THE HOLY GRAIL ~ CANTOR DISPROOF
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From: Transfer Principle on 24 May 2010 12:46 On May 22, 4:22 pm, Herc7 <ozd...(a)australia.edu> wrote: > On May 22, 2:45 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > I liked it better back when Herc was trying to disprove Cantor > > rather than prove Genesis because Cantor, unlike Genesis, is > > actually pertinent to sci._math_. > OK then. Thank you. Here on sci.math, I want to see _math_. > If the computable reals And of course, this gives it away. Notice that standard theory ZFC does not dispute that there exist only countably many _computable_ reals. > can be shuffled to fit any randomly > generated diagonal, how can you prove ANYTHING from the value of the > diagonal given the information entropy of a random number is 0. Ah yes, I sort of remember that old "shuffling" argument given by Herc/Cooper. In particular, the statement to be proved was that given a list of reals and another real number, there exists a permutation of that list such that the diagonal of the permuted list is the given real. To put it slightly more rigorously, since the proof used ternary notation, Herc/Cooper is saying that for a certain function: F : omega^2 -> 3 and for every function: g : omega -> 3 (Recall that here 3 denotes the von Neumann ordinal 3, i.e., {0,1,2}). there exists a bijection: h : omega -> omega such that for every natural number n, F(h(n),n) = g(n). I forgot how computability was related to this old proof. Notice that F(m,n) can't be any arbitrary function, since if, say, F(m,1) = 0 for every m, then there evidently is no function h such that F(h(1),1) = 1. Most likely, this list is supposed to be a list of _computable_ reals -- that is, if the function c : omega -> 3 is computable, then there exists a natural number m such that for every natural n: c(n) = F(m,n) Although I'm not sure whether Herc/Cooper's proof is valid in ZFC, the conclusion that there exist only countably many computable reals is not disputed in ZFC. What ZFC does refute is the existence of countably many _uncomputable_ reals. And so let's go back to the three cases that I've agreed consider when I see an anti-Cantor thread: Case 1. Herc/Cooper is trying to propose a new theory in which only computable reals count as reals in the proposed theory. Case 2. Herc/Cooper knows that standard theory proves the existence of uncountably many reals, but doesn't like this fact. Case 3. Herc/Cooper believes that the standard theory proves that there are only countably many reals, and that he has the proof. In this case, we might say that Herc/Cooper is wrong. In another thread, some posters have suggested that I add a fourth case. In Case 4, we say that Herc/Cooper is "not even wrong," because his claim is so outrageous.
From: Herc7 on 24 May 2010 16:40 On May 25, 2:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On May 22, 4:22 pm, Herc7 <ozd...(a)australia.edu> wrote: > > > On May 22, 2:45 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > I liked it better back when Herc was trying to disprove Cantor > > > rather than prove Genesis because Cantor, unlike Genesis, is > > > actually pertinent to sci._math_. > > OK then. > > Thank you. Here on sci.math, I want to see _math_. > > > If the computable reals > > And of course, this gives it away. Notice that > standard theory ZFC does not dispute that there > exist only countably many _computable_ reals. > > > can be shuffled to fit any randomly > > generated diagonal, how can you prove ANYTHING from the value of the > > diagonal given the information entropy of a random number is 0. > > Ah yes, I sort of remember that old "shuffling" > argument given by Herc/Cooper. In particular, > the statement to be proved was that given a > list of reals and another real number, there > exists a permutation of that list such that the > diagonal of the permuted list is the given real. > > To put it slightly more rigorously, since the > proof used ternary notation, Herc/Cooper is > saying that for a certain function: > > F : omega^2 -> 3 > > and for every function: > > g : omega -> 3 > > (Recall that here 3 denotes the von Neumann > ordinal 3, i.e., {0,1,2}). > > there exists a bijection: > > h : omega -> omega > > such that for every natural number n, > > F(h(n),n) = g(n). > > I forgot how computability was related to this > old proof. Notice that F(m,n) can't be any > arbitrary function, since if, say, F(m,1) = 0 > for every m, then there evidently is no function > h such that F(h(1),1) = 1. > > Most likely, this list is supposed to be a list > of _computable_ reals -- that is, if the function > > c : omega -> 3 > > is computable, then there exists a natural number > m such that for every natural n: > > c(n) = F(m,n) > > Although I'm not sure whether Herc/Cooper's proof > is valid in ZFC, the conclusion that there exist > only countably many computable reals is not > disputed in ZFC. What ZFC does refute is the > existence of countably many _uncomputable_ reals. > > And so let's go back to the three cases that I've > agreed consider when I see an anti-Cantor thread: > > Case 1. Herc/Cooper is trying to propose a new > theory in which only computable reals count as > reals in the proposed theory. > Case 2. Herc/Cooper knows that standard theory > proves the existence of uncountably many reals, > but doesn't like this fact. > Case 3. Herc/Cooper believes that the standard > theory proves that there are only countably many > reals, and that he has the proof. In this case, > we might say that Herc/Cooper is wrong. > > In another thread, some posters have suggested > that I add a fourth case. In Case 4, we say > that Herc/Cooper is "not even wrong," because > his claim is so outrageous. Translation, "No because I told you so". I have made no claim in this post that there are only countable reals, I merely show that Cantor's proof by diagonalization is inconsistent with probability theory. Try answering the question I put forth to you 3 days ago. If the computable reals can be shuffled to fit any randomly generated diagonal, how can you prove ANYTHING from the value of the diagonal given the information entropy of a random number is 0? Herc
From: Transfer Principle on 24 May 2010 19:38 On May 24, 1:40 pm, Herc7 <ozd...(a)australia.edu> wrote: > On May 25, 2:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > Case 1. Herc/Cooper is trying to propose a new > > theory in which only computable reals count as > > reals in the proposed theory. > > Case 2. Herc/Cooper knows that standard theory > > proves the existence of uncountably many reals, > > but doesn't like this fact. > > Case 3. Herc/Cooper believes that the standard > > theory proves that there are only countably many > > reals, and that he has the proof. In this case, > > we might say that Herc/Cooper is wrong. > > In another thread, some posters have suggested > > that I add a fourth case. In Case 4, we say > > that Herc/Cooper is "not even wrong," because > > his claim is so outrageous. > I have made no claim in this post that there are only countable reals, > I merely show that Cantor's proof by diagonalization > is inconsistent with probability theory. OK then, we can safely eliminate Case 1. > Try answering the question I put forth to you 3 days ago. > If the computable reals can be shuffled to fit any randomly > generated diagonal I have yet to see a valid proof in ZFC that this holds. To me, this _conjecture_ is a bit strange. After thinking about it for a while, I'm starting to wonder whether it's even plausible. Notice that Herc/Cooper claims to have proved it, and even once gave a link to a website which demonstrates the conjecture by providing a random list and diagonal, then reorders the list such that the diagonal matches the given real. But this was flawed for two reasons: 1. The list and diagonal real were both _finite_. 2. I was able to stump the computer by choosing my own real for the diagonal, and the computer wasn't able to find a permutation of the list for it. Of course, Herc/Cooper was discussing infinite lists, and so a counterexample for a finite list doesn't disprove the infinite case. But, after thinking about it for a while, I believe to have found an infinite counterexample after all. Let's say we want the diagonal to be: D = .020202... This is the rational number 1/4 (in ternary). So according to Herc/Cooper, there exists a permutation of the list of computable reals such that this number lies on the diagonal. This list contains every computable real, so in particular, it contains: R = .111111... This is the rational number 1/2, which, being computable, must be on the list. But where is it? It can't be the first number, since the first digits of D and R don't match. It can't be the second number, since the second digits of D and R don't match. It can't be the third number, since the third digits of D and R don't match. And so on. Thus the computable reals can't be shuffled to fit the diagonal 1/4. Therefore, the conjecture is false. QED And so we can snip the rest of Herc's question, since its premise is false. The computable reals can't be shuffled to fit any diagonal.
From: Herc7 on 24 May 2010 20:04 On May 25, 9:38 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On May 24, 1:40 pm, Herc7 <ozd...(a)australia.edu> wrote: > > > > > > > On May 25, 2:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > Case 1. Herc/Cooper is trying to propose a new > > > theory in which only computable reals count as > > > reals in the proposed theory. > > > Case 2. Herc/Cooper knows that standard theory > > > proves the existence of uncountably many reals, > > > but doesn't like this fact. > > > Case 3. Herc/Cooper believes that the standard > > > theory proves that there are only countably many > > > reals, and that he has the proof. In this case, > > > we might say that Herc/Cooper is wrong. > > > In another thread, some posters have suggested > > > that I add a fourth case. In Case 4, we say > > > that Herc/Cooper is "not even wrong," because > > > his claim is so outrageous. > > I have made no claim in this post that there are only countable reals, > > I merely show that Cantor's proof by diagonalization > > is inconsistent with probability theory. > > OK then, we can safely eliminate Case 1. > > > Try answering the question I put forth to you 3 days ago. > > If the computable reals can be shuffled to fit any randomly > > generated diagonal > > I have yet to see a valid proof in ZFC that this holds. > > To me, this _conjecture_ is a bit strange. After > thinking about it for a while, I'm starting to > wonder whether it's even plausible. > > Notice that Herc/Cooper claims to have proved it, > and even once gave a link to a website which > demonstrates the conjecture by providing a random > list and diagonal, then reorders the list such that > the diagonal matches the given real. But this was > flawed for two reasons: > > 1. The list and diagonal real were both _finite_. > 2. I was able to stump the computer by choosing my > own real for the diagonal, and the computer wasn't > able to find a permutation of the list for it. > > Of course, Herc/Cooper was discussing infinite > lists, and so a counterexample for a finite list > doesn't disprove the infinite case. > > But, after thinking about it for a while, I believe > to have found an infinite counterexample after all. > > Let's say we want the diagonal to be: > > D = .020202... > > This is the rational number 1/4 (in ternary). So > according to Herc/Cooper, there exists a permutation > of the list of computable reals such that this number > lies on the diagonal. This list contains every > computable real, so in particular, it contains: > > R = .111111... > > This is the rational number 1/2, which, being > computable, must be on the list. But where is it? It > can't be the first number, since the first digits of > D and R don't match. It can't be the second number, > since the second digits of D and R don't match. It > can't be the third number, since the third digits of > D and R don't match. And so on. > > Thus the computable reals can't be shuffled to fit the > diagonal 1/4. Therefore, the conjecture is false. QED > > And so we can snip the rest of Herc's question, since > its premise is false. The computable reals can't be > shuffled to fit any diagonal. You are selecting specific diagonals based on the list. The probability of fitting a random diagonal to the list of computable numbers resulting in a different set of numbers is 0. Herc
From: William Hughes on 24 May 2010 20:39
On May 24, 9:04 pm, Herc7 <ozd...(a)australia.edu> wrote: > On May 25, 9:38 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > > > On May 24, 1:40 pm, Herc7 <ozd...(a)australia.edu> wrote: > > > > On May 25, 2:46 am, Transfer Principle <lwal...(a)lausd.net> wrote: > > > > Case 1. Herc/Cooper is trying to propose a new > > > > theory in which only computable reals count as > > > > reals in the proposed theory. > > > > Case 2. Herc/Cooper knows that standard theory > > > > proves the existence of uncountably many reals, > > > > but doesn't like this fact. > > > > Case 3. Herc/Cooper believes that the standard > > > > theory proves that there are only countably many > > > > reals, and that he has the proof. In this case, > > > > we might say that Herc/Cooper is wrong. > > > > In another thread, some posters have suggested > > > > that I add a fourth case. In Case 4, we say > > > > that Herc/Cooper is "not even wrong," because > > > > his claim is so outrageous. > > > I have made no claim in this post that there are only countable reals, > > > I merely show that Cantor's proof by diagonalization > > > is inconsistent with probability theory. > > > OK then, we can safely eliminate Case 1. > > > > Try answering the question I put forth to you 3 days ago. > > > If the computable reals can be shuffled to fit any randomly > > > generated diagonal > > > I have yet to see a valid proof in ZFC that this holds. > > > To me, this _conjecture_ is a bit strange. After > > thinking about it for a while, I'm starting to > > wonder whether it's even plausible. > > > Notice that Herc/Cooper claims to have proved it, > > and even once gave a link to a website which > > demonstrates the conjecture by providing a random > > list and diagonal, then reorders the list such that > > the diagonal matches the given real. But this was > > flawed for two reasons: > > > 1. The list and diagonal real were both _finite_. > > 2. I was able to stump the computer by choosing my > > own real for the diagonal, and the computer wasn't > > able to find a permutation of the list for it. > > > Of course, Herc/Cooper was discussing infinite > > lists, and so a counterexample for a finite list > > doesn't disprove the infinite case. > > > But, after thinking about it for a while, I believe > > to have found an infinite counterexample after all. > > > Let's say we want the diagonal to be: > > > D = .020202... > > > This is the rational number 1/4 (in ternary). So > > according to Herc/Cooper, there exists a permutation > > of the list of computable reals such that this number > > lies on the diagonal. This list contains every > > computable real, so in particular, it contains: > > > R = .111111... > > > This is the rational number 1/2, which, being > > computable, must be on the list. But where is it? It > > can't be the first number, since the first digits of > > D and R don't match. It can't be the second number, > > since the second digits of D and R don't match. It > > can't be the third number, since the third digits of > > D and R don't match. And so on. > > > Thus the computable reals can't be shuffled to fit the > > diagonal 1/4. Therefore, the conjecture is false. QED > > > And so we can snip the rest of Herc's question, since > > its premise is false. The computable reals can't be > > shuffled to fit any diagonal. > > You are selecting specific diagonals based on the list. > > The probability of fitting a random diagonal to the list of computable > numbers resulting in a different set of numbers is 0. > > Herc There are however an infinite number of counterexamples. Interestingly this set is uncountable. - William Hughes |