You're saying that, just because you can *write down* the missing sequence (at the beginning, middle or anywhere else in the list), it follows that there *is* no missing sequence. Looks pretty wrong to me.
Cantor's proof disqualifies any candidate enumeration. You respond by saying, "well, here's another candidate!" But Cantor's procedure disqualified *any*, repeat *any* candidate enumeration. Barry Brent On Nov 20, 2007, at 11:42 AM, Torgny Tholerus wrote: > Bruno Marchal skrev: >> But then the complementary sequence (with the 0 and 1 permuted) is >> also well defined, in Platonia or in the mind of God(s) >> >> 0 1 1 0 1 1 ... >> >> But this infinite sequence cannot be in the list, above. The "God" >> in question has to ackonwledge that. >> The complementary sequence is clearly different >> -from the 0th sequence (1, 0, 0, 1, 1, 1, 0 ..., because it >> differs at the first (better the 0th) entry. >> -from the 1th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs >> at the second (better the 1th) entry. >> -from the 2th sequence (0, 0, 0, 1, 1, 0, 1 ... because it differs >> at the third (better the 2th) entry. >> and so one. >> So, we see that as far as we consider the bijection above well >> determined (by God, for example), then we can say to that God that >> the bijection misses one of the neighbor sheep, indeed the "sheep" >> constituted by the infinite binary sequence complementary to the >> diagonal sequence cannot be in the list, and that sequence is also >> well determined (given that the whole table is). >> >> But this means that this bijection fails. Now the reasoning did >> not depend at all on the choice of any particular bijection- >> candidate. Any conceivable bijection will lead to a well >> determined infinite table of binary numbers. And this will >> determine the diagonal sequence and then the complementary >> diagonal sequence, and this one cannot be in the list, because it >> contradicts all sequences in the list when they cross the diagonal >> (do the drawing on paper). >> >> Conclusion: 2^N, the set of infinite binary sequences, is not >> enumerable. >> >> All right? > > An ultrafinitist comment to this: > ====== > You can add this complementary sequence to the end of the list. > That will make you have a list with this complementary sequence > included. > > But then you can make a new complementary sequence, that is not > inluded. But you can then add this new sequence to the end of the > extended list, and then you have a bijection with this new sequence > also. And if you try to make another new sequence, I will add that > sequence too, and this I will do an infinite number of times. So > you will not be able to prove that there is no bijection... > ====== > What is wrong with this conclusion? > > -- > Torgny > > > Dr. Barry Brent [EMAIL PROTECTED] http://home.earthlink.net/~barryb0/ --~--~---------~--~----~------------~-------~--~----~ 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 -~----------~----~----~----~------~----~------~--~---
