Cantor Confusion
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From: David Marcus on 23 Dec 2006 12:41 cbrown(a)cbrownsystems.com wrote: > 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). Ah, but you are forgetting the crank's axiom: "Everything true for finite numbers is also true for infinity." -- David Marcus
From: Virgil on 23 Dec 2006 13:13 In article <1166873125.594545.142580(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > stephen(a)nomail.com schrieb: .. > > > > So that answers your question "what is wrong with WM's mapping?" > > It is not a mapping. > > > > > You need to construct a surjection. > > > No, I don't. > > > > Yes you do, if you want to show that there exists a surjection > > from edges to paths. > > You cannot construct a surjection because irrational numbers do not > exist. As there are no numbers, irrational or otherwise, involved in the structure of an infinite binary tree, questions of their existence are irrelevant. > But we can construct a surjection into the set of all existing > nunmbers Only if one assumes that "the set of all existing numbers" is countable. So WM again argues in circles. > > > >> Showing that each path will accumalate the weight of two edges > > >> has nothing to do with constructing a mapping. > > > But it shows that in an infinite tree there are two edges per path. > > > > No. It shows that in large finite trees there are approximately > > two edges per path. It says nothing about infinite trees. > > 1 + 1/2 + 1/4 + ... + 1/2^n is larger than 1. That does not prove that > 1 + 1/2 + 1/4 + ... > 1? > Yes, set theorist must confess 1 + 1/2 + 1/4 + ... <1. But they need not "confess" that that is in any way relevant to the demonstrated uncountability of the set of paths of am infinite binary tree.
From: Virgil on 23 Dec 2006 13:22 In article <1166873365.794178.241750(a)42g2000cwt.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1166854303.474151.267360(a)h40g2000cwb.googlegroups.com>, > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > Virgil wrote: > > > > In article <1166849530.679744.243900(a)a3g2000cwd.googlegroups.com>, > > > > "Newberry" <newberry(a)ureach.com> wrote: > > > > > > > > > stephen(a)nomail.com wrote: > > > > > > Newberry <newberry(a)ureach.com> wrote: > > > > > > > > > > > > > > > > > No, I don't. What I am showing here is that each path will > > > > > > > accumulate > > > > > > > the weight of two edges as it approaches infinity. > > > > > > > > > > > > That is not a mapping. > > > > > Correct. > > > > > You need to construct a surjection. > > > > > > > > > No, I don't. > > > > > > > > You do if you want to claim that there are "as many" edges as paths, as > > > > that is what such a claim says. > > > > > > > > > > Showing that each path will accumalate the weight of two edges > > > > > > has nothing to do with constructing a mapping. > > > > > > > > > But it shows that in an infinite tree there are two edges per path. > > > > > > > > And the same two edges for infinitely many other paths as well, so your > > > > paths are being used more than once. > > > > > > Is it true that the ratio of edges over paths converges to two as we > > > approach infinity? > > > > > > lim{n-->oo} (2*2^n - 2)/2^n = 2 > > Yes but that would not even be required. Enough to now that > lim{n-->oo} (2*2^n - 2)/2^n > 1. > > Set theorists (or should we say set terrorists?) confirm 1 + 1/2 + 1/4 > + ... < 1. > > > > It is true that the ratio of terminal nodes to paths converges to 1 as > > the path lengths increase towards infinity. So if your logic held every > > endless path would have a terminal node. > > The logic required says: No path separates itself without an additional > edge. Every *existing* separated path has its own edge. Only if that edge is a terminal edge of that path. Which in these infinite binary trees can not occur. If a path through any edge has further edges then all those paths, and there are more than one, through that edge share that edge equally. So that in a binary tree in which no path end, there is NO path at all "having its own edge" in the sense demanded by WM.
From: Virgil on 23 Dec 2006 13:31 In article <1166873954.726498.74250(a)79g2000cws.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > 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. > > As not all natural numbers do exist, the set is potentially infinite, > i.e., it is finite. It has a maximum. L_D. Taking this maximum and > adding 1 or takin L_D ^ L_D or so yields another maximum. In any cae > there is a maximum. Which is always exceeded by a larger maximum. > > > 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. > > Here is anothe proof: > The set of all even natural numbers contains at least one number larger > than its cardinality. Can WM name that number? If not, how can he prove its existence? > This proves that there cannot be an infinite > cardinal number of tis set. It does not prove anything until it has itself been proved. And WM cannot prove it. > This proves that there cannot be an > infinite cardinal numbers of N. Circular argument again. WM cannot prove any of the claims in this circle without assuming one of them. > > There is no largest existing natural number. > > False. Try to find a nunmber larger than he largest existig number. You > will fail. Which "largest existing natural number" is that? You name one, and we will find a larger one. > > > > Assume it exists. Call it N_L. > > > > It is easy to see that the set > > A={1,2,3,...,N_L} exists. > > If we do not take into accoun that there are physical constraints, yes. What physical constraints limit non-physical existence? > > > > 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. > > Of course it does exist. For every N_L, there is an N_L + 1. > It is N_B unless you find a larger one. For every N_B there is an N_B + 1.
From: Virgil on 23 Dec 2006 13:37
In article <1166874463.316054.202290(a)73g2000cwn.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > 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. > > Not directly but indirectly. But the printing allowance originally > issued shows that such an allowance was required. > > > > But the hardliners were not mathematical hardliners or astronomical > > hardliners. They were religious hardliners. > > The same as today. WM and HdB, and their ilk, certainly preach a "gospel", but we who do not insist on, or rely on, such absolute truths, nor accept the preachers' unfounded assumption of infallability, reject it. |