An uncountable countable set
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From: MoeBlee on 3 Oct 2006 19:27 MoeBlee wrote: > > I reject the von Neumann ordinals as an inadequate model of the naturals. > > They are the carrier set of a Peano system. Of course, I meant the finite von Neumann ordinals. MoeBlee
From: Dik T. Winter on 3 Oct 2006 20:45 In article <b3dc0$45221476$82a1e228$24601(a)news1.tudelft.nl> Han de Bruijn <Han.deBruijn(a)DTO.TUDelft.NL> writes: > Dik T. Winter wrote: > > In article <1159726829.184763.303470(a)i3g2000cwc.googlegroups.com> > > Han.deBruijn(a)DTO.TUDelft.NL writes: .... > > > Obviously they exist, because > > > how can we approach e.g. the Continuum Hypothesis by employing limts? > > > > I have no idea about the meaning of this statement, but off-hand I ould > > say that there is no way. > > Now we are getting somewhere. In applicable mathematics, approaching the > infinite via the limit concept is the _only_ viable way. Apparently some do disagree. But what is the connection with the continuum hypothesis? In what way is CH "approaching the infinite"? -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: Tony Orlow on 3 Oct 2006 21:54 Virgil wrote: > In article <4522e14e(a)news2.lightlink.com>, > Tony Orlow <tony(a)lightlink.com> 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. > > 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. >>> 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 "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. >>> >>>>> 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. sum(x=1->aleph_0: 1)=aleph_0. > > 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. > >>>>> (2) since this N can be proved to be well ordered, >>>>> Therefore, every subset of N with no first element must be empty. >>>> My N is not "well ordered" any more than the reals. >>> Then it is not vN, and does not satisfy the Peano properties, and is >>> irrelevant in any discussion of natural numbers. >> Not really. > > Really. That also makes TO irrelevant. Not half as irrelevant as yourself.
From: Tony Orlow on 3 Oct 2006 21:58 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?
From: Tony Orlow on 3 Oct 2006 22:14
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. > > 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. > > 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. > > 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. > > 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. > > 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 reject the von Neumann ordinals as an inadequate model of the naturals. > > They are the carrier set of a Peano system. All Peano systems are > isomorphic. All your nonsense about this reduces simply to your > preference that the word 'natural number' be used to mean something > different. It doesn't matter. We could just as well call them > 'schmatural numbers' and you'd have nothing to complain about. But you > would not be able to use 'natural number' in YOUR sense and still be > within the common notion of plain ordinary counting numbers, since your > sense does not provide for a Peano system (the notion of a Peano system > DOES capture the common notion of plain ordinary counting numbers). > What does it lack as a Peano system? If each has a successor and predecessor except for 0, and each is either before of after any other, why does it faila s a model |