THE HOLY GRAIL ~ CANTOR DISPROOF
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From: Transfer Principle on 27 May 2010 00:10 On May 24, 5:39 pm, William Hughes <wpihug...(a)hotmail.com> wrote: > On May 24, 9:04 pm, Herc7 <ozd...(a)australia.edu> wrote: > > 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. > There are however an infinite number of counterexamples. Hmmm. Let's look at Herc/Cooper's post again. > > The _probability_ of fitting a _random_ diagonal to the list of computable > > numbers resulting in a different set of numbers is 0. (emphasis mine) Probability? Random? We know that in standard analysis, probability is usually defined in terms of the Lebesgue measure. > Interestingly this set is uncountable. But what's the Lebesgue measure of this set? Is it zero? If so, then we can vindicate Herc/Cooper's claim after all. So let us restate the claim. Question: Let X be the subset of [0,1] such that a real number x is in X iff no permutation of the set of computable reals, written in ternary, can produce that number on the diagonal. What is the Lebesgue measure of X? Or we may write this even more formally: Let F : omega^2 -> 3 (the von Neumann ordinal 3) be a function such that for every function g : omega -> 3, there exists a unique natural number m such that F(m,n) = g(n) for all n if and only if g is a computable function. Then let X be the subset of [0,1] such that for every real number x, x is in [0,1] iff for every h : omega -> omega bijective, the sum of F(h(n),n)/3^(n+1) as n ranges 0 to infinity is not x. What is the Lebesgue measure of X? William Hughes tells us that X is uncountable, but we are interested in its measure. There are three cases: Case 1. X has zero Lebesgue measure. Case 2. X has positive Lebesgue measure. Case 3. X is not Lebesgue measurable. But I doubt that Case 3 applies, for otherwise we would have produced a nonmeasurable set X without AC -- which would render ZF+~AC (hence ZF by Cohen) inconsistent. So either Case 1 or Case 2 applies. If Case 1 holds, then Herc/Cooper would be right to say that the probability of choosing a real that can't be on the diagonal of the list of computable reals is _zero_. But I'm not sure how to decide this question. Let us start out by looking at some elements of X. From the argument involving 1/2 (0.111... in ternary) above, we know that any real whose ternary expansion contains only zeros and twos is in X. This is a well known set, the Cantor set, and it is uncountable yet has measure zero. For every computable real, the set of all reals with no digit in common with the given real (and thus must be in X) is also an uncountable null set (and the proof of this is similar to that for the Cantor set). There are only countably many computable reals, and the union of countably many null sets is still null. Thus, so far, the reals found to be in X still form a set of Lebesgue measure zero. Notice that X contains all computable reals. This is because for every computable real x, there exists another computable real y such that x and y have no digits in common. If x is in [0,1/2], consider y = x+1/2 or x+0.111..., where we add either 1 to each digit of x, or 2 if there's a carry. If x is in [1/2,1], we can consider y = x-1/2 = x-0.111... instead. But for the remaining reals (all of which must be uncomputable) having at least one digit in common with every real, there's still no guarantee that we can find a permutation of the list such that the target real lies on the diagonal. Suppose there's some real which has only one digit in common with a computable real -- let's call the diagonal d, call the computable real x, and say that the first digit is the common digit. Then x must appear first on the list. But then we can find another real with only the first digit in common with d -- namely y, where y = x+1/6 = x+0.0111... (unless this causes a carry in the first digit, in which case we consider y = x-1/6 = x-0.0.111... instead). Then y must also appear first on the list -- but evidently x and y can't both be first on the list! So every real with has only one digit in common with some computable real must be in X as well. Indeed, I suspect that every real which has only finitely many digits in common with some computable real must be in X as well. But by now, there might be so many reals in X that it is no longer null.
From: William Hughes on 27 May 2010 10:32 On May 27, 1:10 am, Transfer Principle <lwal...(a)lausd.net> wrote: > On May 24, 5:39 pm, William Hughes <wpihug...(a)hotmail.com> wrote: > > > On May 24, 9:04 pm, Herc7 <ozd...(a)australia.edu> wrote: > > > 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. > > There are however an infinite number of counterexamples. > > Hmmm. Let's look at Herc/Cooper's post again. > > > > The _probability_ of fitting a _random_ diagonal to the list of computable > > > numbers resulting in a different set of numbers is 0. > > (emphasis mine) > > Probability? Random? There are _however_ an infinite number of counterexamples (emphasis added). My comment was not intended to contradict the probability statement. It is true that the set of counterexamples may have measure 0 in the usual Lebesgue measure. Indeed, I suspect this, though I cannot see a simple way of demonstrating it. However, the set of counterexamples is non-empty and uncountable. [Note that the idea that an infinite number of reals can have Lebesgue measure 1 already requires uncountability. If you insist on trying to make Herc's statements consistent you will need a theory of probability in which a countable infinity of elements, each with the same probability, has a probability of 1.] - William Hughes
From: Daryl McCullough on 27 May 2010 11:15 LauLuna says... >You are overlooking the possibility that 'this cannot be proven by >Roger Penrose' express no proposition, that it be paradoxical just >like the Liar sentence. But it obviously is *not* paradoxical. Assuming it is true does not lead to a contradiction. Assuming it is false does not lead to a contradiction. That's very different from the liar paradox, in which both assumptions lead to contradictions. >Penrose will surely know that if he proves it, then it is not true; so >he must know he cannot prove it; No, that doesn't follow. That only follows if you *also* assume that Penrose never proves anything false. (There is actually a technical difficulty in applying the notion of "what is provable" to a human being, as opposed to a formal theory, and that's this: for a formal theory, if the theory proves A, and B logically follows from A, then the theory proves B. Humans don't live long enough to prove all the logical consequences of what they've already proved. And furthermore, once a human spots a contradiction in what they've done, they will typically try go back and fix their assumptions to eliminate the contradiction. They don't just go ahead proving everything.) >in fact he has proved that the sentence cannot be proven by Roger Penrose; No, he hasn't. He's proved that *if* he never proves anything false, *then* he will never prove that sentence. He hasn't proved that he will never prove that sentence. On the other hand, let's assume the following facts about Penrose: (1) Penrose believes that he never proves anything false. (2) Penrose proves everything that is a logical consequence of things that he believes. In that case, since "this cannot be proven by Roger Penrose" is a logical consequence of "Penrose never proves anything false", then Penrose *will* prove the sentence "this cannot be proven by Roger Penrose". It follows that the sentence is both false and provable by Penrose. So for the "Penrose statement" we have the following: 1. If Penrose never proves anything false, then the Penrose statement is true, but not provable by Penrose. 2. If Penrose can prove the Penrose statement, then the Penrose statement is false (and Penrose proves false statements). It's exactly the case with Godel's theorem: The theory is either unsound (proves false things) or incomplete (fails to prove true things). -- Daryl McCullough Ithaca, NY
From: Transfer Principle on 27 May 2010 15:52 On May 25, 10:11 am, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Transfer Principle <lwal...(a)lausd.net> writes: > > Notice that Herc/Cooper claims to have proved it, > Notice that Herc/Cooper claims to be both Genesis Adam Back to Genesis Adam again? I definitely do _not_ accept Herc/Cooper's claim that he is Genesis Adam, since, to say the least, Adam is _dead_ according to Genesis (for having eaten the forbidden fruit). How can someone be a specific person mentioned in a book if according to that book, that person is dead? (Unless that book is wrong, of course but Herc/Cooper also states belief that the book is true.) Therefore, just because I'm entertaining one of Herc/Cooper's claims (about Cantor), it doesn't mean that I necessarily accept any of his other claims (especially not those about Adam). Indeed, I convinced Herc/Cooper to start this thread about Cantor in order to avoid discussing Genesis on sci.math, but Genesis and Adam have still turned up in this thread.
From: Transfer Principle on 27 May 2010 16:07
On May 25, 12:21 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > Don Stockbauer <donstockba...(a)hotmail.com> writes: > > Now, now, now. Let's not fall victim to the "Argument Against the > > Man" (er, Person, nowadays). > In this instance the information that Herc appears to be clinically > insane is salient. Recall the nature of lwalker's valiant quest. He > wants to vindicate people who get called (mathematical) cranks I made a resolution that I would avoid five-letter insults such as the word "crank" unless at least one other poster in the thread has already used the word. Here Aatu has used the word "crank," and in the plural, to boot. This means that use of the word "crank," in both singular and plural, is now open to me. Also, Aatu refers to Herc/Cooper as "insane." I don't accept that most so-called "cranks" are "insane." Of course, much of the evidence supporting Aatu's claim of his insanity are non-mathematical (Genesis Adam and all that). So I should reject all of Herc/Cooper's mathematics just because of his insane claim to be Adam? This issue has also come up with Archimedes Plutonium and his Atom Totality theory. I'd like to consider some interesting mathematical theories yet be able to separate these "cranks'" mathematical ideas from their more controversial theories about Adam and Atom. Of course, if only Herc/AP would stick to alt.atheism and sci.physics, respectively, when discussing their non-mathematical ideas, it would be easier, for me as well as everyone else, to read about their ideas for math here on sci.math. |