Cantor Confusion
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From: Sebastian Holzmann on 30 Oct 2006 13:25 MoeBlee <jazzmobe(a)hotmail.com> wrote: > mueckenh(a)rz.fh-augsburg.de wrote: > Probably what he is trying to get across to you is that dropping the > axiom of infinity does not entail that all sets are finite. Dropping > the axiom of infinity only entails that it is undecided whether there > are infinite sets. You can't infer from Z without the axiom of infinity > that there are no infinite sets. To prove that there are no infinite > sets in a Z set theory requires adopting an axiom that there are no > infinite sets not just dropping the axiom of infinity. Unfortunately, it is impossible to formulate such an axiom in first order predicate logic.
From: MoeBlee on 30 Oct 2006 14:35 Sebastian Holzmann wrote: > MoeBlee <jazzmobe(a)hotmail.com> wrote: > > mueckenh(a)rz.fh-augsburg.de wrote: > > Probably what he is trying to get across to you is that dropping the > > axiom of infinity does not entail that all sets are finite. Dropping > > the axiom of infinity only entails that it is undecided whether there > > are infinite sets. You can't infer from Z without the axiom of infinity > > that there are no infinite sets. To prove that there are no infinite > > sets in a Z set theory requires adopting an axiom that there are no > > infinite sets not just dropping the axiom of infinity. > > Unfortunately, it is impossible to formulate such an axiom in first > order predicate logic. What I have in mind is not the model-theoretic sense I think you are referring to. What I mean is just an axiom such as ~Ex x is infinite. I'm not claiming that this would preclude that the theory has models that have as elements of the universe infinite sets. I'm just saying that the theory would have ~Ex x is infinite.as a theorem. That is, if one wants the theory to have the theorem ~Ex x is infinite, then ZFC without the axiom of infinity does not provide that theorem. MoeBlee
From: Sebastian Holzmann on 30 Oct 2006 14:56 MoeBlee <jazzmobe(a)hotmail.com> wrote: > Sebastian Holzmann wrote: >> Unfortunately, it is impossible to formulate such an axiom in first >> order predicate logic. > > What I have in mind is not the model-theoretic sense I think you are > referring to. What I mean is just an axiom such as ~Ex x is infinite. > I'm not claiming that this would preclude that the theory has models > that have as elements of the universe infinite sets. I'm just saying > that the theory would have ~Ex x is infinite.as a theorem. That is, if > one wants the theory to have the theorem ~Ex x is infinite, then ZFC > without the axiom of infinity does not provide that theorem. I just wanted to add some additional point. You are, of course, right in your argument.
From: MoeBlee on 30 Oct 2006 15:06 mueckenh(a)rz.fh-augsburg.de wrote: > > With the axiom of choice, it is possible to well-order the reals. > > But it can also be shown that such a well-ordering can not be defined > > by a formula in ZFC. > > How then can the well-ordering be accomplished? Zermelo proved tat it > could be accomplished. The word 'accomplished' is not in set theory. ZFC proves the sentence 'AxEr r well orders x'. But there is no assertion about 'accomplished' since 'accomplished' is not in the language of the theory. ZFC proves that any set has a well ordering, but ZFC doesn't show us how to describe anything about that well ordering. ZFC just tells us that such a well ordering exists. Period. That is exists. Indeed, that is quite understandably unsatisfying for many people. But it is not in itself a contradiction. Yes, the axiom of choice may be regarded as unintuitive; but being unintuitive does not in itself entail contradiction. On the other hand, it may be unintuitive NOT to adopt the axiom of choice. For example, let me ask you this: Suppose we have a function f from x onto y. So f "covers" every member of y with a member of x. So, intuitively, there must be at LEAST as many members of x as there are of y, since every member of y gets mapped to (is "covered") by at LEAST (maybe more than one) member of x. So we would think that there must be an injection from y into x. That is, we would think there must be a function that maps each member of y to one of the members of x that y is mapped to by the function f. At least, at first blush, that seems intuitive, right? Okay, but try to prove it without the axiom of choice. MoeBlee
From: MoeBlee on 30 Oct 2006 15:12
mueckenh(a)rz.fh-augsburg.de wrote: > Get it right: It is nonsense to talk about infinite sets if there is no > axiom of infinity and, therefore, no possible definition of infinity. No, YOU need to get it right. You have it completely wrong. We don't need the axiom of infinity to define the predicate 'is infinite'. We need the axiom of infinity to prove that there is an object of which that predicate holds. MoeBlee |