Hi. Bruno could do this better, but I like the practice. I guess you're trying to demonstrate that the form of Cantor's argument is invalid, by displaying an example in which it produces an absurd result. Start with a set S you want to show is not enumerable. (ie, there is no one-one onto map from Z^+ to S). The form of the diagonalization argument is as follows: give me any, repeat, any, particular candidate for an enumeration of S. This should be a map from Z^+ into S. (If it isn't such a map, it isn't an enumeration.) I will show you an element D of S that your candidate enumeration omits. (That is, I will show you that your candidate is not onto.) Hence, S is not denumerable. In your first attempt, your D is not an element of your S (= Z^+). So your first attempt doesn't fit the form of the diagonalization argument on this account. More fundamentally, it also fails to fit the form of Cantor's argument because you haven't tried to debunk *any* candidate enumeration, but a particular one. In your second attempt, if I understand you, you start with a map from the primes (all of them!) and then (your work suggests, but I think you'd need more details--what's the image of 4, for example?) from the rest of Z^+, into S = Z^+ again. This example doesn't invalidate Cantor's argument either. Again, you debunk a particular candidate enumeration, not any and all candidate enumerations. So you don't arrive at the absurdity you seem to be after, even if you fill in the details I mentioned. Barry On Dec 16, 2007, at 3:49 AM, Daniel Grubbs wrote:Hi Folks, I joined this list a while ago but I haven't really kept up. Anyway, I saw the reference to Cantor's Diagonal and thought perhaps someone could help me. Consider the set of positive integers: {1,2,3,...}, but rather than write them in this standard notation we'll use what I'll call 'prime notation'. Here, the number m may be written as m = n1,n2,n3,n4,... where ni is the number of times the i'th prime number is a factor of m. Thus: 1 = 0,0,0,0,0,... 2 = 1,0,0,0,0,... 3 = 0,1,0,0,0,... 4 = 2,0,0,0,0,... 5 = 0,0,1,0,0,... ... 28 = 2,0,0,1,0,0,0,... ... If we now apply the diagonal method to this ordered set, we get the number: D = 1,1,1,1,1,... Has this just shown that the set of positive integers is not denumerable? I can see that one may complain that D is clearly infinite and therefore should not be in the set, but consider the following... Let's take the original set and reorder it by exchanging the places of the i'th prime number with that of the number in the i'th position. (i.e. First switch the number 2 with the number 1 to move it to the first position. Then switch 3 with the number -- now 1 -- in the 2nd position. Then 5 with the 1 which is now in the 3rd position. Etc...) Because we are just trading the positions of the numbers, all the same numbers will be in the set afterwards. The set is now: 2 = 1,0,0,0,0,... 3 = 0,1,0,0,0,... 5 = 0,0,1,0,0,... 7 = 0,0,0,1,0,... 11= 0,0,0,0,1,... ... Now instead of adding 1 to each 'digit' of the diagonal, subtract 1. This will ensure that the diagonal number is different from each of the numbers in the set. Thus, D = 0,0,0,0,... But this is the number 1 which we know was in the set to begin with. What happened to it? I would suggest that the diagonal method does not find a number which is different from all the members of a set, but rather finds a number which is infinitely far out in the ordered set. If anyone can find where I've gone wrong, please let me know. Dan GrubbsDr. Barry Brent barrybrent@earthlink.net http://home.earthlink.net/~barryb0/
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