the problem with Cantor
|-|erc
what variety do you get in the decimal expansions?
Every possible sequence of digits to infinite length
is represented as a computable number. You might
argue Omega is not represented in it entirety, but
Omega to 10 decimal places is computable,
Omega to 100 decimal places is computable,
Omega to a googol decimal places is computable,
...
What does that ... mean? It means Omega is computable
to infinite decimal places. OK you argue, show me the
actual computable real where it is apparent to oo decimal
places and in theory there is none. But the set of CR does
contain this sequence of digits. ALL sequences of digits
to oo are computable. And I think that's good enough to
refute the existence of sets that are uncountably larger than oo.
FACT: All sequences of digits are computable to infinite length.
Herc
--
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Robert J. Kolker
Not true. The set of possible infinite sequences is a non-countable set.
The set of computable (i.e. Turing computable) infinite sequences is a
countable set.
Bob Kolker
fishfry
"|-|erc" <h@r.c> wrote:
> When you consider computable reals 0<=cr<1
> what variety do you get in the decimal expansions?
>
> Every possible sequence of digits to infinite length
> is represented as a computable number.
No, that's just not true. There are uncountablyl many such sequences,
but only countably many computable numbers.
You might
> argue Omega is not represented in it entirety, but
> Omega to 10 decimal places is computable,
> Omega to 100 decimal places is computable,
> Omega to a googol decimal places is computable,
> ...
What do you mean here by Omega?
Anyway, your reasoning is flawed. 10 is a finite natural number; and
10^100 is a finite natural number; and 10^googol is a finite natural
number; but the set of all finite natural numbers, N, is not a fiinite
natural number.
You have to be careful when passing to infinity.
>
> What does that ... mean? It means Omega is computable
> to infinite decimal places.
You haven't used any properties of computable numbers here. Whatever you
mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
the order type of N?), you haven't shown anything about its
computability.
>OK you argue, show me the
> actual computable real where it is apparent to oo decimal
> places and in theory there is none. But the set of CR does
> contain this sequence of digits. ALL sequences of digits
> to oo are computable. And I think that's good enough to
> refute the existence of sets that are uncountably larger than oo.
>
> FACT: All sequences of digits are computable to infinite length.
>
Clearly not.
|-|erc
OK, the set of possible infinite sequences are computable to how many digits?
Herc
|-|erc
> In article <JNMjk.23889$IK1....@news-server.bigpond.net.au>,
> "|-|erc" <h@r.c> wrote:
>
> > When you consider computable reals 0<=cr<1
> > what variety do you get in the decimal expansions?
> >
> > Every possible sequence of digits to infinite length
> > is represented as a computable number.
>
> No, that's just not true. There are uncountablyl many such sequences,
> but only countably many computable numbers.
>
>
>
>
> You might
> > argue Omega is not represented in it entirety, but
> > Omega to 10 decimal places is computable,
> > Omega to 100 decimal places is computable,
> > Omega to a googol decimal places is computable,
> > ...
>
> What do you mean here by Omega?
Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
something like that
>
> Anyway, your reasoning is flawed. 10 is a finite natural number; and
> 10^100 is a finite natural number; and 10^googol is a finite natural
> number; but the set of all finite natural numbers, N, is not a fiinite
> natural number.
>
> You have to be careful when passing to infinity.
How many digits is Omega computable to?
>
>
>
> >
> > What does that ... mean? It means Omega is computable
> > to infinite decimal places.
>
> You haven't used any properties of computable numbers here. Whatever you
> mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
> the order type of N?), you haven't shown anything about its
> computability.
>
>
> >OK you argue, show me the
> > actual computable real where it is apparent to oo decimal
> > places and in theory there is none. But the set of CR does
> > contain this sequence of digits. ALL sequences of digits
> > to oo are computable. And I think that's good enough to
> > refute the existence of sets that are uncountably larger than oo.
> >
> > FACT: All sequences of digits are computable to infinite length.
> >
>
> Clearly not.
OK, the set of possible infinite sequences are computable to how many digits?
Herc
Peter Webb
"|-|erc" <h@r.c> wrote in message
news:NePjk.23952$IK1....@news-server.bigpond.net.au...
Any finite number.
Peter Webb
"|-|erc" <h@r.c> wrote in message
That depends upon exactly how Omega is defined, ie the TM rules.
You might be able to work out the first n digits (with n finite) if you can
prove that for all n digits you can use some argument to prove that all of
these definitely halt or don't halt. However, no such proof is possible in
general; at some point in the expansion you will just have to say that you
don't know whether the TM terminates, and at that point you don't know what
the rest of Omega looks like.
>
>
>>
>>
>>
>> >
>> > What does that ... mean? It means Omega is computable
>> > to infinite decimal places.
>>
>> You haven't used any properties of computable numbers here. Whatever you
>> mean by Omega (do you mean lower-case omega, 'w' in ASCII, to represent
>> the order type of N?), you haven't shown anything about its
>> computability.
>>
>>
>> >OK you argue, show me the
>> > actual computable real where it is apparent to oo decimal
>> > places and in theory there is none. But the set of CR does
>> > contain this sequence of digits. ALL sequences of digits
>> > to oo are computable. And I think that's good enough to
>> > refute the existence of sets that are uncountably larger than oo.
>> >
>> > FACT: All sequences of digits are computable to infinite length.
>> >
>>
>> Clearly not.
>
> OK, the set of possible infinite sequences are computable to how many
> digits?
>
> Herc
>
>
As I said before, any finite number.
Mariano Suárez-Alvarez
> "fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote
>
>
>
> > In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
> > "|-|erc" <h...@r.c> wrote:
>
> > > When you consider computable reals 0<=cr<1
> > > what variety do you get in the decimal expansions?
>
> > > Every possible sequence of digits to infinite length
> > > is represented as a computable number.
>
> > No, that's just not true. There are uncountablyl many such sequences,
> > but only countably many computable numbers.
>
> > You might
> > > argue Omega is not represented in it entirety, but
> > > Omega to 10 decimal places is computable,
> > > Omega to 100 decimal places is computable,
> > > Omega to a googol decimal places is computable,
> > > ...
>
> > What do you mean here by Omega?
>
> Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
> something like that
Are you supposed to *know* what Omega is? That's a
very mild requirement for your original post to have
any sense at all...
Now you are asking what the first bit is?
> [ ... yet another instance of the apparently endemic
> sci.math disease causing people to `pass to the limit'
> with no regards to meaningfulness, correctness or even
> manners, snipped. ]
-- m
|-|erc
|-|erc
> > OK, the set of possible infinite sequences are computable to how many
> > digits?
> >
> > Herc
> >
> >
>
> Any finite number.
Bill Goates has so much money in the bank.
If you asked whether he had any finite number of money in the bank
the answer was always yes, how much money total does he have?
Infinite.
Herc
|-|erc
In a binary system of numbers, roughly 1/2 of all computable reals (0<=cr<1)
begin with 1, and the other 1/2 begin with 0.
Say Omega starts with 0.1
1/2 of CR also begin with 0.1.
Say Omega starts with 0.10
1/4 of CR also begin with 0.10
Omega then becomes 0.101
1/8 of CR also begin with 0.101
No matter how many digits of Omega you consider, a percentage of computable
reals will compute that string of digits.
As the number of reals in the list of computable reals approaches infinity,
the number of digits of Omega that are computed approaches infinity.
Considering the entire infinite set of computable reals, it contains the sequence
of digits contained in Omega (to infinite length)
Herc
Virgil
"|-|erc" <h@r.c> wrote:
>
> Say Omega starts with 0.1
What requires "Omega" to have any binary representation at all?
Is |-| uck claiming that omega is a computable real that falls between 0
and 1?
Unless he is, the rest of his post is garbage.
|-|erc
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.
> Is |-| uck claiming that omega is a computable real that falls between 0
> and 1?
no. I'm claiming all the digits of Omega appear in sequence in the set of
computable reals.
>
> Unless he is, the rest of his post is garbage.
True or False?
As the number of reals in the list of computable reals approaches infinity,
the number of digits of Omega that are computed approaches infinity.
Herc
|-|erc
On Jul 29, 10:16 pm, "|-|erc" <h...@r.c> wrote:
> "fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote
>
>
>
> > In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
> > "|-|erc" <h...@r.c> wrote:
>
> > > When you consider computable reals 0<=cr<1
> > > what variety do you get in the decimal expansions?
>
> > > Every possible sequence of digits to infinite length
> > > is represented as a computable number.
>
> > No, that's just not true. There are uncountablyl many such sequences,
> > but only countably many computable numbers.
>
> > You might
> > > argue Omega is not represented in it entirety, but
> > > Omega to 10 decimal places is computable,
> > > Omega to 100 decimal places is computable,
> > > Omega to a googol decimal places is computable,
> > > ...
>
> > What do you mean here by Omega?
>
> Isn't the 1st digit of Omega 1 if the 1st Turing machine halts, 0 otherwise
> something like that
-Are you supposed to *know* what Omega is? That's a
-very mild requirement for your original post to have
-any sense at all...
-Now you are asking what the first bit is?
No we don't need to know any values of Omega. We know by redundancy
that we have computed Omega to n digits because every possible sequence
to n digits has been computed.
Herc
Mariano Suárez-Alvarez
> > "|-|erc" <h...@r.c> wrote:
>
> > > "Peter Webb" <webbfam...@DIESPAMDIEoptusnet.com.au> wrote >
> > > > "|-|erc" <h...@r.c> wrote in message
> > > >news:UhPjk.23954$IK1....@news-server.bigpond.net.au...
> > > > > "fishfry" <BLOCKSPAMfish...@your-mailbox.com> wrote
> > > > >> In article <JNMjk.23889$IK1.3...@news-server.bigpond.net.au>,
There is another option: meaningless.
-- m
Mariano Suárez-Alvarez
> "Mariano Suárez-Alvarez" <mariano.suarezalva...@gmail.com> 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
|-|erc
=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?
Now you get it! What I said was, in this context, is that I can write down your number.
Do you agree that the 1st 10 digits of Omega appear in sequence in the right position,
somewhere in the list of computable reals?
Herc
|-|erc
> True or False?
> As the number of reals in the list of computable reals approaches infinity,
> the number of digits of Omega that are computed approaches infinity.
[m]
-There is another option: meaningless.
So you're saying no digits of Omega are ever computed, by any computer ever?
Can anyone else evaluate whether my proposition above is true or false?
Herc
kunzmilan
I tried to prove something similar.
But, when we define the rational matrix R, r(i,j) = j/i,
we see that there are as many rational numbers lesser than
1, as rational numbers greater than 1. Since in the first row of R,
all natural numbers n are enumerated, rational numbers greater than 1
overflow capacity of natural numbers to count, regardless to the
infinite length of the string of natural numbers. This infinity
increases the size of the rational matrix R, too.
kunzmilan
Peter Webb
"|-|erc" <h@r.c> wrote in message
Any = for all n e N. Which somewhat equates to "all", except in common
English "all" is slightly ambiguous in that it can mean either "any" or "the
set of".
But the deal is, you give me any finite number and I can work it out to that
precision.
Peter Webb
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.
Dave L. Renfro
I believe the issue is the existence of a uniform
computability procedure for each n-digit truncation
of the number. Sure, there is such a procedure
for the first 100 digits, such a procedure for
the first 1000 digits, such a procedure for the
first 10000 digits, and so on, but these procedures
have to be "essentially the same" for all such
n-digit truncations if we're going to use them
to verify the computability of the number. For
computable numbers this can be arranged, but for
non-computable numbers this can't be arranged.
Kind of like uniform continuity vs. continuity,
or any situation in which a statement of the form
(for all x)(there exists y) is true, but not the
corresponding statement where the quantifiers
are switched, (there exists y)(for all x).
Anyway, I don't know much about formal computability,
so maybe someone who does can comment on whether
this general idea has any bearing on the actual
issue at hand.
Dave L. Renfro
Virgil
"|-|erc" <h@r.c> wrote:
> "Virgil" <Vir...@gmale.com> wrote in...
> > > Say Omega starts with 0.1
> >
> > What requires "Omega" to have any binary representation at all?
>
> its a real number.
By what definition of "real number"?
The most commonly accepted definitions of "real number" involve either
Cauchy sequences or Dedekind cuts of rationals, neither of which allow
omega as a real number, so what is your definitions of "real number"
that includes omega?
>
>
> > Is |-| uck claiming that omega is a computable real that falls between 0
> > and 1?
>
> no. I'm claiming all the digits of Omega appear in sequence in the set of
> computable reals.
You first have to establish that omega has digits.
>
> >
> > Unless he is, the rest of his post is garbage.
>
> True or False?
No, garbage
> As the number of reals in the list of computable reals approaches
> infinity,
> the number of digits of Omega that are computed approaches infinity.
Not even false.
Virgil
"|-|erc" <h@r.c> 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?
I do not even agree that Omega has any digits appearing any place.
Virgil
"|-|erc" <h@r.c> wrote:
> [Herc]
> > True or False?
> > As the number of reals in the list of computable reals approaches infinity,
> > the number of digits of Omega that are computed approaches infinity.
>
> [m]
> -There is another option: meaningless.
>
> So you're saying no digits of Omega are ever computed, by any computer ever?
One would be well advised to avoid use of any computer on which any such
digits have been computed.
>
> Can anyone else evaluate whether my proposition above is true or false?
I see questions, but no proposition in this posting.
Mariano Suárez-Alvarez
And he agreed!
-- m
hagman
> When you consider computable reals 0<=cr<1
> what variety do you get in the decimal expansions?
>
> Every possible sequence of digits to infinite length
> is represented as a computable number. You might
> argue Omega is not represented in it entirety, but
> Omega to 10 decimal places is computable,
> Omega to 100 decimal places is computable,
> Omega to a googol decimal places is computable,
> ...
>
> What does that ... mean? It means Omega is computable
> to infinite decimal places. OK you argue, show me the
> actual computable real where it is apparent to oo decimal
> places and in theory there is none. But the set of CR does
> contain this sequence of digits. ALL sequences of digits
> to oo are computable. And I think that's good enough to
> refute the existence of sets that are uncountably larger than oo.
>
> FACT: All sequences of digits are computable to infinite length.
>
> --
> ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Whoa! This looks a bit like the WM tree in disguise!
hagman
Dave L. Renfro
>> 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
Ralf Bader
He has been asked what he means by omega, but didn't give a sensible answer.
Maybe it is something like what is called omega here:
http://plus.maths.org/issue37/features/omega/index.html
But who knows.
Ralf
Virgil
<72ad002f-da47-4eda...@f63g2000hsf.googlegroups.com>,
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.
|-|erc
-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
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
Peter Webb
"|-|erc" <h@r.c> wrote in message
> "Mariano Suárez-Alvarez" <mariano.su...@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.
You can't compute the set of computable reals. If you think you can, give me
an algorithm for doing so.
> If you think I can't,
> then which digit am I going to miss?
>
It depends upon how you define Omega; ie, the rules for the TM that you use.
> Computable reals displays EVERY type of decimal expansion imaginable,
> there is nothing it misses.
>
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.
lwa...@lausd.net
> 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.
Ross A. Finlayson
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
***************************************************************
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
> And so another nonstandard mathematician |-|erc, has
> appeared out of lurkdom to refute Cantor.
Funny that you think Herc is a MATHEMATICIAN.
MoeBlee
ju...@diegidio.name
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
MoeBlee
> 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.
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
ju...@diegidio.name
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
ju...@diegidio.name
> 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?
-LV
Dave L. Renfro
> 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
Virgil
<9a63cf17-a30b-4865...@m44g2000hsc.googlegroups.com>,
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.
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.
> 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 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
<09f99364-6b14-4ebf...@25g2000hsx.googlegroups.com>,
ju...@diegidio.name wrote:
> 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?
Among other things, how a sequence which differs from every member of a
list can still be a member of that list.
ju...@diegidio.name
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
> 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
Balthasar
<renf...@cmich.edu> 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.
>>
And so on in this case means: it differs from ALL entries (i.e
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?!
>>
>> 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.
>>
>> That is to say ...
>>
Yes?
>>
>> that, at the limit, ...
>>
Limit? What limit? Hell!!!
>>
>> 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
_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
>
> 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
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)
MoeBlee
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
Virgil
<4383c973-d11d-44e6...@t54g2000hsg.googlegroups.com>,
ju...@diegidio.name wrote:
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.
Virgil
Balthasar <nomail@invalid> wrote:
> >> 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.
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
wrote:
>
> 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. :-)
ju...@diegidio.name
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
> 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.)
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.
> 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.
Dave L. Renfro
MoeBlee
> 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
MoeBlee
> > 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.
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
ju...@diegidio.name
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
ju...@diegidio.name
> 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)
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?)
> 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? 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
MoeBlee
> 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?)
Not intuitionistic logic ALONE. I said COMBINED. You are contradicting
intuitionistic logic COMBINED with (2):
> > 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?
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).
> 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.
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.
> > 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?
No answer from you.
MoeBlee
Tonico
**************************************************************
"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
Balthasar
wrote:
>>
>> In a fully-computable realm, ...
>>
Do we have a formal definition of /fully-computable realm/, Moe?
Moreover does this idiot ASSUME that the context of our considerations
is a "fully-computable realm"? - Whatever that may be.
>>
>> ... any formalizable property does express a set [...].
>>
Fascinating. One might think that (at least in a "standard realm")
/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.)
>>
>> It is rather the accepted approach that is problematic in this respect.
>>
Crank speak.
lwa...@lausd.net
> > Any finite number.
> Bill Goates has so much money in the bank.
> If you asked whether he had any finite number of money in the bank
> the answer was always yes, how much money total does he have?
> Infinite.
In this post, we can see exactly what Herc is thinking, and
why his thinking is not considered valid in ZFC.
Herc believes that if phi(n) for every natural n then phi(N).
What makes this so-called "infinite induction" seem valid
is that it holds for a few chosen cases.
Suppose phi(n) is the formula "x dominates n" -- that is,
the cardinality of x is at least n. Then phi(N) must also
hold in ZFC (_not_ ZF+~AC, so there's no need for anyone
to bring up Dedekind finite sets) -- "x dominates N," or
the cardinality of x is at least that of N.
And the Bill Gates analogy as basically referring to how
much money he has in the bank -- a cardinality question. So
if phi(n) is the formula "Gates has at least n cents in the bank"
and phi(n) holds for every natural n, then it _does_ hold that
his bank account is infinite.
But the problem is that it doesn't hold for general phi. In
particular, one can't conclude -- at least not in ZFC -- that if
phi(n) is the formula "the truncation of Omega to n decimal
digits" and phi(n) holds for every n (which it does, since each
of these approximations to n is rational hence computable)
then phi(N) -- that Omega is in the list -- must hold as well.
Indeed, if it did, then since the list consists only of computable
numbers and it contains Omega, then one must conclude that
Omega is computable -- but I know that even Herc would
disagree with that conclusion. Herc is trying to refute Cantor,
not Turing here. It's not that Herc doesn't believe in the
existence of uncomputable numbers -- it's that he believes that
only countably many reals exist.
As always, I believe that there can be rigorous theories in
which the claims made by so-called "cranks" such as Herc
can hold. There are two ways to make card(N) = card(R) --
either by making N bigger or by making R smaller.
Perhaps, in the spirit of this thread, we could make R smaller
by insisting that only computable numbers exist. Then there
would be no Omega or any other numbers missing from the
list, and so Herc's claim that there exist only countably many
reals can still hold.
of
Tonico
> On Thu, 31 Jul 2008 15:58:51 -0700 (PDT), MoeBlee <jazzm...@hotmail.com>
***********************************************************
This quote by the almost unbelievable and psychodelic WM should be
framed and hung at the entrance of his office in that rather
suspicious and highly loose german college, or whatever, in which he
teaches...or stuff.
Regards
Tonio
|-|erc
>
> > 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.
***************************************************************
[Tonico]
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...
[Herc]
Well spotted flipside of my claim - the diag argument.
The list will contain every finite prefix of the anti-diagonal.
When the diag argument was invented we imagine something like this.
123
456
789
The diagonal is 159
The antidiagonal is 261
Voila - 261 is not on the list.
But, if the list is the computable reals,
2 is on the list
26 is on the list
261 is on the list
Every finite prefix of the antidiagonal, UP TO OO LENGTH is on the list.
This gives me grave doubts of claims that the sequence is missing.
In a way the sequence is there, every finite prefix to infinite length means
you don't miss a digit. Sorry to have to tell you but you're all fools,
only when Artificial Intelligence arrives in 20 years will you be corrected. :)~
Herc
Tonico
> "Tonico" <Tonic...@yahoo.com> wrote
**********************************************************
Yeah...that sounded like :"you're wrong, you suck, I'm right I win.
Wait till jesus (muhammed, buddha, your-favourite-god) arrives...just
wait! Then you'll see who's the fool!"
Well, some fundies already mention NOW the year 2012, after many
failed dates in the past, and now you've mentioned A.I. in 20 years
more (which I can't fully understand what has to do with this
issue...). We shall see, I supose, and no further debate is worth.
Regards
Tonio
Balthasar
wrote:
>>
>> "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)
>>
> This quote by the almost unbelievable and psychodelic WM should be
> framed and hung at the entrance of his office in that rather
> suspicious and highly loose german college, or whatever, in which he
> teaches...or stuff.
>
Agree.
Also quite nice:
"I can prove that there exists an n which requires more than 10^80 bits
to name it. This n does not exist."
(W. Mueckenheim)
"Remember, there exists an infinite set. But there is no claim that all
its elements exist."
(W. Mueckenheim)
Anyway, he really knows his stuff:
"[Consider] the potentially infinite set N. If we add another element,
the set remains the same."
(W. Mueckenheim)
B.
Balthasar
> On Fri, 1 Aug 2008 00:58:28 -0700 (PDT), Tonico <Toni...@yahoo.com>
> wrote:
>>>
>>> "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)
>>>
>> This quote by the almost unbelievable and psychodelic WM should be
>> framed and hung at the entrance of his office in that rather
>> suspicious and highly loose german college, or whatever, in which he
>> teaches...or stuff.
>>
"My goal is to show that set theory is self contradictory,
in particular Cantor's claim of the infinite set of finite
naturals. This goal is not difficult to achieve. There are
several proofs."
(W. Mückenheim)
B.
ju...@diegidio.name
> On Thu, 31 Jul 2008 15:58:51 -0700 (PDT), MoeBlee <jazzm...@hotmail.com>
> wrote:
>
> >> In a fully-computable realm, ...
>
> Do we have a formal definition of /fully-computable realm/, Moe?
>
> Moreover does this idiot ASSUME that the context of our considerations
> is a "fully-computable realm"? - Whatever that may be.
Now that the discussion is over and all results are clear, i can take
the time for you:
Apart from your absolute cluelessness, you are by far the most
revolting piece of shit around.
Keep enjoying your conversantion.
-LV
MoeBlee
> As always, I believe that there can be rigorous theories in
> which the claims made by so-called "cranks" such as Herc
> can hold.
Not just "so-called".
> There are two ways to make card(N) = card(R) --
> either by making N bigger or by making R smaller.
Since all Peano systems are isomorphic with one another, and since all
complete ordered fields are isomorphic with one another, what you're
asking for is at least a pretty tall order.
Anyway, cranks aren't interested in rigorous theories.
MoeBlee
Balthasar
wrote:
>>
>> There are two ways to make card(N) = card(R) --
>> either by making N bigger or by making R smaller.
>>
>> [lwal..., designated crank]
>>
Does this asshole have a brain? If we "either make N bigger or R
smaller", the resulting set(s) would not be N and/or R anymore.
*sigh* (Clearly there's a REASON why we write "R*" and "N*" in
non-standard analysis instead of "N" and "R".) Hence there's NO WAY "to
make card(N) = card(R)" IF "N" denotes the set of natural numbers, "R"
the set of real numbers, "Card" cardinality (defined in the spirit of
Cantor and Frege) and "=" identity (in the way WE -i.e. non-cranks- know
it).
Right, there are two way to make 1 = 2 --
either by making 1 bigger or by making 2 smaller.
(Actually, there's a third way: by redefining
"=" as =/=.)
Why oh why are mathematical cranks just so one-dimensional?
"[Consider] the potentially infinite set N. If we add another element,
the set remains the same." (W. Mueckenheim)
Balthasar
>>>
>>> There are two ways to make card(N) = card(R) --
>>> either by making N bigger or by making R smaller.
>>>
>>> [lwal..., designated crank]
>>>
> Does this asshole have a brain?
>
And if so - why doesn't he use it?
Virgil
"|-|erc" <h@r.c> wrote:
Who says they must all be written down in order to know that they are
all included in a set?
> 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...
>
Is Huck really claiming to be able to show that there is a number in in
a list and not in that same list? He seems to be challenging WM's
position as King Kook.
>
>
> [Herc]
> Well spotted flipside of my claim - the diag argument.
>
> The list will contain every finite prefix of the anti-diagonal.
>
> When the diag argument was invented we imagine something like this.
>
> 123
> 456
> 789
>
> The diagonal is 159
> The antidiagonal is 261
> Voila - 261 is not on the list.
>
> But, if the list is the computable reals,
> 2 is on the list
> 26 is on the list
> 261 is on the list
>
> Every finite prefix of the antidiagonal, UP TO OO LENGTH is on the list.
Up to but not including oo, which means every member of the list has a
sufficiently long but finite prefix differing from an equally long
prefix of the diagonal.
>
> This gives me grave doubts of claims that the sequence is missing.
Doubt all you want, but until you can support your doubts with better
evidence than you so far have been able to do, they will not convince
anyone else.
> In a way the sequence is there, every finite prefix to infinite length means
> you don't miss a digit. Sorry to have to tell you but you're all fools
But not so foolish as to be convinced by such foolish arguments as you
have been presenting.
|-|erc
**********************************************************
Regards
Tonio
You ignored my argument and focused on the little emoticon ending. No
wonder you want to withdraw now, your superinfinity is looking pretty weak.
You can't even form a unique sequence of digits using diagonalisation on my set
can you?
http://www.freewebs.com/namesort/linux.html
Tell me in your own words what this site does.
There's 3 buttons numbered 1 2 3.
Press 1 2 3 2 3 2 3 and describe what happens.
Then I'll present my proof.
Herc
|-|erc
You dumb fuck, I can still make a new number
[HERC]
I don't care about your new number, give me a new sequence of digits.
Herc
Virgil
"|-|erc" <h@r.c> wrote:
> I don't care
Certainly not about truth.
|-|erc
The truth is there is nothing larger than infinity.
You have failed to give me a new sequence of digits not on the computable reals.
Cantor's proof is an exercise in going round in circles.
Herc
David Bernier
Suppose we have a list LF of the numbers that have a terminating
decimal expansion:
2.3, 3.14159, 2000000, 0.00000000000056, etc.
1/3 = 0.33333.... =
__
0.3 [ 0.3periodic].
The first 5 decimals of 1/3 are the same as those of 0.33333 , and so
on for
5, 7, ... 100, ... 1000 decimals and more.
But if we take a fixed number 'c' on LF and compare it to
__
0.3 [ 0.3periodic] , then either
__
c < 0.3 or
__
c>0.3 .
Don't you think the same thing happens with Omega_{Herc} and LF?
After all, you're only comparing a finite (but arbitrarily large) number
of decimals.
In other words, what works for
Omega_{Herc} and LF should work for
__
0.3 and LF ...
David Bernier
|-|erc
You've shattered the illusion that Omega is on some list but the problem
with Cantors proof still stands.
Every finite prefix of possible sequences is computable, so the original
diagonalisation technique fails to find a new sequence.
123
456
789
In this simple finite list, the diagonal is 159.
The anti-diagonal is 261.
It looks like diagonalisation actually finds new sequences of digits.
but when you apply it to the set of computable reals it doesn't, the
list is already saturated with every possible sequence.
That's the claim, that there is a problem, not that Omega is on the list.
The computable reals contains every possible sequence up to infinite length,
and you tell me with a cloak and digger trick that you can find new sequences.
You can't. All the sequences are computed, up to oo length, end of story.
Have you taken a look at
http://www.freewebs.com/namesort/linux.html
It proves that a list of reals that contains every digit oo times in each column
can be sorted to have any random diagonal. The diagonal is independent of
such lists, e.g. the computable reals. That's another problem with Cantor's proof,
the diagonal does not even depend on the list, it's just a random variable.
Herc
Virgil
"|-|erc" <h@r.c> wrote:
> "Virgil" <Vir...@gmale.com> wrote in...
> > In article <qLLkk.24732$IK1....@news-server.bigpond.net.au>,
> > "|-|erc" <h@r.c> wrote:
> >
> >
> >
> >
> > > I don't care
> >
> > Certainly not about truth.
>
> The truth is there is nothing larger than infinity.
> You have failed to give me a new sequence of digits not on the computable
> reals.
You have failed to give us any listing of all computable reals.
The existence of such a listing would be proof that the computable reals
are countable, and the non-existence of any such listing would be
evidence off that set's uncountability.
> Cantor's proof is an exercise in going round in circles.
It is an eample of clearer thinking than Herkimer has yet demonstrated.
|-|erc
> In article <LkNkk.24748$IK1....@news-server.bigpond.net.au>,
> "|-|erc" <h@r.c> wrote:
>
> > "Virgil" <Vir...@gmale.com> wrote in...
> > > In article <qLLkk.24732$IK1....@news-server.bigpond.net.au>,
> > > "|-|erc" <h@r.c> wrote:
> > >
> > >
> > >
> > >
> > > > I don't care
> > >
> > > Certainly not about truth.
> >
> > The truth is there is nothing larger than infinity.
>
>
> > You have failed to give me a new sequence of digits not on the computable
> > reals.
>
> You have failed to give us any listing of all computable reals.
> The existence of such a listing would be proof that the computable reals
> are countable, and the non-existence of any such listing would be
> evidence off that set's uncountability.
>
>
> > Cantor's proof is an exercise in going round in circles.
>
> It is an eample of clearer thinking than Herkimer has yet demonstrated.
Use a Universal Turing Machine to compute all the reals.
let the jth digit of the ith real = UTM(i,j) mod 10
where UTM has 2 parameters, the TM it is emulating and the input tape to the TM.
UTM(TM, input) = output
It will have holes in it where the UTM doesn't halt, but thanks to multitasking we
can continue on computing every computable real.
Diagonalisation will not produce any new sequence of digits as I have demonstrated.
You are far too stupid to admit that and maintain an argument.
Herc
Virgil
"|-|erc" <h@r.c> wrote:
Then Herkimer must be claiming that the diagonal appears at some finite
position in the list. Which position is it in, dumbkopf?
>
> 123 456 789
>
> In this simple finite list, the diagonal is 159.
>
> The anti-diagonal is 261.
For the above simple finite list, there are 729 different base 10
anti-diagonals.
>
> It looks like diagonalisation actually finds new sequences of digits.
> but when you apply it to the set of computable reals it doesn't, the
> list is already saturated with every possible sequence.
Is any list of all computable reals itself computable? And given any
arbitrary infinite list of computable reals, is the diagonal for that
list computable?
Until Herk the Jerk can definitively answer those, one way or the
other, there are holes in his arguemnts.
> The computable reals contains every possible sequence up to infinite
> length, and you tell me with a cloak and digger trick that you can
> find new sequences.
I don't claim that any such "new" sequence is necessarily computable,
merely that for any list it exists.
> You can't. All the sequences are computed, up to oo length, end of
> story.
There is nothing in Cantor's proof that requires the 'anti-diagonal" to
be computable, merely existing in that non-physical way that numbers
have of existing.
>
> Have you taken a look at http://www.freewebs.com/namesort/linux.html
>
> It proves that a list of reals that contains every digit oo times in
> each column can be sorted to have any random diagonal.
It does not prove anything at all about such lists, since it only works
with finite lists of finite lengths.
> The diagonal is independent of such lists
> That's another problem with Cantor's
> proof, the diagonal does not even depend on the list, it's just a
> random variable.
Nonsense.
Cantor's original 'diagonal' proof applies only to lists of infinite
binary sequences with values in {m,w}.
So anything not valid for that original proof is irrelevant.
|-|erc
You can't correctly define an anti-diagonal. You also can't cite any
sequence of numbers that is new to the list. Oh sure it works for
123
456
789
but it doesn't work for the infinite set of computable reals. The computable reals
contains all finite sequences of digits up to oo length. Every pattern possible is
computable.
> >
> > 123 456 789
> >
> > In this simple finite list, the diagonal is 159.
> >
> > The anti-diagonal is 261.
>
> For the above simple finite list, there are 729 different base 10
> anti-diagonals.
If usenet posts had audio you'd hear clapping right now.
Did you remember the antidiagonals for the 6 permutations of the list?
> >
> > It looks like diagonalisation actually finds new sequences of digits.
> > but when you apply it to the set of computable reals it doesn't, the
> > list is already saturated with every possible sequence.
>
> Is any list of all computable reals itself computable? And given any
> arbitrary infinite list of computable reals, is the diagonal for that
> list computable?
yes computable things are computable
yes there is no contradiction in computing the diagonal somewhere in the list
>
> Until Herk the Jerk can definitively answer those, one way or the
> other, there are holes in his arguemnts.
>
> > The computable reals contains every possible sequence up to infinite
> > length, and you tell me with a cloak and digger trick that you can
> > find new sequences.
>
> I don't claim that any such "new" sequence is necessarily computable,
> merely that for any list it exists.
>
> > You can't. All the sequences are computed, up to oo length, end of
> > story.
>
> There is nothing in Cantor's proof that requires the 'anti-diagonal" to
> be computable, merely existing in that non-physical way that numbers
> have of existing.
Numbers exist because they can be computed. If they couldn't be computed
you wouldn't know what they were.
> >
> > Have you taken a look at http://www.freewebs.com/namesort/linux.html
> >
> > It proves that a list of reals that contains every digit oo times in
> > each column can be sorted to have any random diagonal.
>
> It does not prove anything at all about such lists, since it only works
> with finite lists of finite lengths.
Oh you had a look.
Claim: Let L be a list of numbers in [0,1] such that for each d in
{0,...,9} and for all n >= 1, the set {i| n-th digit of the decimal
representation of L_i is d} is infinite. Let D be a random variable
uniform on [0,1]. Then almost surely there is a permutation L' of L
such that D is the diagonal of L'.
>
>
>
> > The diagonal is independent of such lists
>
>
>
> > That's another problem with Cantor's
> > proof, the diagonal does not even depend on the list, it's just a
> > random variable.
>
> Nonsense.
Did you randomise the diagonal at the website, and produce an equivalent set
in the second list with that new diagonal? What does that mean?
>
> Cantor's original 'diagonal' proof applies only to lists of infinite
> binary sequences with values in {m,w}.
>
> So anything not valid for that original proof is irrelevant.
Anything valid for an equivalent proof is relevant.
This brings up an important point. Virgil, like most of you, believe numbers
and mathematical things exist without being computable? Tell me if you
subtracted the theories of mathematics that are computable from the total
set of theories in mathematics that you believe in, what are you left with?
Herc
Virgil
"|-|erc" <h@r.c> wrote:
> Use a Universal Turing Machine to compute all the reals.
Which of the uncountably many incomputable reals will it compute first?
lwa...@lausd.net
> If we "either make N bigger or R
> smaller", the resulting set(s) would not be N and/or R anymore.
> *sigh* (Clearly there's a REASON why we write "R*" and "N*" in
> non-standard analysis instead of "N" and "R".) Hence there's NO WAY "to
> make card(N) = card(R)" IF "N" denotes the set of natural numbers, "R"
> the set of real numbers, "Card" cardinality (defined in the spirit of
> Cantor and Frege) and "=" identity (in the way WE -i.e. non-cranks- know
> it).
> Right, there are two way to make 1 = 2 --
> either by making 1 bigger or by making 2 smaller.
> (Actually, there's a third way: by redefining
> "=" as =/=.)
In this post, Balthasar clearly compares the formulae
"card(N) = card(R)" and "1 = 2." We know that the
negation of both of these are theorems of ZFC. From
the perspective of Balthasar -- and most standard
mathematicians -- any poster who claims that
"card(N) = card(R)" might as well be claiming that
"1 = 2" -- either claim merits the "crank" label.
But I disagree and feel that there are worlds of
difference between "card(N) = card(R)" and "1 = 2,"
even though "(card(N) = card(R)) <-> (1 = 2)" is a
theorem of ZFC. Formulae involving infinite sets are
open to various interpretations, such as classical,
constructivism, intuitionism -- whereas those
involving finite sets are not subject to these
types of interpretations. Thus I consider "~(1 = 2)"
to be "absolutely true" in a way that the formula
"~(card(N) = card(R))" isn't -- since the former
holds for both classical and constructivist
mathematicians, unlike the latter where there is
some debate as to what N and R actually _are_.
Herc has mentioned computable reals in this thread,
as if he were a sort of constructivist who only
believes in the existence of computable reals.
To a constructivist, Omega simply _isn't_ a real
number -- it doesn't even _exist_. Omega is thus
_not_ an element of R. To someone who only
philosophically believes in the existence of
computable reals, there exist only countably
many reals -- and therefore there's a bijection
between N and R -- found by enumerating every
Turing algorithm that computes a real.
Perhaps, just as Balthasar has intimated with
his example of *N and *R, a constructivist
shouldn't use the symbol R to denote the set
of real numbers that he believes exists, but to
use some other notation such as Rc -- so that
Herc's claim becomes card(N) = card(Rc). Then
the symbol R would be reserved for the full,
classical set R -- anyone who doesn't believe
in the existence of the classical set R would
have no right to use the symbol R in a formula.
Maybe it's too bad that this isn't the case (cf.
the subthread of the WM thread in which
Balthasar, among others, discussed pedantry vs.
abuse of notation re: definition of card).
> Why oh why are mathematical cranks just so one-dimensional?
When I first started posting at sci.math about a
year ago, I came in fully prepared that someone
would call me a "crank." By the standard
mathematicians' definition of "crank," I am
indeed a borderline "crank," simply because I do
not state that Herc and other posters like him
are 100% wrong. I may not like the label, but I
cannot deny that I satisfy their definition of
the term.
And I'd rather be labeled a borderline "crank"
than blindly accept ZFC and classical mathematics
as the only theories worth discussing and call
anyone who posts anything that contradicts ZFC
or classical analysis "wrong."
Virgil
"|-|erc" <h@r.c> wrote:
Cantor did, at least for any list of binary sequences (which is all he
used, the decimal model came later and was done by others.
> You also can't cite any
> sequence of numbers that is new to the list.
I can cite a rule which will produce a sequence not in a given list.
>
> but it doesn't work for the infinite set of computable reals.
Then the rule must define an incomputable sequnce
> The computable
> reals
> contains all finite sequences of digits up to oo length. Every pattern
> possible is
> computable.
Since there are uncountably more real numbers than computable real
numbers, you must explain how that works.
>
>
> > >
> > > 123 456 789
> > >
> > > In this simple finite list, the diagonal is 159.
> > >
> > > The anti-diagonal is 261.
> >
> > For the above simple finite list, there are 729 different base 10
> > anti-diagonals.
>
> If usenet posts had audio you'd hear clapping right now.
> Did you remember the antidiagonals for the 6 permutations of the list?
>
> > >
> > > It looks like diagonalisation actually finds new sequences of digits.
> > > but when you apply it to the set of computable reals it doesn't, the
> > > list is already saturated with every possible sequence.
> >
> > Is any list of all computable reals itself computable? And given any
> > arbitrary infinite list of computable reals, is the diagonal for that
> > list computable?
>
> yes computable things are computable
There are things that one can speak of which are not, like the set of
all Dedekind cuts of the rationals or the set of all equivalence classes
of Cauchy sequences mod the null sequences.
Either of which makes the set of reals to computable and ssome members
of that set also not computable.
> yes there is no contradiction in computing the diagonal somewhere in the list
>
> >
> > Until Herk the Jerk can definitively answer those, one way or the
> > other, there are holes in his arguemnts.
> >
> > > The computable reals contains every possible sequence up to infinite
> > > length, and you tell me with a cloak and digger trick that you can
> > > find new sequences.
> >
> > I don't claim that any such "new" sequence is necessarily computable,
> > merely that for any list it exists.
> >
> > > You can't. All the sequences are computed, up to oo length, end of
> > > story.
There can, and must be be uncomputable sequences, at least unless there
are uncountably many computable sequences.
> >
> > There is nothing in Cantor's proof that requires the 'anti-diagonal" to
> > be computable, merely existing in that non-physical way that numbers
> > have of existing.
>
> Numbers exist because they can be computed.
> If they couldn't be computed
> you wouldn't know what they were.
That's what uncomputable numbers are like, you don't know exactly what
they are.
>
>
> > >
> > > Have you taken a look at http://www.freewebs.com/namesort/linux.html
> > >
> > > It proves that a list of reals that contains every digit oo times in
> > > each column can be sorted to have any random diagonal.
> >
> > It does not prove anything at all about such lists, since it only works
> > with finite lists of finite lengths.
>
> Oh you had a look.
>
> Claim: Let L be a list of numbers in [0,1] such that for each d in
> {0,...,9} and for all n >= 1, the set {i| n-th digit of the decimal
> representation of L_i is d} is infinite. Let D be a random variable
> uniform on [0,1]. Then almost surely there is a permutation L' of L
> such that D is the diagonal of L'.
"Almost surely" in any infinite set allows for infinitely many
exceptions.
"Almost surely" in an uncountably infinite set allows uncountably many
exceptions.
>
>
> >
> >
> >
> > > The diagonal is independent of such lists
> >
> >
> >
> > > That's another problem with Cantor's
> > > proof, the diagonal does not even depend on the list, it's just a
> > > random variable.
> >
> > Nonsense.
>
> Did you randomise the diagonal at the website, and produce an equivalent set
> in the second list with that new diagonal? What does that mean?
That each list has its own private diagonal(s) which need not work for
any other list.
>
> >
> > Cantor's original 'diagonal' proof applies only to lists of infinite
> > binary sequences with values in {m,w}.
> >
> > So anything not valid for that original proof is irrelevant.
>
> Anything valid for an equivalent proof is relevant.
>
> This brings up an important point. Virgil, like most of you, believe numbers
> and mathematical things exist without being computable?
I believe that when working with a set of axioms, whatever the axioms
require to exist does exist, at least within that axiom system.
Tell me if you
> subtracted the theories of mathematics that are computable from the total
> set of theories in mathematics that you believe in, what are you left with?
Most of mathematics, except for, say, finite groups.
|-|erc
I meant to say, use a UTM to compute all the computable reals, that's what
you asked for.
Although there are no incomputable reals.
Herc
herbzet
MoeBlee wrote:
> Since all Peano systems are isomorphic with one another ...
Not sure what you mean here. By the upward LS theorem, there are
models of PA of every infinite cardinality.
--
hz
lwa...@lausd.net
> On Fri, 01 Aug 2008 19:31:59 +0200, Balthasar <nomail@invalid> wrote:
> > Does this asshole have a brain?
> And if so - why doesn't he use it?
In Balthasar's opinion, if I were to use my
brain, I would see that Herc is 100% wrong
and that there's no possible interpretation
in which Herc could be right.
Of course, I'm fully aware that in that in
ZFC, Herc's reasoning is invalid. I already
know that the negations of Herc's claims are
theorems of ZFC.
But Herc already has enough people to tell
him that he's wrong. Why should I simply
repeat what others are telling him? Indeed,
I'd claim that for me to keep repeating "Herc,
you're wrong!" over and over again isn't
fully using my brain either.
I believe that repeating "You're wrong"
seldom wins anyone to one's point of view. I
prefer to say "I disagree" instead -- though
I'm not perfect and have probably called
several people "wrong" -- both standard and
"crank" -- during my first year of posting here.
When Herc makes a claim that N and R have the
same cardinality, I point out that his claim
doesn't hold in ZFC, but perhaps there's some
other rigorous theory in which his claim could
possibly hold. In other words, I'm trying to
find the shortest distance between the claim
and something that does hold.
Indeed, Balthasar has already mentioned *N
and *R. The sets *N and *R do have the same
cardinality, namely c. So this is a short
path from the claim to something that does
hold (though I don't believe that Herc
had *N and *R in mind, but this may be
similar to WM's ideas, since WM has mentioned
Robinson in his posts).
Since Herc mentions computable reals, we see
how there are only countably many reals that
are computable. So I use my brain and realize
that to someone who only believes in the
existence of computable reals, there are
exactly as many (computable) reals as there
are naturals.
This may not be _exactly_ what Herc has in
mind, but it's the shortest distance between
his claim that fails in ZFC and something that
does hold.
It's the difference between insulting someone
and constructive criticism. Many standard
mathematicians prefer the former -- since it
is easier -- while I strive for the latter. I
believe that it's better to find a theory in
which Herc's claims hold rather than simply
repeat how he's wrong. And I have found such
a theory -- the theory of computable reals.
Virgil
<157ede31-794c-4177...@27g2000hsf.googlegroups.com>,
lwa...@lausd.net wrote:
I am of the opinion that there is more than one Turing algorithm to
compute any computable real, for some computable reals there are very
likely countably many such algorithms. So that constructability of such
a bijection is, at best, problematical.
What I object to is WM, and others, telling me that I am not allowed to
accept ZFC as a theory worth discussing. I have no objections to someone
preferring not to accept ZFC or whatever, but I do object to their
attempting to control what I am allowed to think about.
Particularly in the case of those like WM who are both mathematically
and logically incompetent.
Virgil
"|-|erc" <h@r.c> wrote:
> "Virgil" <Vir...@gmale.com> wrote in...
> > In article <ZlQkk.24844$IK1....@news-server.bigpond.net.au>,
> > "|-|erc" <h@r.c> wrote:
> >
> >
> > > Use a Universal Turing Machine to compute all the reals.
> >
> > Which of the uncountably many incomputable reals will it compute first?
>
> I meant to say, use a UTM to compute all the computable reals, that's what
> you asked for.
Some of those computable reals will take all the time your UTM has.
So it may never get past the first one.
>
> Although there are no incomputable reals.
What exists outside UTMs is not necessarily computable.
If Herkimer's universe is limited to what a UTM can produce, his life is
pretty lifeless.
|-|erc
> In article <aaSkk.24893$IK1....@news-server.bigpond.net.au>,
> "|-|erc" <h@r.c> wrote:
>
> > "Virgil" <Vir...@gmale.com> wrote in...
> > > In article <ZlQkk.24844$IK1....@news-server.bigpond.net.au>,
> > > "|-|erc" <h@r.c> wrote:
> > >
> > >
> > > > Use a Universal Turing Machine to compute all the reals.
> > >
> > > Which of the uncountably many incomputable reals will it compute first?
> >
> > I meant to say, use a UTM to compute all the computable reals, that's what
> > you asked for.
>
> Some of those computable reals will take all the time your UTM has.
>
> So it may never get past the first one.
> >
> > Although there are no incomputable reals.
>
> What exists outside UTMs is not necessarily computable.
>
> If Herkimer's universe is limited to what a UTM can produce, his life is
> pretty lifeless.
brrrrr bing *pretty*
it does not compute
Herc
Alan Smaill
> Indeed, Balthasar has already mentioned *N
> and *R. The sets *N and *R do have the same
> cardinality, namely c. So this is a short
> path from the claim to something that does
> hold (though I don't believe that Herc
> had *N and *R in mind, but this may be
> similar to WM's ideas, since WM has mentioned
> Robinson in his posts).
I think I asked this before: *N and *R are not uniquely defined in
Robinson's approach.
So what do you mean when you say that *N and *R
have the cardinality c? There are certainly non-standard
models of PA that are countable.
Of course, there are countable models of the reals with their
arithmetic properties also, from Lowenheim-Skolem.
--
Alan Smaill
Horand....@googlemail.com
> What I object to is WM, and others, telling me that I am not allowed to
> accept ZFC as a theory worth discussing. I have no objections to someone
> preferring not to accept ZFC or whatever, but I do object to their
> attempting to control what I am allowed to think about.
Well said!
> Particularly in the case of those like WM who are both mathematically
> and logically incompetent
Or intentionally and maliciously dishonest. I'm still not sure about
WM, but my gut tells me that he knows his position is untenable, and
he gets his jollies out of riling up everyone else.
David Bernier
Let's say a Turing machine is decimal if, starting with
a tape with k consecutive 1s , where k is a positive integer, it
halts and leaves m(k) 1s on the tape, with 0<=m(k)<10 .
This makes precise the idea of a procedure in a programming language
with input = a positive integer and output = one of the 10 digits.
If you want the diagonal and anti-diagonal to be computable
and on the list, since diag and anti-diag depend on the ordering
of the list, I think it's for you to explain how to order the list
of decimal TMs so that diag and anti-diag are
*computable* ; also the list of decimal TMs should
be complete.
Above, you wrote: "The anti-diagonal is 261 " .
How do you know that the entire anti-diagonal is
computable?
David Bernier
|-|erc
The anti-diagonal is not computable, therefore it doesn't exist.
let UTM(real, digit) mod 10 calculate the list of reals. (where real is deceptively an integer)
the antidiagonal is UTM(digit, digit) + 1 mod 10
if this is computable, then some TM computes it and some emulated TM also computes it,
UTM(ad, digit) mod 10 = UTM(digit, digit) + 1 mod 10
when digit = ad,
UTM(ad, ad) mod 10 = UTM(ad, ad) + 1 mod 10
Contradiction!
Therefore antidiag is an invalid formula.
This is a more sensible conclusion than higher infinities exist as I've basically just
rearranged Cantor's proof. One can argue its a valid specification of a sequence
but it doesn't actually compute a new sequence of digits.
Herc
>
>
>
>
Virgil
<fb9df027-90be-450a...@e53g2000hsa.googlegroups.com>,
Horand....@googlemail.com wrote:
WM has too many published papers supporting his position to be doing it
for fun.
If those papers were as publicly discredited as they ought to be, it
could have serious effects on his job.
Virgil
"|-|erc" <h@r.c> wrote:
>
> The anti-diagonal is not computable, therefore it doesn't exist.
In what axiom system? Unless you can state exactly what assumptions you
are making (and you are making quite a few) there is no way to determine
what does or does not exist in the system formed by those assumptions.
>
> let UTM(real, digit) mod 10 calculate the list of reals. (where real is
> deceptively an integer)
>
> the antidiagonal is UTM(digit, digit) + 1 mod 10
Not in any standard decimal system. So the rest of Herk's garbage is
just that.
David Bernier
With the "mod 10", we can avoid talking about decimal TMs. For
argument 'real', I'd put TM_n, the n'th TM that halts on any input.
For 'digit', I'd put k and we get (after rewriting):
UTM(TM_n, k)%10 // UTM a fixed universal TM
(Indeed, TM_n and k will be presented to the UTM as
input or 1s on the tape of the UTM just before it starts).
for the k'th decimal of the n'th TM that halts under
all inputs, with TMs: TM_1, TM_2 , .... and for k = 1, 2, 3 ....
if we are only concerned with reals in [0, 1], and allowing
say 0.5000... to appear for several always halting TMs:
TM_a, TM_b and so on .
If the anti-diag were computable, there would be a contradiction.
I think what's going on is that the unsolvability of the halting
problem prevents us from writing a program that finds
the n'th always-halting TM.
" The anti-diagonal is not computable, therefore it doesn't exist."
There was a long thread starting with:
http://groups.google.com/group/sci.logic/msg/555732b7f182a8dd ,
"Uncomputable numbers are all in your head".
The conclusion I see is that you're not a Platonist.
You could abandon the Power set axiom for
infinite sets and the axiom of choice.
Constructivists don't draw conclusions using
the law of the excluded middle.
>
> the antidiagonal is UTM(digit, digit) + 1 mod 10
>
> if this is computable, then some TM computes it and some emulated TM also computes it,
>
> UTM(ad, digit) mod 10 = UTM(digit, digit) + 1 mod 10
>
> when digit = ad,
>
> UTM(ad, ad) mod 10 = UTM(ad, ad) + 1 mod 10
>
> Contradiction!
>
> Therefore antidiag is an invalid formula.
>
> This is a more sensible conclusion than higher infinities exist as I've basically just
> rearranged Cantor's proof. One can argue its a valid specification of a sequence
> but it doesn't actually compute a new sequence of digits.
Then we can say you're a computist.
David Bernier
> Herc
Tonico
**********************************************************
In the case of WM, I honestly think he embraces all four talents:
mathematically and logically incompetent, and also intentionally and
maliciously dishonest.
I've had a hard time swallowing the fact that he's actually
""teaching"" people stuff. A pity.
Regards
Tonio
Tonico
> [VIRGIN]
> You dumb fuck, I can still make a new number
>
> [HERC]
> I don't care about your new number, give me a new sequence of digits.
>
> Herc
***************************************************************
I answer here you prior post since for some rather misterious and
completely unreachable for me reason that post doesn't accept any
answer...
1) Well, let's play along: I entered your link, pressed 1232323 and
got a left list of some 60 numbers, then an apaprently randomly
generated numbers, and a second list which apparently contains the
randly gen. number in its main diagoal entries...so? I don't
understand what is this supposed to show.
2) What is "superinfinity"? And why is it mine?
3) I did not ignore your argument: I did address it in my first post,
unless what I wrote there was NO your argument and thus I
misunderstood. You ignored this addressing of mine.
4) "no wonder I want to withdraw now"....from what? A debate? If it
indeed is a debate, what path is left for a debate after you write
"Sorry to have to tell you but you're all fools, only when Artificial
Intelligence arrives in 20 years will you be corrected."?
It seems not only you've utterly made up your mind about this, but
you've also decided we all are fools, you only are wise, we're wrong,
you're right...and we all shall see this in 20 years more!
Well, you've wrapped it all tight and dandy: YOU closed the door.
Regards
Tonio
Tonico
> In article
> <fb9df027-90be-450a-891f-11b4d54f2...@e53g2000hsa.googlegroups.com>,
>
>
>
>
>
> Horand.Gassm...@googlemail.com wrote:
> > On Aug 2, 2:37 am, Virgil <Vir...@gmale.com> wrote:
>
> > > What I object to is WM, and others, telling me that I am not allowed to
> > > accept ZFC as a theory worth discussing. I have no objections to someone
> > > preferring not to accept ZFC or whatever, but I do object to their
> > > attempting to control what I am allowed to think about.
>
> > Well said!
>
> > > Particularly in the case of those like WM who are both mathematically
> > > and logically incompetent
>
> > Or intentionally and maliciously dishonest. I'm still not sure about
> > WM, but my gut tells me that he knows his position is untenable, and
> > he gets his jollies out of riling up everyone else.
>
> WM has too many published papers supporting his position to be doing it
> for fun.
>
Where has WM publish any of his papers on this stuff? Is there any
reachable link where one can acceed to some of them?
Thanx
Regards
Tonio