Cantor Confusion
in [Math]
|
Prev: Pi berechnen: Ramanujan oder BBP
Next: Group Theory
From: MoeBlee on 27 Nov 2006 14:22 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > That is not the question any longer. The question is: Can a set > > > theorist admit that she is in error? It seems impossible. They all are > > > too well trained in defending ZFC. > > > > In error as to what? Set theorists admit mistakes. It is not uncommon > > for books to have errata sheets attached. > > Would you see a contradiction in these two statements? > 1) "The cardinality of omega is |omega| not omega." > 2) "The cardinality of omega, also written as |omega|, is omega". If you give me the context of those remarks, I might have something to say about them. MoeBlee
From: Randy Poe on 27 Nov 2006 14:25 MoeBlee wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > > MoeBlee schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > That is not the question any longer. The question is: Can a set > > > > theorist admit that she is in error? It seems impossible. They all are > > > > too well trained in defending ZFC. > > > > > > In error as to what? Set theorists admit mistakes. It is not uncommon > > > for books to have errata sheets attached. > > > > Would you see a contradiction in these two statements? > > 1) "The cardinality of omega is |omega| not omega." > > 2) "The cardinality of omega, also written as |omega|, is omega". > > If you give me the context of those remarks, I might have something to > say about them. In its original context, statement #1 was about the NOTATION for "cardinality of omega". In its original context, statement #2 was about a theorem that |omega| = omega. This has already been explained several times to Mueck. He either doesn't understand the distinction, or pretends not to. - Randy
From: William Hughes on 27 Nov 2006 15:00 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Virgil schrieb: > > > > > > > > > > > > By induction one can even prove that always a finite number of words > > > > > has been mapped on the natural numbers. > > > > > > > > By induction one can also map an infinite set of words onto the natural > > > > numbers. > > > > > > Then you think the natural numbers are covered by induction? You read > > > my heart. I knew they were not actually infinite. > > > > > > > > > > > > > Now, after you have "completed" this finite list of words, take it and > > > > > construct a diagonal number. > > > > > > > > I prefer to take the infinite set of words. > > > > > > I know, but you don't get it by induction. > > > > > > > > > This is always a finite number, as the list is finite. > > > > > > > > Maybe yours is, mine isn't. > > > > > > > > > Fine. > > > > > So we have proved that there is a finite word which was missing in the > > > > > "complete" list. > > > > > > > > You haven't proved anything, > > > > > > If the set 1,2,3,...,n is finite, then the set 1,2,3,..., n+1 is finite > > > too. > > > Therefore, by induction we (that is: those who can think logically) > > > prove that the generated set is always finite. > > > > The two statements: > > > > The elements of N satsify property X > > and > > The set N satisfies property X > > > > are independent.. Induction proves that the elements of N are finite. > > Induction does not show that N is finite. > > Of course it does not. Induction is a sensible method. By induction > only sensible assertions can be proved. The *set generated by > induction* is always finite. > Using your interpretation of "set generated by induction" all sets of natural numbers are finite. if all sets of natural numbers can be generated by induction, So all we have to do to prove that all sets of natural numbers are finite is to assume that all sets of natural numbers can be generated by induction. No set of natural numbers without a largest element can be generated by induction. So what we have to do to prove that all sets of natural numbers are finite is to assume that there is no set of natural numbers without a largest element. - William Hughes
From: MoeBlee on 27 Nov 2006 15:05 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > If some other theory is consistent, then it has a countable model. > > > > If a FIRST ORDER theory of predicate logic HAS AN INFINITE model, then > > it has a countable model. > > Of course, I implied these properties. One cannot repeat always > everything. But "HAS AN INFINITE" is superfluous, at least the > capitals, because if the theory has not an infinite model, then it has > to have a countable model too. (Every finite model is per definition > countable. The meaning of "countable set" covers "finite set" as well > as "denumerable infinite set". Didn't you study set theory and its > definitions?) What I meant, and admittedly was not clear to say, was that if a theory has an infinite model then it has a DENUMERABLE (countably infinite) model. It's trivial that a theory having a finite model entails that the theory has a countable model. So what is at stake with Skolem's paradox is not JUST that a consistent theory has a countable model (which is a theorem, of course), but more particularly that if a theory has an infinite model then it has a denumerable model. > > > > But ZFC is not > > > > inconsistent simply for Skolem's paradox nor is ZFC even rendered > > > > prohibitively counterintutitive by Skolem's paradox. > > > > > > That's why Skolem liked ZFC soo much? > > > > Yes, Skolem was critical of set theory. > > What do you think, why? He gives reasons in his paper. By the way, Skolem's remarks pertain to Zermelo's set theory, and are not addressed to ZFC itself, since that came later, and the axiom schema of replacement in ZF remedies one of Skolem's criticisms as well as, if I'm not mistaken, by the time we get to ZF, Skolem's own recommendation for using formulas instead of 'definite properties' had been incorporated. > Was he too stupid to recognize this big > advantage of mathematics? Right now, I'm not in the mood to answer questions that are THAT childish. > > language is. > > > > > Why do you think that our > > > arithmetic (which is not a model of ZFC) does contain such a bijection? > > > > I don't know what bijection you are referring to. Exactly what > > bijection in exactly what set do you claim existence? > > No bijection in any infinite set. But why do you think that there is a > bijection N <--> Q? No bijection IN any infinite set? In set theory, 'in' usually means 'member of'. Of course there are bijections that are members of infinite sets. > But why do you think that there is a > bijection N <--> Q? The reason I believe it's a theorem of Z set theory that there is a bijection between the set of natural numbers and the set of ordered pairs of natural numbers is because I've studied at least one proof. (It's not more than routine then to show that there is a bijection between the set of natural numbers and the set of rational numbers.) Your questions seem pointless. MoeBlee
From: MoeBlee on 27 Nov 2006 15:20
mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > Countable means: There exists a bijection with N. After having > > > recognized that this simple definition leads to a contradiction, > > > > We've been over this and over this and over this, and you still persist > > incorrectly. You've not shown a statement P in the language of a Z set > > theory such that both P and ~P are theorems of the theory. > > > > > people > > > defined countable in and outside differently. > > > > I don't know who does that. I don't. > > It is necessary to avoid Skolems antinomy. If you have the same > definition of countability inside and outside the countable model, then > you will have the same result. But that would be a desaster. You're mixed up. The definition of 'is countable' does not change. x is countable <-> Ef(f is a bijection between x and omega v f is a bijection between x and a member of omega). Some sets have as a member such a bijection and other sets don't have as a member such a bijection. That doesn't entail that the definition has changed. > > > And if they will > > > recognize that there is really a contradiction in ZFC, then they will > > > define contradiction differently. I am sure, they will not fail. > > > > If there is shown a contradiction in ZFC, then I won't change any > > definitions; I'll just recognize that ZFC is inconsistent. But you've > > not shown a contradiction in ZFC. > > Yes, the same track should have been followed by set theorists after > Russell had shown his antinomy. But I am sure (and in fact I know) that > you would not agree that a contradiction is a contradiction. And you > are right. That's the simplest way to avoid learning that many of your > efforts have been wasted in vain. Just show a sentence P in the language of set theory such that both P and ~P are theorems of set theory. Then you'll have provided something to talk about. MoeBlee |