Cantor Confusion
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From: G. Frege on 4 Feb 2007 11:10 On Sun, 4 Feb 2007 02:36:47 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: > > [...] I am now halfway chapter 8 and only found one serious error (in > chapter 7). And one place where I have serious doubts (in chapter 8, but > I have to look thoroughly at that). But until this point it is an > excellent review about the history about the thinking about the infinite. > "Man soll den Tag nicht vor dem Abend loben." F. -- E-mail: info<at>simple-line<dot>de
From: mueckenh on 4 Feb 2007 11:14 On 3 Feb., 03:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170413330.341310.269...(a)p10g2000cwp.googlegroups.com> mueck....(a)rz.fh-augsburg.de writes: > > > On 2 Feb., 02:19, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > > > In article <1170348341.624257.130...(a)p10g2000cwp.googlegroups.com> mueck.= > ... > > > If 4 is the set of all existing (whatever that does mean) sets with 4 > > > elements and 5 is the set of all existing sets with 5 elements, we find > > > immediately that 4 is *not* a subset of 5. With this definition of > > > numbers as sets subsetting does not give what you wish. > > > > 4 is the set of all sets with 4 elements. 4 is also every set with 4 > > elements. > > That can not be true. 4 can not be all kinds of different things at once, > at least not in mathematics. What is 4 in mathematics? > With this kind of definition 4 contains 4. Of course. Cardinal number = ordinal number. 4 contains 4 elements (even in set theory). But 4 contains IIII --- not 0,1,2,3. Henry VIII had 7 predecessors --- only including him they were 8 Henries. > > > > Therefore we have to investigate whether such a representation exists. > > > > Note that I use the definition of a number by Peano in order to look > > > > for the existence of that number. > > > > > > What does the latter sentence *mean*? But indeed, for most numbers some > > > representations do not exist, while other representations do exist. The > > > existence of a number is independent of the existence of a representation. > > > > This point of view has lead to the present mess-math. > > > > > For instance, for most rational numbers a decimal representation does not > > > exist. > > > > Correct, for instance for 1/7. > > And for computable numbers some representation does exist. But this representation does not necessarily enable us to determine the trichotomy relation with numbers which are really numbers. > > > > > The definition of a number is given as non-circular as possible: p. 3: > > > > 1) 1 ist eine nat=FCrliche Zahl. > > > > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. > > > > > > in N is wrong here; N is not defined. > > > > N is already in (1), because (1) is identical to "1 is in N". > > N is not *defined*. Whether you use it in (1) is irrelevant, what is > relevant is that it is not defined. > > > > A more proper version is: > > > 2) Jede natürliche Zahl a had einen bestimmten Nachfolger a', und > > > dass ist auch ein natürliche Zahl. > > > > What is every natural number if N is not defined? > > This is a recursive definition of natural numbers. By (1) we have one > natural number, by (2), from that single natural number we get a lot of > other natural numbers. Why do you say N is wrong in (2) but not in (1)? > > > This is getting phylosofical. In mathematics a definition is circular if > in the definition one of the deciding features is the term you want to > define. So a definition as you gave: > 3 is the set of all sets of 3 elements > is a perfect example of a circular definition. !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IT IS NOT A DEFINITION !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! > > I think that everybody able to read and understand these lines will > > know what "+ 1" means while the successor is not immediately clear. > > (The successor of n coul be n+2 or 2*n or 10*n or ....) > > Yes, indeed, and that is the crux. Abstraction. When you use the > '+ 1' notation you are already assuming the existence of addition. > When you do not, you can properly define addition (and all other > operations) using the Peano axioms. You are *defining* the natural > numbers, presumably without any knowledge about what natural numbers > even are. And I may note that with '2*n' you can get a set that is > isomorphic (with respect to all operations) to the natural numbers, > only the naming is different. Consider the set of powers of 2, > (your 2*n case) define: > a '+' b = 2^[ log_2(a) + log_2(b) + 1 ] > a '*' b = 2^[ log_2(a) * log_2(b) + log_2(a) + log_2(b) ] > Call this set K. You may verify that the rings: > R(N, +, *) and R(K, '+', '*') > are isomorphic. I do know that. But I don't want some isomorphic sets. I want to define *the natural numbers* which are I II III .... > > > > And *in this context* the definition of 3 is: > > > succ(succ(1)) > > > no set at all. And the existence of 3 follows from the first set of > > > axioms you gave. > > > > The existence does not follow from axioms. Axioms state something in > > an arbitrary way. They can even state the famous pink elephant. > > The non-existence of a pink elephant in the realm of the axiom that > states that there is a pink elephant is not mathematics but philosophy. > Axioms state what things exist (or do not exist) in their realm. No. This belief is the core of mathmess > > Whether it exists remains to be investigated. > > Your existence is not a mathematical existence. This form of existence is the only possible existence. > > > > Your definition is circular, this one is not. > > > > I do not give a definition but I look for the existence of the number > > already defined. > > So when I ask you for a definition you do not give a definition? Yes, > I have been thinking that all along. I gave two definitions: Peano and that with "+1" which is very close to Dedekind's attitude. (I don't know whether he actually created it, but I know that he would have liked it with "+1" as a primitive). But the axioms do not establish any existence, in particular not when you apply Dedekind's definition of what a number is. Regards, WM
From: mueckenh on 4 Feb 2007 11:24 On 3 Feb., 04:12, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > In article <1170413742.648825.136...(a)l53g2000cwa.googlegroups.com> mueck...(a)rz.fh-augsburg.de writes: > > > On 2 Feb., 02:42, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: > ... > > > > Frankly, I do not see much difference to: The union of finite paths > > > > contains an infinite path. > > > > > > But that is true. > > > > So you say: > > The union of finite trees contains an infinite tree. > > The union of finite chains contains an infinite chain. > > The union of finite paths contains an infinite path. > > Indeed. > How can a union of finite elements contain an infinite element? > > > > > What I state again, again and again, but what you > > > misread each and every time is: "the union of sets of finite paths > > > does not contain an infinite path". > > > > So you say the union of finite paths > > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > contains the infinite path p(oo) = {0,0,0,...}. > > > > But the union of finite sets of finite paths > > {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... > > does not contain the union of finite paths > > p(0) U p(1) U p(2) U ... = {0} U {0,0} U {0,0,0} U ... > > Again, indeed. But that is incorrect! The union of finite sets of finite paths {p(0)} U {p(1),q(1)} U {p(2), q(2), r(2),s(2)} U ... is the set of all finite paths {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } which obviously contains the union of finite paths of the special form p(0), p(1), p(2), .... as a subset. If you say, as you do above, that the union of finite paths contains an infinite path, p(0) U p(1) U p(2) U ... contains the infinite path p(oo) then this is also true for this subset. It is impossible that you loose p(oo) because the set of all finite paths {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contains some more paths than he union of special finite paths p(0), p(1), p(2), .... > > I would say: The union of all finite sets of finite path contains all > > finite paths. > > Right. As I have stated all the time. And so the set of paths in the > complete tree is *not* a subset of the union of all finite sets of > finite paths, something you have stated repeatedly. The union of all finite paths of a special kind contains the due infinite path. The union of all finite paths contains all infinite paths. > > > The union of these finite paths can be consructed (it is > > a set of nodes) and is realized in the binary tree by its nodes. It > > contains all infinite paths of the infinite tree. > > > > The tree contains the union of all finite sets of finite path as well > > as the union of these paths. > > Right. But you *use* that the union of all finite sets of finite paths > *contains* the union of these paths. If the union p(0), p(1), p(2), ....contains p(oo) then the union p(0), p(1), q(1), p(2), q(2), r(2),s(2), ... cannot miss it. > > > > And it is not contained in any of the finite trees. Why is it said to > > > > be contained in the union of all finite trees? > > > > > > Because an infinite path is a union of paths. And when you construct > > > unions of *sets* of paths you are not constructing unions of paths. > > > > The union of a set of sets of elements is a set of elements, not a set > > of sets of elements. > > What definition of union do you use? Ah, I am seeing now. The axiom > of sumsets. But I am *not* talking about a single set, I am talking > about *multiple* sets. To rephrase your wording so that they are > applicable: > > The union of sets of sets of elements is not a set of elements, but a set > > of sets of elements. > To know why the axiom of sumsets is also the axiom of union, you have to > consider the following: > given two sets A and B > by the axiom of pairing there exists a set {A, B} > by the axiom of union there exists a set U{A, B} consisting of the > elements of A and B. That set is normally called the union of A and B. > I allow that the terminology is a bit confusing. But U applied to a > single set of sets yields the set of the elements. But U applied to > multiple sets does not. And you are applying it to multiple sets, so > it does not. Then apply it twice. First you get the single set of all paths, second you get the set of elements. The tree guarantees the existence of these unions. > > If there is an infinite pathlength, then there must be an infinite > > number as well as an infinite segment of natural numbers. > > Why must there be an infinite number? Because a union of finite pathlengths cannot lead to an infinite pathength. The infinite union of pathlengths 1,1,1,... 2,2,2,... 3,3,3 4,4 5 yields a finite pathlength, namely 5. > But rest assured, there may be an > infinite segment (depending on how you define segment). It is necessary. As an infinite natural number would be necessary to obtain an actually infinite set of numbers. > > You say, the former does not exist in N while the latter does exist in > > the form of the whole set N. > > > > That is a contradiction. You see that there is no infinite pathlength > > possible without an infinite pathlength. > > This is gibberish. I only state that infinite paths have infinite pathlength, > and that there is no natural number that maps to. What is the contradiction? There cannot be an infinite pathlength. There cannot be an infinite segment {1,2,3,...n} with a last element unless the last element is infinite. > > > It all boils down to the same story. Let A be the set of natural numbers > plus omege. A is infinite (I think you agree). Remove omega from A. > What do you have now? I state that you now have N. Apparently you > disagree. Why? I also state that because A is infinite, N is also > infinite. But for some reasons you state that N is finite. For what > reasons? I state that N is potentially infinite. There is no number countaing all natural numbers. But if there was a number counting all natural numbers, then it could not be infinite unless there waqs an infinite natural number because the natural numbers count themselves. As long as we are always counting another finite number, the counted set of nunmbers is finite. > > > > > But you said that there is an infinite pathlength corresponding to an > > > > infinite set of paths. > > > > > > But that set of paths is *not* the union of the P(i). > > > > Above you said that the set of all paths p(n) contains the infinite > > path p(oo). > > Yes. But not that the union of the P(i) contains an infinite path. The union of the P(i) is the set of all paths. It includes the set of all paths of the form p(n). Why should it not contain p(oo) if the set of all paths of the form p(n) alone contains p(oo)? > > > > The union of sets of finite trees would neither contain, nor establish, > > > an infinite tree. It would not contain one, because none of the > > > constituent sets contains an infinite tree. It would not establish > > > one, because the union is a set of trees, not a tree. > > > > The union of a set of sets of elements is a set of elements, not a set > > of sets of elements. > > But I am not talking about the union of a set. I am talking about the > union of set*s*. Lets analyse. Given two sets A and B: > by pairing {A, B} exists > by union U{A, B} exists. > The definition of A U B is U{A, B}. And that is a single set containing every element which is in A or in B or in both. The union of some sets of paths is one set of paths. The union of all paths is one set of elements. The tree unions all sets of paths. > > > The union of a set of trees, as I defined it, is a tree. > > I was talking about the union of set*s* of trees. How many trees must be in the sets? The union of two trees is a tree. "Two trees" is a set of two trees. The union of tree paths and 12 paths and 17 paths is a set of nodes, perhaps this set includes even one or more paths. The union of {three paths} and {12 paths} and {17 paths} is a set of paths. These paths are unioned in the tree to give a set of nodes, perhaps including even one or more paths. > > > > > > > > Wrong. Infinite paths do not have a path-length that is a > > > > > > > natural number. > > > > > > > > > > > > Pathlengths ARE natural numbers. > > > > > > > > > > For finite paths. > > > > > > > > In all cases, by definition. > > > > > > *What* definition? > > > > Definition: Map the pathlength x on the number x. > > Fine, I do the mapping. How do I get that x is a *natural* number? Because there are no other numbers of elements (nodes). What you call omega is nothing but the possibility to extend the path of n nodes by another node to n+1 nodes. Regards, WM
From: mueckenh on 4 Feb 2007 11:31 On 2 Feb., 21:01, Virgil <vir...(a)comcast.net> wrote: > > Why should it? But it can be able. For instance it can prove that > > every set of natural numbers is finite while the size of sets of > > natural numbers is unbounded from above. > > Not so. Induction can only prove every set of naturals which is bounded > above by a natural is finite, but that does not, according to the axiom > of infinity, exhaust the possible sets of naturals. There is no natural number in this set which is not covered by induction. So, which number is missing to exhaust N? > What Induction says is precisely: > Given a set, S, of natural numbers such that > (a) S contains the first (smallest) natural, and > (b) whenever a particular natural is a member of S, > the successor of that natural is also a member of S, > then every natural is a member of S. > > And induction says nothing else. Induction says: If P holds for n then it holds for n+1. This is the main property of an infinite set as the set described by the axiom of infinity originally was. Only the dubious interpretation of esoteric infinity removes this set from the realm of induction and from the realm of mathematics too. > > Of course every set of natural numbers is finite, as proved > > by induction over all initial segments. > > That only proves that such initial segments as have a last element are > finite. It says nothing at all about any initial segment which does not > have a last element, and there is one. Every finite natural defines a finite segment. An infinite segment can be defined by an infinite number only, and there is none. ================================== > The set of all sets with exactly 4 elements is not a subset of the set > of all sets with exactly 5 elements since the set of all sets with > exactly 5 elements does not contain any sets with less that five > elemeNts. 4 is also every set with 4 elements. Every set with 4 elements is a subset of that set where one further element has been added. > WM must have completely lost touch with reality to think otherwise. Contrary. Just because I have not lost touch with reality. > The "mess" is only in the minds of those who imagine it into existence. > For those of us who perceive no mess there is no mess. That is an excellent description of your position. > In Peano, you do not have to know what any of the things are, all you > have to know is that they work according to the rules stated. You do not have to know (or to be able to get to know at least) whether or not one natural number is larger than the other? > If you don't know at all what you're looking for, you can't ever tell > whether you've found it or not. Therefore I accept and reported in my book the definitions by Peano and Dedekind. ========================================= > but { p(n) : n in N } is a set of paths and not a path. And the union of the paths which belong to this set is a set of nodes which is realized in the tree. Regards, WM
From: Carsten Schultz on 4 Feb 2007 11:33
mueckenh(a)rz.fh-augsburg.de schrieb: > On 4 Feb., 03:36, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: >> In article <1170470698.824513.309...(a)m58g2000cwm.googlegroups.com> cbr...(a)cbrownsystems.com writes: >> >> > On Feb 2, 5:12 pm, Carsten Schultz <cars...(a)codimi.de> wrote: >> ... >> > > From that it follows that there are exactly four sets with four >> > > elements, since these are the elements of 4. > > Is this intended to be a serious statement? Of course I am serious in pointing out that this is a consequence of what you have written, and had you not snipped your statement from which this is derived, evryone could see this. >> > > It also follows that there >> > > is only one set with four elements, namely four. > > With respect to the property "set having elements" one can in fact > take this position. There is only one set with four elements. So for example {1,2,3,4} = {7,42,666,1984}? > Therefore "a set of four elements" is the same as "every set of all > sets of four elements". Good to know. > >>> > You should >> > > write a book about this. > > That book is available, Who would have guessed this! And in it you prove that 1=4? > for instance from Amazon or from the German > National Library or directly from the publisher > http://www.shaker.de/Online-Gesamtkatalog/details.asp?ID=1471993&CC=21646&ISBN=3-8322-5587-7 I also continuously get paper spam by this publisher. Now I know what kind of customers they have. -- Carsten Schultz (2:38, 33:47) http://carsten.codimi.de/ PGP/GPG key on the pgp.net key servers, fingerprint on my home page. |