Cantor Confusion
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From: William Hughes on 21 Dec 2006 18:38 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > William Hughes schrieb: > > > > > > > > Please give a counter example for a finite number n. > > > > > > > > No such counterexample exists. The claim is that despite > > > > the fact that given any n we can find a single line, L(n), that works > > > > for this n, there is no single line, L_D, that will work for all n. > > > > > > This claim is obviously false. At least my counter claim cannot be > > > disproved by an example.> > > > > see the example that disproves your counter claim below. > > (by mistake you snipped it) > > You are in error. You gave no example showing that not all natural > numbers can be in one line. > > > > > > > > > > > Remember, even if there is not a last line, every n eps N is finite. > > > > > > > > Which means that for any n we can find a single line L(n) that works > > > > for this n. It does not mean that that we can find a single > > > > line L_D, that works for all n. > > > > > > That is impossible, beause there is no "all n". > > > > "It does not mean that that we can find a single > > line L_D, that works for all n." written out in full is > > > > It does not mean that there exists an L_D with the property that: > > if a natural number n can be shown to exist in the diagonal > > n must be an element of L_D. > > If all natural numbers would exist, then, of course, also this L_D > would exist. If L_D does not exist, then we have the proof that not all > natural numbers do exist. No. L_D does not exist whether or not we assume that all natural numbers exist. The proof that L_D does not exist makes no assumption about whether all natural numbers exist. All we need is that the natural numbers are not bounded. This is true whether or not all the natural numbers exist. > > > > There is no claim that "all n" exists. > > > > Restoring the example you snipped > > (and expanding the language slightly) > > > > There exists a linearly ordered set of finite elements that > > does not support quantifier reversal. > > > > Consider the (potentially infinite) set of natural numbers. > > This is linearly ordered. Every element is finite. > > > > For any natural number n that can be shown to > > exist, a natural number m(n), > > such that m(n) > n can be shown to exist. > > > > However it is not true that: > > > > There exists a natural number m, such that for any > > number n that can be shown to exist, m>n. > > There exists a natural number m, such that for any n that can be shown > to exist, m >= n. > It is a number which is at least as large as the largest existing > natural number. > There is no largest existing natural number. Assume it exists. Call it N_L. It is easy to see that the set A={1,2,3,...,N_L} exists. The natural numbers are a potentially infinite set. For any set of natural numbers B that exists, we can find a natural number that is not in B. A exists, therefore there must be a natural that is not in A. Call it N_B. N_B > N_L. Contradiction. Therefore N_L does not exist. -William Hughes
From: William Hughes on 21 Dec 2006 19:26 Han de Bruijn wrote: > William Hughes wrote: > > > Han de Bruijn wrote: > > > >>William Hughes wrote: > >> > >>>The statements are > >>> > >>> A "potentially infinite set" is a function on sets. > >>> > >>> A potentially infinite set takes on the values true and false. > >>> > >>>Any claim that the putative domain and range are not clear > >>>is playing word games. > >> > >>On the contrary. The domain "on sets" of your function is not defined. > > > > Word games. Clearly you had no difficulty determining what > > was intended. > > > >>>If you have a substantive objection (e.g. one cannot define > >>>such a function because you can only define functions on sets, > >>>and the set of all sets of natural numbers does not exist) > >>>then the answer is "No". > >>> > >>>If, however, you wan't to answer this question No, but refuse > >>>to do so because you feel that answering it No would make > >>>you look stupid, I do not feel for you. > >>> > >>>Does a potentially infinite set exist? > >> > >>I have another definition that I can handle, but I can not handle yours. > > > > More word games. For some reason you do not > > wish to give an actual reply like: > > > > No, this would require defining a function on all sets which cannot > > be done. My preferred definition is... > > the Naturals Construction Set, as I've indicated before: "the Naturals Construction Set" is not a definition, it is a property that a set of natural numbers might or might not have. It also sheds no light on the question of whether a potentially infinite set of natural numbers exists as it is trivially obvious that both bounded and unbounded sets can have "the Naturals Construction Set" property. The question is not longer "are you trying to avoid the question" but "why are you avoiding the question". However, the answer to this lies in the psychology of the crank and is not of present interest. - William Hughes > > http://groups.google.nl/group/sci.math/msg/13795822737a77ca?hl=en& > > BTW, you seems to know very well what my objections would be. > > Han de Bruijn
From: Dik T. Winter on 21 Dec 2006 19:47 In article <1166728786.050409.162940(a)48g2000cwx.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > > > But, indeed, if you pay for it, anything can be published. But > > > > that was already true in Galilei's time. > > > > > > You are in error. > > > > It is certainly true, also at that time there were publishers enough that > > would publish anything. And indeed, in 1637 a book by Galilei was > > published in Leiden showing his views: "Discorsi, e dimostrazioni > > mathematiche, intorno � due nuoue szienze". And I do not think that > > Galilei even paid for that publication. The manuscript was smuggled > > from his house (where he was in detention) to the publisher. > > Correct. Why had it to be smuggled? Because hardliners wanted to > maintain their false worldview! > > In 1624 Pope Urban VIII (former Cardinal Barberini) had allowed the > printing of the dialogue about the two word systems. No smuggling was > necessary. That was a different dialogue (and according to my sources it was published in 1632): "Dialogo di Galileo Galilei sopra i due Massimi Sistemi del Mondo Tolemaico e Copernicano". And as a result of that he got home detention. But the hardliners were not mathematical hardliners or astronomical hardliners. They were religious hardliners. > > And, > > yes, in the low countries almost anything could be published. Many > > important books have been published in those times in the Netherlands > > (while it was still in war with Spain). In the first republic ever in > > the western world, there was a lot of tolerance. > > This role has now been taken by institutions like Shaker Verlag. It appears to be necessary in the physics comminity (with the page rates as they are). I do not see so much the necessity in the mathematics community (where no page rates are used). At that time Galilei's book could not be printed in many countries because it would have been forbidden by the RC church. And although, formally, the Spanish king had still everything to say in the Netherlands, he had no power in what actually was the republic, which was at that time mostly Calvinist but with great tolerance to other religions (later that tolerance deteriorated quite a bit). > Merry Christmas. Fr�hliche Weihnachten. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: cbrown on 21 Dec 2006 20:02 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > cbrown(a)cbrownsystems.com schrieb: > > > > > > > > > > > No. But I need no proof in order to see that > > > > > 1 + 1/2 + 1/4 + ...+ 1/2^n + ... >= 2 - 1/2^n > 3/2 for n > 2. > > > > > > > > > > > > > Yes, it appears that you need no proof to make any of your assertions. > > > > You simply "see" them; or "feel" them. That's not what I would call > > > > "really correct mathematics". > > > > > > Call it whatever you want. > > > > I call it "mathiness". > > > > > I can calculate 1 + 1/2. > > > > Good for you! > > Bad for you, if you can't. > > > > > But considering the > > > fact that you would call the following result "correct mathematics", I > > > prefer to stay with my version. > > > > > > Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n > > > different paths and E(n) = 2*2^n - 2 different edges. So we find > > > lim{n-->oo} E(n)/P(n) = 2, or, if we are unable to determine limits, > > > E(n)/P(n) >= 1. Nevertheless "E(n)/P(n) < 1 in the actual infinity". > > > > It is /not/ a fact that I would call this "result" correct mathematics. > > I would call it yet another example of your devotion to "mathiness" > > whenever you claim to provide a "contradiction". > > > There can't be a contradiction in ZFC because everything looking like a > contradiction is mathiness. Assertions don't "look" like contradictions, as much as you may "feel" that they do. They must be /proved/ to be contradictions, which is to say, they must be proved /in ZFC/ to be a contradiction /in ZFC/. > > While "E(n)/P(n)" is certainly well-defined when both E(n) and P(n) are > > real numbers, it is not well-defined otherwise. You simply "feel" that > > Cantor's result must imply "E(n)/P(n) < 1 in the actual infinity"; > > without considering what that assertion /means/. > > > I know that 2^aleph0 > aleph0 is a result of Cantor's theory. Great! > If you > don't know that, you should try to inform you. > And I have proved that E(n) > P(n) for every n and for the limit n --> > oo. > Yes, you have. > > You apply "mathiness" to the problem. You "know" (need no proof) that > > "2 < 3" implies that "2/3 < 1". > > I know that 2/3 < 1 is a convenient way to express 2 < 3. For heaven's sake. "2/3 < 1" is not just "a convenient way" of expressing "2 < 3". One statement is /derived/ from the other: /Because/ 2/3 < 1, it follows that 2 < 3; and /because/ 2 < 3, it follows that 2/3 < 1. How can these statements be derived? Because we have /defined/ what "x/y" means and what "x < y" means in this context. > But if you > don't know that, then we can stay with 2 < 3 as well. > > > You "see" the assertion "|N| < |R|", > > This is a provable theorem in ZF > Yay! > > and because it "looks" like "2 < 3", you "feel" that you can therefore > > conclude "|N|/|R| < 1". > > > > Instead, the statements "|N|/|R| < 1" and "E(n)/P(n) < 1 in the actual > > infinity" are simply mathematical nonsense: neither true nor false. > > Are you unable to see that the formal use of fractions like |N|/|R| < 1 > is completely irrelevant but only a brief way to express |N| < |R| ? "The formal use of fractions like ...". What on earth is "formal" about it? Why not simply /say what you mean/ instead of using misleading language? Suppose you had simply said: Down to level n (for n = 1, 2, 3, ...) we can distinguish P(n) = 2^n different paths and E(n) = 2*2^n - 2 different edges. So we find lim{n-->oo} E(n)/P(n) = 2. Nevertheless, |E(n)| < |P(n)| in the actual infinity. Then I would agree: the fact that E(n)/P(n) = 2 in the limit for natural n indeed has no particular relevance to the provable fact that |E(n)| < |P(n)| for infinite n; and there really is nothing surprising about this. E(n)/P(n) -> 2 is a feature only of finite n. |E(n)| < |P(n)| is a feature only of infinite n. /Many/ things that can be said in the finite case are not true in the infinite case; for example E(n) is finite for natural n, and not finite for infinite n. It is when you try to argue "lim n-> oo E(n)/P(n) = 2 contradicts E(n)/P(n) < 1 in the infinite case" that your confusing notation "looks" like a contradiction (to you; to me it looks like nonsense). Cheers - Chas
From: cbrown on 21 Dec 2006 20:08
Virgil wrote: > In article <1166737382.495151.299530(a)i12g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > but the question is > > > whether you can generate a direct contradiction of the type P & ~P in > > > ZFC. > > > > I do not think that this is of any interest after it can easily be > > shown that the results of ZFC are inacceptable. > > Except that the only demonstration of such "inacceptability" would be > the discovery of some internal contradiction within ZFC or NBG. > You are discounting the fact that for WM, an assertion is unacceptable iff it "feels" unacceptable. Cheers - Chas |