Re: Bijections (was OM = SIGMA1)

From: Bruno Marchal <>
Date: Fri, 16 Nov 2007 12:50:14 +0100

Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit :

> Bruno Marchal skrev:Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit :
>>> What do you mean by "..."?
>> Are you asking this as a student who does not understand the math, or
>> as a philospher who, like an ultrafinist, does not believe in the
>> potential infinite (accepted by mechanist, finistist, intuitionist,
>> etc.).
> I am asking as an ultrafinitist.

Fair enough.
I am not sure there are many ultrafinitists on the list, but just to
let John Mikes and Norman to digest the bijection post, I will say a
bit more.
A preliminary remark is that I am not sure an ultrafinitist can really
assert he is ultrafinitist without acknowledging that he does have a
way to give some meaning on "...".
But I have a more serious question below.

>> I have already explained that the meaning of "...'" in {I, II, III,
>> mystery.
> Do you have the big-black-cloud interpretation of "..."?  By that I
> mean that there is a big black cloud at the end of the visible part of
> universe,

Concerning what I am trying to convey, this is problematic. The word
"universe" is problematic. The word "visible" is also problematic.

> and the sequence of numbers is disappearing into the cloud, so that
> you can only see the numbers before the cloud, but  you can not see
> what happens at the end of the sequence, because it is hidden by the
> cloud.

I don't think that math is about seeing. I have never seen a number. It
is a category mistake. I can interpret sometimes some symbol as
refering to number, but that's all.

>>>> For
>>>> example, the function which sends x on 2*x, for each x in N is such
>>>> a
>>>> bijection.
>>> What do you mean by "each x" here?
>> I mean "for each natural number".
> What do you mean by "each" in the sentence "for each natural
> number"?  How do you define ALL natural numbers?

By relying on your intuition of "finiteness". I take 0 as denoting a
natural number which is not a successor.
I take s(0) to denote the successor of 0. I accept that any number
obtained by a *finite* application of the successor operation is a
I accept that s is a bijection from N to N \ {0}, and things like that.

>>> How do you prove that each x in N has a corresponding number 2*x in
>>> E?
>>> If m is the biggest number in N,
>> There is no biggest number in N. By definition of N we accept that if
>> x
>> is in N, then x+1 is also in N, and is different from x.
> How do you know that m+1 is also in N? 

By definition.

> You say that for ALL x then x+1 is included in N, but how do you prove
> that m is included in "ALL x"?

I say "for all x" means "for all x in N".

> If you say that m is included in "ALL x", then you are doing an
> illegal deduction, and when you do an illegal deduction, then you can
> prove anything.  (This is the same illegal deduction that is made in
> the Russell paradox.)

? (if you believe this then you have to accept that Peano Arithmetic,
or even Robinson arithmetic) is inconsistent. Show me the precise

>>> then there will be no corresponding
>>> number 2*m in E, because 2*m is not a number.
>> Of course, but you are not using the usual notion of numbers. If you
>> believe that the usual notion of numbers is wrong, I am sorry I cannot
>> help you.
> I am using the usual notion of numbers. 

You are not. By definition of the usual natural numbers, all have a

> But m+1 is not a number. 

This means that you believe there is a finite sequence of "s" of the

A =
s(s(s(s( ....s(0)))))))))))))))))))))...)

where "..." here represents a finite sequence, and which is such that
s(A) is not a number.

> But you can define a new concept: "number-2", such that m+1 is
> included in that new concept.  And you can define a new set N2, that
> contains all natural numbers-2.  This new set N2 is bigger than the
> old set N, that only contains all natural numbers.

  Torgny, have you followed my "fairy tale" which I have explain to Tom
Caylor. There I have used transfinite sequence of growing functions to
name a big but finite natural number, which I wrote F_superomega(999),

My "serious" question is the following: is your "biggest number" less,
equal or bigger than a well defined finite number like

If yes, then a big part of the OM = SIGMA_1 thread will be accessible
to you, except for the final conclusion. Indeed, you will end up with a
unique finite bigger universal machine (which I doubt).

If not, let us just say that your ultrafinitist hypothesis is too
strong to make it coherent with the computationalist hypo. It means
that you have a theory which is just different from what I propose. And
then I will ask you to be "ultra-patient", for I prefer to continue my
explanation, and to come back on the discussion on hypotheses after.

Actually, my conversation with Tom was interrupted by Norman who fears
people leaving the list when matter get too much technical; but I was
about to introduce a second fairy, capable of non constructive
reasoning. I recall that the first fairy asked for a big constructively
nameable number. The second fairy was supposed to ask the same except
that she drops the constructivity requirement. In that case you can use
even higher infinities to name (non constructively) a natural number.
This gives a way to name finite number still *vastly* bigger than
In case your bigger natural number is bigger than such monstruosity,
then the whole of the thread will be understandable, although comp can
be made explicitly wrong (in that case) but somehow approximable in
practice (even in the local comp practice).

Of course if your bigger natural number is little than
100^(100^(100^(100^(100^(100^(100^(100^100)], which is already far
beyond any empirical numbers motivated by the observable empirical
world, then we can only conclude that we have very different theory.

BTW, do you agree that 100^(100^(100^(100^(100^(100^(100^(100^100)],
and 100^(100^(100^(100^(100^(100^(100^(100^100)] +1 are numbers? I am
just curious,


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Received on Fri Nov 16 2007 - 06:50:25 PST

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