Re: Bijections (was OM = SIGMA1)

From: Torgny Tholerus <torgny.domain.name.hidden>
Date: Thu, 15 Nov 2007 14:45:24 +0100
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.

I have already explained that the meaning of "...'" in {I, II, III, 
IIII, IIIII, IIIIII, IIIIIII, IIIIIIII, IIIIIIIII, ...}  is *the* 
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, 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.


  
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?


  
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?  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"?

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.)


  
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.  But m+1 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 Tholerus

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Received on Thu Nov 15 2007 - 08:45:41 PST

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