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 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. ========== 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? -- 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 -~----------~----~----~----~------~----~------~--~--- |

- Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Quentin Anciaux
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Quentin Anciaux
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- The big-black-cloud-interpretation. Torgny Tholerus
- Re: The big-black-cloud-interpretation. Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) meekerdb
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Mirek Dobsicek
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Re: Bijections (was OM = SIGMA1) Torgny Tholerus
- Re: Bijections (was OM = SIGMA1) Bruno Marchal
- Cantor's Diagonal Bruno Marchal