and primitives, as well as our
EXPLICIT definitions. What are yours, including your explicit RIGOROUS
definition of "the last sequence", "the last place", and "the limit
sequence"?

There is NO last sequence. There is NO last place. And 'limit' is
meaningless unless you define the specific sense of taking such limits
(by some topology, metric, ordering, or other explicit definition). It
is meaningless (and typically CRANK) to just wave one>s hand and to
speak of "the limit case" when one has not given an actual
mathematical definition of such a limit.

MoeBlee
[/quote]
I have given an argument based on the basic properties of the natural
numbers, and on induction to give a very specific and again basic
sense to the limit case.

Here I have furtherly given a very specific and unambiguous sense to
the diagonal procedure up to a definition for the class of "omega-
sequences" for _any given *rule* to enumerate_ the original, complete,
infinite, and putative enumerable collection (was, a bit ambiguously,
"list": just inherited from the previous discussion).

No such *rule*, no diagonal function to build!
No such *collection* (the "list") to begin with... not even the
computables!?

So that your objections, in general, are a contradiction to your own
language and tools, and not even near to be real objections.

(Why do you keep with this reminding me of *your* axioms? You are just
pointing out the *problems* with the accepted approaches, while the
objections to the above are really immaterial: you deny your own
language and tools when you deny that the above makes any sense as it
stands. In the name of your axioms. And that is all your objections
keep amounting to.)

-LV

Dave L. Renfro · Guest

ju...@diegidio.name wrote:

[quote]We can then call all possible diagonal sequences for any
enumeration rule of such putative list: the class of
"omega-sequences" for that rule. (And this is now a very
well defined notion.)
[/quote]
You can do this, but that>s like saying we can consider
a basket of apples when someone points out that the
rocketship you were describing doesn>t exist.

Anyway, for what it>s worth, this has been studied:

Robert Gray, "Georg Cantor and transcendental numbers",
American Mathematical Monthly 101 #1 (November 1994),
819-832.

[quote]That such complete infinite list itself exists, might
finally be a matter for an axiom.
[/quote]
Why should this require an axiom? It>s a subset of the
collection of all sequences of elements from some fixed
set, so it>s existence follows from the subset selection
axiom.

Dave L. Renfro

MoeBlee · Guest

On Jul 31, 2:41 pm, ju...@diegidio.name wrote:
[quote]On 31 Jul, 22:26, MoeBlee <jazzm...@hotmail.com> wrote:

On Jul 31, 1:06 pm, ju...@diegidio.name wrote:

On 31 Jul, 21:02, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 31, 12:58 pm, ju...@diegidio.name wrote:

I guess the argument goes like: "IF there is such a list, THEN no
sequence can ever escape it."

Let>s assume we have such a list. Then we can apply a suitable
diagonalisation function and get a sequence that differs from the
first sequence in the first place, from the second sequence in the
second place, and so on.

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.

"the last sequence", "the last place", and "the limit sequence". All
undefined nonsense.

It is not undefined nonsense:

"That is to say that, at the limit, any diagonal sequence is still a
sequence in the list, namely the limit sequence."

I am not trying to be ambiguous: what is unclear?

Please READ what I wrote:

We give you our EXPLICIT axioms, rules, and primitives, as well as our
EXPLICIT definitions. What are yours, including your explicit RIGOROUS
definition of "the last sequence", "the last place", and "the limit
sequence"?

There is NO last sequence. There is NO last place. And 'limit' is
meaningless unless you define the specific sense of taking such limits
(by some topology, metric, ordering, or other explicit definition). It
is meaningless (and typically CRANK) to just wave one>s hand and to
speak of "the limit case" when one has not given an actual
mathematical definition of such a limit.

I have given an argument based on the basic properties of the natural
numbers, and on induction to give a very specific and again basic
sense to the limit case.
[/quote]
Your "argument" is NONSENSE. It>s been explained to you.

[quote]Here I have furtherly given a very specific and unambiguous sense to
the diagonal procedure up to a definition for the class of "omega-
sequences" for _any given *rule* to enumerate_ the original, complete,
infinite, and putative enumerable collection (was, a bit ambiguously,
"list": just inherited from the previous discussion).

No such *rule*, no diagonal function to build!
No such *collection* (the "list") to begin with... not even the
computables!?

So that your objections, in general, are a contradiction to your own
language and tools, and not even near to be real objections.

(Why do you keep with this reminding me of *your* axioms? You are just
pointing out the *problems* with the accepted approaches, while the
objections to the above are really immaterial: you deny your own
language and tools when you deny that the above makes any sense as it
stands. In the name of your axioms. And that is all your objections
keep amounting to.)
[/quote]
No, you contradict two principles combined:

(1) Intuitionisitc logic (weaker even than classical logic) combined
with (2) The principle that for any formalizable property P (not
mentioning D) and for any set S there is a set D that is the set of
members of S that have property P.

THEN, my argument is not that you are not allowed to contradict those,
but rather that if you DO contradict those, then what principles do
you offer in their place?

But you have no appreciation really of that question, since you have
no understanding of what is involved in formulating a sufficient body
of logical and mathematical principles to prove the theorems that are
used in calculus, which mathematics for the sciences and technology.

And you contradict yourself: You said you wanted to learn. But
instead, your posting reflects only someone doggedly committed to
repeating his most basic misconceptions.

MoeBlee

MoeBlee · Guest

On Jul 31, 2:41 pm, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
[quote]ju...@diegidio.name wrote:

That such complete infinite list itself exists, might
finally be a matter for an axiom.

Why should this require an axiom? It>s a subset of the
collection of all sequences of elements from some fixed
set, so it>s existence follows from the subset selection
axiom.
[/quote]
If I understand what he>s saying, yes, it>s a theorem that the set
exists. But, contrary, to his description, it is not listable
(enumerable, i.e. countable).

MoeBlee

Guest · Posted: Thu Jul 31, 2008 10:05 pm Post subject: Re: the problem with Cantor

On 31 Jul, 22:57, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 31, 2:41 pm, ju...@diegidio.name wrote:
On 31 Jul, 22:26, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 31, 1:06 pm, ju...@diegidio.name wrote:
On 31 Jul, 21:02, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 31, 12:58 pm, ju...@diegidio.name wrote:

I guess the argument goes like: "IF there is such a list, THEN no
sequence can ever escape it."

Let>s assume we have such a list. Then we can apply a suitable
diagonalisation function and get a sequence that differs from the
first sequence in the first place, from the second sequence in the
second place, and so on.

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.

"the last sequence", "the last place", and "the limit sequence". All
undefined nonsense.

It is not undefined nonsense:

"That is to say that, at the limit, any diagonal sequence is still a
sequence in the list, namely the limit sequence."

I am not trying to be ambiguous: what is unclear?

Please READ what I wrote:

We give you our EXPLICIT axioms, rules, and primitives, as well as our
EXPLICIT definitions. What are yours, including your explicit RIGOROUS
definition of "the last sequence", "the last place", and "the limit
sequence"?

There is NO last sequence. There is NO last place. And 'limit' is
meaningless unless you define the specific sense of taking such limits
(by some topology, metric, ordering, or other explicit definition). It
is meaningless (and typically CRANK) to just wave one>s hand and to
speak of "the limit case" when one has not given an actual
mathematical definition of such a limit.
I have given an argument based on the basic properties of the natural
numbers, and on induction to give a very specific and again basic
sense to the limit case.

Your "argument" is NONSENSE. It>s been explained to you.

Here I have furtherly given a very specific and unambiguous sense to
the diagonal procedure up to a definition for the class of "omega-
sequences" for _any given *rule* to enumerate_ the original, complete,
infinite, and putative enumerable collection (was, a bit ambiguously,
"list": just inherited from the previous discussion).

No such *rule*, no diagonal function to build!
No such *collection* (the "list") to begin with... not even the
computables!?

So that your objections, in general, are a contradiction to your own
language and tools, and not even near to be real objections.
(Why do you keep with this reminding me of *your* axioms? You are just
pointing out the *problems* with the accepted approaches, while the
objections to the above are really immaterial: you deny your own
language and tools when you deny that the above makes any sense as it
stands. In the name of your axioms. And that is all your objections
keep amounting to.)

No, you contradict two principles combined:

(1) Intuitionisitc logic (weaker even than classical logic) combined
with (2) The principle that for any formalizable property P (not
mentioning D) and for any set S there is a set D that is the set of
members of S that have property P.

THEN, my argument is not that you are not allowed to contradict those,
but rather that if you DO contradict those, then what principles do
you offer in their place?

But you have no appreciation really of that question, since you have
no understanding of what is involved in formulating a sufficient body
of logical and mathematical principles to prove the theorems that are
used in calculus, which mathematics for the sciences and technology.

And you contradict yourself: You said you wanted to learn. But
instead, your posting reflects only someone doggedly committed to
repeating his most basic misconceptions.

MoeBlee
[/quote]
Now you are getting just and nise yourself.

I am glad you have given something explicit I can work upon.

I am simply not (yet?) able to explicitly state the list of "my
axioms". I keep telling you: what you get is what you see, natural
numbers plus induction. If anything else is needed, then let>s name
it.

Anyway, I>ll tell you what I can get from your argument above.

-LV

Guest · Posted: Thu Jul 31, 2008 10:41 pm Post subject: Re: the problem with Cantor

On 31 Jul, 22:57, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]On Jul 31, 2:41 pm, ju...@diegidio.name wrote:
I have given an argument based on the basic properties of the natural
numbers, and on induction to give a very specific and again basic
sense to the limit case.

Your "argument" is NONSENSE. It>s been explained to you.

Here I have furtherly given a very specific and unambiguous sense to
the diagonal procedure up to a definition for the class of "omega-
sequences" for _any given *rule* to enumerate_ the original, complete,
infinite, and putative enumerable collection (was, a bit ambiguously,
"list": just inherited from the previous discussion).

No such *rule*, no diagonal function to build!
No such *collection* (the "list") to begin with... not even the
computables!?

So that your objections, in general, are a contradiction to your own
language and tools, and not even near to be real objections.
(Why do you keep with this reminding me of *your* axioms? You are just
pointing out the *problems* with the accepted approaches, while the
objections to the above are really immaterial: you deny your own
language and tools when you deny that the above makes any sense as it
stands. In the name of your axioms. And that is all your objections
keep amounting to.)

No, you contradict two principles combined:

(1) Intuitionisitc logic (weaker even than classical logic)
[/quote]

I am afraid I cannot see where I am contradicting intuitionistic
logic. Could you be explicit? (Can *I* really be the one to perform
such comparisons?)

[quote]combined
with (2) The principle that for any formalizable property P (not
mentioning D) and for any set S there is a set D that is the set of
members of S that have property P.
[/quote]

How can I ever be contradicting that one? In a fully-computable realm,
any formalizable property does express a set, and viceversa. It is
rather the accepted approach that is problematic in this respect.

-LV

[quote]THEN, my argument is not that you are not allowed to contradict those,
but rather that if you DO contradict those, then what principles do
you offer in their place?[/quote]

MoeBlee · Guest

On Jul 31, 3:41 pm, ju...@diegidio.name wrote:
[quote]On 31 Jul, 22:57, MoeBlee <jazzm...@hotmail.com> wrote:
No, you contradict two principles combined:

(1) Intuitionisitc logic (weaker even than classical logic)

I am afraid I cannot see where I am contradicting intuitionistic
logic. Could you be explicit? (Can *I* really be the one to perform
such comparisons?)
[/quote]
Not intuitionistic logic ALONE. I said COMBINED. You are contradicting
intuitionistic logic COMBINED with (2):

[quote]combined
with (2) The principle that for any formalizable property P (not
mentioning D) and for any set S there is a set D that is the set of
members of S that have property P.

How can I ever be contradicting that one?
[/quote]
Because (2) combined with intuitionistic logic, entails that there is
no set that can map onto its power set. So when you claim that there
is a set that maps onto its power set then you are contradicting
intutionistic logic combined with (2).

[quote]In a fully-computable realm,
any formalizable property does express a set, and viceversa. It is
rather the accepted approach that is problematic in this respect.
[/quote]
Whatever you mean by "accepted approach", you are not coming to grips
with this fact:

Intutionistic logic and the axiom schema of separation prove that no
set maps onto its power set.

So if you think there is a set that maps onto its power set then you
must discard either intutionistic logic or the axiom schema of
separation.

[quote]THEN, my argument is not that you are not allowed to contradict those,
but rather that if you DO contradict those, then what principles do
you offer in their place?
[/quote]
No answer from you.

MoeBlee

Tonico · Guest

On Jul 31, 10:58 pm, ju...@diegidio.name wrote:
[quote]On 31 Jul, 09:24, Tonico <Tonic...@yahoo.com> wrote:

On Jul 31, 8:32 am, lwal...@lausd.net wrote:

On Jul 29, 3:25 pm, "|-|erc" <h...@r.c> wrote:

When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?

And so another nonstandard mathematician |-|erc, has
appeared out of lurkdom to refute Cantor.

I>ve only glanced at this thread, but it appears that
|-|erc>s argument is another one based on the
assumption (that doesn>t hold in ZFC) that if for every
natural number n, phi(n) holds, then we must have
phi(N) holding as well.

***************************************************************

I can>t be sure (pretty confussing with all those non-defined terms
and stuff), but at the bottom line it seems |-|erc tries somehow to
"turn over" the argument used in Cantor>s Diagonal proof and he says:

"OK, I can write down all the computable numbers and that way I get
ALL the possible real numbers, [quote: Computable reals displays EVERY
type of decimal expansion imaginable, there is nothing it misses.].
Proof? Very simple: tell me which number would I miss doing this! Oh,
but if you can point such a number then it is computable, and thus I
would have written it...taraaaaan!"

Well, there seems to exist a rather huge logical flaw up there: how
can you know a priori whether you>ve written down all the conmputable
real numbers?
Or even better, and "turning over the tortilla" once again: what if
you present me the list (because it will be a list...right??) of ALL
computable real numbers, and then I use Cantor>s Diagonal argument and
pinpoint a real number which is NOT in that list? Because believe me:
in any list of real numbers there will be a number I can pinpoint and
prove it is NOT in that list...

I guess the argument goes like: "IF there is such a list, THEN no
sequence can ever escape it."

Let>s assume we have such a list. Then we can apply a suitable
diagonalisation function and get a sequence that differs from the
first sequence in the first place, from the second sequence in the
second place, and so on.

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.

-LV
[/quote]
**************************************************************

"Last sequence"? Where is that?
"At the limit"? At the limit of what and when??
"Limit sequence"? Limit of what?
I really don>t understand what you meant.

Regards
Tonio

Virgil · Guest

In article
<9a63cf17-a30b-4865-9de5-d5817df52bb3@m44g2000hsc.googlegroups.com>,
julio@diegidio.name wrote:

[quote]I guess the argument goes like: "IF there is such a list, THEN no
sequence can ever escape it."

Let>s assume we have such a list.
[/quote]
If one assumes a falsehood, one can prove anything, including that that
assumption is false.

So that if assuming that every sequence is in the list still allows us
to prove that there is a sequence not in the list, as it does, then it
is not in the list.

[quote]Then we can apply a suitable
diagonalisation function and get a sequence that differs from the
first sequence in the first place, from the second sequence in the
second place, and so on.

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.
[/quote]
Just what in hell is that supposed to mean?
To be "in" the list means that it has a finite position in the list, so
there is no such thing as a limit member of such a list.

Virgil · Guest

In article
<09f99364-6b14-4ebf-a01f-462df1e3c73d@25g2000hsx.googlegroups.com>,
julio@diegidio.name wrote:

[quote]On 31 Jul, 21:02, MoeBlee <jazzm...@hotmail.com> wrote:
On Jul 31, 12:58 pm, ju...@diegidio.name wrote:

The original list is infinite. By extending the argument: Any such
diagonal sequence differs from the "last sequence" in the "last
place". That is to say that, at the limit, any diagonal sequence is
still a sequence in the list, namely the limit sequence.

"the last sequence", "the last place", and "the limit sequence". All
undefined nonsense.

It is not undefined nonsense:

"That is to say that, at the limit, any diagonal sequence is still a
sequence in the list, namely the limit sequence."

I am not trying to be ambiguous: what is unclear?
[/quote]
Among other things, how a sequence which differs from every member of a
list can still be a member of that list.

Balthasar · Guest

On Thu, 31 Jul 2008 14:05:07 -0700 (PDT), "Dave L. Renfro"
<renfr1dl@cmich.edu> wrote:

[quote]
Let>s assume we have such a list. Then we can apply
a suitable diagonalisation function and get a sequence
that differs from the first sequence in the first place,
from the second sequence in the second place, and so on.

And so on in this case means: it differs from ALL entries (i.e[/quote]
sequences) in the list. And that in turn means, it is not itself a
sequence in the list.

Period.

Though of course in the strange worlds of cranks there are always
additional "possibilities":

"For every line of Cantor>s list it is true that this line
does not contain the diagonal number. Nevertheless the
diagonal number may be in the infinite list." (WM, sci.logic)

Of course, this contradicts ANY form of logic, but who cares?!

[quote]
The original list is infinite. By extending the argument:
Any such diagonal sequence differs from the "last sequence"
in the "last place".

Errr... what "last sequence"? Since the considered list does not have a[/quote]
last sequence, it>s hard to see what this idiot might mean here.

[quote]
That is to say ...

Yes?[/quote]

[quote]
that, at the limit, ...

Limit? What limit? Hell!!![/quote]

[quote]
any diagonal sequence is still a sequence in the list,
namely the limit sequence.

What EXACTLY _is_ /the limit sequence/ (->definition). And where>s the[/quote]
_proof_ that this /limit sequence/ is 1. identical with the "diagonal
sequence" and 2. a sequence in the list? (->proof)

"Unproven statements carry little weight in the world of
mathematics." - Amir D. Aczel

[quote]
Just what do you mean by "a list"? If you mean a correspondence
with 1, 2, 3, ... (what one would expect if you>re using a
diagonal argument with the positive integers), then there
is no last place since there is no last positive integer,
something most 12-year olds could have told you.

Trying to ARGUE with a crank? :-o[/quote]

B.

--

"For every line of Cantor>s list it is true that this line does not
contain the diagonal number. Nevertheless the diagonal number may
be in the infinite list." (WM, sci.logic)

Virgil · Guest

In article
<4383c973-d11d-44e6-8eab-e62ee8299cb0@t54g2000hsg.googlegroups.com>,
julio@diegidio.name wrote:

[quote]On 31 Jul, 22:05, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
ju...@diegidio.name wrote:
Let>s assume we have such a list. Then we can apply
a suitable diagonalisation function and get a sequence
that differs from the first sequence in the first place,
from the second sequence in the second place, and so on.

The original list is infinite. By extending the argument:
Any such diagonal sequence differs from the "last sequence"
in the "last place". That is to say that, at the limit,
any diagonal sequence is still a sequence in the list,
namely the limit sequence.

Just what do you mean by "a list"? If you mean a correspondence
with 1, 2, 3, ... (what one would expect if you>re using a
diagonal argument with the positive integers), then there
is no last place since there is no last positive integer,
something most 12-year olds could have told you.

Dave L. Renfro

Just and nise as always.

"IF there is such a list, THEN no sequence can ever escape it."

We can then call all possible diagonal sequences for any enumeration
rule of such putative list: the class of "omega-sequences" for that
rule. (And this is now a very well defined notion.)

That such complete infinite list itself exists, might finally be a
matter for an axiom.

[/quote]
But requiring the addition of such an axiom to most set theory axiom
systems allows proof of "P /\ ~P", which characteristic is quite
generally regarded as being deleterious to an axiom system>s worth.

Balthasar · Guest

On Thu, 31 Jul 2008 14:30:54 -0700 (PDT), MoeBlee <jazzmobe@hotmail.com>
wrote:

[quote]
So if you make it an axiom, you>ll need to
tell us what the rest of your axioms are.

WHY?! Since he>s a crank, he needn>t. :-)[/quote]

B.

--

"For every line of Cantor>s list it is true that this line does not
contain the diagonal number. Nevertheless the diagonal number may
be in the infinite list." (WM, sci.logic)

Virgil · Guest

In article <9ka4945kfb8bklpec9s9veaa7qgmfcrkuk@4ax.com>,
Balthasar <nomail@invalid> wrote:

[quote]The original list is infinite. By extending the argument:
Any such diagonal sequence differs from the "last sequence"
in the "last place".

Errr... what "last sequence"? Since the considered list does not have a
last sequence, it>s hard to see what this idiot might mean here.
[/quote]
And to compound the idiot>s idiocy of a "last" sequence in a list
without any "last" sequence, he refers to a "last" place in a set of
sequences none of which have "last" places.

At least doubly idiotic.

Balthasar · Guest

On Thu, 31 Jul 2008 15:58:51 -0700 (PDT), MoeBlee <jazzmobe@hotmail.com>
wrote:

[quote]
In a fully-computable realm, ...

Do we have a formal definition of /fully-computable realm/, Moe?[/quote]

Moreover does this idiot ASSUME that the context of our considerations
is a "fully-computable realm"? - Whatever that may be.

[quote]
... any formalizable property does express a set [...].

Fascinating. One might think that (at least in a "standard realm")[/quote]
/being not an element of itself/ is a formalizable property:

x !e x.

Still it>s hard to see how in a "fully-computable realm" the Russell-set
can exist, i.e. a set y such that

Ax(x e y <-> x !e x).

Note that this "property" (in a class theory) does "express" (determine)
a proper class. (Though in NF, where there are only sets, the predicate
"x !e x" is not admissible, since it>s not stratified.)

[quote]
It is rather the accepted approach that is problematic in this respect.

Crank speak.[/quote]

B.

--

"For every line of Cantor>s list it is true that this line does not
contain the diagonal number. Nevertheless the diagonal number may
be in the infinite list." (WM, sci.logic)


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