Bruno Marchal wrote:
> Le 26-mai-06, à 19:35, Tom Caylor a écrit :
> >
> > Bruno,
> > You are starting to perturb me!  I guess that comes with the territory
> > where you're leading us.
>
> You should not worry too much. I confess I am putting your mind in the
> state of mathematicians before the Babbage Post Markov Turing Church
> discovery. Everything here will be transparently clear.
>
> > But of course being perturbed doesn't
> > necessarily imply being correct.  I will summarize my perturbation
> > below.  But for now, specifically, you're bringing in transfinite
> > cardinals/ordinals.
>
> Only transfinite ordinal which are all countable, and even nameable,
> for example by name of growing computable functions as I am
> illustrating.
>
> Be sure you understand why G is a well defined computable growing
> function, and why it grows faster than each initial Fi. If you know a
> computer programming language, write the program!
>
> >  This is where things get perverse and perhaps
> > inconsistent.  For instance, couldn't I argue that G is also infinite?
>
> In which sense? All functions are infinite mathematical object.
> Factorial is defined by its infiinite set of inputs outputs: {(0,1)
> (1,1)(2,2) (3,6) (4,24) (5,120) ...}.
>
> > Take n = some fixed N>1.  Then F1(N) > 1, F2(N) > 2, F3(N) > 3, ...
> > and Fn(N) > n, for all n.  So each member of the whole sequence F1, F2,
> > F3 ... G is greater than the corresponding member of the sequence 1, 2,
> > 3, ... aleph_0 (countable infinity).  Thus, G >(=) countable infinity,
> > even for a fixed n=N>1.
>
> You are right but G is a function. Actually it just does what it has
> been programmed to. I don't see any problem here.
>
OK.  I see that so far (above) there's no problem.  (See below for
where I still have concern(s).)  Here I was taking a fixed N, but G is
defined as the diagonal, so my comparison is not valid, and so my proof
that G is infinite for a fixed N is not valid.  I was taking G's
assignment of an ordinal of omega as being that it is in every way
larger than all Fi's, but in fact G is larger than all Fi's only when
comparing G(n) to Fn(n), not when comparing G(Nfixed) to Fi(N) for all
i's.

> >> Oh Oh Oh Oh Oh .... A new pattern emerge (the Ackerman Caylor one, at
> >> a
> >> higher level).
> >>
> >> F_omega,
> >> F_omega + omega
> >> F_omega * omega
> >> F_omega ^ omega
> >> F_omega [4]  omega (omega tetrated to omega, actually this ordinal got
> >> famous and is named Epsilon Zéro, will say some words on it later)
> >>
> >> F_omega [5] omega
> >> F_omega [6] omega
> >> F_omega [7] omega
> >> F_omega [8] omega
> >> F_omega [9] omega
> >> F_omega [10] omega
> >> F_omega [11] omega
> >>
> >> ...
> >>
> >> In this case they are all obtained by successive diagonalzations, but
> >> nothing prevent us to diagonalise on it again to get
> >>
> >> F_omega [omega] omega
> >>
> >> OK, I think the following finite number is big enough:
> >>
> >> F_omega [omega] omega (F_omega [omega] omega (9 [9] 9))
> >>
> >>
> >> Next, we will meet a less constructivist fairy, and take some new kind
> >> of "big leap".
> >>
> >> Be sure to be convinced that, despite the transfinite character of the
> >> F_"alpha" sequence, we did really defined at all steps precise
> >> computable growing functions ... (if not: ask question please).
> >>
> >
> > It seems to me that you are on very shaky ground if you are citing
> > transfinite numbers in your journey to showing us your ultimate
> > argument.
>
> Please Tom, I did stay in the realm of the finitary. Even intutionist
> can accept and prove correct the way I named what are just big finite
> number. I have not until now transgressed the constructive field, I
> have not begin to use Platonism! There is nothing controversial here,
> even finitist mathematician can accept this. (Not ultra-finitist,
> though, but those reject already 10^100)
>
> >  I also think that if you could keep your arguments totally
> > in the finite arena it would less risky.
>
> I have. You must (re)analyse the construction carefully and realize I
> have not go out of the finite arena. Ordinals are just been used as a
> way to put order on the successive effective diagonalizations. Those
> are defined on perfectly well defined and generable sequence of well
> defined functions. I have really just written a program (a little bit
> sketchily, but you should be able to add the details once you should a
> programming language).
>

OK.  I think you are just throwing me off with your notation.  Do you
have to use transfinite ordinals (omega) to do this?  Couldn't you just
stay with successively combining and diagonalizing, like below, without
using omegas?

G(n) = Fn(n)+1
Gi(n) = G(...G(n)), G taken i times
Then instead of using more and more additional letters, just add
subscripts...
H1(n) = Gn(n)+1
H1i(n) = H1(...H1(n)), H1 taken i times
H2(n) = H1n(n)+1
H2i(n) = H2(...H2(n)), H2 taken i times

Then the subscripts count the number of diagonalizations you've done,
and every time you do the Ackermann thing, instead of adding an omega
you add another subscript.   Then it continues ad infinitum.  You can
do the Ackermann thing with the *number* of subscripts, i.e. do the
Ackermann thing on the number of times you've done the Ackermann
thing... etc.

This may just be a technical point, but it doesn't seem precise to do
very much arithmetic with ordinals, like doing omega [omega] omega,
because you're just ordering things, and after a while you forget the
computations that are being performed.  I can see that it works for
just proving that you can continue to diagonalize and grow, which is
what you're doing.  I just don't want to be caught off guard and
suddenly realize you've slipped actual infinities in without me
realizing it.  I don't think you have.

> > For instance, I think it is
> > very risky to use things like the axiom of choice as part of a theory
> > of everything.
>
> Tom, I am trying to be 100% serious. I am giving you the "real thing".
> If you think I am using the axiom of choice, you better tell me where.
> I don't.
>
> > If you have Arithmetical Platonism as one of your
> > hypotheses, then you need to stick to parts of math that are generally
> > agreed to be a part of the Platonic world and not a man-made tool.
>
> I was not doing any piece of philosophy. Just a part common in science
> and enginering: writing a program which, if ever it would be able to
> run, would stop on a (very big) finite number. It is a problem in
> computer science, and I have illustrated a "modern" solution, if you
> want.
> It is indeed the effectivity (mechanizability, programmability, ...) of
> the diagonalization which makes possible to write programs encoding
> tranfinite ordinals, and here I show how to use that fact to solve a
> "concrete" problem: naming a *very* big number in a program having some
> reasonable length.
>
> >  Of
> > course you'll counter that even man-made tools, in fact anything that
> > we can think of (that's consistent), is part of reality (assuming the
> > multiverse).  But the problem is consistency.  In the realm of the
> > infinite your playing with truth that is not necessarily provable, or
> > part of the universe.
>
> I insist, I did not leave the realm of the finite era, at all. There is
> nowadays nothing philosophical here. Usual Analysis textbook used far
> more less finitary notion.
>
> Please try to circumscribe more clearly the point where you feel we
> have leave the finite era, because if you will miss what will happen
> later were indeed we will leave the finite or effective era. At some
> point, we will be obliged to leave the finite era, and it will be a key
> step, but until now we haven't: we have just written a program naming a
> very big number.
>
> >
> > So you've got me stuck right here.  I think that at this point, you're
> > assuming your conclusion:  everything is number.
>
> But we are not even thinking on things like that. People could even
> think I am leading to the contrary conclusion. Look the tricky problem
> below:
>
> >
> >> Tricky Problem: is there a sequence in which all growing computable
> >> functions belong? Is it possible to dovetail on all computable growing
> >> functions, ...
>
> Diagonalization seems to illustrate the impossibility of sequencing
> effectively all computable growing functions: having a sequence like F1
> F2 F3 F4 ..., we can find by diagonalization a new function not in the
> list. But dovetailing seems to necessitate such a sequence, so it seems
> we can't dovetail on all growing functions, but then how could a
> universal dovetailer exist?
>
> This is an exercise! Not a philosophical question. It is a crucial
> exercise the solution of which will eventually justify the
> differentiation between the logic G and G*.
>
> Before tackling this problem, which has been tricky at a time of course
> (but is transparently clear today for the experts) please be sure we
> have not yet leave the finite and effective era. Don't worry I will
> present other diagonalizations. It is really the "heart of the matter".
>
> Bruno
>
> http://iridia.ulb.ac.be/~marchal/

OK.  So we haven't left the finite behind yet.  It makes intuitive
sense to me that you can diagonalize till the cows come home, staying
within countability, and still not be done.  Otherwise infinity
wouldn't be infinite.

On the tricky question, it also makes intuitive sense that you can
sequence effectively on all computable growing functions.  This is
because the larger the growing function gets, the more uncovered space
(gaps) there are between the computable functions.  Any scheme for
generating growing functions will also leave behind every-growing
uncomputed gaps.  Very unmathematical of me to be so vague, but you've
already given us the answer, and I know you will fill in the gaps.  :)

Tom


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