An uncountable countable set
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From: Virgil on 5 Sep 2006 14:40 In article <1157471262.974912.294510(a)b28g2000cwb.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > In article <1157367209.318653.75760(a)p79g2000cwp.googlegroups.com>, > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > > > > Counting is the most primitive version of addition. > > > > Counting is not addition at all. > > > So you cannot even count, in your infinite unphysical mind? Counting precedes adding, in that one may count when unable to add. But how is one able to add without being able to count? A primitive version of counting, from which the word "calculus" allegedly derives, is Greek shepherds in classical times counting their flocks by pairing off their sheep with pebbles (calculi).
From: MoeBlee on 5 Sep 2006 14:44 Tony Orlow wrote: > Indeed, or we may NOT allow such things, if we adopt a rule of the form > x = S: A P P(x) = P e S. We don't need to go to second order to disallow urelements. The plain axiom of extenstionality disallows urelements. You don't even understand the axiom of extensionality. You're overflowing with opinions about set theory but don't even know what the axiom of extensionality says. > That is, every object is a set of values which > each property has when applied to that object. The truth value of each > statement that may be made about x is EQUAL to the truth value of that > property's membership in x. While some restrictions need to be made > regarding what constitutes a property, this general notion seems sound, no? No. What I don't understand is why you think you wouldn't benefit from understanding what other people (such as those who have studied the subject and written books on it) have come up with so that you can contribute or even dissent on an informed basis rather than flounder in your ignorance. > That's nice. Is there a difference between an abstract object like a > set, and its definition? Yes. > If you change the definition, does that also > change the set? That a different definition may be of a different set does not imply that the definition IS the set. > >> But, that requires the > >> universal quantification over properties and/or sets, and so cannot be > >> stated in first order logic, right? > > > > Right. But it's worse. > > Worse than what? Worse in the sense that there are even more problems with you mixed up idea than that of moving to second order. > That's nice. You're playing Grandpa trying to show the six year old how > to program the VCR like it's a Victrola. No, in this latest matter, it's more like you can't distinguish the picture on the screen from the buttons on the remote control that define which picture will be on the screen. The set defined is not its own definition any more than the picture of the cowboy on the screen is the sequence of buttons on the remote that brought that picture on. MoeBlee
From: Tony Orlow on 5 Sep 2006 14:45 MoeBlee wrote: > Tony Orlow wrote: >> I am NOT accepting the axioms. The axioms are artificial statements >> which can be made to work together, but which do not follow from >> elementary logic the way they should. Every axiom should be justifiable >> based on fundamental concepts. I don't see that here. > > You keep avoiding what I've told you several times. To get an adequate > amount of mathematics, you have to adopt axioms that are not derivable > from pure logic alone. So whatever axioms you produce will be what you > call "artificial statements" just as you call the axioms of set theory > "artificial statements." Perhaps I ignore that statement because you have not given any justification for it. I am not convinced that it's necessary to invent rules which do not follow from more elementary fundamentals. > >> Yes, I am working on that, or was trying to until I got stuck with child >> care duty the last several weeks. It's not far off, but it's not at all >> like ZFC. I have tried to present several axioms, such as N=S^L and IFR, >> and the axioms of internal and external infinity. > > Those are not statements of pure logic, hence they're "artificial > statements". Or, if they are pure logic, then they're inadequate. Inadequate why? N=S^L is inductively provable, and inductive proof is derived directly from logic by forming an infinite loop. That's all very elementary. IFR is justified primarily geometrically, and is also very straightforward. Together they provide means for measuring the relative sizes of infinite sets of the symbolic and the quantitative variety, respectively. The axioms of internal and external infinity are constructive axioms like Peano's, and therefore somewhat "artificial", but justified as a construction. Were Peano's axioms inadequate? > > MoeBlee > Hmmmmm.... Tony
From: MoeBlee on 5 Sep 2006 15:03 Tony Orlow wrote: > The difference between = and <-> disappears when logical truth values > are quantities from 0 through 1, so I don't see that as any better, but > equivalent. You say, in the absence of having specified a syntax for a language in which this all happens. >> > 'equality' would be a better word than 'equivalence' here, I think. > > I suppose, though the same applies to "equivalence classes" doesn't it? > No matter. No, that is the point. There is a difference between members of an equivalence class and the equivalence class itself. > >> So, it's not that a->b -> b->a, but that a=b <-> > >> b=a. > > > > That part seems messed up. a=b <-> b=a is just the symmetry of > > identity. > > Yes, it's that simple. If the object IS the unique set of logical values > applied to all properties, then each unique set of logical values for > each statement about an object IS a unique object. :) Whatever that means, I doubt it is the principle of the symmetry of identity, which is that a=b <-> b=a, which makes no mentions whatsoever of "logical values" or "properties". > >> and the > >> inability to discern two objects by their properties makes them equal, > >> at least until some property is discovered which can discriminate > >> between the two. > > > > That's going to make the theory subjective - depending on discoveries. > > Why don't you look at how different mathematical theories handle > > identity? > > Ummm.... Isn't each isolated theory "subjective" in terms of the > properties that it explores? A theory is a set of sentences closed under entailment. Theories are not made subjective for our reasons for interest in them. The subjectivity is in our deciding to study one theory and not another, but as a set of sentences closed under entailment, the theory itself is not affected by whether we are interested in it or not or by our reasons for interest or disinterest in it. > Two objects are equal only if there exists no way to distinguish them. See, that is what is subjective (or epistemological). We don't define equality by "way to distinguish" but rather by FORMULAS. > How do we know if this is the case? By enumerating all possible > properties of each. Can we do that? No. We can only say that, given the > set of properties under discussion in any given theory, the two are not > distinguishable, within that theory. We cannot say that they are > absolutely the same object. No, we may do better than that in theories in which there are only finitely many primitive predicate symbols, such as set theory. I told you all about that already. > >>> It depends on the specific theory. In a first order theory with > >>> infinitely many primitive predicate symbols, we have no theorem schema > >>> for doing what you suggest. But set theory has only two primitive > >>> predicates (one if you take equality as defined) so we can state such a > >>> theorem schema. However, we don't need to do that since the axiom of > >>> extensionality allows us to prove x=y merely by proving Az(zex <-> > >>> zey). > >> And what is z besides one of the set of properties which defines the > >> sets x and y? > > > > That's a confused view of the axiom of extensionality and the role of > > variables. > > The perception of confusion would appear to be a subjective and rather > relative phenomenon. True. You might not really be confused about the axiom of extensionality and the role of variables, but rather only pretending to be. > >> The distinction between elements and properties is rather > >> tenuous. Is it not a property of y that z e y? > > > > For any PARTICULAR z, it's a property of y that z is or is not a member > > of y. That doesn't entail that y IS the set of properties that y has. > > Consider each object in the universe to be a bit. Does each unique set > correspond to a unique bit string, where each object's bit posiiton is a > 1 if the object is a member, and 0 if it is not? I thought a 'bit' is a 0 or 1. In that case, in set theory, it is not the case that each object is a bit. > The set y IS the set of > objects which are members of y, no more, and no less. That's correct regarding set theory, and it conflicts with your notion that a set is the set of its defining PROPERTIES. A set is the set of its MEMBERS, and it is not the set of defining PROPERTIES. > > In second order, it would be a=b <-> AP(P(a) <-> P(b)), and P=Q <-> > > Aa(P(a) <-> P(a)). > > If you prefer, you may use <-> instead of =. > > > > > But those don't ential that, for example, b = {P | P is a defining > > property of b}. > > Really, they do. Because you say so. But you couldn't demonstrate that entailment in a system. > > Please just read a book on logic. > > Please just think hard. :) Yes, If I just think hard enough, in a blaze of enlightenment I'll see that you and you alone have the answers. MoeBlee
From: Tony Orlow on 5 Sep 2006 15:04
Dik T. Winter wrote: > In article <1157367467.816725.158560(a)i3g2000cwc.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > > > > > > Then such numbers like 1/3 do not exist (in that representation). > > > > > > Indeed, in your tree with terminating edges, such numbers do not exist. > > > > > > > But > > > > the same holds for Cantor's list. > > > > > > I do not see how you can jump to that conclusion. > > > > Cantor's list has as many lines as my tree. The diagonal has as many > > digits as each path of my tree. The only difference is that the paths > > split while the diagonal does not. > > And the difference is that given the list we can immediately state what > the n-th digit of the diagonal is for arbitrary n. You can not state what > the n-th path is for arbitrary n. The H-riffic number system indeed can. :) > > > > > More to the topic: Do you know how Cantor's diagonal is constructed > > > > from a list of reals? And how this list is constructed? > > > > > > Again, the second proof was *not* about the reals. > > > > Please spare your nonsense. Cantor did not consider anything else than > > the reals. If today the proof concerns some wider range then this is > > not due to Cantor. > > Please spare your nonsense. The second part of that article is *definitely* > not about the reals. It uses the reals only "beispielsweise". Then later it was applied to the reals to show them uncountable. It confuses symbolic combinatorics with the real continuum, which is a mistake. > > > > But I know how the > > > diagonal is constructed. I have no idea how the list is constructed. > > > But again (it appears that I have to repeat myself a lot in this > > > discussion), there is *no* construction of the list given. The proof > > > is along the lines of: "given *any* list, I can show such". So the > > > proof is valid without regards to the actual construction of the list. > > > > > The same is true for my tree. > > What is the tenth path in your tree? I think it's 1/2^sqrt(1/2). It depends what order one enumerates in. > > > > > > > There is no process of construction. If you have difficulties to > > > > > > comprehend that: it is the same as with Cantor's list. The tree is > > > > > > defined once and for all. That's it. > > > > > > > > > > Wrong. But I am not going to explain that again. > > > > > > > > What is the difference between my tree and Cantor's list. There was no > > > > explanation and there is none, because there is no difference. > > > > > > You give a construction process for your tree. In the case of the list > > > there is no construction process involved. The proof is about "given > > > *any* tree". > > > > Forget the construction process of the tree. Take any tree which > > contains all reals. As you say: My proof is about "given *any* > > infinite tree". > > What is the tenth path of that tree? > > > > The list is not a list that is constructed, but a list that is given, or > > > assumed. No construction involved. Whether the list contains finite > > > binary expansions or not is completely irrelevant. The process with Cantors > > > proof is more like the following: > > > give me a list of infinite sequences of 0's and 1's, and I show that there > > > is a sequence that is not in the list. > > > Do you have problems with proofs by contradiction? > > > > Give my a tree of infinite paths consisting of 0's and 1's, and I show > > that there are not less edges than paths. > > Indeed. If all edges terminate, also all paths terminate, and both are > countable, and 1/3 is not in the tree. If edges do *not* terminale, > also paths do not terminate, and both are not countable, and 1/3 is > in the tree. Every path must be countable. Do you have uncountable bit string lengths? Indeed a countably infinite number of paths pass through each node, but a countably infinite number of nodes are on each infinite path. There are half as many paths as edges, since a new path is established for every pair of edges added to the tree. Any other logic regarding the infinite binary tree is flawed at the root. > > > > > Please give a reference, or better: copy and paste your explanation. > > > > > > I have repeated my argument above. But I think you will not read it. > > > 1/3 is not in your tree because all edges (and hence paths) are > > > terminating. The diagonal is not in the list because it is defined > > > in such a way that it can not be in the list. There is no relation > > > between the two. > > > > The list, if existing, contains a diagonal (before anything is > > exchanged). Doesn't it? > > Yup. > > > > If you want to do the same with 1/5 > > > the case is similar. The problem appears when you want to count *all* > > > edges. At each level, the number of edges is 2^n - 1 or somesuch. But > > > also at each level we still have paths that need to be distinguished. > > > > You know that sets of order 2^omega are countable? > > No. Can you prove it? It is precisely Cantor's diagonal proof that > shows that it is not countable. By "not countable", all that is meant is "bigger than countably infinite". That's significant, but "uncountable" is a misnomer. :) Tony |