Cantor Confusion
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From: mueckenh on 24 Dec 2006 07:25 Franziska Neugebauer schrieb: > >> The number of sums (left to the "=") and the number of n's (right to > >> the "=") is not bounded. What precisely is your problem? > > > > The infinitely many sumands in > > > > 1 > > 1+1 > > 1+1+1 > > ... > > 1. Every line contains finitely many summands. > 2. Every n e N is finite. Do you agree? Of course. Every set of lines is finite. > > > yield an infinitude of sums but not an infinitude of summands. > > 3. Every single sum contains only finitely many summands. Of course. And finitely many different sums are less than infinitely many different sums. But I think, here we have a problem which is analogous to the problem of the tree: Here we have only finitely many summands, nevertheless they realise infinitely many sums. > That is your problem? Not really. I know that it is not a problem but only the relegious confession of some matheologists. =================================== > I do not claim that there are two lines each of which mutually lacks a > number the other one does not lack. Hence I do not have to show what > you have asked for. But that would be the only possibility to need more than one line for all natural numbers. ================================ > Linguistic Question: Is it meaningfull to speak of an "approximation" if > one denies the existence of the thing which is approximated (the > irrational number)? Not as long as not all deny its existence. Regards, WM
From: Franziska Neugebauer on 24 Dec 2006 08:09 mueckenh(a)rz.fh-augsburg.de wrote: > Franziska Neugebauer schrieb: > [...] the problem of the tree [...] Today there is only one "finite" tree I care for. A green one. >> That is your problem? > > Not really. I know that it is not a problem but only the relegious > confession of some matheologists. I thought you want to discuss _your_ problems. >> I do not claim that there are two lines each of which mutually lacks >> a number the other one does not lack. Hence I do not have to show >> what you have asked for. > > But that would be the only possibility to need more than one line for > all natural numbers. Why don't you stay on topic? Where is your proof of (*)? >> Linguistic Question: Is it meaningfull to speak of an "approximation" >> if one denies the existence of the thing which is approximated (the >> irrational number)? > > Not as long as not all deny its existence. To me this reads As long as all deny the existence of the entity to be approximated it is not meaningful to speak of an "approximation". Is this correct? I do not deny its existence. Do you? What meaning does have an "approximation" to you if the approximated entity (irrational number) is supposed to not exist? F. N. -- xyz
From: Newberry on 24 Dec 2006 10:37 mueckenh(a)rz.fh-augsburg.de wrote: > Newberry schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Newberry schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > > > > 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? > > > > > > Yes. But that is another topic. (In short: There are less than 10^100 > > > bits in the universe. Therefore, no irrational number can be > > > approximated better than to about 10^-100. The Cauchy criterion for > > > converging series fails.) > > > > What cannot be approximated? > > You know the Cauchy criterion. For every epsilon > 0 you must be able > to find an n_0 such that for n > n_0 we have |a_n - pi| < epsilon. > > If you have less than 10^100 bits available, then you cannot realise an > a_n which reproduces more than the first 10^100 bits of a number like > pi. (Unless pi was a number showing some pattern, like the number > 0.111.., which would make it possible to express every a_n using less > than 10^100 bits.) > > You cannot find such a number a_n without having pi already. But this > number shall be used to establish the existence of pi. This is > impossible. pi does not exist *as a number* (it exists as an idea So it does exist after all. BTW, an algorithm exists for calculating ALL its digits. and > in form of close approximations). > > Regards, WM.
From: Newberry on 24 Dec 2006 10:53 Virgil wrote: > 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. Why does it not matter? The cardinality of the inexes in the limit is aleph0, and the cardinality of the nodes in any infinite path is aleph0. It means that in calulating the limit lim{n-->oo} (2*2^n - 2)/2^n = 2 we transversed all the infinite paths. > > 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: David R Tribble on 24 Dec 2006 12:16
muecken h wrote: >> Not yet a number, unless you can specify it such that one (at least >> you) can decide whether q < n or q = n or q > n for any natural number >> n given to you. Only saying that it is not equal to any natural number >> given is not enough. > David R Tribble schrieb: >> I did not say that q is not equal to any given natural number. >> I said that q is not equal to any natural number that has been >> "brought into existence". Trichotomy is well defined for q. >> For any natural n given, if n has been "brought into existence", >> then q > n. >> >> Since you believe that only those numbers that have been "brought >> into existence" exist, you should readily grasp the definition of q. > mueckenh wrote: > Yes. You mean a number q which exists but is larger than any existing > number. > [...] > It is not my assumption that a non existing number does exist. Could > well be an axiom of set theory. Well, now I'm confused. Could you provide an example of a natural number that does not exist? |