Cantor Confusion
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From: MoeBlee on 9 Nov 2006 13:54 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > MoeBlee schrieb: > > > > > > > > > > > > > > > > For ordinals, > > > > > > > > > > > > x<y <-> xey > > > > > > > > > > > > where 'e' is the epsilon membership symbol. > > > > > > > > > > That is identical with my definition. Take for instance Zermelo's > > > > > definition of the naturals or Cantor's own (Collected works, p. > > > > > 289-290), then you can see it. Your definition has been simply > > > > > translated from Cantor's. Please don't conclude from your own ignorance > > > > > on mine. > > > > > > > > You wrote: > > > > > > > > "My question is : Do you maintain omega > n for all n e N? I know that > > > > modern set theory says so. If something can be larger than a number, > > > > then it must be a number." > > > > > > > > Given that "n < omega" stands for "n e omega", and given that N is > > > > omega, what point is there in asking anyone whether they maintain that > > > > n < omega for every n e N? You're asking someone whether he or she > > > > maintains that n e omega for every n e omega. What is the point of > > > > asking that? > > > > > > > > > If omega > n then we cannot have a diagonal with omega digits in a > > > matrix the lines of which have only n (= less than omega) digits, > > > because the digits of the diagonal are digits of the lines. > > > > That's just another reenactment of your dogmatic pronouncements. I gave > > you a proof in set theory, from axioms using only first order logic. > > You give an argument that does not use any recognizable logic and from > > no stated axioms. > > The diagonal cannot have more positions than every line. In order to > see this you need not study set theory or logic or whatever. It is > simply the fact which everybody accepts who has not deliberately been > blinded by assuming the existence of a whole number larger than any > natural number but counting them. > > This number does not exist. The diagonal has not omega positions. So I guess the answer to my question is that you are not prepared to show that my proof has anything more than first order logic applied to Z set theory nor are you prepared to demonstrate a contradiction in Z set theory. MoeBlee
From: William Hughes on 9 Nov 2006 13:58 mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > David Marcus schrieb: > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > David Marcus schrieb: > > > > > > > > > > > The set of natural numbers is an infinite set that contains only finite > > > > > > numbers. > > > > > > > > > > Please do not assert over and over again this unsubstantiated nonsense > > > > > (this word means exactly what you think) but give an example, please, > > > > > of a natural number which does not belong to a finite sequence. If you > > > > > cannot do so, then it is obviously unnecessary to consider N as an > > > > > infinite sequence, because all its members belong to finite sequences. > > > > > > > > I didn't say anything about sequences, finite or otherwise. So, your > > > > request is irrelevant to my statement. > > > > > > The sequence of natural numbers is not comprehensible in ZFC? Neither > > > is the sequence of partial sums of a converging series? Nor are the > > > finite sequences which are called (initial) segments of sequences which > > > are ordered sets. Also the expression "extended sequence" for an > > > uncountable ordered set is new to you? > > > > Non sequitor. > > ? > I did not yet conclude anything but asked some questions. > > > > Let's make it simple. I'll give a statement and you say whether you > > think it is provable in ZFC. Is > > > > The set of natural numbers is infinite > > > > provable in ZFC? Please answer "Yes" or "No". > > > > > > Don't you think that you should label all your posts as > > > > "NON-STANDARD MATHEMATICS"? > > > > > > Cantor invented omega and defined omega as a whole number. > > > Who changed this standard meaning? > > > Why do you think this meaning was changed? > > > When do you think the contrary meaning became standard? > > > What is the contrary meaning? > > > Do you agree that A n: n < omega is incorrect? > > > If not, why do you complain about non-standard meaning on Cantor's > > > definition of omega as a whole number? > > > > Since Cantor predates axiomatic set theory, if you write anything that > > uses Cantor's definitions without checking whether the definitions are > > still standard is "Non-Standard Mathematics". > > Therefore I put above list of questions in order to find out what your > understanding of the standard is. If you say: My position is standard, > that is fine for you, but it is not sufficient to show anything but > orthodoxy. > > > If you want to discuss > > history, that is fine, but you should label your posts as such. This is > > simple courtesy. If you use words without defining them, readers assume > > you are using them in their current meanings. If you are using > > historical meanings, then either say so or use a different word. > > > > As for the current definition of omega, Kunen's book is a good > > reference. According to Kunen, omega is not a natural number. > > That is out of any question. > > > I'm > > guessing that by "whole number" you mean natural number, but I really > > don't know, since you seem to have your own language for everything and > > you never give definitions for any of the words that you use. > > "Whole number" is Cantor's name for his creation. > > My question is : Do you maintain omega > n for all n e N? I know that > modern set theory says so. If something can be larger than a number, > then it must be a number. No. It must be something that can be compared with a number. (Ususall, the term number to include both omega and the elements of N, however if you want to restrict the term number to the elements of N, knock yourself out. However, be careful to point out that your use of the term number is not the usual one.) Omega is not an element of N. However you can compare omega with any element of n. - William Hughes
From: MoeBlee on 9 Nov 2006 14:17 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > Wrong.. You have not understood the meaning. We have: The set of all > > > sets does not exist. But if all mathematical entities including all > > > sets do exist in a Platonist universe, how can it be that the set of > > > all sets does not exist? > > > > That is NOT Fraenkel, Bar-Hillel, and Levy's point at all! > > You admitted that you do not understand their philosophical remarks. So > let it be. No, please do not misparaphrase me. I said that I admit that I don't fully MASTER their philosophical remarks. But I said that I do know basically what they're talking about and what they're saying. And what they're saying is definitely NOT that sets themselves grow or are indeterminate, but rather, as I explained for you, that we may make choices as to which axioms to adopt as those choices will determine how inclusive or exclusive is any universe of sets for those axioms, which is roughly put by me, but good enough along with a recommendation that the reader read for himself the section in the book. > > So you have not made ANY point regarding your own views by quoting > > Fraenkel, Bar-Hillel, and Levy. > > > > > > That does not suggest that > > > > Fraenkel, Bar-Hillel, and Levy consider that a set itself can grow or > > > > have different members at different times or anything like that. > > > > > > They do not consider it for themselves but they report that some > > > mathematicians could adhere to that point of view. As far as I > > > remember, they do *not* say hat this point of view is wrong or > > > illogical or silly. > > > > And they don't advocate it at all. > > They mention this point of view as a possible one, and, as far as I > remember, as the only possibility for a platonist. WHERE? Where in that book do they mention the point of view that a set can grow or have different members at different times and that that is the only possible view for a platonist? You've got it completely backwards. It is very much a platonist view that sets are NOT as just described. > >If you want a theory in which there > > are growing sets, then, by all mean, go ahead and develop one. > > Not necessary. It has already been there. It is the well known > mathematics without set theory. What are the axioms, primitives, and definitions of this theory that has "already been there" and that has growing sets? > > > > For > > > > such axiomatic set theories, membership in a set is definite and sets > > > > are determined strictly by membership. And Fraenkel, Bar-Hillel, and > > > > Levy do not dispute that. > > > > > > If sets are strictly determined by membership, and if all sets do > > > exist, why then doesn't just that set exist the members of which are > > > sets with no further specification. > > > > Because no one who works with Z set theory claimed that for any > > possible description there is a set that has the properties described. > > It is not important whether one claims that or not. It is CRUCIAL whether one claims that or not. > > And what do you mean "all sets exist"? Do you mean all sets that exist > > do exist? (Which is of course true.) > > No, it is wrong, because the set of all those sets does not exist. That in Z set theories there is no set S that has the property that all other sets are members of S does not entail that it is is not the case that all sets that exist do exist. And you didn't answer my question. > > Or do you mean that for any > > specification of properties, there exists a set having those > > properties? > > Not even all sets with a well-defined specific property do exists. In Z set theories, we do NOT claim that for any property expressed as a formula that there is a set that has as members all and only those objects that have the property. > Already the definition of the natural numbers n > 0 by all sets with > cardinality n fails, Yes, there is not such a definition in Z set theory. So? > because the set of all sets equinumerous to n does > not exist. Yes, in Z set theories, for n not equal to 0, there is no set whose members are all and only those sets equinumerous with n. So? So, since you seem in the last couple of your remarks only to be reiterating Z set theory, what is your point? What is your answer to my question, which is: What do you mean when you say "all sets exist"? MoeBlee
From: MoeBlee on 9 Nov 2006 14:31 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > As I said, I don't care about (3). > > My mathematical notions include the fact that the diagonal of a matrix > cannot have more elements than every line. If in your opinion this > position is not valid in ZFC or in your logic, then we should stop > here. Then why did you even START to comment on my proof and raise all kinds of specious objections when all your objections only boil down to the fact that the axiom of infinity is used? I said it is a proof in Z set theory, nothing more and nothing less. If you had simply STARTED by saying, "But your proof uses the axiom of infinity", then I would have said, "Yes, of course it does, just as I said AT THE OUTSET that my proof is in Z set theory" and that would be that. > > > The diagonal is assumed to exist such that each of its digits exists, > > > actually. This is established by the mapping on a column. > > > > No it is not. > > > > > But it cannot > > > be established by the mapping on any line. You should recognize that > > > the following bijection between columns and lines shows a > > > contradiction, because one element is missing: > > > > > 1 <--> 1 > > > 1,2 <--> 2 > > > 1,2,3 <--> 3 > > > ... > > > 1,2,3,...n <--> n > > > ... > > > 1,2,3,... <--> omega > > > > 1, 2, 3 ... is NOT a line. > > But it is the whole column which would correspond to the line omega > (which does not exist). > > > > And it is NOT a contradiction that 1, 2, 3, .. omega is equinumerous > > with omega+1. > > Of course, the set 1,2,3... is equinumerous to the set 2,3,4,..., 1. > Nevertheless the diagonal of a matrix cannot have more elements than > every line. If in your opinion this position is not valid in ZFC or in > your logic, then we should stop here. Again, NOW you say that, after I already said so many times that my proof is in Z set theory. > > > > I answered that already. And the cardinality of the diagonal is not > > > > assumed, but is proven, to be omega. > > > > > > It is *assumed* by stating the axiom of infinity. Without this > > > assumption the length was not omega. > > > > It is PROVEN from the axiom of infinity. > > But the existence of this axiom is assumed. The existence of the AXIOM is assumed? The axiom exists, obviously. You can read it for yourself. It exists as a formula, which is what it is - a certain kind of formula. You can see a typographical representation of the formula whenever you like. Are you now saying we should doubt the existence of formulas? If that is not what you're saying, then in what sense is relevent to say that the existence of the axiom is assumed? > > since I would not have bothered to even post > > a proof about denumerable sequences and talk with you about it for so > > long if I accepted any condition that I can't use the axiom of > > infinity. > > That is *not my condition* but it is the result which follows from the > fact that the diagonal cannot be longer than every line. If it is not a condition, then what is the sense of your objection? OF COURSE I used the axiom of infinity. I said my proof is in Z set theory. > If I have the > choice either to accept the axiom of infinity with the condition that a > diagonal can be longer than every line, or to drop both notions, then I > choose the second. Fine! Don't accept the axiom of infinity. But that's irrelevent to my claim, which is that my proof is a proof, in Z set theory, of the denumerability of the diagonal. > If you can live with the contrary, then try to do > it. I wish you nice dreams. It's not a matter of whether I accept the axiom of infinity. I said mine is a proof in Z set theory. And mine is a proof in Z set theory, whether you or even *I* accept the axiom of infinity or not. MoeBlee
From: MoeBlee on 9 Nov 2006 14:49
mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > About a couple of weeks ago you presented an argument about trees, and > > as you presented your argument, as you described it, it was clearly > > reasonable to regard you as intending that as a proof in set theory (it > > would have been UN reasonable to think the contrary). But later you > > switched, so that your argument was not to be taken as in set theory. > > Why don't you try to find out how it could be presented in set theory? > Or why that cannot be done? That is completely beside the point. You represented that you were giving us a set theoretic proof, then you switched away from that. It's not MY job to vindicate your argument as a set theoretic proof. Yet, STILL, I am open to seeing your argument as set theoretic if you would just cooperate. I can't even begin to evalute your argument in terms of set theory if you won't give me a mathematical definition of the relation you first mention in your argument. And from there, I cannot even begin to evaluate your argument without confirming that your use of mathematical terminology is based on the same definitions I would have. Moreover, definitions in graph theory and tree theory differ even among different authors. I am interested in knowing the precise mathematics of your argument. And I humbly represent myself as not an expert in graph theory or trees, which is all the MORE reason that the definitions need to be clear to me. > > Now, when I say that I am giving a proof in set theory, and after I > > discuss that with you while having reminded you of that so many times, > > you switch not your OWN argument this time, but you switch to make the > > terms of MY argument subject to criticism for USING AN AXIOM OF SET > > THEORY! > > You misunderstood. See below. > > > > I'm really very curious what satisfaction you get from such games you > > play. > > > I hope that there are some lurkers who will learn what are the > consequences of accepting ZFC. By switching around the way you do, you don't prove anything about consequences of ZFC. > > There can't be any intellectual satisfaction in such mindless games. So > > what is it get from them? > > I do not prohibit to use the axiom of infinity, but I show that the > consequences of its use are absurd. Now you're switching AGAIN! Previously, you objected that I got my conclusion only because I availed myself of the axiom of infinity. NOW you're switching to saying that you don't begrudge use of the axiom of infinity but that now you point to (what you claim to be) an absurdity that results from using the axiom of infinity. But what absurdity? Not a contradiction IN set theory. But rather a conflict with YOUR own informal, non-axiomatic, mathematical notions. So, again, were back to (1) on that list I gave. You've shown nothing in my proof that is not first order logic applied to the axioms of set theory. You've shown no contradiction in set theory. You've only remarked that set theory conflicts with your notions (you also claim that your notions are correct, of course). > It leads to such results as a > diagonal which is longer than any line of a matrix or a vase which > contains zero numbers at noon or Tristram Shandy getting ready with his > diary. In the binary tree it leads the result that there are not less > edges than paths. In my opinion that is enough to prove ZFC being > wrong. If you prefer to remain within this wrong system, good luck. MoeBlee |