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From: Marchal <marchal.domain.name.hidden>

Date: Mon Nov 8 03:48:04 1999

Hal Finney wrote:

*>Marchal <marchal.domain.name.hidden> writes:
*

*>> Step n owns 2^(n-1) initial segments.
*

*>>
*

*>> Now, could you give me a bitstring which is not generated
*

*>> by this countably infinite process ?
*

*>
*

*>Are you familiar with Cantor's diagonalization argument which proves that
*

*>the real numbers are greater in cardinality than the integers? It
*

*>directly disproves your statement.
*

*>
*

*>The real number 2/3, = .10101010... is never output by your procedure.
*

*>If you disagree, tell me at what step it is output.
*

Gosh! Please Hal, I have explicitely explained how and why my

procedure cannot be disproved by Cantor diagonalisation, nor does

my procedure contradicts the uncountability of the reals.

I have also explicitely said that I don't ask for ma procedure to

output the reals at some step. (but that is the case for any

procedure giving the complete expansion of a real!).

What I say is that the UD generates the entire decimal or binary

expansion of all the real: 2/3, = .10101010... is generated at

the first, second, third, ... steps.

Note that any procedure outputing the expansion of a real does that.

The fact is that my procedure generates also non-computable reals.

There is no contradiction with diagonalisation, because at each step

the procedure generates a set of decimal expansion and I don't

provide (and I cannot provide!) a way to pick a particular

expansion as the one being the expansion of a non computable real.

Exemple:

Suppose for simplicity that 0.00010110011110110... is the binary

expansion of an uncomputable real (like Chaitin's number for exemple).

The my procedure generates, 0,at the first step 00, at the second

(among 01,10,11), 000 at the third step (among the 8 others),

and so one.

What I am saying is incredibly trivial, even if it looks a little

paradoxical.

You know, I can write a procedure which gives me with

certainty the answer for any well defined mathematical question.

It looks crazy? Take the question ``is Goldbach conjecture

true or false"?

Here is my procedure: generate the set {true, false}. I know that

the answer is in the set. Of course I don't know which one it is

but I have never been pretending knowing that.

In the same way, it is easy to realise that during his infinite

running the UD generates the binary expansion of each reals

including the non-computable one. (Exercice: show that any attempt

to build a LIST of all the reals using my procedure will fail. Hint:

diagonalisation!).

NOW, the important and perhaps less trivial point is the following one:

1) With comp we survive self-duplication (at a unknown level!, but here

I will not insist although it is fundamental in other part of the proof).

2) From the first person point of views the delays of virtual

reconstitution in the UD does not change the way of quantifying the

indeterminism.

SO we must take the generation of non computable reals into account

in the search for a measure on the computational histories.

Bruno

PS 1) I send this message friday. For obscur reasons it has not been

send as I discover in the archive. Does someone knows if something

happens friday with the mailing list ?

Received on Mon Nov 08 1999 - 03:48:04 PST

Date: Mon Nov 8 03:48:04 1999

Hal Finney wrote:

Gosh! Please Hal, I have explicitely explained how and why my

procedure cannot be disproved by Cantor diagonalisation, nor does

my procedure contradicts the uncountability of the reals.

I have also explicitely said that I don't ask for ma procedure to

output the reals at some step. (but that is the case for any

procedure giving the complete expansion of a real!).

What I say is that the UD generates the entire decimal or binary

expansion of all the real: 2/3, = .10101010... is generated at

the first, second, third, ... steps.

Note that any procedure outputing the expansion of a real does that.

The fact is that my procedure generates also non-computable reals.

There is no contradiction with diagonalisation, because at each step

the procedure generates a set of decimal expansion and I don't

provide (and I cannot provide!) a way to pick a particular

expansion as the one being the expansion of a non computable real.

Exemple:

Suppose for simplicity that 0.00010110011110110... is the binary

expansion of an uncomputable real (like Chaitin's number for exemple).

The my procedure generates, 0,at the first step 00, at the second

(among 01,10,11), 000 at the third step (among the 8 others),

and so one.

What I am saying is incredibly trivial, even if it looks a little

paradoxical.

You know, I can write a procedure which gives me with

certainty the answer for any well defined mathematical question.

It looks crazy? Take the question ``is Goldbach conjecture

true or false"?

Here is my procedure: generate the set {true, false}. I know that

the answer is in the set. Of course I don't know which one it is

but I have never been pretending knowing that.

In the same way, it is easy to realise that during his infinite

running the UD generates the binary expansion of each reals

including the non-computable one. (Exercice: show that any attempt

to build a LIST of all the reals using my procedure will fail. Hint:

diagonalisation!).

NOW, the important and perhaps less trivial point is the following one:

1) With comp we survive self-duplication (at a unknown level!, but here

I will not insist although it is fundamental in other part of the proof).

2) From the first person point of views the delays of virtual

reconstitution in the UD does not change the way of quantifying the

indeterminism.

SO we must take the generation of non computable reals into account

in the search for a measure on the computational histories.

Bruno

PS 1) I send this message friday. For obscur reasons it has not been

send as I discover in the archive. Does someone knows if something

happens friday with the mailing list ?

Received on Mon Nov 08 1999 - 03:48:04 PST

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