Bruno Marchal wrote:
> Hi,
>
> OK, let us try to name the biggest natural (finite) number we can, and
> let us do that transfinite ascension on the growing functions from N to
> N.
>
> We have already build some well defined sequence of description (code)
> of growing functions.
>
> Let us choose the Hall Finney sequence to begin with (but the one by
> Tom Caylor can be use instead).
>
> F1 F2 F3 F4 F5 ...
>
> With F1(n) = factorial(n), F2(n) = factorial(factorial n), etc.
>
> Note this: Hal gave us a trick for getting from a growing function f, a
> new function growing faster, actually the iteration of the function.
> That is, Hal gave us a notion of successor for the growing function.
> Now the diagonalization of the sequence F1 F2 F3 F4 ..., which is given
> by the new growing function defined by
>
> G(n) = Fn(n) + 1
>
> gives us a growing function which grows faster than any Fi from Hal's
> initial sequence. Precisely, G will grow faster than any Fi on *almost
> all* number (it could be that some Fi will grow faster than G on some
> initial part of N, but for some finite value (which one?) G will keep
> growing faster. Technically we must remember to apply our growing
> function on "sufficiently big input' if we want to benefit of the
> growing phenomenon. We will make a rough evaluation on that input
> later, but let us not being distract by technical point like that.
> The diagonalization gives an effective way to take the "limit" of the
> sequence F1, F2, F3, ...
>
> G grows faster than any Fi. Mathematician will say that the order type
> of g, in our our new sequence F1 F2 F3 ... G,  is omega (the greek
> letter).
>

Bruno,
You are starting to perturb me!  I guess that comes with the territory
where you're leading us.  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.  This is where things get perverse and perhaps
inconsistent.  For instance, couldn't I argue that G is also infinite?
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.

> But G is just a well defined computable growing function and we can use
> Hall Finney "successor" again to get the still faster function, namely
> G(G(n)).
>
> The order type of G(G(n)) is, well, the successor of omega: omega+1
>
> And, as Hall initially, we can build the new sequance of growing
> functions (all of which grows more than the preceding sequence):
>
> G(n) G(G(n)) G(G(G(n))) G(G(G(G(n)))) etc.
>
> which are of order type omega, omega+1, omega+2, omega+3, omega+4, etc.
>
> Now we have obtained a new well defined infinite sequence of growing
> function, and, writing it as:
>
> G1, G2, G3, G4, G5, G6, ...  or better, as
>
> F_omega, F_omega+1, F_omega+2, F_omega+3
>
> just showing such a sequence can be generated so that we can again
> diagonalise it, getting
>
> H(n) = Gn(n) + 1, or better
>
> H(n) = F_omega+n (n) + 1
>
>
> Getting a function of order type omega+omega: we can write H =
> F_omega+omega
>
> And of course, we can apply Hall's successor again, getting
> F_omega+omega+1
> which is just H(H(n), and so we get a new sequence:
>
> F_omega+omega+1, F_omega+omega+2, F_omega+omega+3, ...
>
> Which can be diagonalise again, so we get
>
> F_omega+omega+omega,
>
> and then by Hal again, and again ...:
>
> F_omega+omega+omega+1, F_omega+omega+omega+2, F_omega+omega+omega+3
>
> ...
>
> Oh Oh! a new pattern emerges, a new type of sequence of well defined
> growing functions appears:
>
> F_omega, F_omega+omega, F_omega+omega+omega, F_omega+omega+omega+omega,
>
> And we can generated it computationnaly, so we can diagonalise again to
> get:
>
> F_omega * times omega,
>
> and of course we can apply Hal's successor (or caylor one of course)
> again, and again
>
> ....
>
> 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.  I also think that if you could keep your arguments totally
in the finite arena it would less risky.  For instance, I think it is
very risky to use things like the axiom of choice as part of a theory
of everything.  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.  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.

So you've got me stuck right here.  I think that at this point, you're
assuming your conclusion:  everything is number.

Tom

> Tricky Problem: is there a sequence in which all growing computable
> functions belong? Is it possible to dovetail on all computable growing
> functions, ...
> 
> I let you think,
> 
> Bruno
> 
> http://iridia.ulb.ac.be/~marchal/


--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups 
"Everything List" group.
To post to this group, send email to everything-list@googlegroups.com
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at 
http://groups.google.com/group/everything-list
-~----------~----~----~----~------~----~------~--~---

Reply via email to