Cantor Confusion
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From: Dik T. Winter on 4 Feb 2007 22:33 In article <JCx39B.C4K(a)cwi.nl> "Dik T. Winter" <Dik.Winter(a)cwi.nl> writes: > In article <1170470698.824513.309540(a)m58g2000cwm.googlegroups.com> cbrown(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. It also follows that there > > > is only one set with four elements, namely four. So 4=1. You should > > > write a book about this. > > > > Dik is even reviewing it, I think. > > I am. 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. Having read chapter 8 now completely, there are two errors there. But it remains an excellent review about the history of thinking. I just started chapter 9, and, well, I will not say anymore now. I need a few days for the remaining chapters. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 4 Feb 2007 22:44 In article <80kcs2p92giuv70hehcbifcoild1d1uac8(a)4ax.com> G. Frege <nomail(a)invalid> writes: > On Sun, 04 Feb 2007 21:22:31 GMT, Michael Press <rubrum(a)pacbell.net> > wrote: > >> > >> "Man soll den Tag nicht vor dem Abend loben." > >> > > I looked this up and found the meaning. > > How does mid-sentence capitalization work? > > > The substantives in German are usually written this way. > > > Is this verse? > > > A saying. And indeed, it is also a saying in Dutch, when translated. And this was quite appropriate as a response to my earlier posting (which it was). But we do not have such capitals mid-sentence. In Dutch: "Men moet de dag niet prijzen voor het avond is." -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 4 Feb 2007 23:10 In article <1170605669.990641.151350(a)h3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > On 3 Feb., 03:07, "Dik T. Winter" <Dik.Win...(a)cwi.nl> wrote: .... > > > 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? succ(succ(succ(succ(0))), according to the Peano axioms. > > With this kind of definition 4 contains 4. > > Of course. Cardinal number = ordinal number. 4 contains 4 elements > (even in set theory). Depends on the model. > But 4 contains IIII --- not 0,1,2,3. What *is* IIII. You never have defined it. You really do not like definitions, as they pin down the real meaning. > Henry VIII had 7 predecessors --- only including him they were 8 > Henries. You are shifting position again? When I asked you about what basic way, III c IV c V, you answered that I had to continue with IIII, IIIII, etc. > > > 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. Perhaps. How do you establish trichotomy between 1/13 and 1/64? Are you really going to base-26 to establish that? That would be pathetic. > > > > > 1) 1 ist eine nat�rliche Zahl. > > > > > 2) Jede Zahl a in N hat einen bestimmten Nachfolger a' in N. .... > > 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)? Where in (1) is N? I do not see N at all. > > 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 !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! You stated that when I asked you for a definition. So what is happening here? > > > 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 > .=2E. And elsewhere you are using 1, 2, 3, 4, VIII, etc. Pray remain consistent. > > 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 Ah, so the parallel axiom, and all Euclidean axioms are 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. As you do not define your form of existence, it is impossible to talk about it, at least mathematically. > > > > 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. Existence is a mathematical thing when you can establish it by axioms or through theorems based on axioms. Anything else is merely phylosophy. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Dik T. Winter on 5 Feb 2007 00:04 In article <1170606249.100767.104120(a)j27g2000cwj.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > 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: > > > 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? Where in the above is there a union of finite elements? Unions are about *sets*, not about elements. > > > 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) ... } Indeed. > which obviously contains the union of finite paths of the special form > p(0), p(1), p(2), .... > as a subset. Wrong. There is no obviously here. The union of sets of elements does *not* contain the union of elements. This is abundantly clear from the first eight chapters of your book. > 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. Let's get back to the finite. A = {{a, b}, {a, c}, {a, d}} and B = {{a, d}, {b, d}, {c, d}}. A U B consists of: {{a, b}, {a, c}, {a, d}, {b, d}, {c, d}}. It does *not* contain unions of the sets that are elements of A or B. > 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), .... In what way does {p(0), p(1), q(1), p(2), q(2), r(2),s(2) ... } contain: p(0) U p(1) U p(2) U ...? > > > 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. *Sets* of paths. Remember? > > 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. What *sets* of paths are you using? > > 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. Which is a set of nodes. So you are not taking the union but the union of the union. Let us analyse. A set of paths is a set of sets of nodes. So let us consider the basic two trees (level 1 and level 2). The sets of paths are: L1: {{0, 1}, {0, 2}} L2: {{0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. Two sets. Unite them once: {{0, 1}, {0, 2}, {0, 1, 3}, {0, 1, 4}, {0, 2, 5}, {0, 2, 6}}. Unite this one, and we get: {0, 1, 2, 3, 4, 5, 6}. I do not see a path there. > > > 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. Yes. Infinite unions can give finite things. But that is not a contradcition. > > 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 think so, but provide no proof. > > 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. Oh. > There cannot be an infinite segment {1,2,3,...n} with a last element > unless the last element is infinite. Sure. But again that is not in contradiction to anything I state. > > 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. There is no *natural* number containing all natural numbers. How you conclude from this that there is no number containing all natural numbers escapes me. Pray give a proper mathematical definition of "number". > 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. Note how you sometimes use "natural number" and othertimes "number" as if they were the same? > > > 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)? I think we are talking at cross purposes. And are misunderstanding each other about what the meaning is of the term "the set of all paths p(n)". > > > 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. Right. So the union of all paths is a set of nodes. But I do not think that set is a path. That set is a tree (by your definition). > > > 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. I was talking about the union of set*s* of trees. The union of a set containing (as elements) two trees and a set containing (as element) two other trees, is a set containing (as element) two to four 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. This does not make sense at all. > > > 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. Makes no sense again. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Fuckwit on 5 Feb 2007 01:48
On Mon, 5 Feb 2007 04:10:21 GMT, "Dik T. Winter" <Dik.Winter(a)cwi.nl> wrote: >> >> What is 4 in mathematics? >> > succ(succ(succ(succ(0)))), according to the Peano axioms. > Right. @WM: In axiomatic set theory we (usually) have the definitions: 0 := {} , 1 := {0} , 2 := {0,1} , 3 := {0,1,2} , 4 := {0,1,2,3} , and succ(x) := x u {x} . Hence 4 = succ(succ(succ(succ(0)))) , as required. F. |