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|-|erc · Guest

"Mariano Suárez-Alvarez" <mariano.suarezalvarez@gmail.com> wrote
On Jul 30, 7:02 am, "Peter Webb"
<webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
[quote]So...

Say that I pick a number between 1 and 10, but
I do not tell you which. Then you make a list of all
thenumbers between 1 and 10. Do you think that after
making that list you know which number I picked?

-- m

********************
Nicely argued.
[/quote]
-And he agreed!

No, I said I could *write down* your number (by writing down all 10)

Just like I can write down all the digits of Omega in sequence,
by computing the set of all computable reals. If you think I can>t,
then which digit am I going to miss?

Computable reals displays EVERY type of decimal expansion imaginable,
there is nothing it misses.

Herc

|-|erc · Guest

"Virgil" <Virgil@gmale.com> wrote ...
[quote]In article
72ad002f-da47-4eda-8781-7ef8b13776da@f63g2000hsf.googlegroups.com>,
"Dave L. Renfro" <renfr1dl@cmich.edu> wrote:

|-|erc wrote:

Do you agree that the 1st 10 digits of Omega appear
in sequence in the right position, somewhere in the
list of computable reals?

Virgil wrote:

I do not even agree that Omega has any digits appearing any place.

I believe he>s talking about Chaitin>s omega.

http://www.google.com/search?q=Chaitin>s-omega

Dave L. Renfro

If he refuses to say which one, the he is hardly in a position to object
to my counter based on an omega rather more relevant to Cantor.
[/quote]

I>m using Herc>s Omega, I defined it already.

its a real number. if the 1st Turing machine halts, Omega begins 0.1,
if the 1st Turing Machine runs forever, Omega begins 0.0. Similarly
for the second digit and so on, giving it a binary representation.

Herc

Guest · Posted: Thu Jul 31, 2008 5:32 am Post subject: Re: the problem with Cantor

On Jul 29, 3:25 pm, "|-|erc" <h...@r.c> wrote:
[quote]When you consider computable reals 0<=cr<1
what variety do you get in the decimal expansions?
[/quote]
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.

Peter Webb · Guest

"|-|erc" <h@r.c> wrote in message
news:XD7kk.24268$IK1.5635@news-server.bigpond.net.au...
[quote]"Mariano Suárez-Alvarez" <mariano.suarezalvarez@gmail.com> wrote
On Jul 30, 7:02 am, "Peter Webb"
webbfam...@DIESPAMDIEoptusnet.com.au> wrote:
So...

Say that I pick a number between 1 and 10, but
I do not tell you which. Then you make a list of all
thenumbers between 1 and 10. Do you think that after
making that list you know which number I picked?

-- m

********************
Nicely argued.

-And he agreed!

No, I said I could *write down* your number (by writing down all 10)

Just like I can write down all the digits of Omega in sequence,
by computing the set of all computable reals.
[/quote]

You can>t compute the set of computable reals. If you think you can, give me
an algorithm for doing so.

[quote]If you think I can>t,
then which digit am I going to miss?

[/quote]
It depends upon how you define Omega; ie, the rules for the TM that you use.

[quote]Computable reals displays EVERY type of decimal expansion imaginable,
there is nothing it misses.

[/quote]
That statement may or may not be true, depending upon how you define the
word "imaginable". Give me a definition, and I will tell you if your
statement is correct.

Ross A. Finlayson · Guest

lwalke3@lausd.net wrote:
[quote]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.
[/quote]
Where ZF>s universe is the "set of all sets that don>t contain
themselves", which it is in some other theory or ZF would be complete,
which via Goedel>s incompleness would have it inconsistent, then ZF>s
universe is the Russell set. Skolemize, it>s countable: bijecting to
the natural numbers while containing itself, so they contain themselves.

It>s the contrapositive (else it>s not), it does hold in ZFC because if
it didn>t then ZFC would yield a contradiction, so it does:
contradiction: ZFC is inconsistent.

That article by Chaitin that Bader notes about the sampling of the
programs is quite remarkable.

http://plus.maths.org/issue37/features/omega/index.html

For example, he suggests that a program of length n can be sampled by
sampling n-many bits at random, but, then he has sampled n-many
different programs, of length 1, ..., n. Consider sampling a real
number, uniformly from the unit interval, where "like any number" it>s
sampled by successive bits of its expansion. In sampling a rational
number a particular rational is sampled infinitely many times, in an
irrational, each distinctly once. In the consideration of infinite
programs, eg loops, there are many ways to consider statically why a
program, even if it never halts (which basically has that it loops in
place, the instruction at address x is: goto x) instead loopingly halts
or it can be compressed.

Borel vs. Combinatorics, anyone?

There>s only one theory with no axioms, consistent and complete, it>s a
theory of nothing.

Coincidentally, that>s everything.

Regards,

Ross F.

Tonico · Guest

On Jul 31, 8:32 am, lwal...@lausd.net wrote:
[quote]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.
[/quote]
***************************************************************

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...

Of course, it could be |-|erc didn>t actually mean the above...

Regards
Tonio

Well

MoeBlee · Guest

On Jul 30, 10:32 pm, lwal...@lausd.net wrote:

[quote]And so another nonstandard mathematician |-|erc, has
appeared out of lurkdom to refute Cantor.
[/quote]
Funny that you think Herc is a MATHEMATICIAN.

MoeBlee

Guest · Posted: Thu Jul 31, 2008 7:58 pm Post subject: Re: the problem with Cantor

On 31 Jul, 09:24, Tonico <Tonic...@yahoo.com> wrote:
[quote]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...
[/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. 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]Of course, it could be |-|erc didn>t actually mean the above...

Regards
Tonio

Well[/quote]

MoeBlee · Guest

On Jul 31, 12:58 pm, ju...@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. 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]
"the last sequence", "the last place", and "the limit sequence". All
undefined nonsense.

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"?

MoeBlee

Guest · Posted: Thu Jul 31, 2008 8:02 pm Post subject: Re: the problem with Cantor

On 31 Jul, 20:58, 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.
[/quote]

In this sense, I guess we might call all possible diagonal sequences
for any enumeration rule of the list: the class of omega sequences for
that rule.

-LV

[quote]
-LV

Of course, it could be |-|erc didn>t actually mean the above...

Regards
Tonio

Well[/quote]

Guest · Posted: Thu Jul 31, 2008 8:06 pm Post subject: Re: the problem with Cantor

On 31 Jul, 21:02, MoeBlee <jazzm...@hotmail.com> wrote:
[quote]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.
[/quote]

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?

-LV

[quote]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"?

MoeBlee[/quote]

Dave L. Renfro · Guest

ju...@diegidio.name 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.

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 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

Guest · Posted: Thu Jul 31, 2008 9:20 pm Post subject: Re: the problem with Cantor

On 31 Jul, 22:05, "Dave L. Renfro" <renfr...@cmich.edu> wrote:
[quote]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
[/quote]
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.

-LV

MoeBlee · Guest

On Jul 31, 1:06 pm, ju...@diegidio.name wrote:
[quote]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?
[/quote]
Please READ what I wrote:

[quote]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"?
[/quote]
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

MoeBlee · Guest

On Jul 31, 2:20 pm, ju...@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]
You mean the existence of a list (an ENUMERATION) of the set of anti-
diagonals of all lists of computable reals? Then make that an axiom
along with WHAT OTHER axioms for mathematics? It>s inconsistent with
the axioms of Z set theory. So if you make it an axiom, you>ll need to
tell us what the rest of your axioms are.

MoeBlee


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