Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: Virgil on 23 Dec 2006 13:45 In article <1166895046.650593.195620(a)a3g2000cwd.googlegroups.com>, "Newberry" <newberry(a)ureach.com> wrote: > 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 It does not matter. There is a theorem in analysis that says Lim_{n->oo}(f(n)/g(n)) = (Lim_{n->oo}f(n))/ (Lim_{n->oo}g(n)) PROVIDED all 3 limits converge to finite values There is no theorem in analysis, or anywhere else, that says what WM was trying to imply by his limit statement.
From: stephen on 23 Dec 2006 13:54 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: >> >> >> >> >> >> > 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? Yes. Just as when we approch infinity there are twice as many nodes as there are leaf nodes. But an infinite tree has an infinite number of nodes, and zero leaf nodes. > The cardinality of the indexes n is aleph0, correct? Yes, but irrelevant. You cannot just claim that something that is true "in the limit" is true for the infinite case. Do you agree that an infinite tree has 0 leaf nodes? Do you agree that for large finite trees the total number of nodes is approximately twice the number of leaf nodes? Do you agree that an infinite tree has an infinite number of nodes? Do you think that 2*0 = oo? Stephen
From: Lester Zick on 23 Dec 2006 14:01 On Sat, 23 Dec 2006 02:28:35 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >In article <1pkoo2dd7u1tcqeoidr3u9jdl0iio3gr31(a)4ax.com> Lester Zick <dontbother(a)nowhere.net> writes: > > On Thu, 21 Dec 2006 15:22:56 -0700, Virgil <virgil(a)comcast.net> wrote: > > >In article <1166737132.026953.46380(a)48g2000cwx.googlegroups.com>, > > > mueckenh(a)rz.fh-augsburg.de wrote: > > >> Dik T. Winter schrieb: > > >> > > >> > Indeed. No axiom is necessarily true. > > > > Fortunately nothing you say is necessarily true. > >Of course not. Neither what you state. Well there's a subtle distinction here, Dik. You don't speak for me. You can only speak for yourself. I don't claim what I say isn't true whereas you and others do claim what you say isn't true. So I can and do claim what I say is true whereas you've already admitted what you say isn't true. So unless you can prove what I say isn't true I suggest you confine yourself to subjects like modern mathematics and set "theory" where truth is irrelevant. > > >> How then should the truth of AC be proved? > > > > > >One does not prove axioms, one either assumes them or does not assume > > >them. > > > > Unless unlike neomathematikers one is not too lazy or stupid to prove > > them true. > >Did you prove the first postulate of Euclid? Or are you too lazy or >stupid to do so? I'm too lazy or stupid to do it for mathematikers who are too lazy or stupid to do it for themselves because I don't claim to be the kind of mathematician who appears to be too lazy or stupid to do mathematics while nonetheless pontificating about mathematics in terms he doesn't claim to be true. ~v~~
From: Lester Zick on 23 Dec 2006 14:04 On Sat, 23 Dec 2006 12:41:49 -0500, David Marcus <DavidMarcus(a)alumdotmit.edu> wrote: >cbrown(a)cbrownsystems.com wrote: [. . .] >Ah, but you are forgetting the crank's axiom: "Everything true for >finite numbers is also true for infinity." So concern for "truth" would be the hallmark of crackpots but not modern mathematikers, David? ~v~~
From: Newberry on 23 Dec 2006 14:16
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: > >> >> >> > >> >> >> > 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? > > Yes. Just as when we approch infinity there are twice as many nodes > as there are leaf nodes. The nodes are countable and there are twice as many of them than paths .... But an infinite tree has an infinite number > of nodes, and zero leaf nodes. > > > The cardinality of the indexes n is aleph0, correct? > > Yes, but irrelevant. You cannot just claim that something that is true > "in the limit" is true for the infinite case. Hmm ... Why not? What is the cardinality of the nodes in an infinite path? > > Do you agree that an infinite tree has 0 leaf nodes? Yes > Do you agree that for large finite trees the total number of nodes > is approximately twice the number of leaf nodes? Yes > Do you agree that an infinite tree has an infinite number of nodes? Yes > Do you think that 2*0 = oo? No What is supposed to follow from all this? > > Stephen I think it all boils down to this: something that is true "in the limit" is NOT true for the infinite case. |