An uncountable countable set
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From: MoeBlee on 3 Oct 2006 22:14 Tony Orlow wrote: > > The property of not being an infinite natural holds for the first > > natural, and holds for the successor of each non-infinite natural, so > > that it must hold for ALL naturals. > > It holds for all finite naturals, but if there are an infinite number of > naturals generating using increment, then there are naturals which are > the result of infinite increments, which must have infinite value. That's just Tonymathsoundingspeak nonsense. Drop the undefined verbiage "generated using increment" and 'infinite increments", look instead at axioms and proofs, and what is left is the theorem that there are no infinite natural numbers, or an opportunity for you to propose alternative axioms. > > If a property is true for the first and also for the such successor of > > every one for which it is true, then it is true for ALL naturals without > > exception. > > > > And being finite is such a property. > > > > So that TO's system violates the inductive property up front. > > No, the inductive proof of an equality applies to all n, finite or > infinite. But "is finite" is an inequality, equivalent to "<oo". > lim(n->oo: n)=oo, not <oo. You can only increment a finite value a > finite number of times before you get infinite values out of it. Then get some axioms and definitions for which that notation and verbiage is defined. > > All versions of the standard set of naturals prohibit infinite naturals > > by essentially the same argument. > > I reject it as circular. sum(x=1->aleph_0: 1)=aleph_0. You can reject it as being any shape you want to think it is and with as much of your own personal undefined notation you want to bring in. But first order logic does not permit the fallacy of circular argument and there is nothing but first order logic applied to axioms in the proof that no natural number is infinite. > Each of my T-riffics has a successor and predecessor. No discontinuities > there, although there may be uncountable portions of the strings. If every T-riffic has a predecessor and a sucessor, then your set and operation do not provide for a Peano system and do not capture the everyday notion of the non-negative integers. Moeblee
From: Virgil on 3 Oct 2006 23:21 In article <45231438(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Virgil wrote: > > The property of not being an infinite natural holds for the first > > natural, and holds for the successor of each non-infinite natural, so > > that it must hold for ALL naturals. > > It holds for all finite naturals It holds for ALL naturals, as that " inductiveness" is an essential part of the definition of the naturals. Anything without that property, whatever it may be, is not the set of naturals, though it may contain the naturals as a proper subset. , but if there are an infinite number of > naturals generating using increment, then there are naturals which are > the result of infinite increments, It does not follow that having an endless supply of something means that any of them are infinite. TO has repeated that falsehood, many times but it remains false, and will continue to remain false however many times TO repeats it in future. According to every standard definition of natural numbers, the first one is finite, and the successor of any finite natural is also a finite natural, so that by induction (by definition, a necessary property of naturals) EVERY natural is finite. TO may choose to ignore this irrefutable proof that all naturals are finite, but he cannot refute it. > > >>> ERGO: If the first natural is finite and the successor of a finite > >>> natural is a finite then every natural is finite. > >>> > >> Only for a finite number of successive increments. > > > > The inductive property says FOR EVERY NATURAL, idiot. The induction > > property does not exempt any naturals. > > > > If a property is true for the first and also for the such successor of > > every one for which it is true, then it is true for ALL naturals without > > exception. > > > > And being finite is such a property. > > > > So that TO's system violates the inductive property up front. > > No, the inductive proof of an equality applies to all n, finite or > infinite. But in the process proves that none of them can be infinite. > But "is finite" is an inequality, equivalent to "<oo". It is a property, which is all that is needed. > lim(n->oo: n)=oo, not <oo. You can only increment a finite value a > finite number of times before you get infinite values out of it. I can. Nonsense is not a refutation of logic, TO. ANY PROPERTY which holds for the first natural and which holds for the successor of every natural which has it, is a property that EVERY natural has. Being finite, or being less that 'oo', or not being infinite, are all such properties of EVERY natural, regardless of TO's misrepresentations. The inductive proof is an inherent part of the definition of naturals, and has been so ever since Peano. It will not go away simply because TO's intuition balks at it. > > >>> > >>>>> Von Neumann's N is such that > >>>>> (1) {} is a member of N > >>>> {} is the empty set, not a natural number. > >>>> > >>>>> (2) If x is a member of N then so is x \union {x} > >>>>> (3) If S is an subset of N such that > >>>>> (3.1) {} is a member of S, and > >>>>> (3.2) if x is a member of S then so is x \union {x} > >>>>> then S = N. > >>>>> > >>>>> And that is, by general consent, a satisfactory model for N. > >>>> You may be satisfied with it. > >>> Von Neumann was, and the vast majority of those who followed have been. > >>> > >>> That TO is not only bolsters my confidence in it. > >> Good. > >> > >>>>> But that model cannot contain any of TO's allegedly infinite naturals. > >>>> Well, it can. > >>> Let S be the all finite naturals in vN , the set of von Neumann > >>> naturals, then > >>> (1) {} is a member of S > >>> (2) if x is a member of S then so is its successor > >>> Thus S = vN, and there is no room in vN for TO's illusionaries. > >>>>> (1) there cannot be a first infinite natural, and, > >>>> No, there can't, but you have a fallacious first infinite ordinal, so > >>>> why complain? > >>> Because TO is trying to put objects into vN that cannot be there. > >> I reject the von Neumann ordinals as an inadequate model of the naturals. > > > > The same result follows from the Peano rules, or any other set of rules > > by which a standard set of naturals is defined. > > > > All versions of the standard set of naturals prohibit infinite naturals > > by essentially the same argument. > > I reject it as circular. I reject TO as circular. > > > > > So TO will have to find some other set of rules which does not > > explicitely and absolutely prohibit infinite naturals the way that all > > standard systems do. > > Each of my T-riffics has a successor and predecessor. No discontinuities > there, although there may be uncountable portions of the strings. But TO numbers do NOT satisfy this inductive property of naturals: If the first natural has a property, and if the successor of every natural having it also has it, then all naturals have that property. That requirement, which TO cannot escape, proves him wrong.
From: Virgil on 3 Oct 2006 23:24 In article <45231542$1(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > Dik T. Winter wrote: > > In article <40ef7$452210c2$82a1e228$23007(a)news1.tudelft.nl> Han de Bruijn > > <Han.deBruijn(a)DTO.TUDelft.NL> writes: > > ... > > > We can say that the number of balls Bk at step k = 1,2,3,4, ... is: > > > Bk = 9 + 9.ln(-1/tk)/ln(2) where tk = - 1/2^(k-1) for all k in N . > > > And that's ALL we can say. > > > > Why the obfuscation? Why not simply Bk = 9 + 9.(k - 1) = 9.k? > > > > > And that's ALL we can say. The version of the problem used here is > > > the first experiment in: > > > > Not only of the first, but also of the second experiment. > > > > But strange as it may appear, the two experiments give different results. > > In the first experiment there is no ball that escapes from being taken > > out. In the second experiment there are quite a few balls that are never > > taken out. The whole point is that you can not use limits to determine > > what is the valid answer. > > Hi Dik. I think the point that WM, Han, myself and others are trying to > make is that limits gives a more reasonable answer than transfinite set > theory. Why is it more credible to have the balls disappear due to > labeling, than to apply the infinite series and see that it diverges? That the "series" diverges means that there is no such thing as a limit, so that method does not say anything about the result. TO argues that the longer the balls stay in the vase the fewer are in the vase at noon.
From: MoeBlee on 3 Oct 2006 23:31 Tony Orlow wrote: > MoeBlee wrote: > > 1. Sets, countability, and uncountability, and even natural numbers are > > not mentioned in first order PA. The language of first PA does not > > include symbols - either primitive or defined - for those. > It mentions the set, as containing 0 NO IT DOES NOT. Please listen to me for a change. First order PA has no defined predicate for 'is a set' nor for 'containing' nor 'member' nor 'natural number'. > and also the successor of any > member. It doesn't mention any limits on the number of iterations, nor > the operation that constitutes successor. It certainly doesn't mention > any predecessor discontinuities besides the initial one at 0. Yes, you're correct that first order PA DOES have 'S' which is primitive in the language of the theory and is read as 'successor of'. But, as to "limits on the number of iterations, nor the operation that constitutes successor. It certainly doesn't mention any predecessor discontinuities," whatever you mean by that, it is a theorem of first order PA that every object except has a unique immediate predecessor (as 'immediate predecessor' is defined) and that there is no object between an object and its immediate predecessor nor between an object and its successor, and the theory doesn't have to address "limits on the number of iterations" for the theory to entail the theorems we've mentioned. > > 2. The structure <w 0 S + *> is a model of first order PA, where w is > > the set theoretic set of natural numbers, 0 is the set theoretic empty > > set, S is the set theoretic successor operation on natural numbers, + > > is the set theoretic addition operation on natural numbers, and * is > > the set theoretic multiplication operation on natural numbers. > > Okay > > > > > 3. By Lowenheim-Skolem-Tarski, first order PA has uncountable models, > > and Ax(x = 0 v Ey x=Sy) is a theorem of first order PA. But it is NOT > > required that the model regard 'S' as standing for the usual set > > theoretic successor function. So first order PA does not "define the > > inductive set". > > It defines an inductive structure for the set, but not the meaning of > successor. Still, there is no reason why we cannot apply the normal unit > increment uncountable times. Your "inductive structure for the set" and "apply the normal unit increment uncountable times" is just T-mathsoundingspeak. It's not something you have shown to be statable in first order PA or in any particular mathematical theory. > > 3. In set theory, we define 'Peano system' in a way that captures our > > intuitive notion of the counting numbers starting with zero. And we > > prove that all Peano systems are isomorphic. In particular, all Peano > > systems have a denumerable carrier set. So even though first order PA > > has uncountable models, such uncountable models are not part of any > > Peano system. > > It sounds like convention more than anything. We define 'Peano structure'. I've already explained to you too many times what a mathematical definition is. The definition of 'Peano structure' does capture the intuitive notion of the counting numbers. The rest of what I mentioned in that paragraph are theorems. > > 4. [I think the following is correct, and hope to be corrected if it is > > not correct:] In Z set theory without the axiom schema of replacement, > > it is underdermined whether there exist inductive sets that are > > uncountable (where an inductive set is one that has 0 as a member and > > is closed under the successor operation), but with the axiom schema of > > replacement, there are proven to be inductive sets that are > > uncountable. [I think this item 4. is is correct, and hope to be > > corrected if it is not correct:] In any case, again, first order PA > > does not define "the inductive set", since there may be many inductive > > sets. w (which is the LEAST inductive set) is the carrier set of ONE OF > > the models of first order PA, but w is defined in set theory, not in > > first order PA. > > Yes. > > > > > 5. But, per item 3, no uncountable set, whether or inductive or not, is > > a carrier set for a Peano system. That is to say, no, it is not true > > that it is possible to have a set for a Peano system (note 'system', > > not 'axioms') that has a successor operation "applied" an uncountable > > number of times. And the notion of such a thing reduces to just another > > variation on the unsupported and axiom-less confusion (suffered by many > > cranks) that a limit ordinal, such as the first infinite ordinal, can > > somehow (or must somehow) also be a successor. > > I am not talking about limit ordinals, but about an uncountable > successorship. I understand that. But my point, even though you are blind (due to self-blindfolding) to this, is that, as far as set theory is concerned, your remarks reduce to just what I said they do. Aside from set theory, if your remarks are to be anything but babble, then please for godssakes tell us just what theory you are talking about. > > 6. The first order PA axioms are all proven in set theory. In > > particular, the induction axiom schema of first order PA is proven as a > > theorem schema of set theory. > > Using the axiom of infinity I presume? Yes, we use the axiom of infinity to prove the existence of a carrier set for the Peano axioms. But as to the particular matter of proving an induction schema, we can prove an induction schema for natural numbers even without the axiom of infinity. > >>> ERGO: If the first natural is finite and the successor of a finite > >>> natural is a finite then every natural is finite. > >>> > >> Only for a finite number of successive increments. > > > > No such qualification is needed. Giving such a qualification only > > reveals a lack of understanding of what the induction schema is. What > > the poster said is correct, without any qualification about "finite > > number of sucessive increments" needed. > > Inductive proof such as this are taken only to hold for finite n, since > that's all there is in N, but if that's the case, all it proves is that > finite n are finite. It says nothing, in the standard understanding, > about what happens after an infinite number of increments, say, one for > each real in [0,1]. I've already told you too many times that we can have inductive proof for MANY DIFFERENT kinds of sets. A
From: Virgil on 3 Oct 2006 23:51
In article <45231907(a)news2.lightlink.com>, Tony Orlow <tony(a)lightlink.com> wrote: > MoeBlee wrote: > > Tony Orlow wrote: > >> Virgil wrote: > >>> In article <452279ce(a)news2.lightlink.com>, > >>> Tony Orlow <tony(a)lightlink.com> wrote: > >>> > >>> > >>>>>>> David R Tribble wrote: > >>>>>>>>> So I ask again, where are those infinite naturals and reals you keep > >>>>>>>>> talking about? It's obvious they are not in N. > >>>>>>> Tony Orlow wrote: > >>>>>>>> [No] it's not. > >>>>>>> Every member of N has a finite successor. Can you prove that your > >>>>>>> "infinite naturals" are members of N? > >>>>>>> > >>>>>> Yes, if "finite successor" is the only criterion. > >>>>> But it is not. In addition, N must be minimal with respect to closure > >>>>> under successorship, which excludes TO's "infinite naturals". > >>>> Which of Peano's axioms states such a requirement? > >>> The last one: > >>> If a property holds for the first natural, and holds for the successor > >>> of every natural number for which it holds, then the property holds for > >>> all natural numbers. (This axiom of induction is also called > >>> "mathematical induction". Its point is to limit the natural numbers to > >>> just those which are required by the other axioms.) > >> Its point is to define the inductive set. There is no indication there > >> that an uncountably infinite number of successions can't occur. > > > > We need to be clear just what we're talking about. > > > > 1. Sets, countability, and uncountability, and even natural numbers are > > not mentioned in first order PA. The language of first PA does not > > include symbols - either primitive or defined - for those. > > It mentions the set, as containing 0 and also the successor of any > member. It doesn't mention any limits on the number of iterations, nor > the operation that constitutes successor. It certainly doesn't mention > any predecessor discontinuities besides the initial one at 0. There is no "predecesssor discontinuity" at 0, because there is no continuity in N at all. N is discretely ordered, not coninuously ordered. > > > > 3. By Lowenheim-Skolem-Tarski, first order PA has uncountable models, > > and Ax(x = 0 v Ey x=Sy) is a theorem of first order PA. But it is NOT > > required that the model regard 'S' as standing for the usual set > > theoretic successor function. So first order PA does not "define the > > inductive set". > > It defines an inductive structure for the set, but not the meaning of > successor. Still, there is no reason why we cannot apply the normal unit > increment uncountable times. There is certainly no reason to suppose we can. The 5th PA says : If a property holds for 0, and holds for the successor of every natural number for which it holds, then the property holds for all natural numbers. At least according to http://en.wikipedia.org/wiki/Peano_axioms#The_axioms which prohibits uncountably incrementing, at least for the naturals. > > > > 3. In set theory, we define 'Peano system' in a way that captures our > > intuitive notion of the counting numbers starting with zero. And we > > prove that all Peano systems are isomorphic. In particular, all Peano > > systems have a denumerable carrier set. So even though first order PA > > has uncountable models, such uncountable models are not part of any > > Peano system. > > It sounds like convention more than anything. Try reading it instead of only listening to it and you might learn something to your advantage, TO. > > > > > 4. [I think the following is correct, and hope to be corrected if it is > > not correct:] In Z set theory without the axiom schema of replacement, > > it is underdermined whether there exist inductive sets that are > > uncountable (where an inductive set is one that has 0 as a member and > > is closed under the successor operation), but with the axiom schema of > > replacement, there are proven to be inductive sets that are > > uncountable. [I think this item 4. is is correct, and hope to be > > corrected if it is not correct:] In any case, again, first order PA > > does not define "the inductive set", since there may be many inductive > > sets. w (which is the LEAST inductive set) is the carrier set of ONE OF > > the models of first order PA, but w is defined in set theory, not in > > first order PA. > > Yes. > > > > > 5. But, per item 3, no uncountable set, whether or inductive or not, is > > a carrier set for a Peano system. That is to say, no, it is not true > > that it is possible to have a set for a Peano system (note 'system', > > not 'axioms') that has a successor operation "applied" an uncountable > > number of times. And the notion of such a thing reduces to just another > > variation on the unsupported and axiom-less confusion (suffered by many > > cranks) that a limit ordinal, such as the first infinite ordinal, can > > somehow (or must somehow) also be a successor. > > I am not talking about limit ordinals, but about an uncountable > successorship. Which is equally as nonsensical as TO's notions about limit ordinals. > > > > > 6. The first order PA axioms are all proven in set theory. In > > particular, the induction axiom schema of first order PA is proven as a > > theorem schema of set theory. > > Using the axiom of infinity I presume? > > > > >>> ERGO: If the first natural is finite and the successor of a finite > >>> natural is a finite then every natural is finite. > >>> > >> Only for a finite number of successive increments. For all of them, however many there may be for the naturals. TO's repetitive denial of what is patently true in ZF, or NBG, is boring, as are many of his other essays into building castles in air. > > > > No such qualification is needed. Giving such a qualification only > > reveals a lack of understanding of what the induction schema is. What > > the poster said is correct, without any qualification about "finite > > number of sucessive increments" needed. > > Inductive proof such as this are taken only to hold for finite n, since > that's all there is in N, but if that's the case, all it proves is that > finite n are finite. It also says that, at least in N, there are only finite n. If TO wants some infinite things of some sort to play his silly games with, he must go elsewhere to find them as they do not exist in N and ca |