# Re: Bijections (was OM = SIGMA1)

 Bruno Marchal skrev: Le 15-nov.-07, à 14:45, Torgny Tholerus a écrit : But m+1 is not a number.  This means that you believe there is a finite sequence of "s" of the type A = s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s(s( ....s(0)))))))))))))))))))))...) where "..." here represents a finite sequence, and which is such that s(A) is not a number. Yes, exactly.  When you construct the set of ALL natural numbers N, you have to define ALL these numbers.  And you can only define a finite number of numbers.  See more explanations below. 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, Yes, I agree.  All explicitly given numbers are numbers.  The biggest number is bigger than all by human beeings explicitly given numbers. 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. There is some sort of "temporal" distinction between the numbers of different type.  You have to "first" have all numbers of type 0, "before" you can construct the numbers of type 1.  And you must have all numbers of type 1 "before" you can construct any number of type 2, and so on. The construction of numbers of type 1 presupposes that the set of all numbers of type 0 is fixed.  When the set N of all numbers of type 0 is fixed, then you can construct new numbers of type 1. 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. -- Torgny --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. To post to this group, send email to [EMAIL PROTECTED] To unsubscribe from this group, send email to [EMAIL PROTECTED] For more options, visit this group at http://groups.google.com/group/everything-list?hl=en -~----------~----~----~----~------~----~------~--~---