An uncountable countable set
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From: Lester Zick on 3 Oct 2006 18:19 On 3 Oct 2006 13:46:18 -0700, imaginatorium(a)despammed.com wrote: >Lester Zick wrote: >> On 3 Oct 2006 10:13:54 -0700, imaginatorium(a)despammed.com wrote: >> >> >Lester Zick wrote: >> > >> ><snip lesser maundering...> >> > >> >> But if the limit concept is the only sensible way to approach the >> >> infinite, one is left to ponder whether the limit concept is infinite >> >> and if not how a limits to the limit concept can be approached? >> > >> >Beautifully put, Lester, but surely the only way would be by >> >regression? >> >> Naturally by finite tautological regression to self contradictory >> alternatives. Or subdivision. > >Hmm. Not sure what self-contradictory alternatives are. "Contradiction of contradiction" for one is the self contradictory tautological alternative to "contradiction". "Different from differences" for another is the self contradictory tautological alternative to "differences". And "not not" for yet another is the self contradictory tautological alternative to "not". > How can >something be alternative to something that's self-contradictory? Tautologically take what is "not" what's generally self contradictory. However you have to be sure of what's self contradictory in general scientific terms because there are general and particular self contradictions. If one simply asserts "Susan is not Susan" the self contradiction is particular and not general. So just denying it results in regression to at best a particular truth and possibly another particular self contradiction and not a general truth since "Susan not Susan" and "not (Susan not Susan)" don't necessarily exhaust all possibilities for truth between them. > Oh, >but I see, we could just use subdivision. Wow - that really would >_never_ have occurred to me. But I'm sorry, I don't quite see what we >would be subdividing; is there any chance you could clarify? (Hope the >answer won't entail any of these self-contradictory alternatives...) Not if you employ subdivision to begin with, Brian. One can subdivide indefinitely and infinitesimally as through the bisection of an angle or straight line segment without violating containment and "infinity" is the number of those infinitesimals whereas one can never increment a number without either producing another finite number or violating containment. ~v~~
From: Virgil on 3 Oct 2006 18:45 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. > > > > > 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. > > > > > > >>> 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. 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. > > >>> (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.
From: Ross A. Finlayson on 3 Oct 2006 18:48 Lester Zick wrote: > On 3 Oct 2006 13:46:18 -0700, imaginatorium(a)despammed.com wrote: > > >Lester Zick wrote: > >> On 3 Oct 2006 10:13:54 -0700, imaginatorium(a)despammed.com wrote: > >> > >> >Lester Zick wrote: > >> > > >> ><snip lesser maundering...> > >> > > >> >> But if the limit concept is the only sensible way to approach the > >> >> infinite, one is left to ponder whether the limit concept is infinite > >> >> and if not how a limits to the limit concept can be approached? > >> > > >> >Beautifully put, Lester, but surely the only way would be by > >> >regression? > >> > >> Naturally by finite tautological regression to self contradictory > >> alternatives. Or subdivision. > > > >Hmm. Not sure what self-contradictory alternatives are. > > "Contradiction of contradiction" for one is the self contradictory > tautological alternative to "contradiction". "Different from > differences" for another is the self contradictory tautological > alternative to "differences". And "not not" for yet another is the > self contradictory tautological alternative to "not". > > > How can > >something be alternative to something that's self-contradictory? > > Tautologically take what is "not" what's generally self contradictory. > However you have to be sure of what's self contradictory in general > scientific terms because there are general and particular self > contradictions. If one simply asserts "Susan is not Susan" the self > contradiction is particular and not general. So just denying it > results in regression to at best a particular truth and possibly > another particular self contradiction and not a general truth since > "Susan not Susan" and "not (Susan not Susan)" don't necessarily > exhaust all possibilities for truth between them. > > > Oh, > >but I see, we could just use subdivision. Wow - that really would > >_never_ have occurred to me. But I'm sorry, I don't quite see what we > >would be subdividing; is there any chance you could clarify? (Hope the > >answer won't entail any of these self-contradictory alternatives...) > > Not if you employ subdivision to begin with, Brian. One can subdivide > indefinitely and infinitesimally as through the bisection of an angle > or straight line segment without violating containment and "infinity" > is the number of those infinitesimals whereas one can never increment > a number without either producing another finite number or violating > containment. > > ~v~~ Hi, Tony, the "Factorial/Exponential Identity, Infinity", that was last year on sic.math. In it, I showed an expression that was obviously untrue in the finite, but appeared to be so in the infinite, for some input values. I just read it, and let's see, here, hypermatrices, Stirling's numbers, density of infinite bit strings, basically it is about density of semi-infinite bit strings. I see a variety of avenues for research. We get to talking about the real numbers, which I call R. Sometimes you want to talk about them as continuous, other times, basically discrete. So, there are optional umlaut and bar accents, two dots for one, and the bar for the other, and both for both, R bar umlaut, R bar, R umlaut, blah blah blah, R dot dot, the bar is horizontal. Then, there are the natural numbers, for example N, the symbol used to denote the set of numbers {0, 1, 2, 3, ...}. I call the natural numbers N. N E N! That's says that the set N contains itself. It also says it is contained by itself. Thus the direct product of infinitely many copies is zero. That's why ZF is insufficient to be the foundation for some mathematics, because it is so easy to comprehend. That's a pun about comprehension. Mathematics works that way. I read Cornelius Lanczos' "Linear Differential Operators", let me tell you, he doesn't much talk about linear differential operators. I wrote about the ball and vase, basically the faster they move the harder it is to see them. Hi Lester, hey how's it going. As usual, we're sitting here on sci.math. You want a countable uncountable set? Look at the rationals. 2^|X|, they go. Ross
From: Dik T. Winter on 3 Oct 2006 19:20 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. -- dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
From: MoeBlee on 3 Oct 2006 19:22
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. 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. 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". 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. 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. 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. 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. > > 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. > 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). > > Then it is not vN, and does not satisfy the Peano properties, and is > > irrelevant in any discussion of natural numbers. > > Not really. Yes, really. Your set is not a carrier set for a Peano system. Your half-formed notion is not isomorphic with any Peano system. Your notion does not correspond to the common notion of plain ordinary counting numbers. It's fine that you want to develop some different mathematical notions. But it is just inane the way you carry that out as if you are CORRECTING other mathematics and as you claim that your very personal notion is the one to capture, correct, or supplant the everyday common notion of counting numbers. Moreover, you are inane when you claim that your notions are somehow "true of the world" (or however you put it) or are in accord with some correct logic that standard mathematics is not accord with, especially since you can't even begin to articulate what your method of logic is, not even informally, let alone formally. MoeBlee |