Cantor Confusion
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From: mueckenh on 23 Dec 2006 06:47 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 à 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. > 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. Correct. That's why I said that print allowance was required. Regards, WM
From: stephen on 23 Dec 2006 10:44 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: >> >> >> >> > 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. >> >> >> 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? Yes. Just as the number of leaf nodes per node converges to 1/2 as we approach infinity. Yet there are 0 leaf nodes in an infinite tree. Stephen
From: Newberry on 23 Dec 2006 12:30 Virgil wrote: > 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. What about the ratio of all the edges to all paths? Does it converge to 2? lim{n-->oo} (2*2^n - 2)/2^n = 2 So if your logic held every > endless path would have a terminal node.
From: Newberry on 23 Dec 2006 12:32 mueckenh(a)rz.fh-augsburg.de wrote: > 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. Are you saying that x^2 = 2 does not have a solution? > 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: Newberry on 23 Dec 2006 12:41
stephen(a)nomail.com wrote: > 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: > >> >> > >> >> > 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. > >> > >> >> 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? > > Yes. Just as the number of leaf nodes per node converges to 1/2 > as we approach infinity. So when we approach infinity there are twice as many edges as paths as this limit shows, lim{n-->oo} (2*2^n - 2)/2^n = 2 correct? The cardinality of the indexes n is aleph0, correct? Yet there are 0 leaf nodes in an infinite tree. > > Stephen |