Cantor Confusion
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From: Ross A. Finlayson on 4 Nov 2006 22:53 Virgil wrote: .... > > In such matters, ZF does not allow quantified membership by unstated > implication, but requires an explicit and pre-existing set to be for > every quantification. > > http://en.wikipedia.org/wiki/ZFC#The_axioms > 3) Axiom scheme of separation (also called the Axiom scheme of > comprehension): If z is a set and is any property which may be > possessed by elements x of z, then there is a subset y of z containing > those x in z which possess the property. > > > > > > > Again, depends on the model. > > > > Of course. But the usual set theory without yet specifying any models > > starts from this point. > > Everything is a set is true in theories like ZF, with restricted > > comprehension. Unrestricted comprehension requires a universal set. > > And "unrestricted comprehension " also allows contradictions like > Russell's paradox, which is why sensible mathematicians avoid it. For that there must be another and for that another ad infinitum, there's the cumulative hierarchy. So, where's the cumulative hierarchy? There is none in ZF. The cumulative hierarchy is not a set, in ZF. ZF is thus obviously incomplete, and because you can't quantify over anything/everything: inconsistent. Everything is a set means everything is a set. Ross
From: MoeBlee on 5 Nov 2006 21:57 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > As > > to set theory, for the tenth time: Without the axiom of infinity it is > > UNDETERMINED whether every set is finite. > > For the eleventh time: Infinity is NOWHERE. To assume the existence of > actual infinity is one of the greatest errors of human mind. Therefore, > without explicitly assuming this notion, it cannot be anywhere. Wonderful! But you still can see that my point stands. ZFC without the axiom of infinity does not have as a theorem that there do not exist infinite sets (equivalently, ZFC without the axiom of infinity does not have as a theorem that all sets are finite). > > Whatever your point, you won't be able to show that merely dropping the > > axiom of infinity from the Z axioms entails that there are only finite > > sets. > > Whatever your point, you won't be able to show that merely dropping the > > axiom of rabbithood from the Z axioms entails that there are only > non-rabbit > sets. Your analogy just agrees with what I said. You don't seem to have a point. > To put it in other words: It simply an imbecile nonsense to talk about > finished infinity without explicitly stating that it was not. I don't know what you mean by "stating that it was not". I don't get you. Aren't you the one who is reading Fraenkel, Bar-Hillel, and Levy? How can you read in that book and not understand that the axiom of infinity is completely independent of a ZFC without the axiom of infinity? You understand, e.g., that the axiom of choice is completely independent from ZF, right? That is, neither the axiom of choice nor the negation of the axiom of choice is a theorem of ZF. Same thing with the axiom of infinity per ZF without the axiom of infinity. With ZF minus the axiom of infinity, neither the axiom of infinity nor the negation of the axiom of infinity is a thoerem. That is all I'm saying. I don't know why in the world you won't just say you recognize this simple point. > > > Modern set theory simply cannot describe developing sets as > > > it apparently cannot describe sets with limited contents of > > > information. > > > > Whatever your definition of "developing sets" and 'limited contents of > > information", the fact remains that dropping the axiom of infinity does > > NOT entail that there are no infinite sets. > > > > > These things are unknown to the slaves of formalism. Read a good book > > > like Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: > > > "Foundations of Set Theory", North Holland, Amsterdam (1984). There you > > > will find more about that topic. > > > > Oh please, I've read more in that book than you have. > > That is strange. I read all of it, but you read more. This simple > sentence alone would prove that you must be a set-theorist. Okay, touche on that one. You scored a point and your remark was funny. But the point now is to wonder how you could have read all of that book and understand so little that is in it. Neither the axiom of infinity nor its negation are theorems of ZF without the axiom of infinity. Get it? MoeBlee
From: MoeBlee on 5 Nov 2006 22:13 mueckenh(a)rz.fh-augsburg.de wrote: > MoeBlee schrieb: > > > > > Es ist sogar erlaubt, sich die neugeschaffene Zahl omega als Grenze zu > > > denken, welcher die Zahlen n zustreben, wenn darunter nichts anderes > > > verstanden wird, als daß omega die erste ganze Zahl sein soll, welche > > > auf alle Zahlen n folgt, d. h. größer zu nennen ist als jede der > > > Zahlen n. (p. 195) > > > > > > This definition has been conserved up to our days: Limesordinalzahl or > > > limit ordinal number. > > > > I regret that I don't read German. > > A disadvantage for a set theorist. I'm not a set theorist. But it is a disadvantage not to read German anyway. > Here is my translation: It is allowed to understand the new number > omega as limit to which the (natural) numbers n grow, if by that we > understand nothing else than: omega shall be the first whole number > which follows upon all numbers n, i.e., which is to be called larger > than each of the numbers n. Okay. I don't see any problem with that. Would you please refresh the context by saying what point it is that you draw from that quote? Also, I stress that formalized Z set theories are not required to answer to nor conform to any particular formulation, writing, or notion of Cantor. > > But I'd like to know what you might > > propose as a mathematical definition of 1/omega in Z set theory. > > Notice, I'm not asking what Cantor wrote. I'm asking what is the > > definition of 1/omega specifically in Z set theory. > > If omega is larger than 2, then 1/omega is smaller than 1/2. Now > replace 2 by n and let n-->oo. That's not a coherent set theoretical definition. Roughly put, to define an operation symbol -such as '\' is being used as a 2-place operation symbol with a conditional that the arguments are ordinals - is to first prove for some formula F, that for any two ordinals, n and m, there exists a unique x such that F(n m x). Then you can define: n/m = the unique x such that F(n m x). Otherwise, just throwing a bunch of math-sounding language together, such as "If omega is larger than 2, then 1/omega is smaller than 1/2. Now replace 2 by n and let n-->oo." is not to make a definition in set theory. More simply, your "n --> oo" for ordinals is not even defined. By the way, Suppes has one of the best chapters on mathematical definition I've seen in an introductory textbook, in his 'Introduction To Logic' and an abridgement of that chapter is in his 'Axiomatic Set Theory'. MoeBlee
From: MoeBlee on 5 Nov 2006 22:28 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 it cannot be a fraction because ZF does > > > not yet know how to divide elements, > > > > In ZF we define various operations of division. As far as I know, there > > is not a dvision operation for sets in general. By the way, I do see that there are definitions of subtraction, division, and logarithms for ordinals. But they are not anything like your non-definition of ordinal division. > Why then do you not understand how an edge of the binary tree can be > divided? Haven't I asked you a few times already not to effectively put words in mouth? I never said I don't understand how an edge can be divided. I never said anything about "divided edges". > > > then it can only be a whole > > > number, I would guess.# > > > > These problems you're having are of fitting set theory to your own > > system of terminology. To work in set theory, we don't need to care > > about your own system of terminology. > > "Whole number" is not my terminology. It is AS you are using it here. > Please don't conclude from your own ignorance on mine. You are ignorant of the basics of set theory. That's not so bad onto itself. It's just that you have such a big mouth about set theory while you are ignorant of its basics. > Thank you. You're welcome for the time I've spent trying to explain to you what you should learn from a basic textbook by yourself. MoeBlee
From: MoeBlee on 5 Nov 2006 23:54
mueckenh(a)rz.fh-augsburg.de wrote: > Fraenkel, Abraham A., Bar-Hillel, Yehoshua, Levy, Azriel: >"Foundations of Set Theory", North Holland, Amsterdam (1984). Here's a post by mueckenh(a)rz.fh-augsburg.de from a thread that is now not open to reply: [begin post] David R Tribble schrieb: > Yes, I can see now that these are all finite sets. > And which are proper subsets of infinite sets. The set of all naturals > that have been written now, for example. Obviously it's an ever > growing set as time goes on, and will never contain the entire set > of naturals that are possible. So it's simply a finite subset of N, > and always will be. > Somehow you are using this fact to "prove" that N can't exist, perhaps > employing some marvelous mathematical logic that has not been > tainted by mainstream teachings. You show several finite sets. > How do they prove anything about infinite sets? "a reasonable way to make this conform to a platonistic point of view is to look at the universe of all sets not as a fixed entity but as an entity capable of 'growing', i.e. we are able to 'produce' bigger and bigger sets" [A.A. FRAENKEL, Y. BAR-HILLEL, A. Levy: Foundations of Set Theory, 2nd ed., North Holland, Amsterdam (1984) p. 118]. Why should simple infinite sets exist in another way? Just because there is an axiom which cannot be satisfied like the axiom that there be a straight bent line? Regards, WM [end post] If I undestand WM correctly, he's arguing that infinite sets can only exist as described - capable of "growing". If that is not his argument and if that is not the purpose of his adducing this quote by Fraenkel, Bar-Hillel, and Levy, then I welcome him correcting any misunderstanding I have about what WM is trying to say. I admit that I do not fully master the philosophical remarks by Fraenkel, Bar-Hillel, and Levy, but I think I get the general idea, and it is not suggestive that the sizes of sets themselves are variable. The context of the quote is a discussion of the pros and cons of an axiom of restriction, which would, roughly speaking, stipulate that there are no sets except those whose existence follows from the axioms. And the context includes discussions of various proposed axioms to "restrict" the universe of sets. So different axioms give us various possible larger or smaller universes, or some universes that include some sets and other universes that exclude certain of those sets but include certain others. So that is the sense of "growing" or "producing larger and larger", which is to say that we can decide which axioms we are to adopt and such decisions will determine whether or not the universe includes or excludes certain sets. 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. 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. MoeBlee |