Cantor Confusion
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From: Dik T. Winter on 26 Oct 2006 19:42 In article <1161856793.990116.183680(a)e3g2000cwe.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > Within the *real* numbers the limit does exist. And a decimal number is > > nothing more nor less than a representative of an equivalence classes. > > So we are agian at this point: The real numbers do exist. For the real > numbers we have LIM 10^(-n) = 0. Therefore, i ther limit n--> oo tere > is no application of Cantor's argument. And again we are back at this point. You do not comprehend what I am writing. The reason real numbers exist is that there are sequences of rationals that come arbitrarily close to each other. The reason that a decimal expansion is a representative of a real number is because the sequence of finitely terminations of that number is a sequence of rationals that comes arbitrarily close to other such sequences and so falls in an equivalence class. *No* limit is involved in all of this. > > I do not understand why you want to tie that list in with the definition > > of the reals? > > > Because the real numbers are used in that list. Representatives of the equivalence classes that actually are the real numbers. > > > But you should try to understand. In Cantor's list, there are > > > limits of sequences, not equivealence classes. > > > > The list is a list of equivalence classes. In the decimal case > > representatives of those equivalence classes are used. > > And that are limits. No, they are not, they are sequences of rational numbers. > > > > Where in the construction above did I use the limit omega? > > In using any infinite sequence already Oh. I did not know that I did use the limit omega there. > > > You will need it in order to construct a real number and its decimal > > > representation for a Cantor list. > > > > No. By the theory, each decimal number is a representative of an > > equivalence class of sequences of rational numbers. By the construction > > we get another decimal number that is also a representative of an > > equivalence class of sequences of rational numbers. No omega is needed. > > Why then do you think omega is needed at all in mathematics? Because it comes in handy on many occasions. What *is* needed is the axiom of infinity, because that guarantees that you can even talk about infinite sequences. (And ultimately about limits.) Without that axiom there is no way to prove that infinite sets do or do not exist. > Real > numbers without the axiom of infinity would be a nice construction. I would not know how to do that. Perhaps that is possible, but in that case you would need to reformulate the limit concept. > Omega is introduced only by the axiom of infinity. No. It is *defined* using properties obtained through the use of that axiom. In some fields of mathematics you do not need omega at all, but only the axiom of infinity (analysis, and I think also algebra, number theory, and a host of other fields). Try to set up some form of analysis without that axiom, and consider that Leibniz acknowledged that he did not know how to *define* his infinitesimals. But go on forward with what Leibniz had done if you want to use undefined things. -- 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 26 Oct 2006 20:12 In article <1161860460.122685.294590(a)m73g2000cwd.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > Dik T. Winter schrieb: > > > > > > > lim [n-->oo] {-1,0,1,2,3,...,n} = N > > > > > > > > > > is obviuously wrong too. > > > > > > > > Depends on how you define the limit. > > > > You did not state why that one was obviously wrong. > > You think it must be proved that -1 is not a natural number? As it must > be proved that an always increasing positive function does not converge > to zero? No, you must tell me what that limit is, until now you have not provided any definition of a limit of sets so I can verify whether the above is true or false. Until now you have defined the limit of exactly *one* sequence of sets, but that definition is not sufficiently general to determine whether the above statement is true or false. > > > > > Therefore lim [n-->oo] {1,2,3,...,n} = N. > > > > > > > > Depends on how you define the limit. > > > > > > That *is* the definition of the this limit. It is a wrong definition > > > only in case N does not actually exist. > > > > You state "therefore", in general definitions are not conclusions from > > earlier arguments. But N obviously exist; it is simply a letter. > > "N" is simply a letter. N, or better |N, as we understand it by > convention in mathematics is a set most instructively expressed by lim > [n-->oo] {1,2,3,...,n} = N. I think that the |N notation is prevalent amongst German matematicians, but not so much so with Anglosaxon mathematicians. They do in general use a bold-face N. (And I have no idea what the French use, because I currently do not have a French textbook on this subject in front of me.) But what that definition states is that you denote the limit of that sequence by the letter N. What it is is not in the definition, so I have no idea. (Of course, I have, but you formulate your definitions so badly that they can be very confusing at times.) And I may note that this definition does *not* rules out that lim{n -> oo} {-1, 0, 1, ..., n} = N > > > Look here: lim [n-->oo] {1,2,3,...,n} := N. > > > > Finally. A definition. So you define the notation lim for a particular > > sequence of sets. But can you, in that case, use that definition in your > > arguments? This is not the only sequence of sets for which you did use > > the limit notation. > > Not the set is here the important thing with my definition. The > definition of the operator "lim [n-->oo]" covers every case where n > grows without upper bound such that either it comes never to a end or > such that a non-natural number, namely omega, is reached. But this is again different, you state here that N is the set of natural numbers, and you define the limit of a particular sequence of sets as being that set N. But it does *not* cover arbitrary sequences of sets. What, according to that definition, is: lim {n -> oo} {-1, 0, 1, ..., n} the definition does not make that clear. So the definition is not about the "operator" but about a particular sequence of sets. > In the first > case, we have 1/n < epsilon for every positive epsilon and we may > *define or put* > lim [n-->oo] 1/n = 0. > In the second case we have without further ado > lim [n-->oo] 1/n = 0. > That is the difference between potential and actual infinity. Well, in mathematics the first form is valid, the second is not valid. But now you are talking in analysis where limits are properly defined (because there is a topology and a metric). > > > > That requires proof. > > > > > > LOL. Idle discussion. > > > > Perhaps, but you use it as argument, and I want to know whether it is a > > valid argument, and so I want to see a proof. > > How do you know that the infinite set omega = N, which contains {n} or > n+1 if it contains n, does have a positive number of elements? How do > you know that it has more than one element? You do not know that (and can not prove it), until you have *defined* the ordering relations. Once you *have* the ordinal numbers, you can define the ordering relations, and a valid definition would be the following (assuming the letters are ordinal numbers): (1) a = b if there is an order preserving bijection between a set with ordinal a and a set with ordinal b (2) a <= b if there is an order preserving injection from a set with ordinal a and a set with ordinal b from these you can define: (3) a != b iff not a = b (4) a < b iff a <= b and a != b (5) a > b iff b < a (6) a >= b iff b <= a Now you have an ordering. You have to show that it defines an ordering relation indeed (not so very difficult). And as there is an injection from {} to N, we know that {} <= N (from 2), and as there is no bijection from {} to N, we know that {} != N (from 1 and 3), and so by (4), 0 < omega. > If you can prove that, then > the positive contents of the vase at noon, i.e., for t --> oo, is > proved too. Why? You are still assuming that the existence of a limit (which does not exist in this case) coincides with continuity. As I did tell you before, given the function: f(x) = 0 when x != 0 and f(0) = 1 the limit of f(x) for x going to 0 does exist, and is 0, but f(0) = 1. Give me a citation of any textbook on analysis where that is *not* the case. -- 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 26 Oct 2006 20:29 In article <1161861368.657796.161130(a)k70g2000cwa.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > Dik T. Winter schrieb: > > > As you allow constructions by lists, and as there are uncountably many > > lists, there are also uncountably many WM-constructible numbers. > > Where are those list? How are uncountably many different lists are > constructed by countably many words? Any infinite sequence of decimal digits is a list. > > Yes, that is precisely what I wrote, quite sometime ago already, about > > the computable numbers. It is easy to show that the computable numbers > > are countable. But the theory has been developed a bit since then. > > The problem with what K=F6nig wrote is that not every finite definition > > also gives a number. > > If you mean that there are more finite definitions than numbers, you > are right. Therefore we have a countable set of constructable numbers. Pray, refrain from using non-standard terminology. WM-construcatble = computable. > > When you substitute "computable" for "constructable", both are indeed > > right. But Cantor was wrong. It is only after Turing that this was > > solved. The finite definitions are countable, but not every finite > > definition gives a number. So there exists a list of finite definitions > > (this has been formalised using Turing machines), but this is not a > > list of computable numbers, because there will be non-halting Turing > > machines in the list, and you do not know how to separate them from > > the rest. Look up the halting problem. There is no computable list > > of computable numbers. > > Why do you insist on this obvious fact? It does not support your > position. There is a list of all finite constructions or definitions > (encoded by numbers, G?del). This is the definition of countablity. I thought that G?del encoded theorems? But it is trivial (you do not need G?del for that) to find that the number of finite definitions is countable. > The diagonal number of this list shows the listed numbers are > uncountable. Contradiction. But indeed, there exist such a list of finite definitions, but not every finite definition yields a computable number. Some of those definitions do not yield a number at all, so you can not even take the diagonal of the numbers defined with such a list. Getting back to the definition of computable number: a computable number is encoded by a Turing machine that, when it is given a natural number n, responds with the n-th digit of the number encoded by it. Taking the diagonal means that you can ask the n-th Turing machine the n-th digit of the number it represents. But how can you find that digit if the n-th Turing machine does not halt when asked for the n-th digit? -- 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 26 Oct 2006 20:36 In article <1161885287.076956.190220(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > jpalecek(a)web.de schrieb: > > > > The model is that simple that any student in the first semester could > > > understand it. Every paths which branches into two paths necessarily > > > needs two additional edges for this sake. It is only your formalistic > > > attitude that blocks your understanding. But you must not think that > > > anybody is blocked like you. > > > > This looks like a proof by induction. Indeed, you can prove your > > formula > > by induction for FINITE paths. > > Thanks for admitting your understanding so far. > > > But for infinite ones, you must do one > > more transfinite step. > > No. Any real number has only finite digit positions, according to Dik, > who rigorously denies that 0.111... cannot be indexed completely by > natural indexes. Now you see that this opinion leads to induction and > to a contradiction. So you hurry to switch to transfinity. Pray. No, you are wrong. 0.1 is rational, so is 0.11, and so on, so for every *finite* n, 0.1...1 (n-times) is rational. This does *not* show that 0.111... is rational. > Indeed, i is not simple. Do we need transfinite induction for all the > reals? Then, in fact, the natural numbers are not sufficient. Then > some digits of 0.111... are undefined. Or do we need transfinity only > in special cases, namely then when otherwise set theory would be > smashed? You need transfinity when you want to show that something that holds in the finite case also is valid in the infinite case. Induction will not show that 0.111... is rational, it can only show that all the finite initial parts are rational. And I again note that the notation 0.111... (in the decimals) has only meaning due to the definition of that notation. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: William Hughes on 26 Oct 2006 21:26
mueckenh(a)rz.fh-augsburg.de wrote: > Dik T. Winter schrieb: > > > > As you allow constructions by lists, and as there are uncountably many > > lists, there are also uncountably many WM-constructible numbers. > > Where are those list? How are uncountably many different lists are > constructed by countably many words? > > > Yes, that is precisely what I wrote, quite sometime ago already, about > > the computable numbers. It is easy to show that the computable numbers > > are countable. But the theory has been developed a bit since then. > > The problem with what König wrote is that not every finite definition > > also gives a number. > > If you mean that there are more finite definitions than numbers, you > are right. Therefore we have a countable set of constructable numbers. > > > When you substitute "computable" for "constructable", both are indeed > > right. But Cantor was wrong. It is only after Turing that this was > > solved. The finite definitions are countable, but not every finite > > definition gives a number. So there exists a list of finite definitions > > (this has been formalised using Turing machines), but this is not a > > list of computable numbers, because there will be non-halting Turing > > machines in the list, and you do not know how to separate them from > > the rest. Look up the halting problem. There is no computable list > > of computable numbers. > > Why do you insist on this obvious fact? It does not support your > position. There is a list of all finite constructions or definitions > (encoded by numbers, Gödel). This is the definition of countablity. > The diagonal number of this list shows the listed numbers are > uncountable. Contradiction. There is indeed a list of all possible finite constructions or definitions. Call this list A. However, this list must contain things that look like definitions but are not because the method given to produce a number does not halt. We cannot get a diagonal number from A, because some of the members of A do not give numbers. So there is no contradiction. So let's take list B. We get list B from A by taking only the true definitions from A (that is we throw away anything that will does not halt). So now we have the list B, so by definition B is countable. Now we can get a diagonal number, d. This number cannot be a member of B. But note. d cannot have a finite definition. A finite definition of d would include not only the diagonal procedure (finite description) but also the list B (infinite description) or a finite description of a way to get the list B from the list A (i.e. a solution to the halting problem). So d is not a member of B. If we do not take a contrucivist viewpoint this is not a contradiction, there are lots of things that are not a member of B. But let's say we take a constructivist viewpoint. d is not computable, so d does not exist. (Nothing that is not a member of B exists.) We still do not have a contradiction because if we take a constructivist viewpoint the list B does not exist so we cannot use it to form the diagonal d. -William Hughes |