Obections to Cantor's Theory (Wikipedia article)
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From: David Kastrup on 19 Jul 2005 17:23 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> > David Kastrup said: >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> >> >> > There is no reason to expect the natural number corresponding to >> >> > the subset to be a member of that subset. >> >> >> >> There is no such expectation. The only expectation is that _every_ >> >> natural number is _either_ a member of its corresponding subset, or >> >> not. >> > Okay. >> >> _Depending_ on that, the constructed subset will either _not_ or >> >> _do_ contain the number, respectively. >> > Redundant, but okay. >> >> Sure it is redundant. But you should be the last person to complain >> about your need to get this hammered bit by bit into your skull. >> >> >> This constructed subset then does not correspond to _any_ natural >> >> number. >> >> > This is where it goes wrong. >> >> Uh yes. That's the conclusion: that the assumption of the mapping >> goes wrong, since it fails to cover the constructed subset. >> >> > The consructed subset corresponds to the natural number denoted by >> > the binary string constructed from the right to the left, where each >> > successive bit is a 1 if the successive natural is a member, and 0 >> > if it is not. >> >> You are assuming a particular mapping, the proof holds for _any_ >> mapping. So let's see where your mapping takes us. Since 2^n>n >> always, the subset we get is simply N itself, at least as long as we >> assume that N only contains finite numbers. However, that is an >> assumption that you are not willing to make, so let is call this >> particular subset of N by the name Q. >> >> So what natural number corresponds to Q according to your logic? I'll >> take a daring guess at your muddled thoughts and suppose 111...111 or >> something like that. Is 111...111 itself a member of Q? > Well, that's an interesting question. yes, it is 111...111, and > 111...111 is also a member of N, but those two are a little > different. The first is a map of membership, and has N number of > 1's. The second, is a digital number, and presumably denotes > N-1. Therefore, it has log2(N) 1's, rather than N. To represent 2^N > reals, you need N digits, but to represent N naturals, you only need > log2(N) digits. So is would not seem like the "111...111 that is a map of membership" could itself be contained in the "111...111 that is a map of membership" since then you'd need "2^N digits" instead of just the "N" you have. Of course, this is complete and utter bullshit, since N is not a natural number to start with, but even if we dive into your utter bullshit with a vengeance, it does not hold water. >> > It is so basic, I cannot even quite see what erroneous assumption >> > you are making. It seems like you are assuming subset number X >> > must contain X as a member. If so, how do you justify this >> > assumption? >> >> I don't. But I assume that for every X, subset number X must >> either contain number X as a member, or doesn't. And if it does, >> subset Q will not contain number X as a member, and if it doesn't, >> subset Q will contain number X as a member, and so subset Q differs >> from every subset X at the position of number X. > > This is essentially the diagonal argument in disguise. You are > assuming you have the same number of numbers as digits, which is > clearly a false assumption when using any base greater than 1. Fine. So you agree that there can't be a surjection of the naturals onto the subsets of N. Thanks for ceding that point. Now was this so hard? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: G. Frege on 19 Jul 2005 17:24 On 19 Jul 2005 21:02:29 GMT, Chris Menzel <cmenzel(a)remove-this.tamu.edu> wrote: > > Pretty clearly, you aren't terribly well-educated in set theory. Don't > you think you should understand a field before you try to point out its > flaws? > Isn't that rather typical for a m o d e r n "anti-cantorian"? :-) F.
From: david petry on 19 Jul 2005 17:25 Jesse F. Hughes wrote: > Anyway, what is "Cantor's Theory"? The use of the term "Cantor's Theory" has a long history. For example, it appears in the quote from Kronecker that I included in the article. > > These "anti-Cantorians" see an underlying reality to mathematics, > > namely, computation. > > Petry's own particular nonsense and not a broad program. It represents my attempt to understand the arguments of the anti-Cantorians. > Science does not proclaim that all truth has observable implications. So you say. I'd bet most scientists would disagree with you. > > The artists see the requirement that mathematical statements must > > have observable implications as a restriction on their intellectual > > freedom. > > No editorializing in that excerpt, huh? No. It's a very objective summary of things that mathematicians have said in this newsgroup. > > It is plausible that in the future, mathematics will be split > > into two disciplines - scientific mathematics (i.e. the science > > of phenomena observable in the world of computation), and > > philosophical mathematics, wherein Cantor's Theory is > > merely one of the many possible "theories" of the infinite. > > Bullhonkies. It is plausible only in Petry's feeble imagination. Are you seriously saying that it is inconceivable to you that in the future, mathematics will be split into the two disciplines, as I have stated? You seem to lack imagination.
From: David Kastrup on 19 Jul 2005 17:26 Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: > David Kastrup said: >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> > David Kastrup said: >> >> Tony Orlow (aeo6) <aeo6(a)cornell.edu> writes: >> >> >> >> > David Kastrup said: >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: >> >> >> >> >> >> > David Kastrup <dak(a)gnu.org> wrote: >> >> >> > >> >> >> >> Alec McKenzie <mckenzie(a)despammed.com> writes: >> >> >> >> > It has been known for a proof to be put forward, and >> >> >> >> > fully accepted by the mathematical community, with a >> >> >> >> > fatal flaw only spotted years later. >> >> >> >> >> >> >> >> In a concise 7 line proof? Bloody likely. >> >> >> > >> >> >> > I doubt it had seven lines, but I really don't know how many. >> >> >> > Probably many more than seven. >> >> >> >> >> >> It was seven lines in my posting. You probably skipped over it. It >> >> >> is a really simple and concise proof. Here it is again, for the >> >> >> reading impaired, this time with a bit less text: >> >> >> >> >> >> Assume a complete mapping n->S(n) where S(n) is supposed to cover all >> >> >> subsets of N. Now consider the set P={k| k not in S(k)}. Clearly, >> >> >> for every n only one of S(n) or P contains n as an element, and so P >> >> >> is different from all S(n), proving the assumption wrong. >> >> >> >> > I still do not get this. You have a set of naturals {0,1,2,3...}, >> >> > and a set of binary numbers {0,1,10,11,100,101,110,111,....}. Surely >> >> > there is a bijection between these two sets. >> >> >> >> Fine. >> >> >> >> > So, what is the problem if one interprets the binary numbers (with >> >> > implied leading zeroes) as being a map of each subset, where each >> >> > successive bit represents membership in thesubset by each successive >> >> > natural number? >> >> >> >> Ok, so let's construct P. It is actually easy enough, since n<2^n, >> >> and so P=N. So what number corresponds to N itself in your mapping? >> >> >> > An infinite string of 1's: 1111.....1111. This is (2^aleph_0)-1. >> > >> > Infinite whole numbers are required for an infinite set of whole >> > numbers. >> >> And what number corresponds to the subset of N without the number >> 1111.....1111 in it? >> >> > What is this, 20 questions? It's 01111....1111. Duh! And now if you also take out 01111....1111, what do you get then? -- David Kastrup, Kriemhildstr. 15, 44793 Bochum
From: Chan-Ho Suh on 19 Jul 2005 17:27
In article <1121807658.886274.146310(a)g49g2000cwa.googlegroups.com>, david petry <david_lawrence_petry(a)yahoo.com> wrote: > >The truth of the matter is that the article you wrote constitutes > >*original research* on your part, despite your attempt to ascribe your > >views to the "anti-Cantorians", which is, as quasi pointed out, not a > >well-defined group. Thus, it is not acceptable for inclusion into > >Wikipedia. > > I'm objectively describing, as best I can, a debate which has > appeared in these newsgroups. That's not against any rules I > know about. It's not "original research". You see what you wrote in that manner, but it appears to me like "original research" and I'm sure it would to others (or at least enough others). If you are really seeking to summarize a debate, do so, and it may be of interest to others reading Wikipedia. But as I said, if it doesn't get deleted, I expect you will be very unhappy with the result after many edits. The consensus, as far as I can tell, on mathematical Wikipedia is not so different than that among sci.math regulars. > You are free to edit the > article any way you want, and I will be free to do likewise. > May the best man win :-) No, seriously, I doubt I will edit; it would take too much time anyway. But there are others with more time on their hands...you've been warned :-) |