Re: Bijections (was OM = SIGMA1)

From: Bruno Marchal <marchal.domain.name.hidden>
Date: Tue, 20 Nov 2007 10:46:17 +0100

Le 19-nov.-07, à 17:00, Torgny Tholerus a écrit :

> Torgny Tholerus skrev: If you define the set of all natural numbers
> N, then you can pull out the biggest number m from that set.  But this
> number m has a different "type" than the ordinary numbers.  (You see
> that I have some sort of "type theory" for the numbers.)  The ordinary
> deduction rules do not hold for numbers of this new type.  For all
> ordinary numbers you can draw the conclusion that the successor of the
> number is included in N.  But for numbers of this new type, you can
> not draw this conclusion.
>>
>> You can say that all ordinary natural numbers are of type 0.  And
>> the biggest natural number m, and all numbers you construct from that
>> number, such that m+1, 2*m, m/2, and so on, are of type 1.  And you
>> can construct a set N1 consisting of all numbers of type 1.  In this
>> set there exists a biggest number.  You can call it m1.  But this new
>> number is a number of type 2.
>>
>> It may look like a contradiction to say that m is included in N, and
>> to say that all numbers in N have a successor in N, and to say that m
>> have no successor in N.  But it is not a constrdiction because the
>> rule "all numbers in N have a successor in N" can be expanded to "all
>> numbers of type 0 in N have a successor in N".  And because m is a
>> number of type 1, then that rule is not applicable to m.
>
> You can comapre this with the Russell's paradox.  This paradox says:
>
> Construct the set R of all sets that does not contain itself.  For
> this set R there will be the rule: For all x, if x does not contain
> itself, then R contains x.
>
> If we here substitute R for x, then we get: If R does not contain
> itself, then R contains R.  This is a contradiction.
>
> The contradiction is caused by an illegal conclusion, it is illegal
> to substitute R for x in the "For all x"-quantifier above.
>
> This paradox is solved by "type theory".  If you say that all
> ordinary sets are of type 0, then the set R will be of type 1.  And
> every all-quantifiers are restricted to objects of a special type.  So
> the rule above should read: For all x of type 0, if x does not contain
> itself, then R contains x.
>
> In this case you will not get any contradiction, because you can not
> substitute R for x in that rule.





This points on one among many ways to handle Russell's paradox. Type
Theories (TT) are nice, but many logicians prefer some untyped set
theory, like ZF, or a two types theory like von Neuman Bernays Godel
(VBG). Or Cartesian closed categories, toposes, etc. But set theory is
a bit out of the scope of this thread.
All such theories (ZF, VBG, TT) are example of "Lobian Machine", and my
goal is to study all such machine without choosing one in particular,
and using traditional math, instead of working really "in" some
particular theories.

Another solution for many paradoxes consists in working with
constructive objects. Soon, this is what we will do, by focusing on the
set of "computable functions" instead of the set of all functions. The
reason is not to escape paradoxes though. The reason is to learn
something about machines (which are finite or constructive object).
Just wait a bit. I will first explain Cantor's diagonal, which is
simple but rather "transcendental".



> Compare this with the case of the biggest natural number:
>
> Construct the set N of all natural numbers.  For this set N there
> will be the rule: For all x, if N contains x, then N contains x+1.
>
> Suppose that there exists a biggest natural number m in N.  If we
> substitute m for x, then we get: If N contains m, then N contains
> m+1.  This is a contradiction, because m+1 is bigger than m, so m can
> not be the biggest number then.
>
> But the contradiction is caused by an illegal conclusion, it is
> illegal to substitute m for x in the "For all x"-quantifier above.
>
> This paradox is solved by "type theory".  If you say that all
> ordinary natural numbers are of type 0, then the natural number m will
> be of type 1.  And every all-quantifiers are restricted to objects of
> a special type.  So the rule above should read: For all x of type 0,
> if N contains x, then N contains x+1.
>
> In this case you will not get any contradiction, because you can not
> substitute m for x in that rule.
>
> ===========
>
> Do you see the similarities in both these cases?
>


Except that naive usual number theory does not lead to any paradox,
unlike naive set theory.
*you* got a paradox because of *your* ultrafinistic constraint.
So you are proposing a medication which could be worst that the disease
I'm afraid.
Very few people have any trouble with the potential infinite N = omega
= {0, 1, 2, 3, 4, 5, ...}. With comp it can be shown that you don't
need more at the ontological third person level. What will happen is
that infinities come back in the first person point of views, and are
very useful and lawful. Now, sound Lobian Machines can disagree on the
real status of some of those infinities, but this does not concern us,
at least, not urgently, I think.

Bruno



http://iridia.ulb.ac.be/~marchal/

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Received on Tue Nov 20 2007 - 04:46:38 PST

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