Cantor Confusion
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From: Virgil on 24 Dec 2006 14:05 In article <1166963142.411543.257670(a)a3g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > 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. Wm's objections are the religious (purely faith based) claims of a fanatical anti-mathematician. > > =================================== > > > 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. One needs more than one line for two natural numbers, if the lines are not to be identical in content. So that WM claims to need fewer for more? > > ================================ > > > 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. So when WM speaks of approximations to something which he claims does not exist, he admits he speaks meaninglessly? > > Regards, WM
From: Dik T. Winter on 24 Dec 2006 19:45 In article <1166962708.594095.140560(a)f1g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > In article <1166907306.745536.315650(a)h40g2000cwb.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: .... > > > It was Dik who insisted that a well ordering of the reals can be done > > > (not only be proved), if AC is true. Here is the discssion: > > > > Indeed, it can be done *when AC is true*. But perhaps we have a different > > view on the meaning of the sentence "can be done"? > > It has been proved that there is *no definable well-ordering* of the > reals. So it cannot be *done*. The only thing that can be done is to > prove that a well-ordering exists. Zermelo did it. Oh well, I was slightly incorrect. You need also V=L. "...; it is also possible to show that the ZFC axioms are not sufficient to prove the existence of a definiable (by a formula) well-order of the reals. However it is consistent with ZFC that a definable well- ordering of the reals exists -- for example, it is consistent with ZFC that V=L, and it follows from ZFC+V=L that a particular formula well-orders the reals, or indeed any set." So it has *not* been proven that there is *no definable well-ordering* in ZFC. What has been proven was that ZFC is insufficient to *prove* such. > Now we know that this well-odering never can be found. In order to keep > contradictions far away from ZFC, we accept that situation. A definiable well-ordering is *not* inconsistent with ZFC. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: cbrown on 24 Dec 2006 20:42 mueckenh(a)rz.fh-augsburg.de wrote: > cbrown(a)cbrownsystems.com schrieb: > > > What "authorities" do you claim I am quoting when I state "the function > > f(x) = 2*x is a bijection between the set of all naturals, and the set > > of all even naturals"? > > > > I am simply applying specific definitions of the terms "function", > > "bijection", "the set of all naturals" and "the set of all even > > naturals", using the usual logic. > > This bijection between sets (initial segments {1,2,3,...n} and > {2,4,6,...2n}) is only valid for finite sets. Suppose I instead counter-claim that this bijection is obviously "not valid" for finite sets; only for infinite sets. Is that what you call a "correct mathematical argument"? Simply claiming something is valid or not valid? How is it different from your own argument? > > > > > 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). > > > > > > Then you really do accept 1 + 1/2 + 1/4 + ... < 1 ? > > > > Of course not. How do you infer that from my statements? > > From your statement that the same calculation, in case of edges, yields > less than edge per path. > What? I claim that there is /no/ calculation "|N|/|R|" yielding a real number; because division is not defined unless both |N| and |R| are real numbers. Thus "|N|/|R| < 1" is nonsense; as I previously stated quite clearly. > The number of edges accumulated by one path up to level n is a function > like that above: > f(n) = 2 - 1/2^n. Very easy. Yes; when you write it that way, it "looks" just like a valid statement about the real numbers. So, you "feel" that it is a both a sensible and a valid statement about things which are not real numbers; when instead it is nonsense. You might as well state that |N| > sqrt(|R|) with equal (lack of) sense. Cheers - Chas
From: Newberry on 24 Dec 2006 21:47 Virgil wrote: > In article <1166952651.859923.213150(a)i12g2000cwa.googlegroups.com>, > mueckenh(a)rz.fh-augsburg.de wrote: > > > stephen(a)nomail.com schrieb: > > > > > Newberry <newberry(a)ureach.com> wrote: > > > > > > > > > What is supposed to follow from all this? > > > > > > That what is true "in the limit" is not necessarily true for the infinite > > > case. > > > > For instance the diagonal number of Cantor's proof differs from any > > list entry of a finite list. But it certainly does not differ from any > > list entry of an infinite list. (Compare the case 1.000...1 = 0.999...) > > The Cantor construction rule produces for any countably infinite list > a number which necessarily differs from every number in the list. Well, but two different strings can represent the same number e.g. 0.9999999999999999999999999... 1.0000000000000000000000000... Not sure what the implicationsa are for Cantor's argument.
From: Dik T. Winter on 24 Dec 2006 21:49
In article <1166962337.912882.88060(a)42g2000cwt.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: .... > > > No, I don't. What I am showing here is that each path will accumulate > > > the weight of two edges as it approaches infinity. > > No, you do not. In the finite case you do, approximately. In the > > infinite case you do not because *all* parts you pass to a path are > > reduced to size 0. > > All parts? Can you state *any* part of an edge that does not have size 0 that is passed on to an infinite path? > The diagonal digits a_nn of Cantor's argument have to be > multiplierd by 10^-n in order to yield the diagonal number SUM a_nn * > 10^-n. But there is no problem with yielding zero for n --> oo? > --- Remarkable. Apparently you do not understand the working of limits. It has been proven that *every* Cauchy sequence yields a real number. And it is easy to see that *every* decimal expansion is in fact a Cauchy sequence. 10^-n is *never* zero. You need limits. > > Consider what we pass on to the paths that go > > left only. Look at the following: > > level 1: 1 edge at level 1 > > level 2: 1/2 edge at level 1 + 1 edge at level 2 > > level 3: 1/4 edge at level 1+ 1/2 edge at level 2 + 1 edge at level 3 > > etc. for each finite n, but what about the infinite case? > > level 'oo': 1/oo edge at level 1 + 1/oo edge at level 2 + ... > > Where do you see the 2 coming up? It is certainly *not* the series > > 1 + 1/2 + 1/4 + ..., because there is *no* edge that is passed in > > full, there is also *no* edge of which 1/2 is passed, etc. > > There is no edge in the *finite* domain which is "passed in full". In > the infinite series there is an edge passed in full. Which edge is passed in full to the path that goes at each step to the left? > Otherwise the path > does not exist at all. That is what you are trying to prove. But you can not just state it. You hav to *prove* it. > The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no > smallest term 2^-n. Indeed. > It can be applied to calculate the edges per path. You should know that > absolutely converging series can be reordered. You should know that you can *not* reverse absolutely converging series. And that is what you are doing. Because that would result in a series without start. > =============================== > > I now understand what you were trying to do. > > It lasted rather a while. Consider it a complaint about your clarity and lack of definitions. > > But in the complete tree > >the parts of edges that are assigned to an infinite path are not 1, > >1/2, 1/4 etc. There is *no* edge that is completely assigned to a > > path, > > Then the tree is incomplete. You fail to understand infiniteness > already on this low level? Circular reasoning abounds here. Care to give a *proof* that the tree is not complete in that case? This statement is completely similar to the statement that the completed set of natural numbers should have a last element. For which you also fail to give a proper proof. Please state when you intend to no longer reason within ZF (especially the axiom of infinity). As long as you are reasoningin in ZF, there is *no* largest element of the (completed) set of natural numbers. > > Note that in each finite tree, the complete edge assigned to a path is > > the last edge on the path. As there is no last edge in an infinite > > path, this does not hold in the infinite tree. > > So there are different paths without different edges? No. I never did state that. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/ |