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From: Virgil on 21 Oct 2005 17:17 In article <MPG.1dc2ed008cb30cc798a527(a)newsstand.cit.cornell.edu>, Tony Orlow <aeo6(a)cornell.edu> wrote: > stephen(a)nomail.com said: > > That means that imagining any completed bijection also leads to a > > contradiction. w has just as much existence as your bijection. > > If the bijection is complete, so is w. > > The bijection is never "completed". Whatever it is, it is not a bijection until it is completed. Thus TO concedes that while there may be "partial bijections", there can never be a total one. But every two sets allow "parial bijections". So TO's "partial bijections are irrelevant to the issue. > There is always a natural for every subset, but you cannot define the entire > set so that any such natural can be identified for it. When you speak of a > mapping to the entire set, you include the implicit identification of the > last > element and the completed set. Sets are static objects, not processes, at least everywhere except in the wilderness of TOmatica. There is no such thing as an incompleted set except in that strange land. > It's unending. There is no point you can stop and say it's over. No > matter what set you have, there is a number corresponding to it that > is greater. No matter what number you have, there is a set containing > it further down the bijection. If it isn't all there then it isn't a set at all.
From: David R Tribble on 21 Oct 2005 17:24 Tony Orlow: >> f(0) = ...000 = {} >> f(1) = ...001 = {0} >> f(2) = ...010 = {1} >> f(3) = ...011 = {0,1} >> etc. Any questions? > David R Tribble: >> Where are the infinite subsets, such as the set of even numbers? > Tony Orlow: >> Well, infinitely far down the list of course! >> . >> . >> . >> f(N/3) = 0:010101.....010101 = {0,2,4,6,8,....} >> f(2N/3)= 0:101010.....101010 = {1,3,5,7,9,....} > David R Tribble: >> All of the sets you listed, finite and infinite, are composed of >> only finite naturals. Which means that they are all subsets of N. > Tony Orlow: > Incorrect. See the digits to the left of the ellipses? Those are in > infinite positions, representing infinite evens and odds. See the elements within the braces? They are all finite. All those sets are members of P(N). So you're mapping from *N to P(N), which like I said before, is easy. You'd have to use numbers like your 1:000...001 [*] before you map to any subsets containing infinite naturals. But then you'd be outside the members of *N. [*] I think. Who knows?
From: David R Tribble on 21 Oct 2005 17:33 David R Tribble said: >> The infinite natural x in *N that maps f(x) to the infinite set of >> primes is: >> x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... >> >> But like all the other subsets your f mapping defines, the set of >> primes contains only finite numbers, which means that it is one of >> the subsets of N. > Tony Orlow wrote: > If there are no infinite primes, then the subset would correspond to a > finite value. Really? So x (above) must be a finite number? What about n = 1+1+1+1+..., which is obviously less than x? Is it also finite?
From: David R Tribble on 21 Oct 2005 17:37 David R Tribble said: >> The infinite natural x in *N that maps f(x) to the infinite set of >> primes is: >> x = 2^2 + 2^3 + 2^5 + 2^7 + 2^11 + 2^13 + ... >> >> But like all the other subsets your f mapping defines, the set of >> primes contains only finite numbers, which means that it is one of >> the subsets of N. > Tony Orlow wrote: > If there are no infinite primes, then the subset would correspond to a > finite value. > > Not all subsets have to have infinite values. I have only listed finite > numbers here, but do take note of the ellipses. Are there infinite primes? If > not, then there arent an infinite number of them. Well, there are no infinite primes. There can't be, because any infinite natural must have at least a finite number of prime factors. (If N is infinite, then N/2 must also exist, according to your rules. Which means that no infinite N is prime, right?) So since there are no infinite primes, there must only be a finite number of them, according to you. So therefore there must be a largest prime, right? Oh, but I suppose that it's "finite but unidentifiable", right? Euclid must have gotten this one wrong, huh?
From: William Hughes on 21 Oct 2005 22:03
Dave Rusin wrote: > In article <1129901285.084347.34810(a)g49g2000cwa.googlegroups.com>, > William Hughes <wpihughes(a)hotmail.com> wrote: > > >Yes all sets have a cardinality. > > That's your Choice. Am I assuming AC here? My understanding was that, while AC was needed to put a total ordering on the equivalence classes under bijection, one could assert the existence of these classes even without AC. Or are the cardinalities defined not as all possible equivalence classes but as the equivalence classes on the ordinals? -William Hughes |