Cantor Confusion
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From: Virgil on 23 Dec 2006 01:18 In article <1166854110.016734.127930(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > stephen(a)nomail.com wrote: > > Newberry <newberry(a)ureach.com> wrote: > > >> 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. > > Does it converge to two edges per path as we approach infinity? In every finite tree the ratio of terminal nodes to paths is exactly 1. if one could extrapolate that to infinite trees, every infinite path would have a terminal node. Such attempts to carry over from finite to infinite usually mislead.
From: Virgil on 23 Dec 2006 01:20 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 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.
From: mueckenh on 23 Dec 2006 06:25 stephen(a)nomail.com schrieb: > Newberry <newberry(a)ureach.com> wrote: > > > stephen(a)nomail.com wrote: > >> Newberry <newberry(a)ureach.com> wrote: > >> > >> > stephen(a)nomail.com wrote: > >> >> Newberry <newberry(a)ureach.com> wrote: > >> >> > >> >> > >> >> > What is wrong with WM's mapping? You divide each edge into two halves > >> >> > and pass one half to each branch. Then you divide the passed half again > >> >> > and again. > >> >> > >> >> So which edge is mapped to the path that always goes left? > >> >> Remember, you need to uniquely map each path to an edge. > >> > >> > 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. > > 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. But we can construct a surjection into the set of all existing nunmbers, i.e., we can show that no path can separate itself from another path wihout an additonal edge. By this fact we see that even an infinite separated path, if it exists, cannot exist without another edge. The trick, however, is not bad: We state that there are irratioal numbers each of which can be distinguished from all other numbers. But we have no means to make the distincton in form of an edge. And, circumventing straight logic, we state that "in the infinite", there are more paths than edges, i.e., they exist without having necessary separating edges of their own. > > >> 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. Regards, WM
From: mueckenh on 23 Dec 2006 06:29 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. Regards, WM
From: mueckenh on 23 Dec 2006 06:39
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. > 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. This proves that there cannot be an infinite cardinal number of tis set. This proves that there cannot be an infinite cardinal numbers of N. > > > > > > > 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. False. Try to find a nunmber larger than he largest existig number. You will fail. > > 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. > > 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. It is N_B unless you find a larger one. In any case the set includng N_B is a finite set with cardinal number N_B. Regards, WM |